Iit 3 Unlocked

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IIT - MATHS SET - 3

INDEX

1.

VECTOR ALGEBRA ...........................................................................................

2

2.

MATRIX AND DETERMINANTS .......................................................................

24

3.

QUADRATIC EQUATIONS ...............................................................................

52

4.

INDEFINITE INTEGRATION ...........................................................................

78

5.

DEFINITE INTEGRATION ...............................................................................

106

6.

AREAS UNDER CURVES ..................................................................................

128

7.

DIFFERENTIAL EQUATIONS ..........................................................................

158

IIT-MATHS-SET-III

1

VECTOR ALGEBRA

2

VECTORS_THEORY A scalar is a quantity, which has only magnitude but does not have a direction. For ILLUSTRATION time, mass, temperature, distance and specific gravity etc. are scalars. A Vector is a quantity which has magnitude, direction and follow the law of parallelogram (addition of two vectors). For ILLUSTRATION displacement, force, acceleration are vectors. (i)

(ii)

There are different ways of denoting a vector : a or a or a are different ways.    We use for our convenience a, b , c etc. to denote vectors, and a, b, c to denote their  magnitude. Magnitude of a vector a is also written as a . A vector a may be represented by a line segment OA and arrow gives direction of this vector.. Length of the line segment gives the magnitude of the vector. A

O

Here O is the initial point and A is the terminal point of OA

ADDITION OF TWO VECTORS    

 Let OA  a , AB  b and OB  c .  Here c is sum (or resultant) of vectors a and b . It is to be noticed that the initial point of coincides with the terminal point of and the line joining the initial point of to the terminal point of represents vector in magnitude and direction. B c = ab b O

a

A

TYPE OF VECTORS (i)

Equal Vectors Two vectors are said to be equal if and only if they have equal magnitudes and same direction.

3

IIT-MATHS-SET-III

(ii)

A

B

C

D

AB = CD

As well as direction is same

Zero Vector (null vector)

A vector whose initial and terminal points are same, is called the null vector. For ILLUSTRATION.



Such vector has zero magnitude and no direction, and denoted by 0 .

    AB  BC  CA  AA     or AB  BC  CA  0 C

B

A

(iii)

Like and Unlike Vectors Two vectors are said to be (i)

Like, when they have same direction.

(ii) Unlike, when they are in opposite directions. and are two unlike vectors as their directions are opposite, and are like vectors.

(iv)

Unit Vector 4

VECTORS_THEORY A unit vector is a vector whose magnitude is unity. We write, unit vector in the direction of as . Therefore = .

(v)

Parallel vectors Two or more vectors are said to be parallel, if they have the same support or parallel support. Parallel vectors may have equal or unequal magnitudes and direction may be same or opposite. As shown in figure

a

O

b

C

B

c

E

(vi)

A

D

Position Vector If is any point in the space then the vector is called position vector of point , where O is the origin of reference. Thus for any points A and B in the space,

(vii) Co–initial vectors Vectors having same initial point are called co–initial vectors. As shown in figure: Here and are co–initial vectors.

D d

C

c

a

O

b

SOME PROPERTIES OF VECTORS    (i)  (Vector addition is commutative) a b b a

5

IIT-MATHS-SET-III (ii)

      a  b  c  a  b  c (Vector addition is associative)

(iii)

     a  b  a  b , equality holds when a and b are like vectors

(iv)

     a  b  a  b , equality holds when a and b are unlike vectors



 



(v)

   m  na   mna  n  ma  (where m , n are scalars)

(vi)

    m a  b  ma  mb (where m , n are scalars)

(vii)

    m a  b  ma  mb (where m is a scalar)









COLLINEAR VECTORS 

Two vectors are said to be collinear if and only if there exists a scalar m such that a  mb . Thus (i) any vector a and zero vector are always collinear.. (ii) like and unlike vectors are collinear.

   Note that xa  yb  0  x  y  0 if and only if a and b are non–collinear. Thus repre-



sentation of any vector as a linear combination of non–collinear vectors a and b is unique.

COPLANAR VECTORS      are coplanar if there exists a relation  Three vectors a, b, c xa  yb  zc = 0 (where x, y, z are scalars, not all zero) Thus,

 (i) any two vectors a and b and a zero vector are always coplanar..   (ii) if any two of a , b and c are collinear, then a , b and c are coplanar.. (iii) there exists a plane which can contain all coplanar vectors.

  Note that xa  yb  zc  0 

 x = 0, y = 0, z = 0 if and only if vectors a , b and c are

non–coplanar.

 Any vector r , coplanar with non–collinear vectors a and b , can be expressed as a linear  combination of vectors a and b uniquely.. i.e., for same scalars m and n

6

VECTORS_THEORY B

b nb

r O

ma

a

A

Any vector r in space can be written as a linear combination of three non–coplanar vectors    a , b and c uniquely..

 i.e., r  la  mb  nc for some scalars l , m and n .

COLLINEARITYAND COPLANARITY OF POINTS (i)

The necessary and sufficient condition for three points with position vectors and to be collinear is that there exist scalars x, y, z, not all zero, such that , where x + y + z = 0.

(ii)

The necessary and sufficient condition for four points with position vectors and to be coplanar is that then exist scalars x, y, z and u, not all zero, such that , where x + y + z + u = 0.

SECTION FORMULA   and  . Let A, B and C be three collinear points in space having position vectors a, b r Let or,

AC n  CB m m AC  n CB

or,

  m AC  n CB

. . . (i)

(As vectors are in same direction)

    Now, OA  AC  OC  AC  r  a Þ

. . . (ii)

. . . (iii)

  ma  nb  Using (i), we get r = mn

ORTHOGONAL SYSTEM OF UNIT VECTORS Let OX, OY and OZ be three mutually perpendicular straight lines. Given any point P(x, y, z) in space, we can construct the rectangular parallelopied of which OP is a diagonal and OA = x, OB = y, OC = z. 7

IIT-MATHS-SET-III Here A, B, C are (x, 0, 0), (0, y, 0) and (0, 0, z) respectively and L, M, N are (0, y, z), (x, 0, z) and (x, y, 0) respectively. Let î , ˆj, kˆ denote unit vectors along OX, OY and OZ respectively..

    We have r  OP  xiˆ  yjˆ  zkˆ as OA  xiˆ , OB  yjˆ and OC  zkˆ .

   ON  OA  AN

   OP  ON  NP         So, OP  OA  OB  OC NP  OC, AN  OB





  r | r |  OP  x 2  y 2  z 2





 r xiˆ  yjˆ  zkˆ rˆ    = liˆ  mjˆ  nkˆ |r | x2  y2  z 2  r  lr î  mr ˆj  nr kˆ l  cos  

m  cos  

n  cos  

x x2  y 2  z 2

(where  is the angle between OP and x–axis)

y x2  y 2  z 2

(  is the angle between OP and y–axis)

z x2  y 2  z 2

(  is the angle between OP and z–axis)

l ,m,n are defined as the direction cosines of the line OP and x, y, z are defined as direction ratios

of the line OP.

 If P  (x1, y1, z1) and Q  (x2, y2, z2) then PQ = (x2 – x1) î + (y2 – y1) ˆj + (z2 – z1) kˆ 8

VECTORS_THEORY Therefore PQ =

 x2  x1 2   y2  y1 2   z2  z1 2

Hence direction ratios of the line through P and Q are x2 – x1, y2 – y1 and z2 – z1 and its direction cosines are

x2  x1 y2  y1 z2  z1 , and . PQ PQ PQ

SOME PROPERTIES OF DIRECTION COSINES AND RATIOS (i) lr,mr,nr are the projection of r on x, y and z–axis. (ii) r = l î  m ˆj  n kˆ (iii) l 2  m 2  n 2 = 1 l m n   , then a, b, c are the a b c ratios of the line whose direction cosines are l ,m and n .

(iv) If a, b and c are three real numbers such that

direction

SCALAR PRODUCT OF TWO VECTORS (DOT PRODUCT)    The scalar product, a .b of two non–zero vectors a and b is defined as a b cos  , where  is angle between the two vectors, when drawn with same initial point. Note that 0     .

   is defined as zero. If at least one of a and b is a zero vector, then a.b

PROPERTIES (i)

    a. b  b . a

(ii)

   2 a 2  a.a  a  a 2

(iii)

 ma  .b  m  a.b   a. mb 

 

(iv)   cos

(v)

(scalar product is commutative)

 





(where m is a scalar)

    a. b     a .b   

1 

    a.b  0  Vectors a and b are perpendicular to each other.. 

[ a , b are non–zero vectors]. (vi) î . ˆj = ˆj.kˆ  kˆ .iˆ = 0

      (vii) a. b  c  a.b  a.c



9



IIT-MATHS-SET-III (viii)





2

 a  b  . a  b   a 2  b

   a2  b 2

  (ix) Let a  a1 î  a2 ˆj  a3 kˆ , b  b1 î  b2 ˆj  b3 kˆ   Then a. b  a1 î  a2 ˆj  a3 kˆ . b1 î  b2 ˆj  b3 kˆ



*





= a1b1  a2b2  a3b3

Algebraic projection of a vector along some other vector

    a.b ˆ a.b ON  OB cos   b    a b |a|

VECTOR (CROSS) PRODUCT  The vector product of two non–zero vectors a and b , whose module are a and b respectively, is the vector whose modulus is ab sin  , where   0      is the angle between   vectors a and b . Its direction is that of a vector n perpendicular to both a and b , such   are in right–handed orientation. that a, b ,nˆ

 By the right–handed orientation we mean that, if we turn the vector a into the vector b through the angle , then n points in the direction in which a right handed screw would move if turned in the same manner.     Thus a  b  a b sin  n

  If at least one of a and b is a zero vector, then a  b is defined as the zero vector..

PROPERTIES (i)

    ab   b  a

(ii)

 ma   b  m  a  b   a   mb  (where m is a scalar)









 





10

VECTORS_THEORY     (iii) a  b  0  vectors a and b are parallel. (provided a and b are non–zero vectors).

 (iv) î  î  ˆj  ˆj  kˆ  kˆ  0 .









(v) î  ˆj  kˆ    ˆj  î  , ˆj  kˆ  î   kˆ  ˆj ,kˆ  î  ˆj   î  kˆ .

       (vi) a  b  c  a  b  a  c .





 (vii) Let a = a1 î + a2 ˆj  a3 kˆ and b  b1 î  b2 ˆj  b3 kˆ , then î ˆj   a  b  a1 a2 b1 b2

kˆ a3 . b3

= î  a2b3  a3b2   ˆj  a3b1  a1b3   kˆ  a1b2  a2b1 

  a b (viii) sin  =   . (Note : we cannot find the value of  by using this formula) a b

(ix) Area of triangle =

(x)

1 1 1   ap  ab sin   a  b 2 2 2

  Area of parallelogram = ap  ab sin   a  b .

SCALAR TRIPLE PRODUCT   and  is defined as a  b .c The scalar triple product of three vectors a, b c





 Let a = a1 ˆ + a2 ˆj + a3 ˆ , b = b1 ˆ + b2 ˆj + b3 ˆ , c = c1 î + c2 ˆj + c3 ˆ . i i k k k

11

IIT-MATHS-SET-III kˆ a a3  î 2 b2 b3

î ˆj   a  b  a1 a2 Then b1 b2



a    a  b .c  c1 2 b2



a3 a  ˆj 1 b3 b1

a3 a a a  c2 1 3  c3 1 b3 b1 b3 b1

a3 ˆ a1 k b3 b1

a1 a2  b1 b2 c1

a2 b2

a2

a3

b2 c2

b3 c3

                 Therefore a  b .c  b  c .a   c  a  .b   b  a .c   c  b .a    a  c  .b

















         Note that a  b .c  b  c .a  a. b  c , hence in scalar triple product dot and cross are inter      changeable. Therefore we denote a  b .c by  a b c  .

















PROPERTIES    a  b . c represents the volume of the parallelopied, whose adjacent sides are rep  and  in magnitude and direction. Therefore three vectors resented by the vectors a,b c



(i)



   a,b , c are coplanar if and only if

 a 

(ii) Volume of the tetrahedron =

1 6

 b

a1

 c  = 0 i.e., b1 c1

a2

a3

b2 c2

b3 =0 c3

   a b c  .  

        (iii)  a  b c d    a c d   b c d  (iv)

 

 a a b = 0.

VECTOR TRIPLE PRODUCT     The vector triple product of three vectors a ,b and c is defined as a  b  c . If at least   and  is a zero vector or  and  are collinear vectors or  is perpendicular one of a,b c c a b         to both b and c , only then a  b  c  0 . In all other cases a  b  c will be a non–zero













 vector in the plane of non–collinear vectors b and c and perpendicular to the vector a .

         Thus we can take a  b  c   b   c , for some scalars  and  . Since a  a  b  c ,



    a. a  b  c

 









  0    a. b     a. c   0 12

VECTORS_THEORY 

       a. c   ,    a. b  , for same scalar  .



   Hence a  b  c  







  

  

 a. c  b   a.b  c  , for any vectors

   a,b and c satisfying the

conditions given in the beginning.

 In particular if we take, a  b  î , c  ˆj , then  = 1.

         Hence a  b  c   a. c  b  a. b c









RECIPROCAL SYSTEM OF VECTORS  Let a ,b and c be a system of three non–coplanar vectors. Then the system of vectors          a, b and c which satisfies a.a  b .b  c .c = 1 and             a. b  a. c  b . a  b . c  c . a  c . b = 0, is called the reciprocal system to the vectors           a,b , c . In term of a,b , c the vectors a,b, c are given by a =

 b c  ,b    a b c   

  c a  ,c    a b c   

  a b   a b c  .  

PROPERTIES (i) a.b  a.c  b.a  b.c  c.a  c.b = 0 (ii) The scalar triple product [a b c] formed from three non–coplanar vectors a, b, c is the reciprocal of the scalar triple product  a b c formed from reciprocal system.

ANGLE BETWEEN TWO LINES   Let the vector equations of two lines be r  a   b and r  c   d . These two lines are   parallel to the vectors b  a1î  b1 ˆj  c1kˆ and d  a2î  b2 ˆj  c2 kˆ respectively. Therefore,





angle between these two lines is equal to the angle between b and d . Thus, if  is the angle   b. d cos     between the lines, then b d . In Cartesian Form it is given as cos   l1l2  m1m2  n1n2 

a1a2  b1b2  c1c2 a12  b12  c12

a22  b22  c22

, where l1 ,m1 ,n1 and l2 ,m2 ,n2 are direction cosines

and a1 ,b1 ,c1 and a2 ,b2 ,c2 are direction ratios of the given lines. *

13

  If the lines are perpendicular, then b .d = 0.

(vector form)

IIT-MATHS-SET-III i.e., l1l2  m1m2  n1n2 = 0 or a1a2  b1b2  c1c2 = 0 (Cartesian form) *

    If the lines are parallel, then b and d are parallel, therefore b =  d for some scalar  . a1 b1 c1 i.e., a  b  c . 2 2 2

14

VECTORS_THEORY

15

IIT-MATHS-SET-III

ASSIGNMENT

16

VECTOR ALGEBRA

SECTION-B MULTIPLE ANSWER QUESTIONS 1.

The vectors li + j + 2k, i + lj – k and 2i – j + lk are coplanar if a)= -2

2.

d) 

=0

b) –2, -3

c) all x < 0

d) all x > 0

Let a = 2i – j + k, b = i + 2j – k and c = i + j – 2k be three vectors. A vector in the plane of b and c whose projection on a is of magnitude 2 / 3 is a) 2i + 3j – 3k

4.

c) 1 - 3

The values of x for which the angle between the vectors a = xi – 3j – k and b = 2xi + xj – k is acute, and the angle between the vector b and the axis of ordinates is obtuse, are a) 1, 2

3.

b) = 1 + 3

b) 2i + 3j + 3k

c) –2i – j + 5k

d) 2i + j + 5k

Let a = 4i + 3j and b be two vectors perpendicular to each other in xy-plane. The vectors c in the same plane having projections 1 and 2 along a and c are a) -

5.

2 11 i+ j 3 2

b)2 = 1 - 22

c) (8, 8, 9)

If a, b c are three unit vectors such that a x (b x c) =

1 2

d)2 =

1  cos 2 2

d) at the tip of R(2)

b and c being non parallel then

b) angle between a and c is p/4 d) angle between a and b is p/3

1

 2 cos 

b) p = 1  2 cos  , q  1  2 cos  2 cos  d) p = q  1  2 cos 

The vectors ai + 2aj – 3ak, (2a + 1) i + (2a + 3)j + (a + 1) k and (3a + 5) i + (a + 5) j + (a + 2) k are non-coplanar for a in a) {0}

17

2 11 i+ j 3 2

If a, b and c be non-coplanar unit vectors equally inclined to one another at an acute angle q. If a x b + b x r = pa + qb + rc then a) p = r 1 c) r = 1  2 cos 

9.

d)

c) 2 = - cos 2

b) at the tip of R(4)

a) angle between a and b is p/2 c) angle between a and c is p/3 8.

2 11 i+ j 5 5

The point of intersection of the lines l1 : r(t) = (i – 6j + 2k) + t(i + 2j + k) l2 : R(u) = (4j + k) + u (2i + j + 2k) a) at the tip of r(7)

7.

c) -

Let the unit vectors A and B be perpendicular and the unit vector C be inclined at an angle q to both A and B. If C = aA + bB + g(A x B) then a) = 

6.

b) 2i – j

b) (0, )

c) (-, 1)

d) (1, )

IIT-MATHS-SET-III 10.

If K is the length of any edge of a regular tetrahedron then the distance of any vertex from the opposite face is a) 3/2 K

11.

7 75

c)

2 K 3

d)

3K

b) cos-1

26 27

c) cos-1

3 15

2 d) cos-1 3

If the unit vectors a and b are inclined at an angle 2 q and < 1, then if 0 £ q £ p, q lies in the interval   a)  0,   6

13.

2 2 K 3

Two sides of a triangle are formed by the vectors a = 3i + 6j – 2k and b = 4i – j + 3k. Acute angles of the triangle are a) cos-1

12.

b)

 5  ,  b)   6 

   c)  ,  6 2

  5  d)  ,  2 6 

If (a x b) x (c x d) = ha + kb = rc + rd where a and b are non-collinear and c and d are also non-collinear then a) h = [b c d]

b) k = [a c d]

c) r = [a b d]

d) s = [a b c]

18

VECTOR ALGEBRA

SECTION-C PASSAGE TYPE QUESTIONS PASSAGE 1: Let C: r(t) = x(t) i + y(t) j + z(t) k Be a differentiable curve i.e xlim 0

r t  h   r h  exist for all t. The vector h

R’(t) = x’(t)i + y’(t)j + z’(t) k If not O, is tangent to the curve C at the point P(x((t), y(t), z(t)) and r’(t) points in the direction of increasing t. 1.

The point P on the curve r(t) = (1 – 2t) i + t2 j + 2e(t-1) k at which the tangent vector r’ (t) is parallel to the radius vector r(t) is a) (-1, 1, 2)

b) (1, -1, 2)

c) (-1, 1, 2)

d) (1, 1, 2)

2.

A parametrized tangent vector to r(t) = ti + t2j + t3k at (2, 4, 8) is a) R(u) = 2i + 4j + 8k + u(i + j + 4k) b) R(u) = i + 2j + 4k + u(i + 4j + 12k) c) R(u) = i + 4j + 12k + u(2i + 4j + 8k) d) R(u) = 2i + 4j + 8k + u (i + 4j + 12k)

3.

The tangent vector to r(t) = 2t2 i + (1 – t) j + (3t2 + 2) k at (2, 0, 5) is a) 4i + j – 6k

b) 4i – j + 6k

c) 2i – j + 6k

d) 2i + j – 6k

PASSAGE 2: Equation of a line can be obtained as the intersection of two planes, or passing through a point and parallel to given plane. Similarly equation a plane can be obtained having different condition e.g. passing through three points or through a point and perpendicular to two planes. 4.

The line through the point c, parallel to the plane r.n = 1 and perpendicular to the line, r = a + tb is a) r = c + ta x n

5.

b) r = a + t(n1 – n2)

c) r = a + tn2

d) r = a + t(n1 x n2)

b) r . (a – b) = q d) r.((b – a) x n) = [a n b]

The plane which passes through a and is perpendicular to the plane r.n = q and is parallel to the line r = b + tc is a) r .b = [a n c]

19

d) r = a + t(c x n)

The plane which passes through the two points a and b and is perpendicular to the plane r.n = q is a) r.((b – a) x n) = q c) r.((b – a) x n) = [a b n]

7.

c) r = c + tn

The line through the point a and parallel to the planes r.n1 = q1, r.n2 = q2 is a) r = a + tn1

6.

b) r = c + tb x n

b) [r n c] = [a n c]

c) r.a = [b n c]

d) [r c n] = [a n c]

IIT-MATHS-SET-III

SECTION-D MATCHING TYPE QUESTIONS 1.

2.

If a and b are two units vectors inclined at angle a to each other then 2

i) a  b < 1 if ii) a  b  a  b if

a) << 3 b) /2 <   

iii) a  b  2

c) = /2

iv) a  b <

d) 0 £ q < p/2

2

a, b, c, d are given vectors. Match solution of equation in coloumn 1 to its solution in column 2. c.a

i) r x a + (r . b) c = d

a) r = c - a . b b

ii) r = r x a + b

b) r = a . a a + a x

iii) r x b = c x b, r . a = 0

c) r = - b 2 (a x b) + yb, y is a parameter

a .r

1

1  a .b  iv) r x b = a, where a, b are such that a is perpendicular to b  = 1  a 2   a  b  a x b  3.

Given two vectors a = i + j - k, b = i – j + k, c = i + 2j – k i) a vector perpendicular to the vector a and coplanar with a and b

4.

ii) a vector perpendicular to a and the vector in (i)

a) (1/ 2 ) (-i + 2j + 3k) b) 1/ 2 (i + k)

iii) a vector perpendicular to b and c

c)

iv) a vector perpendicular to a and a + c

d) (1/ 2 ) (j + k)

1 6

(2i + j + k)

The area / volume of i) triangle with vertices whose position vector w.r.t O is –i + 2j + 3k, 2i – j – k, i + j - k a) 3 /2 ii) tetrahedra with vertices O, i + j – k, i – j + k, -i + j + k b) 2/3 iii) tetrahedra with vertices –i + k, 2i – j, i + 2j + 5k, i + 2j + k iv) triangle with vertices i, j, i + j + k

5.

c) 89 /2 d) 6

Let a, b, c be any three vector then match the following vectors i) a x (b + c) + b x (c + a) ii) a x (b x c) + b x (c x a) iii) (a x b) x (c x b) iv) (a x b) x (c x a)

a) [c b a ] a b) [a c b] b c) (a x b) x c d) (a + b) x c

20

VECTOR ALGEBRA

SECTION-E WRITE THE ANSWER 1.

If d =  (a x b) +  (b x c) + r(c x a),[a b c] = 1/8 and d. (a + b + c) = 8 then  +  + r is equal to.—————

2.

b = 4i + 3j and c be two vectors perpendicular to each other in the xy-plane. If ri, i = 1,2 ........n are the vectors in the same plane having projections 1 and 2 along b and c respectively n

then  ri i 1

2

is equal to .—————

3.

If A = (1, 1, 1) and C = (0, 1,-1) are given vectors and B is a vector satisfying A x B = C and A.B = 3 then 9 B 2 is equal to.——————

4.

If [b c d] = 24 and (a x b) x (c x d) + (a x c) x (d x b) + (a x d) x (b x c) + ka = 0 then k is equal to.———————

5.

Suppose that a, b, c do not lie in the same plane and are non zero vectors such that a

= 1, b - 2, c = 2, a. b = 1, b. c = 2 and the angle between a – b and b – c is p/6. If d

is any vector such that d. a = d . b = d. c and d 2 = α 2 k for any scalar a, then k is equal to.———————————— 6.

If a = i + 2i – 3k, b = 2i + j – k and u is a vector satisfying a x u = a x b and a. u = 0 then 2 2 u is equal to.————

7.

If the vectors ai + j + k, i + bj + k and i + j + ck, )a ¹ b ¹ c ¹ 1) are coplanar, find the value of 1 1 1   is ——————— 1  a 1  b 1 c

8.

Anon zero vector a is parallel to the line of intersection of the plane determined by the vectors i, i + j and the plane determined by the vectors i – j, i - k. If the acute angle between a and the vector i – 2i + 2k is q find 2 cos q————————

9.

Let OA = a, OB = 10a + 2b and OC = b, where O, A and C are non-collinear points. Let p denote the area of the quadrilateral OABC, and let q denote the area of the parallelogram with OA and OC as adjacent sides. If p = kp find K.—————————

10.

Let a = i – j, b = i + 2j + 2k, c = 2i + j + 2k and d = 2i – j + k. If p is the shortest distance between the lines r = a + t b and r = c + p d find 2p2————————

11.

21

If a, c, d are non-coplanar vectors satisfying d.(a x (b x(c x d))) = k [a c d] and b. d = 14 find k. ——————


A particle is displacement from the point whose position vector is 5i – 7j – 7k to the point whose position vector is 6i + 2j – 2k under the action of constant forces 10i – j + 11k, 4i + 5j + 6k and 2i + j – 9k. Find the total work done .————————

13.

In a triangle ABC, a point P is taken on the side AB such that AP: BP such that CQ : BQ = 2 : 1.If R is the point of intersection of lines AQ and CP. Suppose that the area of the triangle ABC is D, if it is known that the area of triangle BRC is one unit. Find the value of 4 D —— —————————

14.

ABCD is a regular hexagon. If AB = 4 units find AD  EB  FC —————————— —

15.

In a DABC, the median CM is perpendicular to the angle bisector AL and CM and CM : AL = 1: 3 find 85 cos A.——————————

SECTION-B KEY

1

2

3

4

5

6

7

8

a,b,c

b,c

a,c

b,c

a,b,c,d

a,b,c

a,c

a,b,c

9

10

11

12

13

c

a,b

a,b

b,c

b,d

SECTION-C KEY

1

2

3

4

5

6

7

a

d

b

b

d

c

b

22

VECTOR ALGEBRA

SECTION-D KEY

1.

(i)  (a), (ii)  (c), (iii)  (b), (iv)  (d)

2.

(i) (b), (ii) (d), (iii) (a), (iv) (c)

3.

(i)  (c), (ii)  (d), (iii)  (a), (iv)  (b)

4.

(i)  (c), (ii)  (b), (iii)  (d), (iv)  (a)

5.

(i)  (d), (ii)  (c), (iii)  (b), (iv)  (a)

SECTION-E KEY

1

2

3

4

5

6

7

8

64

20

33

48

3

5

1

1

9

10

11

12

13

14

15

6

1

14

97

17

16

77

23

IIT - MATHS - SET - III

2

MATRIX AND DETEMINANTS

24

MATRICES AND DETERMINANTS

Matrix A system of m × n numbers arranged in the form of an ordered set of m rows and n columas is called an m × n matrix. It can be read as m by n matrix.It is represents as A = [aij]m × n and

 a11 a12 . . . a1n    can be written in expanded form as A   a 21 a 22 . . . a 2n        a m1 a m2 . . . a mn 

 2 1 0   1 1 3   is a 3by 3 matrix. e.g.,   6 5 1

DIFFERENT TYPES OF MATRICES (i)

Square Matrix: A matrix for which the number of rows is equal to the number of

columns (each equal to n) is called a square matrix of order n.  1 0 2 3   2 1 4 5   is a square matrix of order 4. e.g. A =   3 2 4 1   1 0 0 2  (ii) Null Matrix: The matrix whose all elements are zero is called null matrix or zero matrix. It is usually denoted by O. (iii) Identity Matrix : A square matrix in which all the elements along the main diagonal (elements of the from aii) are unity is called an identity matrix or a unit matrix. An identity matrix of order n is denoted by In.

1 0 0   e.g. I3 =  0 1 0   0 0 1  (iv) Scalar matrix : A matrix whose diagonal elements are all equal and other entries are zero, is called a scalar matrix k 0 0  e.g. A   0 k 0  0 0 k    (v) Triangular Matrix: A square matrix whose elements above the main diagonal or below the main diagonal are all zero is called a triangular matrix. 25

IIT - MATHS - SET - III Note:

(i)

[aij]n × n is said to be upper triangular matrix if i > j  aij = 0,

(ii) [aij]n × n is said to be lower triangular matrix if i < j  aij = 0. 1 0 e.g.  0  0

2 1 3  2 1 0  is an upper triangular matrix. 0 3 1  0 0 3

(vi) Diagonal Matrix: A square matrix of any order with zero elements every where, except on the main diagonal, is called a diagonal matrix.

3 0 0    e.g.  0 5 0  is a diagonal matrix of order 3.  0 0 1 (vii) Row Matrix: A matrix which has only one row and n columns is called a row matrix of length n e.g., [2 –1 3 0]1×4

is a row matrix of length 4.

(viii) Column Matrix: A matrix which has only one column and m rows is called a column matrix of length m. 1 3 e.g. x    is a column matrix of length 4.  2     1 (ix) Sub Matrix: A matrix obtained by omitting some rows or some columns or both of a given matrix A is called a sub matrix of A.

2 0 4    e.g., If A   5 6 8  , then  3 2 29 

 2 0  5 6  is a submatrix of A which is obtained by omitting  

third row and third column of A.

EQUALITY OF TWO MATRICES Two matrices A = [aij] and B = [bij] are said to be equal if they are of the same order and their corresponding elements are equal. If two matrices A and B are equal, we write A = B.

ADDITION OF MATRICES If A and B are two matrices of the same order m × n, then their sum is defined to be the matrix of order m × n obtained by adding the corresponding elements of A and B. 26


 a11 a12  e.g. If A   a 21 a 22  a 31 a 32

a13   b11  a 23  and B   b 21  b31 a 33 

 a 11  b11 then A  B   a 21  b 21  a 31  b31

a 12  b12 a 22  b 22 a 32  b 32

b12 b 22 b32

b13  b 23  , b33 

a 13  b13  a 23  b 23  a 33  b 33 

PROPERTIES OF MATRIX ADDITION (i)

Matrix Addition is Commutative If A and B are two m × n matrices, then A + B = B + A.

(ii) Matrix addition is associative If A, B, C are three matrices, each of the order m × n, then (A + B) + C = A + (B + C). (iii) Existence of additive identity If O is the m × n null matrix, then A + O = A = O + A for every m × n matrix A. O is called additive identity.

MULTIPLICATION OF A MATRIX BY A SCALAR If A = [aij]m × n and  is a scalar, then  A = [  aij]m × n

2 1 3  e.g. A   2 5 1  – 3A =    0 1 2 

 6 3 9   6 15 3  .  0 3 6   

MULTIPLICATION OF TWO MATRICES Let A = [aij]m× n and B = [bjk]n×p be two matrices such that the number of columns in A is equal to number of rows in B. Then the m × p matrix C = [cik]m× p , n

where cik =  a ij .b jk (where i = 1, 2, 3 ... m, k = 1, 2, 3 ... p), is called the product of the j1 matrices A and B. We have  a11 a  21  ...  a m1

a12 a 22 ... am2

...

a 1n  ... a 2n  ... ...   ... a mn  mn

 b11 b  21  ...   b n1

b12 b 22 ... bn2

... b1p   c11  c ... b 2p  12   ... ... ...    ... b np   c m1 n p n

where Cij = ai1 b1j + ai2 b2j + ... + ain bnj =

27

a k 1

ik

b kj

c12 c 22 ... cm 2

...

c1p  ... c 2p  ... ...  ,  ... c mp  m p

IIT - MATHS - SET - III Properties of Matrix Multiplication (i)

Matrix multiplication is associative i.e.,

(AB)C = A(BC), A, B and C are m × n, n ×p and p × q matrices respectively.

(ii) Multiplication of matrices is distributive over addition of matrices i.e.,

A(B + C) = AB + AC

(iii) Existence of multiplicative identity of square matrices. If A is a square matrix of order n and In is the identity matrix of order n, then A In = In A = A. (iv) Whenever AB and BA both exist, it is not necessary that AB = BA. (v) The product of two matrices can be a zero matrix while neither of them is a zero matrix. 0 1 1 0  0 0  e.g., If A =  and B   then AB      , while neither A nor B is a null ma0 0 0 0  0 0  trix. (vi) In the case of matrix multiplication of AB = 0, then it doesn’t necessarily imply that A = 0 or B = 0 or BA = 0.

TRACE OF A MATRIX Let A be a square matrix of order n. The sum of the diagonal elements of A is called the trace of A. n



Trace (A) =

a

ii

 a11  a 22  ...  a nm .

i 1

TRANSPOSE AND CONJUGATE OF A MATRIX The matrix obtained from any given matrix A, by interchanging rows and columns, is called the transpose of A and is denoted by A  or AT .

1 2   1 4 7   e.g., If A   4 5  , then A   2 5 823  7 8  32 Properties of Transpose of a matrix (i)

(A)  A

(ii)

(A  B)  A  B

(iii)

(A)  A,  being any scalar

(iv)

(AB)  BA 

28

MATRICES AND DETERMINANTS Conjugate of a Matrix The matrix obtained from any given matrix A, containing complex number as its elements, on replacing its elements by the corresponding conjugate complex numbers is called the conjugate of A and is denoted by A .

1  2i 2  3i 3  4i  1  2i 2  3i 3  4i    e.g., A   4  5i 5  6i 6  7i  , then A   4  5i 5  6i 6  7i     8 7  8i 7   8 7  8i 7  Properties of Conjugate of a matrix (i)

A  A

(iii)

 A    A, 

being any scalar

(ii)

 A  B

(iv)

(AB)  A B

AB

Transpose Conjugate of a Matrix The transpose of the conjugate of a matrix A is called transposed conjugate of A and is denoted by A  . The conjugate of the transpose of A is the same as the transpose of the conjugate of A  i.e.,  A    A   A  If A = [aij]m× n , then A   [b ji ]nm , where bji = aij i.e., the (j, i)th element of A  = the conjugate of (i, j)th element of A.

8  1  2i 2  3i 3  4i  1  2i 4  5i     e.g., If A   4  5i 5  6i 6  7i  , then A   2  3i 5  6i 7  8i   8 3  4i 6  7i 7  8i 7  7  Properties of Transpose Conjugate of a matrix (i)

(A  )  A

(iii) (kA)  kA  , k being any scalar

(ii)

(A  B)  A   B

(iv)

(AB)  B A 

SPECIAL MATRICES Symmetric Matrix A matrix which is unchanged by transposition is called a symmetric matrix. Such a matrix is necessarily square

 2 1 3  1 4 1   e.g.,   3 1 5 Thus if A = [aij]m × n is a symmetric matrix then m = n, aij = aji i.e., A   A . 29


Skew Symmetric Matrix A square matrix A = [aij] is said to be skew symmetric, if aij = – aji for all i and j  0 2 3 1   e.g.  2 0 4 3  3 4 0 1     1 3 1 0  Thus if A = [aij]m × n is a skew symmetric matrix, then m = n, aij = – aji i.e., A   A . Obviously diagonal elements of a square matrix are zero.

Orthogonal Matrix A square matrix A is said to be orthogonal, if AA = AA  I , where I is a unit matrix. Note: (i) If A is orthogonal, then A is also orthogonal. (ii) If A and B are orthogonal matrices then AB and BA are also orthogonal matrices. Unitary Matrix A square matrix A is called unitary matrix if AA  A A  I . Idempotent Matrix: A square matrix A is called idempotent provided it satisfies the relation A2 = A.  2 2 4  e.g. The matrix A   1 3 4  is idempotent as  1 2 3     2 2 4   2 2 4   2 2 4  A   1 3 4   1 3 4    1 3 4   1 2 3  1 2 3   1 2 3      2

Involutary Matrix A matrix A such that A2 = I, is called involutary matrix. Nilopotent Matrix A square matrix A is called a nilpotent matrix, if there exists a positive integer m such that Am =O. If m is the least positive integer such that Am = O, then m is called the index of the nilpotent matrix A.

30


DETERMINANT Equations a1x + b1y = 0 and a2x + b2y = 0 in x and y have a unique solution if and only if a1b2 –a2b1  0. We write a1b2 – a2b1 as

a1 a2

b1 and call it a determinant of order 2. b2

Similarly the equations a1x + b1y + c1z = 0, a2x + b2y + c2z = 0 and a3x + b3y + c3z = 0 have a unique solution if a1(b2 c3 – b3c2) + b1 (a3c2 – a2c3) + c1 (a2 b3 – a3 b2)  0

a1

b1

c1

i.e.,   a 2 a3

b2 b3

c2  0 c3

The number ai, bi, ci (i = 1, 2, 3) are called the elements of the determinant. The determinant obtained by deleting the ith row and jth column is called the minor of the element at the ith row and jth column. We shall denote it by Mij. The cofactor of this element is (–1)i+j Mij, denoted by Cij. Let A = [aij]3×3 be a matrix, then the corresponding determinant (denoted by det A or | A |) is

a11 a12 a 21 a 22 a 31

a 32

a13 a 23 . a 33

It is easy to see that | A | = a11C11+ a12 C12 + a13C13 (we say that we have expanded the determinant | A | along first row). Infect value of | A | can be obtained by expanding it along any row or along any column. Further note that if elements of a row (column) are multiplied to the cofactors of other row (column) and then added, then the result is zero.

PROPERTIES OF DETERMINANTS (i) The value of a determinant remains unaltered, if its rows are changed into columns and the columns into rows.

a1

b1

c1

a1

a2

a3

e.g., a 2 a3

b2 b3

c2  b1 c3 c1

b2 c2

b3 . Thus any property true for rows will also be true for colc3

umns. (ii) If all the elements of a row (or column) of a determinant are zero, then the value of the determinant is zero.

31


0 b1 e.g., 0 b 2 0 b3

c1 a1 c2  0, 0

b1 0

c1 0 0

c3

b3

c3

a3

(iii) If any two rows (columns) of a determinant are identical, then the value of the determinant is zero.

e.g.,

a1

a1

c1

a2 a3

a2 a3

c2  0 c3

(iv) The interchange of any two rows (columns) of a determinant results in change of it’s sign

a1 i.e, a 2 a3

b1 b2

c1 b1 c2   b2

a1 a2

c1 c2

b3

c3

a3

c3

b3

(v) If all the elements of a row (column) of a determinant are multiplies by a non-zero constant, then the determinant gets multiplied by that constant.

a1

kb1

c1

a1

b1

c1

a1

b1

c1

ka 1

kb1

kc1

e.g., a 2 a3

kb 2 kb3

c2  k a 2 c3 a3

b2 b3

c2 and k a 2 a3 c3

b2 b3

c2  a 2 c3 a 3

b2 b3

c2 c3

(vi) If each element of a row (column) of a determinant is a sum of two terms, then determinant can be written as sum of two determinant in the following way:

a1

b1

c1  d1

a1

b1

c1

a1

b1

d1

a2 a3

b2 b3

c2  d 2  a 2 c3  d 3 a 3

b2 b3

c2  a 2 c3 a 3

b2 b3

d2 d3

n

n

n

In general  r 1

f (r) g(r) h(r) a2 a3

b2 b3

c2  c3

n

 f (r)  g(r)  h(r) r 1

r 1

a2 a3

b2 b3

r 1

c2 c3

(vii) The value of a determinant remains unaltered under a column operation of the form Ci  Ci  C j  C k ( j, k  i) or R i  R i   R j   R k ( j, k  i).

a

row

operat ion

of

t he

form

32


a1

b1

c1

a1

b1  2a1  3c1

c1

e.g. a 2 a3

b2 b3

c2  a 2 c3 a 3

b 2  2a 2  3c 2 b3  2a 3  3c3

c 2 , obtained after C2  C2 + 2C1 + 3C3. c3

(viii) Product of two determinants

a1

b1

c1

l1

l2

l3

a2 a3

b2 b3

c 2  m1 m 2 c3 n1 n 2

m3 n3

a1l1  b1m1  c1n1

a1l2  b1m 2  c1n 2

a1l3  b1m 3  c1n 3

 a 2l1  b 2m1  c 2 n1 a 2 l2  b 2 m 2  c 2 n 2 a 3l1  b3 m1  c3n1 a 3l2  b3m 2  c3n 2

a 2l3  b 2 m3  c 2 n 3 a 3l3  b3m3  c3n 3 (row by column multiplication)

a1l1  b1l2  c1l3

a1m1  b1m 2  c1m3

a1n1  b1n 2  c1n 3

 a 2 l1  b 2l2  c 2l3 a 3l1  b3l2  c3l3

a 2 m1  b 2 m 2  c 2 m3 a 3m1  b3m 2  c3m3

a 2 n1  b 2 n 2  c 2 n 3 a 3n1  b3n 2  c3n 3 (row by row multiplication)

We can also multiply determinants column by row or column by column. (ix) Limit of a determinant lim f (x) lim g(x) lim h(x) f (x) g(x) h(x) x a xa x a Let (x)  l (x) m(x) n(x) , then lim (x)  lim l(x) lim m(x) lim n(x) , xa x a x a x a u(x) v(x) w(x) lim u(x) lim v(x) lim w(x) x a

x a

x a

provided each of nine limiting values exist finitely. (x) Differentiation of a determinant

f (x) Let

g(x)

(x)  l(x) m(x) n(x) , u(x) v(x) w(x) f (x) g(x)

then (x)  l (x) u(x)

33

h(x)

h(x)

f (x) g(x) h(x) f (x) g(x) h(x) m(x) n(x)  l (x) m(x) n (x)  l (x) m(x) n(x) v(x) w(x) u(x) v(x) w(x) u (x) v(x) w (x)

IIT - MATHS - SET - III (xi) Integration of a Determinant

f (x) g(x) h(x) Let (x)  a l

b m

c n

b

, where a, b, c, l, m and n are constants,

b

b

 f (x)dx  g(x)dx  h(x) dx a

b

then

 (x) dx 

a

a l

a

a

b m

c n

b

Note that if more than one row (column) of  (x) are variable, then in order to find   (x) dx a first we evaluate the determinant (x) by using the properties of determinants and then we integrate it .

SPECIAL DETERMINANTS (i)

Skew symmetric Determinant

A determinant of a skew symmetric matrix of odd order is zero.

0 e.g.,

c

b

b 0 c a

a 0 0

(iii) Circulant Determinant A determinant is called circulant if its rows (columns) are cyclic shifts of the first row (columms).

a e.g.,

(iv)

(v)

b c

b c a . It can be show that its value is – (a3 + b3 + c3 – 3abc) . c a b

1 a

1 b

1 c  (a  b) (b  c) (c  a)

a2

b2

c2

1

1

1

a a3

b b3

c  (a  b) (b  c) (c  a) (a  b  c) c3

34


(iv)

1

1

2

2

a a3

b b3

1 c2 = (a – b) (b – c) (c – a) (ab + bc + ca) c3

INVERSE OF A SQUARE MATRIX Let A be any n–rowed square matrix. Then a matrix B, if exists, such that AB = BA = In, is called the inverse of A. Inverse of A is usually denoted by A–1 (if exists). We have |A| In = A(adjA)  |A| A–1 = (adjA). Thus the necessary and sufficient condition for a square matrix A to possess the inverse is that |A|  0 and then A–1 =

Adj(A) A square matrix A is called non|A|

singular if |A|  0. Hence a square matrix A is invertible if and only if A is non-singular. Properties of Inverse of a Matrix (i)

Every invertible matrix possesses a unique inverse.

(ii) If A and B are invertible matrices of the same order, then AB is invertible and (AB)–1 = B–1 A–1. (iii) If A is an invertible square matrix, then AT is also invertible and (AT)–1 = (A–1)T. (iv) If A is a non-singular square matrix of order n, then |adjA| = |A|n–1 (v) If A and B are non-singular square matrices of the same order, t hen adj (AB) = (adj B) (adj A)

SYSTEM OF LINEAR SIMULTANEOUS EQUATIONS Consider the system of linear non-homogenoeus simultaneous equations in three unknowns x, y and z, given by a1x + b1y + c1z = d1, a2x + b2y + c2z = d2 and a3x + b3y + c3z = d3 ,

 a1 Let A   a 2  a 3

b2 b3

c1  x  d1     c 2  , X   y  , B   d 2  ,  z   d 3  c3 

a1

b1

c1

d1

b1

c1

Let | A | =   a 2 a3

b2 b3

c2 ,  x  d 2 c3 d3

b2 b3

c 2 , obtained on replacing first column of  by c3

B.

35

b1


a1

d1

c1

a1

b1

d1

Similalry let  y  a 2 a3

d2 d3

c 2 and  z  a 2 c3 a3

b2 b3

d2 . d3

It can be shown that AX = B, x    x , y.   y , z   z .

(i)

Determinant Method of Solution We have the following two cases :

Case I If   0, then the given system of equations has unique solution, given by x   x / , y   y /  and z   z / . Case II If   0, then two sub cases arise: (a) at least one of  x ,  y and  z is non-zero, say  x  0. Now in x.    x , L.H.S. is zero and R.H.S. is not equal to zero. Thus we have no value of x satisfying x.    x . Hence given system of equations has no solution. (b)  x   y   z  0. In the case the given equations are dependent. Delete one or two equation from the given system (as the case may be) to obtain independent equation(s). The remaining equation(s) may have no solution or infinitely many solution(s). For example in x + y + z = 3, 2x + 2y + 2z = 9, 3x + 3y + 3z = 12,    x   y   z  0 and hence equations are dependent (infact third equation is the sum of first two equations). Now after deleting the third equation we obtain independent equations x + y + z = 3, 2x + 2y + 2z = 9, which obviously have no solution (infact these are parallel planes) where as in x + y + z = 3, 2x – y + 3z = 4, 3x + 4z = 7,    x =  y   z  0 and hence equations are dependent (infact third equation is the sum of first two equations). Now after deleting any equation (say third) we obtain independent equations x + y + z = 3, 2x – y + 3z = 4, which have infinitely many solutions (infact these are non parallel planes) For let z =   R, then x 

7  4  2 and y  . Hence 3 3

we get infinitely many solutions.

(ii)

Matrix Method of Solution (a)   0, then A–1 exists and hence AX = B  A–1 (AX) = A–1 B  x = A–1 B and therefore unique values of x, y and z are obtained. (b) If   0, then from the matrix [A : B], known as augmented matrix (a matrix of order 36

MATRICES AND DETERMINANTS 3 × 4). Using row operations obtain a matrix from [A : B], whose last row corresponding to A is zero (which is possible as   0 ). If last entry of B in this matrix is non-zero, then the system has no solution else the given equations are dependent. Proceed further in the same way as in the case of determinant method of solution discussed earlier. Aliter of (ii (b)) : We have AX = B  ((adj A)A)X = (adj A)B  X = (adj A)B. If  = 0, then X = 03 × 1, zero matrix of order 3 × 1. Now if (adj A)B = 0, then the system AX = B has infinitily many solution, else no solution. Note : A system of equation is called consistent if it has a least one solution. If the system has no solution, then it is called inconsistent.

Illustration 8: Solve the system of equations x + 2y + 3z = 1 2x + 3y + 2z = 2 3x + 3y + 4z = 1 with the help of matrix inversion. Solution : The given system of equations in the matrix form can be written as

1 2 3  x  1  2 3 2   y    2        AX = B  3 3 4   z   1  1 2 3  x  1 where A   2 3 2  , X   y  and B   2  .  3 3 4   z   1  Now |A| = 1(12 – 6) – 2 (8 – 6) + 3(6 – 9) = 6 – 4 – 9 = –7  0. Hence the given system has unique solution.

 C11 C12 Let C be the matrix of cofactors of elements in |A|. then C  C 21 C 22  C31 C32 37

C13  C23  C33 


Here

3 2 6 3 4

;

C 23  

C12 

2 2  2 3 4

;

C31 

C13 

2 3 3 3 3

;

C32  

C11 

C 21   C 22 



2 3 1 ; 3 4

C33 

1 2 3 3 3

2 3 5 3 2 1 3 4 2 2

1 2  1 2 3

1 3 5 3 4

 6 2 3 C   1 5 3   5 4 1

 6 1 5  Adj A = C  2 5 4   3 3 1



 6 1 5   6 / 7 1/ 7 5 / 7  AdjA 1 A     2 5 4    2 / 7 5 / 7 4 / 7  |A| 7  3 3 1  3/ 7 3/ 7 1/ 7 



 6 / 7 1/ 7 5 / 7  A B   2 / 7 5 / 7 4 / 7   3/ 7 3 / 7 1/ 7 

1

1

1  2     1 

 x   3 / 7   y   8 / 7      ( A–1 B = X)  z   2 / 7 

 x = –3/7, y = 8/7, z = –2/7

SYSTEM OF LINEAR HOMOGENEOUS SIMULTANEOUS EQUATIONS

38

MATRICES AND DETERMINANTS Consider the system of linear homogeneous simultaneous equations in three unknowns x, y and z, given by a1x + b1y + c1z = 0, a2x + b2y + c2z = 0 and a3x + b3y + c3z = 0. In this case, system of equations is always consistent as x = y = z = 0 is always a solution. If the system has unique solution (the case when coefficient determinant  0), then x = y = z = 0 is the only solution (called trivial solution). However if the system has coefficient determinant = 0, then the system has infinitely many solutions. Hence in this case we get solutions other than trivial solution also and we say that we have non-trivial solutions.

39

IIT-MATHS-SET-III

ASSIGNMENT

40

MATRICES

WORKEDOUT ILLUATRATION ILLUSTRATION : 01 With 1 , 2 as cube roots of unity, inverse of which of the following matrices exists? 1   a)  2   

 2 b)  1

1  

 c)  2 

2   1

d) None of these.

Solution : Ans:(d)

1  2   0,  2 1

2  0 Hence inverse does not exist. 1

1   0, 2  

ILLUSTRATION : 02

2 3 4  5  3 8   , then trace of A is, If A =  9 2 16 a) 17

b) 25

c) 8

d) 15

Solution : Ans:(d) [sum of leading diagonal elements is called Trace of matrix] 2   3   16  15

ILLUSTRATION : 03 If A is an orthogonal matrix, then a) |A| =0

b) |A|=  1

c) |A| = 2

Solution : Ans:(b) A A'  I

 A A' | I |  | A| | A|  1 | A|  1

ILLUSTRATION : 04

41

| A' || A|

for any square matrix

d) None of these.

IIT-MATHS-SET-III

 2 2 4   1 3 4   is an idempotent matrix then x = If A =   1 2 x  a) -5

b) -1

c) -3

d) -4

Solution : Ans:(c) Here A2  A

2 16  4x   2 2 4   2  1 3 16  4x    1 3 4    4  x 8  2x 12  x 2   1 2 x  On comparing 16  4x  4  x  3

ILLUSTRATION : 05 If A is non-singular matrix, then Det  A1    1  a) Det  A2   

1

b) Det  A2 

1 c) Det    A

1 d) Det  A

Solution : Ans:(d) 1 1 1 1 det  AA   det  I   det  A det  A   1  det  A   det  A

ILLUSTRATION : 06 1 2 3 2 4 6   is equal to The rank of   3 6 9  a) 1

b) 2

c) 3

d) None of these.

Solution : Ans:(a) 1 2 3  A   2 4 6   3 6 9 

42

MATRICES

R21  2   1 2 3 0 0 0   p A  1 ~     R31  3  0 0 0 

ILLUSTRATION : 07 If A2  A  I  0,then A1  a) A  I

b) I  A

c) I  A

d) None of these

Solution : Ans:(c) 1 2 1 A2  A  I  0  A  A  A  I   A .0  A  A1 A  A1 I  0

  A  I   A 1  0

 A1  A  I

ILLUSTRATION : 08 8  6 If the matrix A=   2

6

2 7 4  is singular, then = 4  

a) 3

b) 4

c) 2

d) 5

Solution : Ans:(a) If matrix A is singular. Then |A| =0

8 | A|  6 2

6

2

7 4  0 8 7   16  6 6   8  2 24  14  0        4 

  3

ILLUSTRATION : 09 If A is a 3 x 3 matrix and det  3A  k det  A  , k  a) 9

b) 6

Solution : Ans:(d) det  3A  k det  A  33 det  A  k det  A  k  27

ILLUSTRATION : 10 43

c) 1

d) 27

IIT-MATHS-SET-III The equations 2x  y  5, x  3 y  5, x  2 y  0  have a) no solution

b) one solution

c) two solutions

d) infinity many solutions

Solution : Ans:(b) 2x  y  5

.........( i )

x  3y  5

.........( ii )

x2y  0

.........( iii )

Solving  1 &  2  , we get x  2 & y  1 Which is satisfied (3).

44

MATRICES

SECTION A ONE ANSWER TYPE QUESTIONS

1.

1 - 1 If A =   2

2

-1  2 then det adj adjA is 1 

2 -1

a) 17 1 2.

b) 17 2

1 2 If the matrix  3

3 4 5

a)-2

4.

5.

sin3    cos 3 d)   - cos3  - sin3

- sin3  - cos3 

b) A2 + B2

c) A2 + 2AB + B2

d) A + B

Let A be a matrix of order 3 and let D denote the value of determinant A. Then det. (-2A) b) -2D

c) 2D

d) 8D

1 0 The element in the first row and third column of the inverse of the matrix  0

2 3 The matrix  - 4 a) symmetric

45

d) -4

If A and B are two square matrices such that B = -A-1BA, then (A + B)2 is equal to

a) –2

8.

c) 2

sin   cos , then A3 = If A =  cos  - sin  cos 3  sin 3   cos 3 sin3  cos 3   a) b)  - sin3 3 3 cos3 c) - sin3   sin  cos   

a) -8D

7.

b)If A is a scalar matrix so is A-1 d) If |A| equals 2, so does |A-1|

  2 8  is singular then l is 10  b) 4

a) 0 6.

d) 17 4

If A is an invertible matrix, which of the following is not true ? a) If A is symmetric so is A-1 c)If A is a triangular matrix so is A-1

3.

c) 17 3

b) 0

-3 2 5

c) 1

2 1 0

-3  2  is : 1 

d) 7

4 - 5  is: 2  b) skew – symmetric c) non- singular

d) singular

IIT-MATHS-SET-III

9.

2 0 If A =  0

0 2 0

a) 5A

10.

0 0  then A5 = 2 b) 10A

c) 16A

d) 32A

 2  If A =  2  and |A3| = 125, then a is equal to   a) ± 1 b) ±3 c) ±4 d) ±5

11.

 0  1 0  If A =  1 1 , B  3 1 , then value of a for which A2 = B is     a) 1

12.

b) 2A 1

a  b

b) a = cos2q, b = sin2q

d) a non zero perfect square

d) null matrix

- b then : a 

c)a = sin2q, b = cos2q

c) | A | n 2 A

b) | A | n A

b) -2

d) none of the above

0 2 1

d) None of the above

2 1 1 1 = 0 is : 0 1

c) 3

d) -3

If D = diag. d 1 , d 2 , d 3 ,............d n  where d i  0 for all i  1,2,................n, then D-1 is equal to a) diag. d 11 d 21 ................d n1 

18.

c) unit matrix

1  The value of x for with [ 1 1 x ] 0 2

a) 2 17.

c) 1

If A is a square matrix of order n x n, then adj(adjA) is equal to : a) | A | n 1 A

16.

2) positive

- tan  1 tan  1 If I =    1  - tan 1  tan

a) a =1, b =1

15.

d) no real values

If A is a square matrix, then adj. AT – (adjA)T is equal to : a) 2 |A|

14.

c) I

Determinant of a skew symmetric matrix of odd order is a) zero

13.

b) –1

b) D

c) I n

d) 0

If A = diag. d 1 , d 2 , d 3 ...........d n  then A n is equal to : a) A 1

b) I n

c) diag.d1n , d 2n ,.............d nn 


46

MATRICES 19. If A, B are two square matrices such that AB = A and BA = B, then : a)A, B are idempotent c) only B is idempotent

20.

1 If A =  1

1 and n  N then An is equal to : 1

a) 2 n A 21.

b) 2 n1 A

c) nA

b) 1

c) 0

b) symmetric

cos x

If A = - sinx  a) sin x cos x

28.

b) skew symmetric

b) skew symmetric

b) AA1 sinx  and A (adjA) = k cosx 

c) a diagonal matrix


1 0 

d) A – A1

0 then the value of k is : 1 

c) 2

d) 3

Each diagonal element of a Hermitian matrix is

1 1 5 2 Let A =  - 2 - 1 a) scalar

47


c) A1A

b) 1

a) a real number c) a non-zero real number

29.

c) a diagonal matrix

Let A be a square matrix. Then which of the following is not a symmetric matrix : a)A + A1

27.

d) skew - symmetric

If A is symmetric matrix and n Î N, then An is a) symmetric

26.

c) I

The inverse of a symmetric matrix is a) symmetric

25.

b) A is a null matrix d) A is a triangular matrix

If A, B are symmetric matrices of the same order, then the matrix AB – BA is : a) 0

24.

d) no real values

If a matrix A is symmetric as well as skew symmetric, then : a) A is a diagonal matrix c) A is a unit matrix

23.

d) none of these

If the system of homogeneous equations x + ay = 0, az + y = 0 and ax + z = 0 has infinite number of solutions, then the value of ‘a’ is a) –1

22.

b) only A is idempotent d) none of the above

b) an imaginary number d) none of the above

3  6  then A is : - 3 b) diagonal

c) nilpotent

d) idempotent


 

If A = aij is a scalar matrix, then trace of A is : n

n

i

31.

b)

j

cost If Rt    - sint

37.

38.

c) R (s) + R( t)

b) A 2  B 2

b) non-singular


c) A 2  2 BA  B 2

d) A 2  AB  BA  B 2

c) symmetric

d) not defined

If A and B are square matrices of the same type, then :

 a1 a If in  2  a3

b1 b2 b3

b)A + B = A – B

c) A – B = B – A

d) AB = BA

c1  c 2  the co-factor of a r is Ar , then c1 A1  c 2 A2  c3 A3 is c 3  b) –D

d) D2

c) D

cos 2  cos 2  cossin   cossin      are two matrices such that the If A =  and B = sin 2  sin 2   cossin  cossin  product AB is a null matrix, hen a - b is :  a) 0 b) an odd multiple of c) multiple of p d) none of the above 2 cos If A =  - sin 

sin   1 n then lim n A is  cos  n

a) a zero matrix

b) an identity matrix

cos F    sin  Let 0

- sin 

 

a) F  1

39.

i

If A is a singular matrix, then adj A is :

a) 0

36.

n

d)  a ii

ij

A and B are square matrices of order n x n, then (A – B)2 is equal to

a) A + B = B + A

35.

a j

b) R( s – t)

a) singular 34.

c)

ij

sint  then R(s) R(t) equals : cost 

a) A 2  2 AB  B 2 33.

a i

a) R ( s + t ) 32.

n

n

a)  a ij

2 If  3

1  - 3 A 2 5

1 0


0 0 where a Î R. Then F  1 is equal to 1 

cos 0 b) F    2  1 = - 3  0

0 c)  - 1

c) F 2 


0 then the matrix A equals : 1 48

MATRICES 1 a)  1

40.

42.

2 1

4/7  1 b)    2/7 1/7 

1 2 If f  x  = x 2  4 x  5 then f  A , where A =   2 b) I

b) [0, 2]

1 c)  2

0 d)  1

2 then the matrix A is : - 7 

4 1 

2/7 1 d)    4/7 1/7 

x

x  x x  is singular is : x x    c) x  

2

d)0

2 2 equals. 1

1 2 c) – I

d) 2I

c) (2,4)

4) (2,3)

b) A = O and B = O d) A ¹ O, B ¹ O

b) one

c) two

d) infinite

Let a., b,c be positive real numbers. The following system of equations in x,y and z

x2 y2 z2 x2 y2 z2 x2 y2 z2    1 , 2  2  2  1 ,  2  2  2  1 has : a2 b2 c2 a b c a b c a) no solution c) infinitely many solutions

49

1 1

The number of solutions of the equations x 2  x3  1 ,  x1  2 x 3  2, x1  2x 2 =0 a) zero

46.

0 1

If the matrix AB = O and one of them is non singular then a) A = O or B = O c) it is not necessary that either A = O or B = O

45.

1 c)  1

sin 1  1 - sin 1 sin  The value of the determinant A, A =  lies in the interval - 1 - sin 1 

a) [0, 4] 44.

1 1

x      x The value of x for which the matrix  x  a) x  3 b) x   3

a) O

43.

1 b)  0

- 1 Let A be an invertible matrix and suppose that the inverse of 7A is  4 1 a)  4

41.

1 0

b) unique solution d) finitely many solutions

IIT-MATHS-SET-III

47.

0 - 5 The matrix  7

5

- 7 0 11  is known as - 11 0 

a) symmetric matrix c) upper triangular matrix 48.

If A5 = O such that A n  I for 1 £ n £ 4, then I  A1 equals : a) A4 c) I +A

49.

b) diagonal matrix d) skew symmetric matrix

b) A3 d) I  A  A 2  A 3  A 4

 cos  sin    An  n n  equals A  lim bij . Then lim  Let A =   sin  cos   Let A  bij2 x 2 .Define lim n  n  n  2x2    n  to  0 1 a) zero matrix b) unitary matrix c)   1 0  d) limit does not exist  





SECTION B ONE ANSWER TYPE QUESTIONS

1.

1 1 1 1 1 1  then If A =  1 1 1 a) A3 = 9A c) A + A = A2

2.

b) A3 = 27A d) A-1 does not exist

 1   For all values of l, the tank of the matrix A =  1  2 a) for l = 2, r (A) = 1 c) for l ¹ 2, -1, r(A) = 3

4

5

 8 8  6  9  4 2  21

b) for l = -1, r(A) = 2 d) none of these

50

MATRICES

KEY 1

2

3

4

5

6

7

8

9

10

11

b

b

b

c

b

a

b

c

d

b

d

16

17

20

21

24

25

26

27 28

b

a

b

a

a

a

d

b

31

32

35

36

39

40

41

42 43

d

d

d

b

a

a

b

46

47

48 49

b

d

d

18 19

c

a

33 34

a

a

22 23

b

d

37 38

a

b

a

KEY

1

a,c,d

51

2

a,b,c

12 13

a

a

d

d

c

14

15

b

c

29

30

a

a

44

45

a

d


3

QUADRATIC EQUATION

52

QUADRATIC EQUATION

Basic concepts An equation of the form ax2 + bx + c = 0, where a  0 and a, b, c are real numbers, is called a quadratic equation over reals. The numbers a, b and c are called the coefficients of the quadratic equation. A root of the quadratic equation is a number  (real or imaginary) such that a 2 + b + c = 0. The roots of the quadratic equation are given by  b  b 2  4ac x 2a

The quantity D (= b2 – 4ac) is known as the discriminant of the equation.

Basic Results (i)

The quadratic equation has real and equal roots if and only if D = 0

(ii) The quadratic equation has real and distinct roots if and only if D > 0 (iii) The quadratic equation has complex roots with non-zero imaginary parts if and only if D < 0. (iv) If p + iq (p and q being real) is a root of the quadratic equation, where i   1 , then p –iq is also a root of the quadratic equation. (v) If p  q is an irrational root of the quadratic equation, then p – q is also a root of the quadratic equation provided that all the coefficients are rational, q not being a perfect square. (vi) The quadratic equation has rational roots if D is a perfect square of a rational number and a, b, c are rationals. (vii) If a = 1 and b, c are integers and the roots of the quadratic equation are rational, then the roots must be integers. (viii) If the quadratic equation is satisfied by more than two distinct numbers (real or imaginary), then it becomes an identity i.e., a = b = c = 0 (ix) Let  and  be two roots of a given quadratic equation. Then  +  =

b c and   . a a

(x) A quadratic equation, whose roots are  and  can be written as (x –  ) (x –  ) = 0 i.e., ax2 + bx + c  a(x -  ) (x –  ).

METHODS OF INTERVALS (wavy curve method) In order to solve inequalities of the form Px  P x   0,  0 , where P(x) and Q(x) are polynomials, we use the following method: Qx  Qx 

53

IIT - MATHS - SET - III If x1 and x2 (x1 < x2) are two consecutive distinct roots of a polynomial equation, then within this interval the polynomial itself takes on values having the same sign. Now find all the roots of the polynomial equations P(x) = 0 and Q(x) = 0. Ignore the common roots and write

x   1 x   2 x   3 . . . . . x   n  Px   f x   x  1 x   2 x   3  . . . . . x   m  , Qx  Where a1, a2, . . . . . an, b1, b2, . . . . . , bm are distinct real numbers. Then f(x) = 0 for x = a1, a2, . . . . . , an and f(x) is not defined for x = b1, b2, . . . . . , bm. Apart from these (m + n) real numbers f(x) is either positive or negative. Now arrange a1, a2, . . . . . , an, b1, b2, . . . . . , bm in an increasing order say c1, c2, c3, c4, c5, . . . . . , cm+n. Plot them on the real line. Draw a curve starting from right of cm+n along the real line which alternately changes its position at these points. This curve is known as the wavy curve.

The intervals in which the curve is above the real line will be the intervals for which f(x) is positive and the intervals in which the curve is below the real line will be the intervals in which f(x) is negative.

QUADRATIC EXPRESSION The expression ax2 + bx + c is said to be a real quadratic expression in x where a, b, c are real and a  0, Let f(x) = ax2 + bx + c, where a, b, c  R (a  0). 2 2   b  4ac  b 2  b  D  f(x) can be re-written as f(x) = a  x  2a   4a 2   a  x  2a  – 4a 2  , where      

D = b2 – 4ac is the discriminant of the quadratic expression. Therefore y = f(x) represents a parabola whose axis is parallel to the y-axis, with vertex at A  b D  , .  2a 4a 

Note that if a > 0, the parabola will be concave upwards and if a < 0 the parabola will be concave downwards and it depends on the sign of b2 - 4ac that the parabola cuts the x-axis at two points 2 2 2 (b – 4ac > 0), touches the x-axis (b –4ac = 0) or never intersects with the x-axis (b - 4ac < 0). This gives rise to the following cases: (i) a > 0 and b2–4ac < 0 54

QUADRATIC EQUATION y = f(x)

 f(x) > 0  x R x

x

–b/2a

In this case the parabola always remains concave upwards and above the x-axis (ii) a > 0 and b2– 4ac = 0 y = f(x)

 f(x)  0  x  R x

x

–b/2a

In this case the parabola touches the x-axis and remains concave upwards. (iii) a > 0 and b2 – 4ac > 0. Let f(x) = 0 has two real roots  and  (  <  ). Then f(x) > 0  x  (–  ,  )  (  ,  ), y = f(x) –b/2a

f(x) < 0  x  (  ,  ) and f(x) = 0 for x {, } .



x

x

In this case the parabola cuts the x-axis at two points  and  and remains concave upwards. –b/2a x

x

(iv) a < 0 and b2 – 4ac < 0

y = f(x)

 f(x) < 0  x  R.

In this case the parabola remains concave downwards and always below the x-axis. (v) a < 0 and b2 – 4ac = 0 – b/2a

x

x

 f(x)  0  x  R. y = f(x)

55


In this case the parabola touches the x-axis and remains concave downwards. (vi) a < 0 and b2 – 4ac > 0 Let f(x) = 0 have two real roots  and  (  <  ). Then f(x) < 0  x  (–  ,  )  (  ,  ), 

f(x) > 0  x  (  ,  ) and f(x) = 0 for x {, } .

x

x

–b/2a

y = f(x)

In this case the parabola cuts the x-axis at two point  and  and remains concave downwards. Notes: (i) if a > 0, then minima of f(x) occurs at x = –b/2a and if a < 0, then maxima of f(x) occurs at x = –b/2a (ii) If f(x) = 0 has two distinct real roots, then a.f(d) < 0 if and only if d lies between the roots and a.f(d) > 0 if and only if d lies outside the roots.

4.

INTERVALS OF ROOTS In some problems we want the roots of the equation ax2 + bx + c = 0 to lie in a given interval. b a

c a

2 For this we impose conditions on a, b and c. Since a  0, we can take f(x) = x  x  .

(i) Both the roots are positive i.e., they lie in (0,  ) if and only if roots are real, the sum of the roots as well as the product of the roots is positive. 

+  =

and  =

b  0 a c > 0 with b2 – 4ac  0. a

Similarly, both the roots are negative i.e., they lie in (–  , 0) if and only if roots are real, the sum of the roots is negative and the product of the roots is positive. i.e.,   (ii)

b 0 a

and  =

c > 0 with b2 – 4ac  0. a

Both the roots are greater than a given number k if and only if the following three conditions are satisfied: b

 k and f(k) > 0. D  0, 2a 56

QUADRATIC EQUATION

–b/2a

x

k –b/2a

x

x

x

k

(iii) Both the roots are less than a given number k if and only if the following conditions are satisfied: b

D  0,  < k and f(k) > 0. 2a (iv) Both the roots lie in a given interval (k1, k2), if and only if the following conditions are satisfied: b

D  0, k1 < < k2 and f(k1) > 0, f(k2) > 0. 2a

–b/2a

x

k1 –b/2a

x

k2

x

k1

k2

x

(v) Exactly one of the roots lies in a given interval (k1, k2) if and only if f(k1). f(k2) < 0.

k2

x

k1 x

k1

x

k2

x

(vi) A given number k lies between the roots if and only if f(k) < 0.

k

x

x

In particular, the roots of the equation will be of opposite signs if and only if 0 lies between the roots  f(0) < 0.

QUADRATIC INEQUATIONS Let f(x) = ax2 + bx + c be a quadratic expression. Then inequations of the type f(x)  0 or f(x)  0 are known as quadratic inequations. The study of these can be easily done by taking the corresponding quadratic expression and by applying the basic results of quadratic expression.

CONDITION FOR COMMON ROOT(S) Let ax2 + bx + c = 0 and dx2 + ex + f = 0 have a common root  (say). Then 2 a  + b + c = 0 and d   e  f  0

57

IIT - MATHS - SET - III Solving for  2 and  , we get

2  1   bf  ce dc  af ae  bd

bf  ce dc  af and    (dc – af)2 = (bf – ce) (ae–bd) ae  bd ae  bd

2 i.e.,  

which is the required condition for the two equations to have a common root. a b c Note: Condition for both the roots to be common is   d e f

THEORY OF POLYNOMIAL EQUATIONS Consider the equation anxn + an–1 xn–1 + an–2xn–2 .... + a1x + a0 = 0

... (1)

(a0, a1 , .... , an are real coefficients and an  0) Let 1 , 2 , .......,  n be the roots of equation (1), Then anxn + an–1 xn–1 + an–2xn–2 + ... + a1x + aa  an ( x  1 ) (x–  2 ) ... (x –  n )

... (2)

Comparing the coefficients of like powers of x, we get a  1 = 1   2   3  .....   n  n 1 an

 1 2  1 2  1 3  1 4  ...   2  3  ...   n 1 n 

a n2 an

............... 1 2 ..... r  ...   n r 1 n r  2 ...  n  (1) r

1 2 ..... n  (1) n

In general

  . 1

2

a nr an

a0 an ...  i  (1)i

a n i an

e.g., If  ,  ,  and  are the roots of ax4 + bx3 + cx2 + dx + e = 0, then     b /a             c / a          d / a

 

e a 58

QUADRATIC EQUATION Remarks 

A polynomial equation of degree n has n roots (real or imaginary).

 If all the coefficients are real then the imaginary roots occur in pairs i.e., number of imaginary roots is always even.  If the degree of a polynomial equation is odd then the number of real roots will also be odd. It follows that at least one of the roots will be real. 

If  is a repeated root, repeating r times of a polynomial equation f(x) = 0 of degree n

i.e., f(x) = ( x   ) r g ( x ) , where g(x) is a polynomial of degree n–r and g ( )  0, then f(  ) = f ( ) = f ( ) = ... = f (r–1) ( ) = 0 and f r ( )  0 and vice versa. Thus polynomial in x of degree n can be factorized into a product of linear/quadratic form.



Remainder Theorem

If we divide a polynomial p(x) in x by (x –  ), then remainder obtained is p(  ). Note that if p(  )  0, then x   is a factor of p(x).

 If a polynomial of degree n has n + 1 roots say x1, x2, ... xn+1 , xi  xj if i  j, then the polynomial is identically zero. i.e., p(x)  0. (In other words, the coefficients a0, ... an are all zero).   2   2   2 < 0, which is not possible if all  ,  and  are reals. So atleast one root i s non-real. As imaginary roots occurs in pair, given cubic equation has two non-real roots and one real root.

59

IIT-MATHS-SET-III

3-A

QUADRATIC EQUATIONS

ASSIGNMENT

60

QUADRATIC EQUATIONS

WORKEDOUT ILLUATRATION ILLUSTRATION : 01

If a and b  0 are the roots of the equation x 2  ax  b  0 , then the least value of x 2  ax  b x  R  is 9 9 1 1 a) b)  c)  d) 4 4 4 4

Ans: (b) Solution : Since a and b are the roots of the equation. x 2  ax  b  0

Therefore, a  b  a and ab = b

 b  0 

Now, ab  b   a  1 b  0  a  1 Putting a=1 in a+b =-a, we get b=-2 Since y  x 2  ax  b is a parabola opening upward. So, ymin  

D 4

[ Using : ymin  

D for y  ax 2  bx  c ] 4a

 a 2  4b  9 =  4    4  

ILLUSTRATION : 02 If , are the roots of the equation ax 2  bx  c  0 , then the value of the determinant. 1 cos      cos  cos     1 cos  cos 

cos 

a) sin    

is

1

c) 1  cos    

b) sin  sin 

d) None of these

Ans: (d) Solution :

We have

1 cos      cos  cos     1 cos  cos 

cos 

1

=

cos 

sin 

0 cos 

sin 

0

cos  1

sin  0

0 cos  0 1

sin  0

0 0

= (0) (0) = 0 for all values of , 61

IIT-MATHS-SET-III

ILLUSTRATION : 03 Let a  cos

2 2  i sin ,   a  a 2  a 4 &   a 3  a 5  a6 . Then the equation whose roots are , is 7 7

a) x 2  x  2  0

b) x 2  x  2  0

c) x 2  x  2  0

d) x 2  x  2  0

Ans: (d) Solution : We have : a  cos

2 2  i sin , 7 7 7

2 2    i sin  a   cos   cos 2  i sin 2  1  0i  1 7 7   7

Now  + = a  a 2  a 3  a 4  a 5  a6

 1  a6  a  a7 a  1 a    1 a  1 a 1 a

 1      1

 a7  1

and  .    a  a 2  a 4  a 3  a 5  a 6   a 4 1  a  a 3  1  a 2  a 3  = a 4 1  a 2  a 3  a  a 3  a 4  a 3  a 5  a6  = a 4  1  a  a 2  3a 3  a 4  a 5  a 6  = a 4  a 5  a6  3a7  a 8  a 9  a10 = 3  a  a 2  a 3  a 4  a 5  a6  a7  1  a 8  a7 a  a, a 9  a7 a 2  a 2 and a 10  a7 a 3  a 3 

 1  a6 3  a  =  1 a

 a  a7 a 1  3  3  1 a 1 a 

= 3  1  2 So, the required equation is

x 2      x    0

or

x2  x  2  0

ILLUSTRATION : 04 Let , be the roots of ax 2  bx  c  0 ;  , be the roots of px 2  qx  r  0 ; and D1 ,D2 the respective discriminants of these equations. If  ,  , and  are in A.P. then D1 : D2 =

a2 a) 2 b

a2 b) 2 p

b2 c) 2 q

d)

c2 r2

62

QUADRATIC EQUATIONS

Ans: (b) Solution : We have    

b c , .  , a a

  

q r and   p p.

Now  ,  , , are in AP         2

2

2

2

                  4       4

b 2 4c q 2 4r b 2  4ac q 2  4rp  2    2 a a p p a2 p2

D1 D2  2  2 a p

D1 a 2   D2 p 2

ILLUSTRATION : 05 If every pair from among the equations x 2  px  qr  0, x 2  qx  rp  0 and

x 2  rx  pq  0 has a

common root, then the sum of the three common roots is a) 2  p  q  r 

c)   p  q  r 

b) p  q  r

d) pqr

Ans: (b) Solution : The given equations are x 2  px  qr  0;

x 2  qx  rp  0

and

x 2  rx  pq  0

Let , be the roots of  i  :  , be the roots of (ii) and  , be the roots of (iii). Since , is a common root of  i  and (ii).   2  p   qr  0 and  2  q  rp  0   p  q   r q  p  0    r Now, , = qr

 r = r = qr

 r=q

Since , and  are roots of (ii). Therefore,   = rp  r  = rp   = p  ++  = q+r+p = p+q+r.. Note: ++  can also be equal to -1/2 (p+q+r) and 0.

ILLUSTRATION : 06 If a  Z and the equation  x  a  x  10   1  0 has integral roots, then the values of a are a) 10,8

Ans: (c) 63

b) 12,10

c) 12,8

d) None of these.

IIT-MATHS-SET-III

Solution : Since a and x are integers. Therefore,  x  a  x  10   1  0 

 x  a  x  10   1



x a  1



x  9



a=8 or a=12.

and x  10  1 or

 x  a  1

and x  10  1

and a  8  or  x  11 and a  12 

ILLUSTRATION : 07 If the equation

 3x 

2

1     27  3 p  15  x  4  0 has equal roots , then p=    

a) 0

b) 2

c) -1/2

d) None of these.

Ans: (d) Solution : The given equation will have equal roots iff Disc = 0 2



 27  3



27  31/ p  27 or 3  31/ p  1 or 31/ p 

1/ p

 15   144  0  27  31/ p  15  12

1 1   0 or  2 9 p

But 1/p cannot be zero. So, p = -1/2.

ILLUSTRATION : 08 The number of solutions of the equation 5 x  5  x  log10 25, x  R is a) 0

b) 1

c) 2

d) infinitely many

Ans: (a) Solution : LHS of the given expression, being the sum of a number and its reciprocal, is greater than or equalto 2 whereas RHS is less than 2. So, the given equation has no solution.

ILLUSTRATION : 09 The integer k for which the inequality x 2  2  4k  1 x  15k 2  2k  7  0 is valid for any x, is a) 2

b) 3

c) 4

d) none of these

Ans: (b)

64

QUADRATIC EQUATIONS

Solution : Let f  x   x 2  2  4k  1 x  15k 2  2k  7  0 . Then,  coeff . of x 2  0 

f  x   0  Disc  0 2

 4  4k  1  4 15k 2  2k  7   0  k 2  6k  8  0  2  k  4.

ILLUSTRATION : 10 The condition that x 3  px 2  qx  r  0 may have two of its roots equal to each other but of opposite signs is b) r  2 p 3  pq

a) r = pq

c) r  p 2 q

d) None of these.

Ans: (a) Solution : Let ,,  be the roots of the given equation such that =-. Then.  +  +  = p   = p. Since  is a root of the given equation, so it satisfies the equation i.e,

 3  p 2  q  r  0  p 3  p 3  pq  r  0  r  pq

ILLUSTRATION : 11 4

4

The number of real roots of  6  x    8  x   16 is a) 0

b) 2

c) 4

d) None of these.

Ans: (b)

Solution : Let y=7-x. Then the given equation becomes

 y  1

4

4

  y  1  16  y 4  6 y 2  7  0

  y 2  1 y 2  7   0  y 2  1  0

 y 2  7  0 

 y  1  7  x  1  x  6 ,8

ILLUSTRATION : 12 The value of a for which the equation  1  a 2  x 2  2ax  1  0 has roots belonging to (0,1) is

65

IIT-MATHS-SET-III a) a 

1 5 2

b) a  2

c)

1 5 a2 2

d) a  2

Ans:(b) Solution : Let f  x    1  a 2  x 2  2ax  1  0 then, f  x   0 has roots between 0 and 1 if. (ii)  1  a 2  f  0   0 and  1  a 2  f  1  0

(i) Disc  0

Now, Disc  0  4a 2  4  1  a 2   0 , which is always true.

1  a  f 0   0 2

  1  a 2   0

a 2  1  0  a  1 or a  1

 and

1  a  f 1  0  1  a  2a  a   0 2

2

2

a  a  1 a  1 a  2   0





a  1

or a  2 or 0  a  1

From (I) and (ii) , we get: a  1

or a  2 .

ILLUSTRATION : 13 If the product of the roots of the equation is 31, then the roots of the equation are real for k equal to a) 1

b) 2

c) 3

d) 4

Ans: (d) Solution : Produce of roots = 31 

2e 2log k  1  31



2k 2  1  31  2k 2  32  k 2  16  k  4.

But k> 0. Therefore, k=4. Now, Disc = 8k 2  8e 2log k  4  8k 2  8e 2log k  4  8k 2  4  0  k Hence, k = 4.

ILLUSTRATION : 14



The roots of the equation a  b



x 2 15



 a b



x 2  15

 2a, where a 2  b  1 , are

66

QUADRATIC EQUATIONS a) 2,  3

b) 4, 14

c) 3,  5

d)  6 ,  20

Ans: (b) Solution : We have, a  b 

 a  b  a  b   a



y

b 1  a b a b

a b

So, by putting a  b



x 2 15

2

 a 2  b  1

 y, the given equation becomes.

1  2a  y 2  2ay  1  0 y 2

 a 2  1  y  a   a2  1



 y  a



ya   b



y  a b  a b



2 x 2  15  1 or x  15  1  x  4, x   14

 a 2  1  b 





x 2  15

 a  b ,a  b

ILLUSTRATION :15 The value of

8  2 8  2 8  2 8 is

a) 10

b) 6

c) 8

d) 4.

Ans:(d) Solution : Let x =

8  2 8  2 8  2 8 . Then,

x  8  2x  x 2  8  2x  x 2  2x  8  0  x 4.

 x  0

ILLUSTRATION : 16







 



2 The harmonic mean of the roots of the equation 5  2 x  4  5 x  8  2 5  0 is

a) 2

b) 4

Ans: (b) Solution : Let , be the roots of the given equtaion. Then, 67

c) 7

d) 8

IIT-MATHS-SET-III

  

4 5 82 5 and   5 2 5 2

Let H be the H.M. of a and b. Then. H

2 16  4 5  4  4 5

ILLUSTRATION : 17 In a traingle PQR, R   / 2 . If tan (P/2) and tan (Q/2) are the roots of the equation ax 2  bx  c  0

 a  0  . then a) a+b=c

b) b+c=0

c) a+c=b

d) b=c

Ans: (a) Solution :  P Q  R   / 2  P  Q  2  2  2  4

 tan P / 2  tan Q / 2 P Q  tan  2  2   tan 4  1  tan P / 2tan Q / 2  

b a  c 1 a 

P Q     tan 2 tan 2     1   b / a and   P Q c tan tan   2 2 a  c b  a a



1



a  c  b



ab  c

ILLUSTRATION : 18 If the roots of the equation x 2  2ax  a 2  a  3  0 are real and less than 3, then a) a<2

b) 2  a  3

c) 3  a  4

d) a  4

Ans: (a) Solution : Let f  x  x 2  2ax  a 2  a  3  0 . Since f  x  has real roots both less than 3. Therefore, Disc >0 and f  3   0

68

QUADRATIC EQUATIONS 

a 2   a 2  a  3   0 and



a  3 and



a  3 and a  2 or a  3



a  2.

a 2  5a  6  0

 a  2   a  3  0

SECTION A - SINGLE ANSWER TYPE QUESTIONS 1.

If  and  are the roots of x 2  x  1  0 then 2 13  3 13 and 3 31  2  31 are the roots of : a) x 2  x  1  0

2.

r  12 r

c) b2/ac

b) b 2  ac

b) 5/2

d) non-negative

 d) b 2  4ac

1 is : 4x  2x  1 c) 13/14 2

d) None of these

b) 2

c) 3

d) 4

b) /3

c) /2

d) /6

b) d/a

c) a/d

d) None of these





The values of a for which the quadratic equation 2 x 2  a 3  8a  1 x  a 2  4a  0 posses roots of opposite signs are given by a) a > 0

69

c) zero

If x 2  x  1 is factor of ax 3  bx 2  cx  d , then the real root of ax 3  bx 2  cx  d  0 is a) -d/a

9.

b) non-positive

If the sum of the squares of the roots of the equation x 2  sin   2x  1  sin    0 is least then  a) /4

8.

d) (1,3)

If the product of the roots of the equation x 2  2 2kx  2e 2 log k  1  0 is 31, then the roots of the equation are real for k equal to a) 1

7.

b)  ,1  2,3

The greatest value of the expression a) 4/3

6.

x2  4 is : x 1 c)  ,1

If r is the ratio of the roots of the equation ax 2  bx  c  0, then a) 1

5.

d) x 2  5 x  7  0

If x, y and z are real numbers, then x 2  4 y 2  9 z 2  6 yz  3 zx  2 xy is always a) Positive

4.

c) x 2  5 x  7  0

The set of values of x which satisfy 5 x  2  3 x  8 and a) 2,3

3.

b) x 2  5 x  7  0

b) a > 5

c) 4 < a < 8.5

d) 0 < a < 4

IIT-MATHS-SET-III 3

10.

The number of positive integral solutions of a) 4

11.

b) 3

12.

13.

b) 4 –5

A solution of the equation

d) 1

x 2  14 x  9 are x 2  2x  3 c) 0, -¥

x  a x  b   x  c x  d  xab

xcd

d) ¥ - ¥

is

c) acb  d   bd a  c 


The integer k for which the inequality x 2  24k  1x  15k 2  2k  7  0 is valid for any x is b) 3

c) 4

d) None of these

If the quadratic expression x 2  2 ax  3a  10  0 ,  x  R then b) |a+ < 5

c) - 5 < a < 2

If  and  be the roots of the equation then the value of a) 0

16.

 0 is

b) cd a  b   abc  d 

a) a > 5

15.

x  55 2 x  7 6

a) ab(c  d )  cd a  b 

a) 2 14.

4

c) 2

The maximum and minimum values of a) 3,1

x 2 3x  4  x  2

b) 1

 2  2  1  2  2   1  is  2  2  b  2  2   b

c) 2

If the roots of the equation ax 2  bx  c  0 be of the form

d) 2 < a < 3

d) 3

  1 2 and then the value of  a  b  c   1 

is : a) b 2  2ac 17.

  2

b)  =  

 2

d) 2b 2  ac

c)   

d)     

If a,b, p,q are non-zero numbers, the two equations 2a 2 x 2  2abx  b 2  0 and p 2 x 2  3 pqx  q 2  0, have a) no common root c) two common roots iff 3 pq  2ab

19.

c) 4b 2  2ac

Let a,b,c be real numbers, a  0. If  is a root of a 2 x 2  bx  c  0 ,  is root of a 2 x 2  bx  c  0 and 0     , then the equation a 2 x 2  2bx  2c  0 has a root  that always satisfies : a)  

18.

b) b 2  4ac

b) one common root iff 2a 2  b 2  p 2  q 2 d)two common roots iff 3qb  2ap

The values of a for which one root of the equation x 2   a  1 x  a 2  a  8  0 exceeds 2 and the 70

QUADRATIC EQUATIONS other is less than 2 are given by a) a >3 20.

b) 0 < a < 10 4

4

c) –2 < a < 3 4

The equation  3  x    2  x    5  2x  has : a) only real roots c) two real & two non-real roots

21.


Given that ax 2  bx  c  0 has no real root and a  b  c  0 , then : b) c >0

c) c <0

d) c 0

If a  b  c  0, then the roots of the equation 3ax 2  4bx  5c  0 are a) positive

24.

c) x   1,   0

b) x  0

a) c =0 23.

b) only non-real roots d) none of the above

x x2  | x |  The solution of is : x 1 | x  1| a) x  0

22.


b) negative

If the roots of the equation

c)real and distinct

d) complex

1 1 1   are equal in magnitude but opposite in sign, than their xa xb c

product is a) 25.

1   a  b2  2

b) 

b) is greater than 61

b) 12,10

Give that for all real x, the expression the expression

1 d) - ab 2

c) is less than 61

d) can not be estimated

c) 12,8

d) None of these

x 2  2x  4 lies between 1/3 and 3. The values between which x 2  2x  4

9.3 2 x  6.3 x  4 lies are: 9.32 x  6.3 x  4

a) 1/3 and 3 28.

1 ab 2

If a  Z and the equation  x  a  x  10   1  0 has integral roots, then the values of a are a) 10,8

27.

c)

If  is a root of the equation x11  x10  x 8  x7  x 5  x 4  x 2  x  20  0 then  12 = a) is equal to 61

26.

1 2  a  b2  2

b) –2 and 0

c) –1 and 1

d) 0 and 2

Let a,b be integers and f  x  be a polynomial with integral coefficients such that f  b   f  a   1 . Then the value of b-a is a) 1

71

b) –1

c) 1, -1



If the equations ax 2  2bx  3c  0 and 3x 2  8 x  15  0 have a common root, where a,b,c are the lengths of the sides of a  ABC , then sin 2 A  sin 2 B  sin 2 C is equal to a) 1

30.

b) 3/2

b) 4

b) 1

c) 2

d) None of these

b) [1,2]

c) [-1,-2]

d)[1, -2]

2

In the equation 4 x 2  2 2 x 3 +48 the value x will be : a) -3/2

34.

d) none of these

x  R :| x  2 | x  = a) [-1,2]

33.

c) 6

2x  3 6 x2  x  6  1  The number of real roots of the equation is x 1 x 1 a) 0

32.

d)2

The number of quadratic equations which are unchanged by squaring their roots is : a) 2

31.

c) 2

b) –2

c) –3

d) 1

If roots of t he equation x n  1  0 are 1, a1 ,a2 ,a3 ............an 1 , then the value of

1  a1 1  a2  1  a3  ........1  an1  will be b) n2

a) n 35.

b) (2,c)

c) (2,9)

d) (1,3)

The value of m for which the equation x 3  mx 2  3x  2  0 has two roots equal in magnitude but opposite in sign is a) 1/2

37.

d) 0

If 9 x  4 x 3 x  2  35  0 then the solution pair is a) (1,2)

36.

c) nn

b) 2/3

If a 

c) 3/4

d) 4/5

cos 2 i sin 2  .Then quadratic equation whose roots are   a  a 2  a 4 and 7 7

  a 3  a 5  a6 is a) x 2  x  2  0

38.

If x  a) 0

39.

b) x 2  x  2  0

c) x 2  x  2  0

d) x 2  x  2  0

1  3 1  2 1   1  5 , then  x  3   5  x  2    x   = x   x   x x 

b) 5

The roots of the equation x

x

c) –5

d) 10

 x x are :

72

QUADRATIC EQUATIONS a) 0 and 4 b) 0 and 1 40.

1  y 1 3

d) 2q

1 1 b)   y  or 3  y   c)  y   3 3 2

2

d)   y  1 2

b) 1

c) 0

d) 3

The number of solutions of the equation sin  e x   5 x  5  x is : a) 0

44.

c) q

The number of real roots of the equation  x  1   x  2    x  3   0 is a) 2

43.

b)1

If y = f(x) = tanx cot3x, then a)

42.

d) 1 and 4

If ,are the roots of the equation x 2  px  p 2  q  0 then the value of 2     2  q is equal to a) 0

41.

c) 0, 1 and 4

b) 1

c) 2

d) infinitely many

The complete set of values of k for which the quadratic equation x 2  kx  k  2  0 has equal roots is given by a) 2  12

45.

b) 2  12 ,2  12 2

2

d) 2  12 2

If a  R and a1 ,a2 ,a3 ...an  R then  x  a1    x  a2   ....   x  an  assumes its least value at x = a) a1  a2  a3  .......  an c) n(a1+a2+a3+...+an)

46.

c) 2  12

b) 2  a1  a2  a3  .....  an  d) None of the above

The ratio of the roots of the equation ax 2  bx  c  0 is same as the ratio of the roots of the equation px 2  qx  r  0. If D1 and D2 are discriminants of ax 2  bx  c  0 and px 2  qx  r  0 respectively, then D1 : D2  a) a2/p2

47.

2 Which statement is true for the equation x 

a) two real roots 48.

x2

 x  1

2


 3?

b) two imaginary roots c) all roots are real

d) nothing can be said

b)   0   |  |

c)     0

d)   0 |  | 

The number of positive integral roots of x 4  x 3  4x 2  x  1  0 are a) 0

73

c) c2/r2

If and      are the roots of the equation x 2  bx  c  0 where c < 0 < b, then a) 0 <   

49.

b) b2/q2

b) 1

c) 2

d) 4


If the roots of the equation x 2  2ax  a 2  a  3  0 are real and less than 3, then a) a<2

51.

b) 2  a  3

b) An 1  bAn  aAn 1

r 7  4 3 p

b)

The equation x 3 / 4log2 x 

2

p 7  4 3 r

 log2 x 

5 4

c) all p and r

b) exactly three real solutions d) non-real roots

b) complex roots

c) purely imaginary roots

d) only one root

If y  2  x   3  3  x  2   5, then [ x  y ] is ([x] denotes the integral part of x) a) 10

56.

d) no p and r

Let a,b and c be real numbers such that 4a  2b  c  0 and ab  0 Then the quadratic equation ax 2  bx  c  0 has a) real roots

55.

d) An 1  bAn  aAn 1

 2 has

a) only one real solution c) exactly one rational solution 54.

c) An 1  aAn  bAn 1

If p,q,r are positive and are in A.P. the roots of the quadratic equation px 2  qx  r  0 are real for : a)

53.

d)a >4

If and  are the roots of the equation x 2  ax  b  0 An   n   n and then which of the following is true ? a) An 1  aAn  bAn 1

52.

c) 3  a  4

b) 15

Let 1

a,b,c

 1  cos x  ax 8

c) 12

be 2

0

non-zero

 bx  c  dx 

2

real

 1  cos x  ax 8

0

d) none of these

2

number

such

that

 bx  c  dx, then the quadratic equation

ax 2  bx  c = 0 has :

a) no root in (0, 2) c) two roots in (0,2) 57.

The equation x  1 a) Unique

58.

b) at least one root in (0,2) d)two imaginary roots



log x1 3  2 x  x 2

  x  3 | x | has  

b) Two solutions

c) No solution

d) More than two

The value of a for which the equation  1  a 2  x 2  2ax  1  0 has roots belonging to (0,1) is a) a  2

b) a  2

c) a 

1 5 2

d)

1 5 a2 2

74

QUADRATIC EQUATIONS

SECTION A - MULTIPLE ANSWER TYPE QUESTIONS 1.

The equation x 1log3 x 2 9 / 2 log3 x 5]  3 3 has A) at least one real solution c) exactly one irrational solution

2.

b) exactly three real solutions d) complex roots

Let f(x) be a quadratic expression which is positive for all real x. If g(x) = f(x) - f’(x) + f”(x), then for any real x. A) g(x) >0

3.

b) a-3/4

b) [0, 1]

b) 4a3 + 3a

3

b) (-1 + i 3 ) / 2

75

d) -a - (1/2)

c) 2 +

3

d) (-1 -i 3 )/2

b) at least one root in (-1, 1) d) none of these

b) at least one root which is negative fraction d) at least one root which is an integer

b) || > 1

c) || < 1

d) none of these

The solution of the equation (6-x)4 + (8-x)4 = 16 are A) 2

11.

c) a - (1/2)

If 0 < a < b < c, and the roots a, b of the equation ax2 + bx + c = 0 are non real complex numbers, then A) || = ||

10.

d) (-¥, ¥)

If A, G and H are the Arithmetic mean, Geometric mean and Harmonic mean between two unequal positive integers. Then the equation Ax2 - |G| x - H = 0 has A) both roots are fractions c) exactly one positive root

9.

c) [1, 2]

If 2a+3b+6c = 0 (a, b, c Î R) then the quadratic equation ax2 + bx + c = 0 has A) at least one in [0, 1] c) at least one root in [0, 2]

8.

d) a-1

The roots of the equation (x2+1) = x(3x2+4x+3), are given by A) 2 -

7.

c) a-1/2

If a is one root of the equation 4x2 + 2x - 1 = 0, then its other root is given by A) 4a3 - 3a

6.

d) g(x) < 0

The real values of l for which the equation, 3x3 + x2 - 7x + l = 0 has two distinct real roots in [0, 1] lie in the interval(s) A) (-2, 0)

5.

c) g(x) £ 0

For a > 0, the roots of the equation logaxa + logx a2 + logx2 a3 = 0, are given by A) a-4/3

4.

b) g(x) ³ 0

b) 4

c) 6

d) 8

The solutions of the equation 1! + 2! + 3! + ..... + (x-1) ! + x! = k2 and k Î l are

IIT-MATHS-SET-III A) 0 12.

c) 2

d) 3

If a, b, c Î R and the equality ax2 - bx + c = 0 has complex roots which are reciprocal of each other then one has A) |b| £ |a|

13.

b) 1

b) |b| £ |c|

c) a = c

d) b £ a

If a, b, c are positive rational numbers such that a > b > c and the quadratic equation (a+b-2c) x2 + (b+c-2a) x+(c+a-2b) = 0 has a root in the interval (-1, 0) then A) b + c > a b) c + a < 2b c) both roots of the given equation are rational d) The equation ax2 + 2b + c = 0 has both negative real roots

14.

How many roots does the following equation possess ? A)

15.

c) 3

d) 4

The equation |x+1| |x-1| = a2 - 2a - 3 can have real solutions for x if a belongs to A) (-, - 1]  [3, ) c) [1- 5 , -1]  [3, 1 +

16.

b) 2

3|x| |2-|x|| = = 1

5]

b) [1- 5 , 1+ 5 ] d) none of these

If a, b, c are rational and no two of them are equal then the equations (b-c)x2 + (c-a) x+a-b = 0 and a(b-c)x2+b(c-a) x+c (a-b) = 0 A) have rational roots c) have exactly one root common

b) will be such that at least one has rational roots d) have at least one root common

76

QUADRATIC EQUATIONS

KEY 1

2

3

4

5

6

7

8

9

10

11

12 13

b

b

d

c

a

d

c

a

d

b

b

d

16

17

20

21

24

25

26

27 28

b

d

c

c

b

b

c

31

32

35

36

39

40

41

d

d

b

b

d

a

b

46

47

48 49

50

51

54

55

56

b

b

b

a

c

a

b

b

9

10

11

18 19

a

c

33 34

a

c

22 23

c

c

37 38

c

a

52 53

a

b

a

b

c

42 43

c

a

14

15

c

b

29

30

d

b

44

45

b

d

14

15

57 58

a

a

KEY 1

2

3

4

5

6

7

8

a,b,c a,b a,c a,b,d a,d a,b,c,da,b,c b,c a,b c,d

16

a,c

77

12 13

b,d a,b,cb,c,d b,c a,c


4

INDEFINITE INTEGRATION

78

INDEFINITE INTEGRALS d [F(x)]  f (x) . Then F(x) is called an integral dx of f(x). Symbolically, it is written as  f (x)dx  F(x) .

Let F(x) be a differentiable function of x such that

Constant of Integration: As the differential coefficient of a constant is zero. We have d d F(x)  f (x) = [F(x)  c]  f (x) dx dx



 f (x)dx  F(x)  c

This constant ‘c’ is called the constant of integration and can take any real value.

PROPERTIES OF INDEFINITE INTEGRATION (i)

 kf (x) dx  k  f (x) dx

(ii)

 [f (x)  f 1

2

(x)  f 3 (x)  .......  f n (x)]dx

=  f1 (x)dx   f 2 (x)dx   f 3 (x)dx .....   f n (x) dx

BASIC FORMULAE Antiderivatives or integrals of some of the widely used functions (integrands) are given below.

d  x n 1  n    x (n  – 1)  dx  n  1 

x n 1  c, n   1 n 1



d 1 (ln | x |)  dx x



 x dx  ln | x |  c



d x (e )  e x dx



 e dx  e

d x (a )  (a x ln a)  dx

79

n   x dx 

1

x

x

c

ax x a dx   c (a  0)   ln a



d (sin x)  cos x dx



 cos xdx  sin x  c



d (cos x)   sin x dx



 sin x dx  – cos x  c



d (tan x)  sec2 x dx



 sec

2

xdx  tan x  c

IIT - MATHS - SET - III 

d (cos ecx)  (  cot x cos ecx) dx



 cosecx cot xdx  – cosecx  c



d (sec x)  sec x tan x dx



 secx tan xdx  secx  c



d (cot x)  – cosec2 x dx



 cos ec xdx   cot x  c

d 1 1 x  dx (sin a )  2 a  x2





d x a (tan 1 )  2 dx a x  a2



x



|x|



d 1 (sec1 x)   dx | x | x 2 1

2

1

x dx  sin 1    c a a2  x2 dx 1 x  tan 1  c 2 a a a

2

1 2

dx  sec 1 (x)  c

x 1

cos x



 cot xdx   sin x dx  ln | sin x |  c



 tan xdx  

 sin x dx = – ln |cosx| + c or ln |secx | + c cos x



 sec xdx  

sec x(sec x  tan x)  x  dx = ln|secx + tanx| + c or ln tan     c sec x  tan x 2 4



 cos ecx dx  

x cos ec x(cot x  cos ecx) dx  ln (cot x  cos ecx)  c or ln tan  c 2 cot x  cos ecx

STANDARD FORMULAE dx











x



a

2

x a

2

dx 2

x a

2

 ln x  x 2  a 2  c

 ln x  x 2  a 2  c

dx 1 x a  ln c 2  a 2a x  a

2

2

dx 1 ax  ln c 2  x 2a a  x

80

INDEFINITE INTEGRALS 







x 2  a 2 dx 

x a2 x 2  a 2  ln x  x 2  a 2  c 2 2



x 2  a 2 dx 

x a2 x 2  a 2  ln x  x 2  a 2  c 2 2



a 2  x 2 dx 

x 2

a2  x2 

a2 x sin  1  c 2 a

METHODS OF INTEGRATION If the integrand is not a derivative of a known function, then the corresponding integral cannot be found directly. In order to find the integral of complex problems, generally three rules of integration are used.  Integration by substitution or by change of the independent variable.  Integration by parts.  Integration using partial fractions.

Integration By Substitution: There are following types of substitutions. 5.1.1 Direct Substitution: If integral is of the form  f (g(x))g(x) dx, then put g(x) = t, provided  f (t)dt exists.

Note: (i)

f (x)

 f (x) dx  ln | f (x) |  c

n (ii)  [f (x)] f (x) dx 

[f (x)]n 1  c, (n   1) n 1

1 (iii) if  f (x) dx  g(x), then  f (ax  b) dx  g(ax  b)  c a

Standard Substitutions

81



For terms of the form x2 + a2 or

x 2  a 2 , put x = a tan  or a cot 



For terms of the form x2 – a2 or

x 2  a 2 , put x = a sec  or acosec 



For terms of the form a2 – x2 or

a 2  x 2 , put x = asin  or a cos 



If both a  x , a  x are present, then put x = acos 

IIT - MATHS - SET - III 

For the type

(x  a) (b  x), put x = a cos2   bsin 2 



For the type



x2  a2  x



n





or x  x 2  a 2 , put the expression within the

bracket = t. 1



For the type  x  a 

1 1 n

(x  b)

1 1 n

1

1  x  b n or  ( n  N, n  1), put  2  x  a  (x  a)

xb t xa 

For

1 , n1 , n 2  N (and > 1), again put (x + a) = t(x + b). (x  a) (x  b) n2 n1

INDIRECT SUBSTITUTION If the integrand is of the form f(x).g(x), where g(x) is a function of the integral of f(x), then put integral of f(x) = t.

INTEGRATION BY PARTS When integrand involves more than one type of functions we may solve them using integration by parts. The rule, its derivation and uses are given below. Derivation of the result We know that

d dv du (uv)  u v dx dx dx

 d(uv) = udv + vdu 

 d(uv)   udv   vdu



 udv  uv –  vdu

If we take u as part 1 and dv as part 2, then the above result can be written as

 (part1) (part 2) = (part 1) [integral of part 2] –  [integral of part 2] [differential of part 1]

PROPER CHOICE OF FIRST AND SECOND FUNCTIONS Usually, the following order is useful in deciding the first part. The function on the left is taken as the first part and the function on right is taken as second part. Inverse, logarithmic, algebraic, trigonometric and exponents are usually taken as first part. The above rule to decide between first part and second part is known as ILATE rule.

82

INDEFINITE INTEGRALS Integrals of the Type

ax

e

sinbx,  eax cosbxdx

Let I   eax sin bx dx Integrate by parts taking eax as the first part, we get

I

eax cos bx   cos bx    aeax   dx b b  

On integrating the second term by parts again, we get 1 a 1 a  I   e ax cos bx   eax sin bx   e ax sin bx dx  b b b b 



1 ax a ax a2 I  – e cosbx  2 e sin bx  2 I b b b

 a 2  e ax   1  b 2  I  b 2 (a sinbx – bcosbx)    I

eax (a sinbx – bcosbx) + c a 2  b2

ax Similarly we can show that  e cos bx 

Integrals of the Type Let I =

e ax (a cos bx + bsinbx) + c a 2  b2

x

 e [f (x)  f (x)]dx

x

 e [f (x)  f (x)]dx

x x  I   e f (x) dx   e f (x) dx

Integrate the first integral on the RHS by parts taking ex as the second function, we get I = ex f(x) –

x

 e f (x) dx   e

x

f (x) dx

 I = ex f(x) Thus to evaluate the integrals of the type  e x g(x) dx , we first try to express g(x) as the sum of the function and its derivative i.e., g(x) = f(x) + f (x) and then we use the result derived above.

INTEGRATION BY PARTIAL FRACTIONS The integrals of rational functions can be evaluated by splitting them into partial fractions. An expression containing a polynomial in numerator and another polynomial in denominator is 83

IIT - MATHS - SET - III called a rational function. polynomial N(x) Rational Function f(x) =  polynomial D(x)

Case-I If the degree of numerator is less than the degree of denominator, we can split the rational functions into simpler fractions according to the factors of denominator. For example: Let f(x) =

x 2x  5x  3 2

x 1 3 = (x  1) (2x  3)  x  1  2x  3 1 3 and are known as partial fractions of f(x). x 1 2x  3

The integral of f(x) can now be evaluated as the sum of the integrals of its partial fractions.   f (x) dx = –log |x – 1| + 3/2 log |2x – 3| + C Case -II If the degree of N(x) is greater than or equal to the degree of D(x), we divide N(x) by D (x) so that the rational function

N(x) R(x) is expressed in the form Q(x) + , where Q(x) and D(x) D(x)

R(x) are polynomials such that the degree of R(x) is less than the degree of D(x). Now as Q(x) is a polynomial, it can be easily integrated and to integrate

R(x) we make use of partial D(x)

fractions as we have done in case-I. The resolution of

R(x) into partial fractions depends upon the nature of the factors of the D(x)

denominator D(x) as discussed below. When denominator D(x) is expressible as the product of non-repeating linear factors. Let D(x) = (x – a1) (x – a2) ..... (x – an). Then we assume that, R(x) A1 A2 An    .....  D(x) x  a1 x  a 2 x  an

where A1, A2 .... An are constants and can be determined by equating the numerator on RHS to the numerator on LHS and then substitute x = a1, x = a2, ..... , x = an respectively.

84

INDEFINITE INTEGRALS

x 1 A B C Let f(x) = (2x  1) (x  2) (x  3)  2x  1  x  2  x  3

A

 x 1 6   (x  2) (x  3)  x  1 35 2

B

 x 1 1   (2x  1) (x  3)  x  2 5

C

 x 1 2   (2x  1) (x  2)  x  3 7 6

dx

1

dx

2

dx

 f (x) dx  35  2x  1  5  x  2  7  x  3 = 

3 1 2 ln| 2x  1|  ln| x  2 |  ln| x  3 |  C 35 5 7

When denominator g(x) is expressible as the product of linear factors such that some of them are repeating. Let D(x) = (x – a)k (x – a1) (x – a2) .... (x – an). Then we assume that B B2 Bn A3 R(x) A1 A2 Ak  1   .....    .....  + 2 3 k x  a1 x  a 2 x  an D(x) x  a (x  a) (x  a) (x  a)

Now to determine constants we equate numerators on both sides. Then from the resulting identity we can find all constants by substituting the various values of x as we have done earlier. When some of the factors of denominator D(x) are quadratic (which can not be factorised further) but not repeating. Corresponding to each quadratic factor of the type ax2 + bx + c, we assume partial fractions Ax  B , where A and B are constants to be determined by comparing ax 2  bx  c coefficients of similar powers of x in the numerator of both sides. (The constants can also be

of the type

determined using methods which we have used earlier.

EVALUATION OF INTEGRALS OF VARIOUS TYPES Integrals of the type

85

1  ax 2  bx  c dx ,



1 ax 2  bx  c

dx,



ax 2  bx  c dx

IIT - MATHS - SET - III Integrals of the type (a)

 ax

px  q dx, (b)  bx  c

2



px  q 2

ax  bx  c

dx , (c)

 ax

2

P(x) dx  bx  c

where P(x) is a polynomial of degree greater than or equal to 2. To evaluate the integrals of type (a) and (b) we express the linear factor in the numerator as px + q = m (derivative of quadratic expression in the denominator) + n To integrate the integrals of type c, we divide the numerator by the denominator and express the integrand as Q(x) +

R(x) , where R(x) is a linear function of x. ax  bx  c 2

 ax

Therefore,

2

P(x) R(x) dx   Q(x) dx   2 dx  bx  c ax  bx  c

Now to evaluate the integral on RHS apply the method discussed in type (a). Integrals of the type

 a sin

2

1 1 dx,  dx, 2 x  b cos x a  b cos 2 x

1

 a  bsin

2

 a  b sin

x

2

dx, 

dx , (a sin x  bcos x)2

 a  bsin

2

1 dx x  ccos2 x

f (tan x) where f(tanx) is polynomial in tanx x  c cos 2 x  d sin cos x

Method of integration 

Divide the numerator and denominator by cos2 x



Put tanx = t  sec2x dx = dt, This substitution will convert the trigonometric Integrals

f (t) , where f(t) is a polynomial in t. This Integral At  Bt  C can be evaluated by the method which is discussed earlier.

into algebraic Integral in the form 

Integrals 1

 a  b cos x dx

of and

the

2

form

of

1

1

 a sin x  b cos x dx,  a  b sin x dx

1

 a sin x  b cos x  c dx

86

INDEFINITE INTEGRALS Method of integration Put sinx =

2 tan x / 2 1  tan 2 x / 2 and cos x  1  tan 2 x / 2 1  tan 2 x / 2

Replace 1 + tan2

x in the numerator by sec2x/2 2

Put tanx/2 = t 

1 x sec2 dx = dt x = dt 2 2

Now, the new integral will be in the form of 

f (t) dt, which can be evaluated by the At  Bt  C 2

methods discussed earlier.

Integrals of the form of



Px 2  qx  r ax 2  bx  c

dx

Method of integration Put Px2 + qx + r = A (ax2 + bx + c) + B

d (ax2 + bx + c) +  dx

Find A and B by comparing coefficients of like powers of x. After that we can proceed as earlier. =

6x 3 1  x 2  sin 1 x  C 2 2

INTEGRATION OF IRRATIONAL ALGEBRAIC FUNCTIONS 

Integral of the type:

 linear

dx linear

To evaluate these type of integrals we substitute linear = t. This substitution will reduce the integral to one of the known forms.



Integral of the type

 (ax

dx 2

 b) cx 2  d

1 To evaluate these type of integrals we substitute x = . This substitution will reduce the t integral to one of the known forms.

Evaluate 

87

dx (x 2  1) x 2  2

IIT - MATHS - SET - III Solution: Let I  

dx 2

(x  1) x 2  2

1 1 Substitute x   dx  2 dt t t

1 dt  tdt t2    1  I=  1 (1  t 2 ) 1  2t 2  2  1 2  2 t  t 

Let 1 + 2 t2 = z2  4t dt = 2z dz



1 zdz dz  2   tan 1 z  C  2 2  z 1  2 z 1 1   z 2  



I



I  tan 1 1  2t 2  C  tan 1 1 


 linear

2 C x2

dx quadratic

1 To evaluate these type of integrals we substitute Linear = . This substitution will reduce the t integral to one of the known forms. 


 f (x, (ax  b)

m1 / n1

, (ax  b) m 2 / n 2 ,....) dx

To evaluate this type of integral we transform it into an integral of rational function by substit u t i n g ax + b = tS, where s is the least common multiple of the numbers n1, n2, ... Integral of the type  x m (a  bx n ) p dx , where m, n and p are rational numbers. To evaluate this type of integral observe the following steps. (i) If p is an integer, the integral reduces to the integral of a rational function by means of the substitution x = tS, where s is least common multiple of the denominator of the fractions m and

88

INDEFINITE INTEGRALS n. (ii) If (m + 1)/n is an integer, the integral can be rationalised by the substitution a + bxn = tS, where s is the denominator of the fraction p. (iii) If (m + 1)/n + p is an integer, substitute ax–n + b = tS, where s is the denominator of the fraction p.

89

AIEEE - CHEMISTRY VOL. - I

ASSIGNMENT

90

ATOMIC STRUCTURE

WORKEDOUT ILLUATRATION ILLUSTRATION : 01 If

3x  4

 x 3  2x  4 dx  log | x  2 | k log f x   C, then (a) K = -

1 2

(b) f x   x 2  2 x  2

(c) f x  | x 2  2x  2 |

(d) k 

1 4

Solution : Ans : (a), (b) and (c)





Since x 3  2x  4  x  2 x 2  2x  2 and the integrand is a proper rational function, we use the method of partial fractions to get 3x  4 A Bx  C   2 x  2x  4 x  2 x  2x  2 3







3x  4  A x 2  2x  2  Bx  C x  2

Putting x = 2, A = 1 and comparing the coefficients of x 2 , we get 0 = A+B, i.e. B = -1. Putting x = 0, we have 4 = 2A - 2C, i.e., C = -1. Thus 3x  4

1

x 1

 x 3  2 x  4   x  2 dx   x 2  2x  2 dx

1 2 = log | x  2 |  log | x  2x  2 |  C 2

1 Hence K =  , so that f x  | x 2  2x  2 | x 2  2x  2, because x 2  2x  2  0 2

ILLUSTRATION : 02 e x 1  sin x   1  cos x dx is equal to (a) log | tan x | C

x x (b) e tan   C 2

(c) sin e x cot x  C

Solution : Ans: (b)

  x  x  e x 1  2 sin   cos    2  2   e x  1  sin x  dx  2 x  1  cos x dx = 2 cos   2

91

x

x 1 x x 1 2 x 2 x x  sec  tan  dx   e sec dx   e tan dx 2 2 2 2 2 2

=

e

=

1 x x x 1 x e sec 2  e x tan   e x sec 2  C  2 2 2 2 2

[integrating by parts]

(d) e x cot x  C

AIEEE - CHEMISTRY VOL. - I x  e x tan  C. 2

ILLUSTRATION : 03

a

2

dx is equal to cos x  b 2 sin 2 x 2

 1  a (a) tan  tan x   C b 

(c)

1 b  tan 1  cot x   C ab a 

(b)

1 b  tan 1  tan x   C ab a 

 1  b (d) tan  tan x   C a 

Solution : Ans : (c) Putting t  tan x , we get

dx sec 2 xdx dt   a 2 cos 2 x  b2 sin2 x  a 2  b2 tan2 x   a 2  b2t 2

1 dt 1 b 1 b  b   2 . tan 1  t   C  tan 1  tan x   C 2  2 ab a  a  = b t 2   a 2  b a . b  

ILLUSTRATION : 04 If

dx

 5  4 cos x  K tan

1

x   M tan   C , then 2 

(a) K = 1

(b) K = 2/3

(c) M = 1/3

(d) M = 2/3

Solution : Ans : (b) and (c) x Since the integrand is a rational function of cosx, we put tan    t . Then 2

I 

= 2

dx  5  4 cos x 

2dt  4 1  t 2 1  t    5  1  t 2 

2

 

 2

dt 5 1  t   4 1  t 2  2

 

dt 2 t 2 x 1  tan 1  C  tan 1  tan   C t 9 3 3 3 2 3 2

Hence K =

2 1 and M = 3 3 92

ATOMIC STRUCTURE

ILLUSTRATION : 05 If

 cos ec2xdx  f  g  x    C , then (a) range g  x     ,  (c) g'  x   sec 2 x

(b) dom f  x     ,    { 0 } 1 (d) f '  x   for all x   0,  x

Solution : Ans : (a), (b), (c). 1 1 2 cos ec2x  cos ec2x  cot 2x  dx = log | cos ec2x  cot 2x | C 2 cos ec2x  cot 2x

 cos ec2dx  2  = Thus f  x  

1 1  cos 2x 1 log  C  log | tan x | C 2 sin 2x 2

1 log | x | and g  x   tan x. Now, range g(x) = ( - , ) and 2

dom f  x   ( - , ) - {0}. Moreover, g'  x   sec 2 x and f '  x  

1 for all x  (0, ). 2x

ILLUSTRATION : 06 The function f whose graph passes through the point (0, 7/3) and whose derivatives is x 1  x 2 is given by  1 2 3/ 2  (a) f  x       1  x   8   3

 1 2 3/ 2  (b) f  x      1  x   8   3

 1 1 (c) f  x       sin x  7   3

(d) None of these

Solution : Ans : (a) f  x    x 1  x 2 dx  C  

1  2x  1  x 2 dx  C  2

 1 2 3/ 2 =    1  x   C  3

So,

7 1   C 3 3

 C=

(Putting 1  x 2  t ) 8 3

 1 2 3/ 2  Therefore, f  x       1  x   8  . 3  

ILLUSTRATION : 07 If I   sec 2 x cos ec 4 xdx  K cot 3 x  L tan x  M cot x  C then (a) K = -1/3

93

(b)L =2

(c) M = -2

(d) none of these


Solution : Ans : (a), (c) Put tanx = t. 2 2

I   1  t

= 

2



1  t  t4

1  t 4  2t 2 dt 2 dt   4   dt   2 dt =  4 t t t

dt 1  t2

1 3 1  tan x  x  tan x  2  tan x   C 4

Therefore, K =

1 3 =  cot x  tan x  2 cot x  C 4

1 ,L  1,M  2 4

ILLUSTRATION : 08 If I =

sin x  sin 3 x  cos 2x dx  Acos x  B log | f  x  | C , then 1 1 2 cos x  1 ,B   , f  x  4 2 2 cos x  1

(b) A 

1 3 ,B   2 4 2

1 3 2 cos x  1 , f  x  (c) A   ,B  2 2 2 cos x  1

(d) A 

1 3 2 cos x  1 ,B   , f  x  2 4 2 2 cos x  1

(a) A 

Solution : Ans : (b), (d) Since the integrand is odd w.r.t sine, we put cosx = t, so cos2x = 2t2 -1 and dt = -sinxdx.

t2  2 1 2t 2  4 1 3 dt dt  dt   dt   2 Thus I =  2 2  2t  1 2 2t  1 2 2 2t  1 1 3 1 . log = 2t 2 2 2

2t  1 1 3  C  cos x  log 2 2t  1 4 2

So,

A

1 3 2 cos x  1 ,B   , f  x  2 4 2 2 cos x  1

or

A

3 1 , f  x  , B 2 4 2

2 cos x  1 C 2 cos x  1

2 cos x  1 2 cos x  1

ILLUSTRATION : 09



dx 2  3x  x 2

 fog  x   C , then

1 (a) f  x   sin x,g  x  

 2x  3 17

1 (b) f  x   tan x, g  x  

 2x  3 17 94

ATOMIC STRUCTURE 1 (c) f  x   sin x,g  x  

 2x  3

(d) none of these

17

Solution : Ans : (c) 2  3  17   x     We have = 2  3x  x    x  3x  2  =  2 4   2

17  3 x   4  2

Therefore, g  x  

3 2  C  sin 1 2x  3  C 17 17 2

x

dx

I 



2

2

 sin 1

 2x  3  17

and f  x   sin 1 x .

ILLUSTRATION : 10 cos 3 x  cos 5 x dx is The value of  sin 2 x  sin4 x (a) sin x  6 tan 1  sin x   C 1

(b) sin x  2  sin x   C 1

(c) sin x  2  sin x   6 tan 1  sin x   C 1

(d) sin x  2  sin x   5 tan 1  sin x   C

Solution : Ans : (c)

R  sin x,cos x  

cos 3 x  cos 5 x sin 2 x  sin4 x

 R  sin x,  cos x  

 cos 3 x  cos 5 x cos 3 x  cos 5 x   sin 2 x  sin 4 x sin 2 x  sin 4 x

Therefore, we substitute sinx = t, so that cosxdx= dt. 2 cos x  1  sin 2 x   1  sin 2 x   cos 3 x  cos 5 x   dx 2 4  sin 2 x  sin 4 x dx   sin x  sin x

1  t  2  t  dt  2



95

2

t t

2

4



2

  1  t

2



6 1  t2

  dt 

=  R  sin x,cos x 

AIEEE - CHEMISTRY VOL. - I 2 1 = t   6 tan t  C t 1

= sin x  2  sin x   6 tan 1  sin x   C

SECTION B ONE ANSWER TYPE QUESTIONS 1.



tan 3 2 x sec 2 x dx equals to 1 sec 3 2 x  3 sec 2x 6 d) None of these

a) sec32x+3sec2x

b)

c) sec32x-3sec2x 2.

 sin 2x  cos 2x dx  a) a =

1 2

5 , b R 4

b) a =

sin x  cos x sin x e cos xdx is equal to 1  sin 2 x



b) esinx-cosx+c

1  x 2



c) esinx+cosx+c

c) logex + 2tan-1x+k If f(0)=f1(0)=0 and f11(x)=tan2x then f(x) is a) log sec x 

6.



e log 11/ x

a)

7.

2

Let

1 2 x 2

1 2 x 2

c) log sec x 

1 2 x 2

d) None of these

dx is

1 tan  x   x 2 



b) logcosx +

b) log e x  tan 1 x  k d) None of these



x 2  1/ x 2

1

d) ecosx-sinx+c

dx is equal to

x  x3

a) logex + loge(1+x2)+k 5.

 5 , b ÎR 4

d) None of these

a) esinx + c

4.



sin 2x  a   b then

 c) a = , b  R 4

3.



1 

x1/ 2 1 x3

dx 

 x 2  1 1 1   log tan  x   c) b)   x 2  x 2 2   

1

1 

 x 2 1  tan 1   2 x 2  

1

d)

2 3 gof (x) + c then

96

ATOMIC STRUCTURE a) f(x) =

8.



b) f x   x 3 / 2

x



e tan

1



ex



b)

x  x2   1 x2 

x 1

a) x e tan

10.

1

x



b) ex tanx + k

1/2

1  2   n 1  x 2   3   C  

x  12

1 tan 1 x e c x

d) None of these

c)

1 x x e tan  k 2 2

x 2 d) e sec

3 /2

1    n 1  x 2   

ax 2  b x c x  ax  b  2

2

2

2

dx

2 +C 3 

3/2

x k 2

1  2   n  1  x 2   3   C  

d) none of these

is equal to

2 2  1  ax  ( b / x )     k sin b)  c  

d) none of these

If  f x dx  F( x) , then  x 3 f x 2 dx equals to a)

ex

c)

1 1  1  2  b) 3  x 

 1  ax  b / x   k c) cos  c  

97

d) 

x 2  1 ln x 2  1  2 ln x  dx is equal to x4

b   ax   xk sin  1  a)  C     

13.

ex x  1

1  sin x x .e dx is equal to 1  cos x

2 1  c)  1  2  3 x 



x  12

b) x 2 e tan 1 x  c

c

1 1  a)  1  2  3 x 

12.

c) 

  dx is equal to  

x x a) e tan   k 2

11.

d) g(x) = sin-1x

e x x  1  /x  1 3 dx equals to

ex a) x  1

9.

c) f x   x 2 / 3

1 2 x Fx 2    Fx 2 dx 2 2





b)

1 2  x Fx 2   Fx 2 dx 2

AIEEE - CHEMISTRY VOL. - I c) 14.

1 2 1  x Fx    Fx 2 dx  2 2 

d) None of these

Suppose f(x) = x  1 2 for x  1 . If g(x) is the function whose graph is the reflection of the graph of f(x) with respect to the line y = x , then  g x dx equals to a) 

2 3 /2 x  x  c, x  0 3

b) 

2 3/2 x  x  c , x  1 3 c)

15.



2 3/2 x  x  c, x  0 3 d)

1 x dx equals to 1 x

x

1 1 2 a) x / 2  1 1  x  sin x 2



c)

16.



1 1 2 b) x / 2  1 1  x  sin x 2



1  x   12 sin 2

1



x



d) None

1





x2 x4 1

3/ 4

dx is equal to

1/ 4

1   a) 1  4   x 

17.

1  c , x  1 x1

1/ 4

c





4

b) x  1

1/ 4

1   c) 1  4   x 

c

If lr means log log log …..x, the log being repeated r times then 

c

x l x l

1   d)  1  4   x  2

x ....l r x 

1/ 4

1

dx is equal

to

l r 1 x  c b) r 1

r+1

a) l (x) + c

c) lr (x) + c

d) None of these

3

18.

2 x Let the equation of a curve passing through the point (0, 1) be given by y =  x .e dx. If the equation

of the curve is written in the form x = f(y) then f(y) is a) 19.

log e 3y  2 

b)

3

log e 3y  2 



c)

3

log e 2  3y 

d) None of these



2 x  4 x 3  .......  2 nx 2 n  1 , 0 < x < 1 then  f x dx is equal to If f(x) = xlim  1 a)  1  x 2

b)

1x

2

c)

1 x 1 2

d)

1 1  x2

98

ATOMIC STRUCTURE 20.

 dx  2  dy  2  If x = f’’(t) cost + f’ (t) . sint, y = -f’(t). sint + f’(t). cost, then   dt    dt       

a) f’(t) + f’’ (t) + c 21.

b) f(t) + f’’ (t) + c

a) A = 1



n 1 n

x 2 x n  1

 cos2 tan

25.

1

x

2 / 3

b)  1  x n 1 /n +c

b)

1  x 

1 / 2 5 / 3

1 2 x +c 2





99

d)

3 2 x +c 2

c) 31  x 1 /2 2 / 3  c d) 21  x 3 3 /2  c

 sin x f xdx is equal to 1

b) x sin 1 x  1  x 2  c d) none of these

tan 4 xdx a tan 3 x  b tan x  x  then

a) a = 1/3



1 3 x +c 3

dx is equal to

x n  xn , 0 < x < 1 , n  N then x n  x n



27.

d) None of these

c)

1 1  x 1 / 2 3 / 2 +c b) 31  x2 /3  c 3

If f(x) = xlim 

If

c)  1  x  n 1 / n  c

1x dx 1  x  equals to

a)  x sin 1 x  1  x 2  c c) constant 26.

d) A = -2

equals to

a) x 2 + c

a)

c) A = 2

dx



24.

d) f’(t) – f’’ (t) +c

  , where u = cotx 

b) A = -1

a) 1  x  n 1 / n  c

23.

dt is equal to

If I n   cot n xdx and I 0  2 I 2  .......  I 8   I 9  I 10  u2 u9  u   .........  = A 2 9 

22.

c) -f’ (t) + f’’(t) +c

1 /2

c) f(x) = x2 + c

2) b =1

1 sin x  3 cos x

d) b = -1/2

dx equals to

a)

1 1  1 log tan  x    2 6  2

c)

1 logsecx   / 6   tan x   / 6  2

b)

1 logcos ecx   / 3  cot x   2

1 d)  logcos ecx   / 3  cotx   / 3 2

AIEEE - CHEMISTRY VOL. - I 28.

 a)

aa

ax

aa

29.



c

d) None of these

 log a 

3

1 whose graph passes through the point (0, 0) is 3  5 sin x  3 cos x

 1  x  cos  2 tan 1  1  x  dx is equal to 

b) ½ x2 +k

c) 1/2x + k

d) None of these

c) log(1+x4)+k

d) None of these

xdx is equal to 1  x4

x5 / 2



1  x7

b)

1 1 2 tan x  k 2

dx is





1 x7  1 log c b) 2 x7  1 d) None of these



x log  1  x 2  dx    x  .log 1  x 2    x   c then

1  x2 a) f(x) = 2

34.

3

1 5 x 1 5 x 1 5 x log 1  tan  b)  log 1  tan  c)  log 1  cot  d) None of these  5 3 2 5 3 2 5 3 2



If

 log a 

ax

2 7/2 7 a) log x  x  1  c 7 c) 2 1  x7 +c

33.

c

b)

a) tan-1x2+k

32.

x

c

a) 1/8 (x2-1) +k 31.

aa

The antiderivative of f(x) =

a)

30.

.a x dx is equal to

x

log a 3 aa

c)

ax

.a

b)   x  

1  x2 2

c)  x   

1  x2 2

d)   x   

1  x2 2

3 x  2  3 sin 2x  The value of  e   dx is  1  cos 2x 

a) e3 x tan 3x  c

b) e3 x cot x  c

c) e3 x tan x  c

d) e3 x sec x  c

100

ATOMIC STRUCTURE 35.

3

Let the equation of a curve passing through the point (0,1) be given by y   x2 e x dx . If the equation of the curve written in the form of x = f(y) then f(y) is

36.

37.

1/ 3

 n 3 y  2

a)

b)   n  3 y  2  

a)

 2



dx equals to 4  5 sin x

c)

c) 

b) 2

x5

c)



 2 tan  x / 2   1  1 d) 3 log  2 tan x / 2  4     

dx equals to

1  x2 1 15

1  x  3x 2

1  x  3x 2

x

4

 x

d) none of these

 2 tan  x / 2   1  b) log  tan x / 2  4     

1 x log tan  5x 3 2

 a)

39.

4

4

 4x 2  8 

b)

 4x 2  8 

1 15

1  x  3x 2

 4 x 2  8 

d) None

x5

dx is equal to 5/ 4

+c

4 1  b)  1  3  5 x 

5/ 4

4  1  c)  1  3  15  x 

dx

  sin x  sin 2x   1 1 2 a) log  1  cos x   log  1  cos x   log 1  2 cos x  6 2 3 2 b) 6 log  1  cos x   2 log  1  cos x   log  1  2 cos x  3

101

4

1/ 4

4  1 a) 1 3  15  x 

40.

d) none of these

1  sin 2 x If the derivative of f(x) w.r.t x is 2 , then f(x) is a periodic function with period f  x

2 tan  x / 2   1 a) 2 tan x. / 2  4  

38.

1/ 3

c)   n  2  3y  

5/ 4

d) None of these

AIEEE - CHEMISTRY VOL. - I 1 2 c) 6 log  1  cos x   log 1  cos x   log  1  2 cos x  2 3 d) None

41.



 1 x  tan 1   dx equals to  1 x 

1  x  2

a) x cos-1x -

42.

If dx 

If



b) B = 35/36

xe x

1  e  x

dx  f  x 

a) f(x) = x+1



47.

c)1/2

f  x  .  x 

1  ex  1

1  ex  1 1  ex  1

d) f  x   2  x  2 

{logf(x) – logf(x)}dx is equal to 2

2

  x     x   1  1    k c) log   k d) None of these b) log 2  f  x   2  f  x  







c) g(x)=

1  ex  1

Let  e x f  x   f 1  x  dx    x  . Then

Let f(x) =

d) A + B = -19/36

1

a) (x) + exf(x)

46.

2

x

 f  x  .  x   f  x  .  x 

  x a) log f x  k  

45.

1  x   d) None

1  e  -2log g(x) +c, then

b) g(x) =

1

44.

2

4e x  6e x 2x x  x =Ax+B loge(9e -4)+c then 9e  4e

a) A = 32/35

43.

1 1 c)  x cos x  2

1  x 

b) x cos-1x+



e x f  x  dx is

b)(x) – exf(x)

c)

1   x   e x f  x   2

d)

1   x   e x f  x   2

x 2 dx

 1  x  1  2



1  x 2 and f(0) = 0. Then f(1) is

a) log(1+ 2 )

 b) log(1+ 2 )4

c)

d) None of these

xn  x n ,x  1 then If f  x   lim n x x  x  n





xf  x  ln x  1  x 2 1  x2

 dx is 102

ATOMIC STRUCTURE





2 a) ln x  1  x  x  c



b)

 



2 2 c) x ln x  1  x  ln x  1  x +c

1 2 x ln x  1  x 2  x 2   c   2





d) none of these

KEY 1

2

3

4

5

6

7

8

9

10

11

d

b

a

d

a

d

d

b

a

a

b

16

17

20

21

22 23

24

25

26

d

b

b

b

c

a

a

31

32

35

36

37 38

39

40

41

b

a

b

c

d

d

a

c

46

47

50

51

54

55

56

b

d

a

d

d

b

a

61

62

63 64

65

66

69

70

71

d

b

d

b

d

c

b

c

103

18 19

b

d

33 34

a

c

48 49

b

b

b

c

b

b

52 53

a

d

67 68

b

b

12 13

14

15

d

b

27 28

29

30

d

b

b

44

45

c

c

59

60

a

a

74

75

d

c

a

a

c

42 43

b

c

57 58

a

b

72 73

a

a


76

77

78 79

b

b

b

91

92

93 94

c

c

b

a

80

81

82

83

84

85

86

a

c

b

c

d

b

d

87 88

a

c

89

90

b

c

d

104

ATOMIC STRUCTURE

105

IIIT - MATHS - SET - III

5

DEFINITE INTEGRATION

106

DEFINITE INTEGRAIS If

d [f (x)] =  (x) and a and b are two values independent of variable x, then dx b b a

 (x) dx  [f (x)]

 f (b)  f (a)

a

is called definite integration of (x) with limits a and b. a is called the lower limit and b is called the upper limit of the integral.

GEOMETRICAL INTERPRETATION OF DEFINITE INTEGRAL b

For f(x)  0 x  [a, b] ,  f (x) dx gives the area ALMB bounded by the curve y = f(x), the x-axis and the lines x = a aand x = b.

y

y = f(x) B

b

i.e.,

 f (x) dx = area ALMB a

++++ A ++++ ++++ x=a + + + + x=b ++++ ++++ x O

L

M

b

In general  f (x) dx represents the algebraic sum of the areas of the figures bounded by the graph of thea function y = f(x), the x-axis and the straight lines x = a and x = b. The area above the x-axis enter in to this sum with a positive sign, while those below the x-axis enter with a negative sign. y B y = f(x) + +

A b

Hence  f (x) dx  area LAP – Area PQR + Area RBM. a

O

++ ++ ++ ++ ++ ++

P R

L

M Q

FUNDAMENTAL THEOREM OF CALCULUS x

If f(x) is a continuous function on [a, b] , then

 f ( t ) d t  F ( x ), a

Where F(x) is any function defined on [a, b] such that F(x)  f (x) ... (1) Therefore without any loss of generality, we can take x

F(x) =

 f (t) dt  c, x [a, b], a

107

... (2)

++ ++ ++ ++ ++ ++

x

IIIT - MATHS - SET - III where c is constant. x

a

Now put x = a in (2)  F(a) =

 f (t) dt  c  c

 F(x) =

a

 f (t) dt  F(a) a

b

 f (t) dt  F(a)

 F(b) =

a

b



 f (t) dt  F(b)  F(a) ... (3) a



(3) is called the fundamental theorem of calculus.

CHANGE OF VARIABLE IN DEFINITE INTEGRATION If the functions f(x) is continuous on [a, b] and the function x = g(t) is continuously differentiable on the interval [t1, t2] and a = g (t1) and b = g (t2), then t2

b

 f (x) dx   f (g(t))g(t) dt a

t1

IMPROPER INTEGRALS Definite integral for which a or b or both are infinite are called improper integrals. These are evaluated in the following manner. 

(i)

b

 f (x) dx  lim  f (x) dx, a

b

a

b

b

 f (x) dx  lim  f (x) dx,

(ii)

x 





a

c



 f (x) dx   f (x) dx   f (x) dx

(iii)





c

Another type of improper integral is that in which the integrand is not defined for a point c  [a, b] . In both type of improper integral we take limiting values as shown in the following example.

BASIC PROPERTIES OF DEFINITE INTEGRALS b

(1)

b

 f ( t) dt =  f (x) dx a

a

108

DEFINITE INTEGRAIS b

(2)

a

 f ( x) dx    f ( x) dx a

b

b

(3)

c

b

f(x) dx  f(x) dx  f(x) dx for any c  R a

a

b

b

c

a

a

(4) .  f ( x ) dx   f (a  b  x ) dx In particular  f (x) dx   f (a  x) dx a

a

a

 a 2 f (x) dx , if f (2a  x)  f (x) 0 f (x) dx  0 f (x) dx  0 f (2a  x) dx   0  0 , if f (2a  x)   f (x) 

2a

(6)

0

 a 2  f (x) dx, if f (x) an even function a f (x) dx  0 f (x) dx  0 f (x) dx   0  0, if f (x)is an odd function  a

(5)

0

a

a

a

nT

(7) If f(x) is a periodic function with period T, then (i)

0

b  nT

(ii)

T

 f (x) dx  n  f (x) dx 0

b

 f (x) dx   f (x) dx , where n  I .

a  nT

a

PIECEWISE CONTINUOUS FUNCTIONS Consider a function f(x) defined on [a, b] which has discontinuity at finite number of points say x1, x2 ... xn. In such a case break the intervals [a,b] in to sub intervals such that f(x) is continuous in each sub interval. Such a function is known as piecewise continuous function.

DEFINITE INTEGRAL AS THE LIMIT OF A SUM Let f(x) be a continuous function on the closed interval [a, b]. Then b

n

 f (x ) dx  lim h  f (a  rh) , where h = a

n 

1

f(x) dx  lim 0

109

n

r0

1 n r  f  n r1  n 

ba . In particular n

IIIT - MATHS - SET - III

ESTIMATING A DEFINITE INTEGRAL b

(1) If (x)  f(x)   (x) for a  x  b, then  (x) dx  a

b

b

 f (x) dx    (x) dx a

a

b

In particular m(b–a) 

 f (x) dx

 M(b – a)

a

where m is the least value, and M is the greatest value of the function f(x) on the interval [a, b]. b

(2)

b

 f (x) dx   | f (x) | dx a

a

DIFFERENTIATION UNDER THE INTEGRAL SIGN (Leibnitz’s Rule) If g is continuous on [a, b] and f1(x) and f2(x) are differentiable functions, whose values lie in [a, b], then f (x)

d 2 g(t)dt  g(f2 (x))f2 (x)  g(f1(x))f1 (x) dx f1 (x)

110

DEFINITE INTEGRAIS

111

IIT-MATHS-SET-III

ASSIGNMENT

112


WORKEDOUT ILLUATRATION ILLUSTRATION : 01

1

Let f (x) = x- x  , for every real number x, where x  is integral part of x. Then  f x  dx is. 1

a) 1

b) 2

c) 0

d) 1/2

Solution : 1

0

1

 x  x dx   x  x dx   x  x dx

1

1

0

=

1

 x  1dx   x  0dx

1

0

0

=

0

1

 x  12   x2        2  1  2  0



1 1   1. 2 2

ILLUSTRATION : 02 x

1

If  f t dt  x   tf t dt, then the value of f(1) is 0

a)

x

1 2

b) 0

d) 

c) 1

1 2

Solution : x

 0

1



f( t )dt  x  tf ( t )dt ,

Þ

x

Þ

f x   1  0  x f ( x )

Þ

f x   1  xf ( x )

x 1   d  d   x  tf( t )dt  f( t )dt    dx   dx  0 x   





[ Using Leibnitz’s Rule] Þ

f x  

1 1  f 1  x 1 2

ILLUSTRATION : 03 x 4 If g (x) =  Cos tdt , then g(x + p) equals 0

a) g(x) + g(p) Solution : 113

b) g(x) - g(p)

c) f(x) g(p)

d) g(x)/g(p)

IIT-MATHS-SET-III

x 4 We have g(x) = cos t dt

 0

 g x    x



=

x



cos 4 t dt  cos 4 t dt 



0

4

 cos t dt 

0

x

 4

g x    

=

4

 cos t dt   cos t dt

= g  x     g    g  x 



0

ILLUSTRATION : 04 /2

The value of

 cosec  x   / 3  cos ec  x   / .6  dx

is

0

a) 2log3

b) -2log 3

c) log3

D) None of these

/2

/2

 cosec  x   / 3  cos ec  x   / 6  dx = 2 

Solution :

0

0

/2

2

 0

      cot  x  3   cot  x  6     /2

        sin  x  3       2 log           sin  x  6     0

/2

   dx 

     2 log     

      sin  x     x    6  3   dx     sin  x   .sin  x   6 3  

Þ 2

 0

1 2 3 2

      log       

      cot  x  3   cot  x  6    

3 2 1 2

   dx 

        2   log 3  log 3   

 4 log 3  2 log 3.

ILLUSTRATION : 05 The value of dx is a) 2

b) 1

c) 0

d) None of these 1

Solution : The function f(x) = x |x| is an odd function. Therefore  x | x | dx  0. 1

114


ILLUSTRATION : 06  /4

If I n 

 tan

n

x dx,n  N , then

0

a)

1 n

b)

1 n 1

c)

1 n 1

d)

1 n2

Solution :  /4

We have, I n  2  I n 



 /4

tan

n 2

xdx 

0

/4

=

 tan

n

xdx

0

 /4

 tan x 1  tan x  dx   tan n

2

0

n

x sec 2 x dx

0

1 n =  t dt ,where t  tan x 0

=

1 n 1

ILLUSTRATION : 07 1

d  1 1   tan  dx is x 1 dx 

The value of  a) p/2 Solution :

b) p/4

c) -p/2

d) None of these

We have, d  1 1  d 1 1  tan    cot x    dx  x  dx 1  x2 1

1

1

1 d  1 1  1 1    1 =  dx  tan x  dx   1  x 2 dx  2  1  x 2 dx = 2 tan x  0  2     2   4 1 1 0

ILLUSTRATION : 08 3

2  1 x 1 x  1   dx The value of the integral is   tan x 2  1  tan x  equal to 1 

a)  Solution : 115

b) 2

c) 4

d) None of these

IIT-MATHS-SET-III 3

3

2  1 x 1 x  1   dx =   tan x 2  1  tan x  1 

x   1 x  cot 1 2 =   tan 2  dx x 1 x 1 1 

3

 dx = 2 1 2

=

ILLUSTRATION : 09 e2

If I1 

2

dx ex and I  2 e log x 1 x dx, then

a) I1  I 2

b) 2I1  I 2

c) I1  2I 2

d) None of these

Solution : e2

= I1 

dx

 log x

Put log x =t

x = et

e

2

2

= I1  

= I1  

et dt t 1

ex dx  I 2 x 1

I1  I 2

ILLUSTRATION : 10 1 2n r  equals 2 n  n r 1 n  r2 lim

a) 1  5

b) 1  5

c) 1  2

d) 1  2

Solution : lim

n 

1 n

2n

 r 1

2n

r 2

n r

2

 lim

n 

 r 1

r 2

r 1  2 n 2

.

1 2 n=

 0

x

2

dx   x 2  1   5  1. 2  0 1 x

116


SECTION A-SINGLE ANSWER TYPE QUESTIONS x 2

1.

The value of

 cot t dt is

Lim 0 x 0

a) 0

x

b) 1 

2.

d) none of these

1  cos 2x dx is 2

The value of the integral



a) -2

b) 2

0

c) -1

c) 0

d) -3 aT

3.

If f(x) is a continuous periodic function with period T, then the integral I 

 f(x)dx

a) equal to 2a

d) none of these

a

b) equal to 3a

c) independent of a

k

4.

k

Let f be a positive function. Let I1  xf x1  x dx, I 2  f x1  x dx where 2k  1  0. 1k 1 k then I1/I2 is



a) 2

b) k



c) 1/2

d) 1

1000

5.

6.

7.

e

x x 

dx is (where [x] is the greatest integer function ) 0 e1000  1 b) c) 1000e  1  /4 e  1 sin   cos  The value of the integral 9  16sin 2 is 0 1 log 3 a) log3 b) log2 c) 20  x log x dx is The value of the integral 2 2 1  x 0 a) 0 b) log7 c) 5log13

The value of e1000  1 a) 1000

e 1 1000

d)

1 log 2 20









8.

d)


 sin mx sin nxdx for m ¹ n (m, n Î I) is

a) 0

b) 

d) none of these



c)  /2

d) 2

c) 3/2 sin1

d) None of these

x

9.

If   x  

 sin  t  dt then  '  1 is equal to 2

1/ x

a) sin1

b) 2sin1 

10.

For any integer n, the integral  e

cos 2 x

cos 3  2n  1 xdx has the value

0

a) 

b) 1

c) 0

 /4

11.

The value of the integral I 

 0

1 b tan 1  a  0,b  0  a) ab a 117

dx a cos x  b 2 sin 2 x is 1 b tan 1  a  0,b  0  b) ab a 2

2

d) none of these

IIT-MATHS-SET-III c)

  a  1,b  1 4 1

12.


d 

 dx  tan

1

1

a) /2

1 a 1 tan 1  ab b ab

d) 1  dx is x

b)/4

c) –/2

d) None of these

c) e 1  1

d) one of these

c) /2

d) /4

c) x = 0

d) x = -1

c) 2  

d) None of these

e

13.

 | log x | dx is


1/ e

a) 1 

 1 b) 2  1    e

1 e  /2

14.

The value of

 0

a) 0

dx 1  tan3 x is

b) 1 x2

15.

t 2  5t  4 The points of extremum of  are 2  e' 0

a) x = -2

16.

b) x=1

The value of the integral 

log 5

0

a) 3+2

ex e x  1 dx is ex  3

b) 4   1

17.

1 x

2

x 2 If I is the greatest of the definite integrals I1   e cos xdx, I 2   e cos xdx 2

0

0

1

1 2

2

I 3   e x dx, I 4   e  x / 2 dx , Then 0

a) I  I 1

b) I  I 2

0

c) I  I 3

d) I  I 4 x

18.

 T T Let f(x) be an odd function in the integral   ,  with a period T. Then F  x    f ( t )dt is  2 2 a

a) periodic c) periodic with period 2T

b) non-periodic d) periodic with period 4T

1

19.

xb  1 The value of the integral  log x dx  b  0  is 0

118

DEFINITE INTEGRATION a) logb

b) 2log(b+1)

c) 3logb

d) None of these

3

20.

The value of  | x  2 |   x  dx (where [x] stands for greatest integer less than or equal to x) 1

a) 7

b) 5

c) 4

sin x  sin 2x  sin 3x sin 2x

21.

3  4 sin x 1  sin x

If f  x   a) 3

sin 3x 4 sin x 1

then the value of

b) 2/3 1

22.

3 sin x

The value of the integral  log

d) 3

 /2  f  x  dx is 0

c) 1/3



d) 0



1  x  1  x dx is

0

a)

1 1   log 2    2 2 4

1  b)  log 2  1   2 2

1  c)  log 4  1   3 8

1  d)  log 3  1   4 2

t2

23.

2 5  4  If, for t > 0 the definite integral  xf  x  dx  t , then f   is equal to 5  25  0

a)

2 5

b) 

2 5

c)

2 5

d) 

2 5

1 n

24.

n If l (m, n) =  t  1  t  dt , then the expression for l (m, n) interms of   m  1,n  1 is 0

2n n    m  1,n  1 a) m 1 m 1

b)

n   m  1,n  1 m1

2n n    m  1,n  1 c) m1 n1

d)

m   m  1,n  1 n 1

x 2 1

25.

If f(x) =



2

e t dt , then the function f  x  decreases in

x2

a) (2,2)

b) (0,¥)

c) no value of x

d)   ,0 

x

bt cos 4t  a sin 4t a sin 4x  1,  0  x   / 4  then a and b are given by dt = 2 t x 0



26.

Let

27.

a) a = ¼, b = 1 b) a = 2 , b = 2 c) a = -1, b = 4 d) a = 2, b = 4 Let f(x), g(x) and h(x) be continuous function on [0,a] such that f(x) = f (a-x), g(x) = -g (a-x), 3h (x) – a

4h (a-x) = 5 then

 f  x .g  x  .h  x  .dx is equal to 0

119

IIT-MATHS-SET-III a) 1

b) 0

c) a

d) –1 a

28.

dx If a continuous function f on [0,a] satisfy f(x). f(a – x) = 1, a> 0 then  1  f x is equal to   0

a) 0 29.

b) a

c) a/2

d) none of these

Let f be a continuous function on R satisfies f(x+y) = f(x) +f(y) for all x, y  R with f(1)=2 and g be a 1

function satisfying f(x) + g(x) = e then the value of the integral  f  x  g(x) dx is x

0

a)

1 4 e

b)

1 e  2 4

c)

2 3

d)

1 e  3 2

2

30.

The values of

  px

3

 qx  r  dx, where p, q, r are constants, depends on the value of

2

a) q

b) r

c) p

d) p and q

x 2

31.

The point of extremum of f  x     t  2   t  1 dt is a 0

a) maximum at x =1 c) minimum at x =1

b) maximum at x = 2 d) minimum at x = 2 2 /4

32.

The value of integral



sin xdx is

0

a) 1

33.

lim 

b) 1/2 x

x4 4

 4t  f  t   dt  x  4

a) 16  f  4 

b) 4  f  4 

If f(x) and g(x) are continuous functions, then

a) depend on 

35.

d) None of these

c) 16  f '  4 

d) 4  f '  4 

is equal to

ln1/ 

34.

c) 3/2



ln 

b) a non zero constant

If f be continuous function of x, then

1 c



bc

ac

 x2  f    f  x   f   x    4  dx  x2  g    g  x   g   x    4  c) zero

d) none of these

 x f   dx is equal to c

120


36.

a)



If



b

a

f  x  dx

n 1

n

b)

c)

f  x  dx  n 2  n,  nI then value of

a) 6 37.

1 b/ c f  cx  dx c a / c

b) 10

Let f : R  R , g : R  R



bc 2

ac

2

b

d) c  f  x  dx

f  x  dx

a

3

 f  x  dx is equal to 3

c) 16

d) 12

be two continuous functions, then the value of the integral

  f  x   f   x    g  x   g   x   dx is a) 

b) 1

3

38.

The value of

c) –1

d) 0

 x

  x

is equal to (where [.] denotes g.i.f. and {.} denotes fractional part)

0

a) 2/3

b) 5/6

c) 1

d) 11/6



39.

40.

cos 2 x The value of  1  a x dx, a  0 is 

a)  If ‘a’

is

a

b) a positive integer,

c) /2 the number

of

d) 2 values of ‘a’

 2  cos 3x 3  a 2  a  cos x  a sin x  20 cos x dx   is     4 4 3   a) 1

b) 2

c) 3

d) 4

x

41.

t The points extremum of F(x) =  e

2

/2

1  t  dt are 2

1

a) x =1

b) x = -1

1

42.

c) x = 2

d) x = -2

c) be  a

d) aeb

c) g  x  g  

g  x d) g 

b

et dt e  t dt  a If  t  1 then  t  b  1 is equal to 0 b 1

a) ae  b

b) ae  b x

43.

4 If g  x    cos tdt then g  x    equals 0

a) g  x   g   

121

b) g  x   g   

 

satisfying


n x  If g  x    | sin t |  | cos t | dt, then g  x   is equal to, where n  N 0 2   a) g  x   g   

 n  b) g  x   g    2 

  c) g  x   g   2

d) None of these

tan x

45.

The value of



1/ e

cot x

dt t dt  + t 1  t 2 is 1 t2  1/ e 

a) 1/2

b) 1 

46.

The value of the integral  

1 a) sin

 

c)  / 4

d) None of these

dx

 x      x 

for    , is

c) sin

b) /2

1

 2

d)  1

47.

If a, b, c be nonzero real numbers such that

  1  cos x  ax 8

2

 bx  c  dx

=

0

2

  1  cos x  ax 8

0

2

 bx  c  dx then the quadratic equation 2 ax  bx  c  0 has

a) no root in (0,2) c) at least one root in (0,1) 48.

b) at least one root in (1,2) d) two imaginary roots x

Consider the function where f  x     t  dt, x>0 and  t  denotes the greatest integer function, then 0

a) is continuous and differentiable for x = 1, 2, 3, ………….. b) f(x) is neither continuous nor differentiable for x = 1, 2, 3………. c) f(x) is continuous but not differentiable for x = 1, 2, 3…….. d) None of these 49.

Let f : R  R such that f  x  2 y   f  x   f  2 y   4xy x, y R and f '  0   0 . If 1

0

2

0

1

1/ 2

I1   f  x  dx,I 2   f  x  dx and I 3   a) I1  I 2  I 3 50.

f  x  dx then

b) I1  I 2  I 3

c) I1  I 2  I 3

d) I1  I 2  I 3

b)

c)

d)

Value of nÎN is equal to a)

122

DEFINITE INTEGRATION 51.

 4 0

The value of 

   n cot   x  4 

a) 0

52.



 /2

0

b) 1  8  7 cos x   7  8 cos x 2 

b) 2/7

d) not defined

c) 3/7

d) 4/7

x

0

If



10

0

a) 50

123

c) 2

Let f  x    | 2t  3 | dt, then f is a) continuous at x=3/2 c) differentiable at x=3/2

54.

dx is :

  dx is equal to  

a) 1/7 53.

  cot   x  4 

b) continuous at x=3 d) differentiable at x=0 10

f  x  dx  5, then

1

  f  K  1  x  dx is equal to K 1

0

b) 10

c) 5

d) none

IIT-MATHS-SET-III

SECTION B-MULTIPLE ANSWER TYPE QUESTIONS 1

1.

  3 t 2 sin 2 t    x  2 dt   2  0 If x satisfies the equation x   2   t 1  t  2 t cos   1 0   3  dt

2

(0) <  < p), then the value of x is

 sin    a) 2    

 sin       

b) -2

 sin   c) 4     

d) none of these

c) b+a

d) none of these

c) less than 1

d) greater than 1

b

2.

x The value of the integral  x dx, a < b is a

a) b-a

b) a-b 1

3.

x2

The value of the integral  e dx is 0

a) less than e

b) greater than e

1

1 2

4.

2

3

2 2

3

4 x x x If I1 =  2 dx, I2  2 dx, I3 =  2 dx and I4 =  2 dx then 0

0

1

a) I1 > I2

1

b) I2 > I1

c) I3 > I4

d) I4 > I3

b) more than 10-7

c) less than 10-6

d) none of these

b) I >

c) I > 0.92

d) I< 0.92

c) I2 > I1

d) I2 = cot-1 x - p/4

c) 3/8

d) 3/8 a2

1

5.

sin x The absolute value of  1  x 8 dx is 0

a) less than 10-7 4

6.

dx

If I =  log x ,then   3

3

a) I < 1

7.

dt and I 2  If I1 =  1  t2 a) I1 = I2

1/ x

dt

 1 t

2

for x > 0, then

1

b) I1 > I2 a  / 2

8.

The value of

  sin

4

x  cos 4 x  dx is

a

a) independent of a

  b) a   2

2

124

DEFINITE INTEGRATION 2

9.

The value a in the interval ([-, 0] satisfying sin  +

 cos 2x dx  0

is



a) -/2

b) -

c) -/3

d) 0

c) I < 7 /2

d) none of these

1

10.

If I = 

1  x dx 3

then

0

a) I < 1

b) I  /2

x/2

11.

The maximum and minimum value of the integral

dx

 1  sin x  2

are

0

a) /4

b) 

c) /2

d) 3/4 b

12.

1 f  x a dx has Let f(a) > 0, and let f(x) be a non decreasing continuous function in [a, b]. Then ba  the a) maximum value f(b

b) minimum value f(a)

c) maximum value of bf(b)

d) minimum value

f a ba

x2

13.

t 2  5t  4 The points of extremum of  dt are 2  et 0

a) x  -2

b) x = 1

c) x = 0

d) x = -1



14.

The values of  which satisfy



sin xdx  sin 2 (   [0, 2]) are equal to

/2

a) /2

125

b) 3/2

c) 7/6

d) 11/6

IIT-MATHS-SET-III

KEY 1

2

b

b

16

17

b

d

31

32

c

d

46

47

d

b

3

4

5

6

7

8

9

10

11

c

c

c

c

a

a

c

c

a,b,c

20

21

24

25

26

a

c

a

d

a

35

36

39

40

41

a

c

d

d

a

d

50

51

52

53

54

d

d

18 19

a

d

33 34

a

c

48 49

c

d

14

15

d

a,b,c,d

29

30

c

b

42 43

44

45

a,b

b

b

b

10

11

12 13

a,c a,d a,c b,c,d a,c

a,c

a,c a,b,c,d a,b,c,d

22 23

b

a

37 38

12 13

c

b

27 28

b

c

a

a a,b,d c

KEY

1

2

3

4

a,b a,b,c a,d a,d

5

a,c

6

7

8

9

14

126


127


6

AREA UNDER CURVES

128

AREA Let f(x) be a continuous non-negative function in the interval [a, b]. The area of the region bounded by the graph of y = f(x), the x-axis and the lines x = a and x = b is given by b

 f (x ) dx a

FORMULAE FOR FINDING THE AREA UNDER BY CURVES 1.

Area ABCDA bounded by the curve y = f(x), x-axis and two ordinates x = a and x = b

b   ydx, if y  0 for x  [a, b] b a is given by  | y | dx   b a    ydx, if y  0 for x [a, b]  a

y = f(x) C

D

A

B x’ x= a

x

x= b

x= a

y = f(x)

x= b

C B

A

x’

D

x

If however y i.e., f(x) changes sign in interval [a, b], say y  0 in [a, c], y  0 in [c, d]and y  0 in [d, b], where a < c < d < b, then area bounded by the curve y = f(x), x-axis and the lines x = a and x = b b

c

d

b

  | y |dx   ydx   ydx   ydx a

a

c

d

= A1 – A2 + A3 , where A1, A2 and A3 are algebraic areas. y = f(x) x= a x’

129

A3

A1 a

c

x

A2

d

x= b b

x

IIT - MATHS - SET - III y A x2 + y2 = 4 O

C

x=1

x x=2

B

2

2

2

Required area = area ABCA  2 ydx  2 1

1

 x 22  x 2 22  1 x 2    2  sin 4  x dx 2 2 2  1

 3  1   0  2sin 1 (1)  2   2sin 1   = 4  3 3 sq. units  2   2 3

2. Area ABCDA bounded by two curves y = f(x), y = g(x) and two ordinates x = a, x = b is given by

 b   (f (x)  g(x) dx, if f (x)  g(x) for a  x  b  a b   | f (x)  g(x) | dx =  b a  (f (x)  g(x)) dx, if f (x)  g(x) for a  x  b  a While using this formula f(x) is taken from the curve which lies above and g(x) is taken from the curve which lies below. If a < c < d < b and f (x)  g(x) for a  x  c f (x)  g(x) for c  x  d f (x)  g(x) for d  x  b

y = g(x)

x= a

x= b

y = f(x)

x

130

AREA

c

d

b

then shaded area =  (f (x)  g(x)) dx   (g(x)  f (x) dx   (f (x)  g(x))dx a c

c d

d

b

  (f (x)  g(x) dx   (f (x)  g(x) dx   (f (x)  g(x) dx a

c

y = f(x)

3.

y = f(x)

y = g(x) y = f(x)

x= a y = g(x) a

d

c

y = g(x) d

b

x

CURVE SKETCHING For the evaluation of area of bounded regions it is very essential to know the rough sketch of the curves. The following points are very useful to draw a rough sketch of a curve. While constructing the graph of f(x, y) = 0, it is expedient to follow the procedure given below: (1) Find the set of permissible values of x (2) Check if the curve is symmetrical about x-axis, y-axis, origin etc (3) Find the period of the curve if it is periodic (4) Find the asymptote(s) of the curve, if any (5) Find the intervals of increase and decrease of the curve. Hence determine the greatest and the least values of the curve, if any. Example 9 : Find the area of the region bounded by the curves y = f(x), y = |g(x)| and the lines x = 0, x = 2, where f and g are continuous functions satisfying f(x + y) = f(x) + f(y) – 8xy for all x, y  R and g(x + y) = g(x) + g(y) + 3xy (x + y) for all x, y  R . also f  (0)  8 and g(0)   4 Solution: Given, f(x + y) = f(x) + f(y) – 8xy for all x, y Putting x = 0 and y = 0, we get f(0) = f(0) + f(0)  f(0) = 0

131

IIT - MATHS - SET - III f (x  h)  f (x) h 0 h

f (x)  lim

f (x  h)  f (x  0) h 0 h

 lim

 lim h 0

 lim h 0

f (x)  f (h)  8xh  f (x)  f (0)  8x.0 h f (h)  f (0)  8x  f  (0)  8x h

Thus f  (x)  8  8x [f  (0)  8] Integrating both sides, we get f(x) = 8x – 4x2 + c

... (2)

Putting x = 0, we get f(0) = 0 + c  c = 0 [ f (0)  0] Hence f(x) = 8x – 4x2 Given g(x + y) = g(x) + g(y) + 3xy (x + y) for all x and y Putting x = y = 0, we get g(0) = 0

... (3) ... (4)

g(x  h)  g(x) g(x  h)  g(x  0)  lim h 0 h0 h h

Now g(x)  lim

 g(h)  g(0)   lim   3x 2  3xh  h 0  h 

 g(0)  3x 2  4  3x 2 [g(0)   4] Thus g  ( x ) = – 4 + 3x2

... (5)

g(x) = – 4x + x3 + k  Putting x = 0, we get g(0) = k 

k = 0 [ g(0)  0]

g(x) = x3 – 4x  For points where y = f(x) and y = g(x) intersect, 8x – 4x2 = x3 – 4x x3 + 4x2 – 12 x = 0  x = 0, 2, – 6  Sign scheme for f(x) i.e., for (8x – 4x2) is

... (6)

132

AREA Y

y = f(x) –

–ve

+ve

0



–ve

2

y = |g(x)| (2, 0)

O

X

Sign scheme for g(x) i.e. for x(x2 – 4) is –

–ve

–2

0 –ve

+ve

2

+ve



f(x) – |g(x)| = 8x – 4x – (4x – x ) ( g(x)  0 in [0, 2]) = x3 – 4x2 + 4x = x(x – 2)2  0 Area bounded by y = f(x) and y = |g(x)| between x = 0 and x = 2 2

2

3

2

  (y1  y2 ) dx   [(8x  4x 2 )  (4x  x 3 )]dx 0

0

2

4   (x 3  4x 2  4x) dx  sq. units . 3 0 Example 10 : x  2x  cos  Find the area enclosed by the circle x2 + y2 = 4, the parabola y = x2 + x + 1, the curve y  sin 4 4  and the x-axis, (where [x] denotes the integral part of x). Solution: Equation of given circle is x2 + y2 = 4 ... (1) – 2  x  2 and – 2  y  2 

Let z = sin2

x x x x  cos = 1 – cos2  cos 4 4 4 4

= 1 + t – t2 , where t = cos Now –



x 4

1 x 1   2 4 2 5 0
x  2x  cos  becomes y = 1 ... (2) for x  [2, 2], curve y = sin 4 4 

Given parabola is

133

IIT - MATHS - SET - III y = x2 + x + 1

... (3)

2

or,

Its axis is x  

1 3  x   y 2 4  1 and vertex is 2

... (4)

 1 3  , .  2 4

Now, we have to find out the area enclosed by the circle x2 + y2 = 4,

y (0, 2) 2

3  1  parabola  y     x   , line y = 1 and x-axis 4  2 

Q A E

P T

S R U O

B D C (2, 0)

x

(–2, 0) – 3 - 1- 1

2

T

S

R

U

D

(–2, 0)

( 3, 0)

(–1, 0)

(–1/2, 0)

(2, 0)

Required area is shown as shaded region in the figure. Hence required area = area OABCO + area PQRS + area RQEAOR + 2 area CBDC. 0

3 1  ( 3  1) 1 

2

2  (x  x  1) dx  2 1

 3



4  x 2 dx

0

2  x3 x 2  x 2 1  x    2 3 1     x  2  4  x  2 sin    2  2  3  1 2  3





  1 1   2 3  1   0      1   2   3 2 





  3 2     (0  )    2 3    

5 2  2 3 1    3 6 3





1  2   3   sq. units . 6  3

134

AREA

SOLVED OBJECTIVE EXAMPLES Example 1 : Area enclosed by the curve |x –2| + |y + 1| = 1 is equal to (a) 4 sq. units

(b) 6 sq. units

(c) 2 sq. units

(d) 8 sq. units

Solution: After shifting the origin at the point (2, –1) the equation of curve becomes, |x| + |y| = 1. This curve will represent a square as shown in the adjacent figure. y x+y=1

y-x=1

A

C O x + y = –1

x–y=1 D

Area of this square is clearly equal to 4 times the area of triangle OAB. Thus required area = 2 sq. units. Example 2 : Area bounded by the curves y = |x| – 2 and y = 1 – |x–1| is equal to (a) 4 sq. units

(b) 6 sq. units

(c) 2 sq. units

(d) 8 sq. units

Solution: Bounded figure ABCD is a rectangle. AB  1  1  2 BC  4  4  2 2

135


A(1, 1) B

Thus, bounded area = ( 2) (2 2 ) = 4 sq. units.

–2

0

–1

1

2

y = |x| –2 y = 1 – |x –1|

D –2 C

Example 3 : Area bounded by the curve y = max{sinx, cosx} and x-axis, between the lines x 

 and x 4

= 2 is equal to (4 2  1)

(a)

2

sq. units

(4 2  1) sq. units 2

(c)

(b) (4 2  1) sq.units

(d) None of these

Solution: Bold lines represents the graph of y = max{sinx, cosx}. Required area,





5 / 4

sin x dx 

 /4

3 / 2



 5 / 4



sin x dx

 2

cos xdx 

 3 / 2

cos x dx y

y = sin x   O /4

 y = cos x



x

136

AREA 

(4 2  1) sq.units 2

Example 4 : Area bounded by the parabola y = x2 – 2x + 3 and tangents drawn to it from the point P(1, 0) is equal to (a) 4 2 sq. units

(b)

4 2 sq. units 3

8 2 sq. units 3

(d)

16 2 sq. units 3

(c)

Solution: Let the drawn tangents be PA and PB. AB is clearly the chord of contact of point P. Thus equation of AB is

1 . (y  0) = x.1 – (2 + 1) + 3 i.e., y = 4 2

x coordinates of points A and B will be given by, x2 – 2x + 3 = 4 

x= 1

i.e., x2 – 2x –1 = 0

2

Thus AB = 2 2 units. y = x2 – 2x + 3

1 Hence  PAB  (2 2).4  4 2 sq. units 2

y=4 A

B

P(1, 0)

Now area bounded by line AB and parabola is equal to 1 2



(4 2  (x 2  2x  3)) dx

1 2

=

137

4 2 sq. units. 3

x

IIT - MATHS - SET - III Thus required area = 4 2 

4 2 8 2  sq. units. 3 3

Example 5 : Area bounded by the curves y = sinx, tangent drawn to it at x = 0 and the line x = 2  4 sq. units (a) 2

(c)

 , is equal to 2

2  4 sq. units (b) 4

2  2 sq.units 4

(d)

2  2 sq. units 2

Solution: The tangent drawn to y = sinx at x = 0 is the line y = x. Clearly the line y = x lies above the graph of y =   sinx x   0,  .  2 / 2

Thus required area  

 0

/ 2

 x2  (x  sin x) dx    cos x   2   0



2  4 sq. units. 4

Example 6 : If A(n) represents the area bounded by the curve y = n. lnx, where n  N and n > 1, the x-axis and the lines x = 1 and x = e, then the value of A(n) + nA(n–1) is equal to (a)

n2 e 1

(b)

(c) n2

n2 e 1

(d) en2

Solution: e

e   e    n  A(n – 1) = (n – 1) l n x dx  n l n x.x  dx A(n) = n  1    1 1  

A(n) + nA(n – 1) = n + n(n – 1) = n2



Example 7 : Value of the parameter a such that the area bounded by y = a2 x2 + ax + 1, coordinate axes and the line x = 1, attains it’s least value, is equal to (a) –

1 4

(b) –

1 2

(c) 

3 4

(d) –1

138

AREA Solution: a2x2 + ax + 1 is clearly positive for all real values of x. Area under consideration 1

   (a 2 x 2  ax  1) dx  0

a2 a  1 3 2

1  (2a 2  3a  6) 6 2   1  3 9  18    2  a 2  a    6   = 1  2  a  3   39  6  2 16  16  6   4 8 

which is clearly minimum for a = –

3 . 4

Example 8 : Area of the region which consists of all the points satisfying the conditions |x – y| + |x + y|  8 and xy  2, is equal to (a) 4(7 – ln8) sq. units

(b) 4 (9 – ln8) sq. units

(c) 2(7 – ln8) sq. units

(d) 2 (9 – ln 8) sq. units

Solution: The expression |x – y| + |x + y|  8, represents the interior region of the square formed by the lines x   4, y   4 and xy  2 represents the region lying inside the hyperbola xy = 2. Required area, y y=4

y=x B

C 4

 2



2

4

  4  x  dx  2  4x  2 l n x 1/ 2

1/ 2

x x = –4 x=4 D y = –4

A

y = –x

= 4(7 – 3 ln2) sq. units Example 9 : A point P moves in xy plane in such a way that [|x|] + [|y|] = 1, where [.] denotes the greatest integer function. Area of the region representing all possible positions of the point P is equal to

139

(a) 4 sq. units

(b) 16 sq. units

(c) 2 2 sq. units

(d) 8 sq. units

IIT - MATHS - SET - III Solution: If [|x|] = 1 and [|y|] = 0 then 1  | x |  2, 0  | y |  1  x  ( 2,  1]  [1, 2), y  ( 1,1)

if [|x|] = 0, [|y|] = 1 y

2

1

Then

–2

x  ( 1, 1), y (2,  1]  [1, 2]

–1

1

2

O

x

–1

–2

Area of required region = 4(2 – 1) (1 – (–1)) = 8 sq. units.

Example 10 : Area enclosed by the curve y = f(x) defined parametrically as x  (a)  sq.units (c)

1 t 2

2t

1 t

1 t2

, y 2

is equal to

(b)  / 2 sq. units

3 sq. units 4

(d)

3 sq. units 2

Solution: Clearly t can be any real number Let t  tan   x  

1  tan 2 

x  cos 2

and y  

1  tan 2 

2 tan 

1  tan 2  x2 + y2 = 1

 sin 2

Thus required area = .12   sq. units.

140

AREA

SUBJECTIVE

LEVEL-I (CBSE LEVEL) REVIEW YOUR CONCEPTS

1.

Find the area bounded by the ellipse

x2 a2



y2 b2

 1 and the ordinates x = ae and x = 0, where

b2 = a2 (1 – e2). Solution : Y x= 0 x = ae a

b e 2 A   a  x 2 dx  ab e 1  e 2  sin 1 e  a0

2.

ae, 0 X

Find the area of the region in the first quadrant enclosed by the x–axis, the line x = 3 y and the circle x2 + y2 = 4.

Solution : Y

1 A 3

3.

3

2

 x dx   0

4  x2 

3

( 3,1)

 3

Find the area of the smaller region bounded by the ellipse x y   1 . (a > 0, b > 0) a b

(2, 0)

x2 a2



y2 b2

y

X

 1 and the straight line

Solution : Y (0, b) a

a

b b A   a 2  x 2 dx   (a  x) dx a0 a0

141

X O

1 x 3

(a, 0)

IIT - MATHS - SET - III 43 (   2) 4

=

Find the area of the region bounded by the parabola y = x2 and the rays given by y = |x|.

4.

Solution :

y 2

1

1

A   (x) dx   x 2 dx  0

2

x =y

0

1 3

x=y y = |x|

y = |x|

(1, 1)

(-1, 1) C O

x

Find the area of the smaller part of the circle x2 + y2 = a2 cut off by the line x =

5.

a . 2

Solution : y A a

A2

x2 + y2 = 4

2

a  a  n dx  (  1) 2 2 2



a/ 2

2

O

C

x=1

x x=2

B

6.

1 Sketch the curves and identify the region bounded by the curves x = , x = 2, y = log x and 2 y = 2x. Find the area of the region.

Solution : Y

2

A



2 x

2 dx 

1/ 2



1/ 2

log ex dx 

(4  2) 5 3  log 2  ] log 2 2 2

(0, 1) x (1, 0)

7.

Find the area of the circle x2 + y2 = 16, which is exterior to the parabola y2 = 6x.

Solution :

142

AREA Y (0, b)

C

2



 8.

2

1  (4) 2  2  16  x 2 dx   6x dx] 2 0 0

(4, 0)

O B (0, 0)

X

(a, 0) E

A

(2, -y)

4 (8  3) 3

Find the area of the circle 4x2 + 4y2 = 9, which is interior to the parabola y2 = 4x.

Solution : 3/ 2 1/ 2  9 A  2   4x dx    x 2 dx  4 0 1/ 2 

Y C

(0, b) (4, 0)

O B (0, 0)

X

(a, 0) A



9.

E

(2, -y)

9 9 1 2  sin 1  8 4 3 6

Find the area of the region in the first quadrant enclosed by the x–axis, the line y = x and the circle x2 + y2 = 32.

Solution : Y 4

A   xdx  0

( 3,1)

4 2



32  x 2 dx  4

4

(2, 0)

10.

X

Find the area of the region enclosed by the circles x2 + y2 = 1 and (x – 1)2 + y2 = 1.

Solution :

143

y

1 x 3

IIT - MATHS - SET - III Y (½,-y) 1 1/ 2  A  2   1  (x  1) 2 dx  1  x 2 dx  1/ 2 0 

O

(1, 0)

X

(½,-y)



2 3  3 2

144

AREA

LEVEL-II

Brush up your Concepts Find the area enclosed by the curves 3x2 + 5y = 32 and y = |x – 2|.

1.

Solution : 2.

Compute the area of the figure bounded by the straight line x = 0 and x = 2, and the curves y = 2x , y = 2x – x2.

Solution : x=2

Y x=0 2

(0, 1)

2

A   2 x dx   (2x  x 2 ) dx 0

3.

0

O

X

 Find the area of the region bounded by the curve y = tanx, the tangent drawn to it at x = 4 and the x –axis.

Solution :

Y

(/4, 1) (/4, 1) /4

A



/4

tan x dx 

0

 0

[1  2x

 ]dx 2

X

O (0, 0)

Q

– Area of OPQ

145

... (1)

IIT - MATHS - SET - III Where area of 

1 Base x height 2

1 1      2 2   1  1   2   2

... (2)

by equation (1) & (2) required area A

4.

1 1 (im2  ) 2 2

Find the area of the region bounded by the curves y = ex ln x and y =

ln x . ex

Solution : s 5.

Sketch the region bounded by the curves y = x2 and y =

2 1 x2

and find its area.

Solution :

6.

Prove that the area bounded by y =

x2  x  1

and the lines x = 0, y = 0 and x = a is a x2  x 1 monotonically increasing function of ‘a’. What is the increase in the area as ‘a’ changes from 2 to 4.

Solution : a

A 0

x2  x 1 dx  A(a) x2  x 1

... (1)

 required area in terms of a Diff. equation (1) w.r.t. ‘a’ dA 0 da

Y x=0

x =a

So, area is a increasing fx of a for change (increase in Area) A  (A a 4  A a  2 )

 2  ln 3 

X y=0

2 1 tan 1 2 4 3

146

AREA

Find the least value of the area bounded by the line y = mx + 1 and the parabola y = x2 + 2x – 3, where m is a parameter.

7.

Solution : x2

x2

A   (mx  1) dx   (x 2  2x  3) dx ` x1

... (1)

x1

(x3 y2) (0, 1) (x4, y1)

where x1, x2 are the values of this equation

(–1, –4) mx + 1 = x2 + 2x – 3  x2 + x (2 – m) – 4 = 0 x1, x2 obtained ... (2)  By equation (1) & equation (2) A

Find the area of the region bounded by the curves y = tanx,  y = cot x,

Solution : Required area / 4

A

and

x = –3 

x=–

x = –/2

x =  x = 

x=0

x = 3 

Y /3

 tan x dx   cot x dx

/6

/4

ln

3 2





147

  x 3 3

 3 x and the x–axis.  6 2

( 

8.

32 3



X


9.

Sketch the region bounded by the curves y = 5  x 2 and y = |x – 1| and find its area.

Solution : 10. Find the area bounded by the curves x2 + y2 = 25, 4y = | 4 – x2| and x = 0 above the x–axis. Solution :

148

AREA

LEVEL-III

CHECK YOUR SKILLS 8a 3 x2 and the curve y = 2 . 4a x  4a 2

1.

Find the area included between the parabola y =

2.

Find the area enclosed by the curve xy2 = a2 (a – x) and the y–axis.

3.

Find the area enclosed by y = x (x – 1)2, y–axis and the line y = 2.

4.

Find the area of the region {(x, y):0  y  x 2  1, 0  y  x 1, 0  x  2} .

5.

Find the area bounded by the curves y = – x2 + 6x – 5, y = – x2 + 4x – 3 and the straight line y = 3x – 15.

6.

e t  e t et  e t For any real ‘t’, let x = 2 + ,y=2+ be a point on the hyperbola 2 2

x2 – y2 – 4x + 4y – 1 = 0. Find the area bounded by the hyperbola and the lines joining the center to the points corresponding to t1 and – t1 . 7.

Sketch the region bounded by the curves y = lnx and y = (ln x)2. Also find the area of the region.

8.

Find the area of the region common to x + y  6, x2 + y2  6y and y2  8x.

9.

Consider the closed figure C given by |x| + |y| = 2 . Let S be the region inside the figure such that any point in it is nearer to the side x + y = 2 then the origin. Find the area of the region S.

10.

1  Let f(x) = max sin x, cox,  , then determine the area of the region bounded by the curves 2 

y = f(x), x – axis , y–axis and x = 2  .

149


PROBLEM ASKED IN IIT-JEE 1.

In what ratio does the x–axis divide the area of the region bounded by the parabolas y = 4x – x2 and y = x2 – x. [1994]

Solution : A 2.

Consider a square with vertices at (1, 1), (–1, 1), (–1, –1), (1, –1). Let S be the region consisting of all points inside the square which are nearer to the origin than to any edge. Sketch the region S and find its area. [1995]

Solution : The equations of the sides of the square are as follows: AB : y = 1, BC : x = –1, CD : y = – 1, DA : x = 1 Let the region be S and (x, y) is any point inside it. Then according to given conditions,

B(-1, 1)

A(1, 1) (0, 1) (0, 1/ 2 )

(–1/ 2,0 )

C(1,– 1)

(1/ 2,0 )

A(1, 1)

y2 = 2x -1 x

G I

E

O

H F

x 2 = 2y - 1 (1, 0)

D(1,– 1)

x 2  y 2  |1  x |,|1  y |,|1  y |

 x2 + y2 < (1 – x)2, (1 + x)2 , (1 – y)2 , (1 + y)2  x2 + y2 < x2 – 2x + 1, x2 + 2x + 1, y2 – 2y + 1, y2 + 2y + 1  y2 < 1 – 2x, y2 < 1 + 2x, x2 < 1 – 2y and x2 < 2y + 1 Now in y2 = 1 – 2x and y2 = 1 + 2x, the first equation represents a parabola (–1/2, 0) with vertex (1/2, 0) and second equation represents a parabola with vertex (–1/2, 0). And in x2 = 1 – 2y and x2 = 1 + 2y, the first equation represents a parabola with vertex at (0, 1/2) and second equation represents a parabola with vertex at (0, – 1/2). Therefore, the region s is the region lying inside the four parabolas y2 = 1 – 2x, y2 = 1 + 2x, x2 = 1 + 2y, x2 = 1 – 2y,

150

AREA where S s the shaded region. Now, s is symmetrical in all four quadrants, therefore, S = 4 × area lying in the first quadrant . Now, y2 = 1 – 2x and x2 = 1 – 2y intersect on the line y = x. The point of intersection is E( 2 – 1, 2 – 1) Area of the region OEFO = area of  OEH + area HEFH



1 2  1)2  2

1/ 2

 2 1

1  2x dx = 1 ( 2  1) 2  (1  2x)3/ 2 . 2 1 ( 1)    2 32

1/ 2

2 1

1 1 (2  1  2 2)  (1  2  2 2) 3/ 2 2 3

=

1 1  (3  2 2)  ( 2  1) 2 ]3/ 2 2 3 1 1  (3  2 2)  [( 2  1) 2 ]3/ 2 2 3 

1 1 (3  2 2 )  ( 2  1)3 2 3

1 1  (3  2 2)  [2 2  1  3 2( 2  1)] 2 3 1 1 1 1  (3  2 2)  [5 2  7]  [9  6 2  10 2  14]  [4 2  5] 2 3 6 6

Similarly, area OEGo =

1 (4 2  5) 6

Therefore, area of S lying in first quadrant =

2 1 (4 2  5)  (4 2  5) 6 3

4 1 Hence, S  (4 2  5)  (16 2  20) 3 3

3.

Let An be the area bounded by the curve y = (tan x)n and the lines x = 0, y = 0 and x =  / 4 . Prove that for n  2, A n  A n  2 

1 1 1  An  and deduce that . n 1 2n  2 2n  2

[1996]

151

IIT - MATHS - SET - III Solution : /4

We have A n 

 (tan x)

n

dx

0

Since 0 < tan x < 1, when 0 < x <  / 4 , we have 0 < (tan x)n + 1 < (tanx)n for each n  N 4

/4

n 1   (tan x) dx < 0

 (tan x)

n

dx

0

 An+1 < An Now for n > 2 / 4

An + An+2 =

/ 4

 ((tan x)

n

 (tan x)

n2

0

]dx =

 (tan x)

n

sec 2 xdx

0

/4

 1  (tan x) n 1  =   (n  1) 0

1  f (x)n 1  n {1  0}   f (x) f (x) dx  =   n  1  (n  1)

Since An+2 < An+1 < An, we get An + An+2 < 2An 



1  2A n n 1

1  2A n n 1

1  An 2n  2 Also for n > 2,

... (1)



An + An < An + An–2  2A n 

=

1 n 1

1 1  An  n 1 2n  2

... (2)

from (1) and (2) 1 1  An  2n  2 2n  2

from (1) and (2) 1 1  An  2n  2 2n  2

152

AREA 4. Let f(x) = max{x2, (1–x)2, 2x (1–x)}, where 0  x  1. Determine the area of the region bounded by the curves y = f(x), x–axis, x = 0 and x = 1. Solution : 5.

Let C1 and C2 be the graph of the functions y = x2 and y = 2x, 0  x  1 respectively. Let C3 be the graph of a function y = f(x), 0  x  1, f(0) = 0. For a point P on C1, let the lines through P, parallel to the axes meet C2 and C3 at Q and R respectively (see the figure). If for every position of P (on C1), the area of the shaded regions OPQ and ORP are equal, determine the function f(x). [1998] (½, 1)

(1, 1)

(0, 1)

C1

C2 Q

P O (0, 0) C3

(1, 0) R

Solution : Refer to the figure in question paper. Let the coordinates of P be (x , x2) where 0  x  1. For the area (OPRO), upper boundary : y = x2 lower limit of x : 0 upper limit of x : x x



x

x

x3 2 t dt  f (t) dt   f (t) dt area (OPRO) =  0 3 0 0

for the area (OPQO) the upper curve : x = the lower curve : x = y/2 lower limit of y : 0 and upper limit of y : x2



153

area (OPQO) =

x2

x2



t dt  

0

0

=

2 3/ 2 x 2 1 2 x 2 [t ]0  [t ]0 3 4

=

2 3 1 4 x  x 3 4

t dt 2

y

IIT - MATHS - SET - III according to the given condition, x

1 3 2 x4 x   f (t) dt  x 3  3 3 4 0

Differentiating both the sides w.r.t. x, we get x2 – f(x) . 1 = 2x2 – x3  f(x) = x3 – x2 , 0  x  1 2x, | x | 1  Let f(x) be a continuous function given by f(x) =  2 . Find the area of the  x  ax  b, | x |  1

6.

region in the third quadrant bounded by the curves x = –2y2 and y = f(x) lying on the left of the line 8x + 1 = 0. [1999] Solution : 2x,| x |  1  f(x) =  2 x  ax  b,| x |  1  f(x) = x2 + ax + b, if x < – 1 = 2x, if – 1  x < 1 = x2 + ax + b, if 1  x f is continuous on R, so f is continuous at – 1 and 1. 

Lim f (x)  Lim f (x)  f ( 1)

x 1

x 1

f (x)  Lim f (x)  f (1)  1 – a + b = – 2 and 2 = 1 + a + b and Lim x 1 x 1 Thus, a – b = 3 and a + b = I a = 2, b = – 1 

 x 2  2x  1 if  f (x)  2x if Hence,  x 2  2x  1 if 

x  1 1  x  1 x1

Next, we have to find the point x = –2y2 and y = f (x) The point of intersection is (–2, –1) 1/ 8





Required area =

2 1/ 8



 2

1

 x   f (x)  dx   2 

x dx   (x 2  2x  1) dx  2 2

x = -2y2x –1/8 y = f(x) –1/8

1/ 8



2x dx

1 1

1/ 8  x 3 2    (  x) 1/ 8  2    x 2  x    [x 2 ]1/1 8 3 2   2  3

2  1 3/ 2 3/ 2   1   8  1      2     1  1     4  2     1 3 2  8    3   64   3

154

AREA 

761 2 5 63 [ 2  2 9 / 2 ]   = 3 3 64 192

7.

Let b  0 and for j = 0, 1, 2, ... n, let Sj be the area of the region bounded by the y–axis and the curve j ( j  1)  y xeay = sin by, . Show that S0, S1, S2, ... Sn are in geometric progression. Also, find their b b sum for a = –1 and b =  . [2001] Solution : Given x = sin by e–ay Now, – 1  sin by  1 – e–ay  sin by  1  –e–ay  x  e–ay  In this case if we take a and b positive, the values –e–ay and e–ay become left bond and right bond of the curve. And due to oscillating nature of sin by, it will oscillate between x = e –ay and x = –e–ay . ( j1)  b

Now S j 



sin by e–ay dy  I =

 sin by e

 ay

dy

/b

Integrating by parts, we get

I

e  ay (a sin by  b cos by) [by Uler’s Integral] a 2  b2

So, S j  

 a ( j1)   aj 1 e  b {a sin ( j  1)  b cos( j  1)}  e b {a sin j  b cos j  2  a b  2

 aj 1   1 ( j1)    S j   a 2  b 2 e b (0  b( 1) j1 )  e b (0  b( 1) j ) 

a  j b

j

 b.(1) e a 2  b2

a  j



b.e b a  1 a    2 2 e b e b 1 a  b







 (–1)j+2 = (–1)2 . (–1)j = (–1)j

be Sj Now, S j1

e

 e

155

 be

1  j b



a

(e b  1) 2 a  b2

a  ( j 1)  b



a

(e b  1) a 2  b2

x = e–ay x = e–ay

S3 S2

a  j b

a  ( j1)  b

y

a

= e b  = constant

S1 O

S0

x

IIT - MATHS - SET - III Sj



S j1 = constant  S0, S1,S2 ... Sj from a G.P..

for a = – 1 and b =  1

.j

.e  .e j 1 Sj  .  (1  e)  (1  2 ) e  1 (1   2 )



n

 Sj 



j 0

.(1  e) (1  2 )



n j

(1  e)

 e  (1   ) (e 2

0

 e1  ...e n )

j 0

(1  e) (e n 1  1) . = (1  2 ) e  1 Find the area of the region bounded by the curves y = x2, y = |2 – x2| and y = 2, which lies to the right of the line x = 1. [2002] Solution : The points in the graph are 8.

A(1, 1), B( 2, 0), C(2, 2), D( 2, 2) 2

2

2 2 Required area =  (x  2(2  x ))dx  1

 (2x

2

 2))dx

2

2



 (2  (x

y =x

2 2

 2)dx 

1

 (4  x

2

)dx

2

y = |2–x2|

2

2

 2x 3   x3    2x    4x   3   3  

2

A

y =1

2

x =1

    20  12 2  square units.  3 

156

AREA

157


7

DIFFERENTIAL EQUATIONS

158

DIFFERENTIAL EQUATION An equation involving independent variable (x), dependent variable (y) and derivative of dependent  dy  variable with respect to independent variable   is called a differential equation e.g.  dx 

(i)

dy  x ln x dx

dy (iii) y = x +a dx

(ii)

dy = cos x dx

(iv)

 d2 y   dy  1   b  2  dx     dx 

4

2

ORDER AND DEGREE OF A DIFFERENTIAL EQUATION The order of a differential equation is the order of the derivative of the highest order occuring in the differential equation. The degree of a differential equation is the degree of the highest order differential coefficient appearing in it, provided it can be expressed as a polynomial equation in derivatives.

FORMATION OF A DIFFERENTIAL EQUATION WHOSE GENERAL SOLUTION IS GIVEN: A differential equation can be derived from its equation by the process of differentiation and other algebraical processes of elimination etc. In order to obtain a differential equation whose solution is f(x, y, c1, c2, . . . , cn) = 0

. . . (i)

where c1, c2, . . ., cn are n arbitrary constants, we have to eliminate the n constants for which we require (n + 1) equations. The given relation along with n more, obtained by successively differentiating it n times, provide us the required (n + 1) relations. The differential equation thus obtained is clearly of the nth order.

SOLUTION OF A DIFFERENIAL EQUATION A solution of a differential equation is an equation which contains arbitrary constants as many as the order of the differential equation and is called the general solution. Other solutions, obtained by giving particular values to the arbitrary constants in the general solution, are called particular solutions. Also, we know that the general integral of a function contains an arbitrary constant. Therefore, the solution of a differential equation, resulting as it does from the operations of integration, must contain arbitrary constants, equal in number to the number of times the integration is involved in obtaining the solution, and this latter is equal to the order of the differential equaiton. Thus we see that the general solution of a differential equation of the nth order must contain n and only n independent arbitrary constants. 159


SOLUTION OF A DIFFERENTIAL EQUATION BY THE METHOD OF VARIABLE SEPARATION If the coefficient of dx is only a function of x and that of dy is only a functinon of y in the given differential equation, then the equation can be solved using variable separation method Thus the general form of such an equation is f(x) dx +  (y) dy

... (i)

Integrating, we get

 f (x)dx   (y).dy  C which is the solution of (i).

SOLUTION OF A DIFFERENTIAL EQUATION OF THE TYPE

Consider the differential equation

dy = f(ax + by + c). dx

dy = f (ax + by + c) dx

. . . (i)

where f(ax + by + c) is some function of ‘ax + by + c’. Let z = ax + by + c.





dz a dz dy ab or dy  dx . dx dx dx b

(i)

dz a dz dz  dx  f (z)   bf (z)  a   dx b dx bf (z)  a

. . . (ii)

In the differential equation (ii), the variables x and z are separted. Integrating (ii), we get

dz

 bf (z)  a   dx  C 

dz

 bf (z)  a  x  C, where z = ax + by + c

This represents the general solution of the differential equation (i).

SOLUTION OF DIFFERENTIAL EQUATION OF THE TYPE dy a1x  b1y  c1 a b c  where 1  1  1 . dx a 2 x  b 2 y  c 2 a 2 b2 c2

160

DIFFERENTIAL EQUATION Consider the differential equation

a b c dy a1x  b1y  c1  where 1  1  1 a 2 b2 c2 dx a 2 x  b 2 y  c 2

. . . (i) Let

a1 b1    (say) a 2 b2



a1  a 2 , b1  b 2



(i) becomes

dy a 2 x  b 2 y  c1  dx a 2x  b2 y  c2



dy   a 2 x  b 2 y   c1  dx a 2x  b2 y  c2

. . . (ii)

Let z = a2x + b2y



dz  a2 z  c1 (ii) becomes dx  b2 z  c2



dz b 2z  b 2c1  a 2 z  a 2c 2  dx z  c2



z  c2 dz  dx  b2  a 2  z  b2c1  a 2c2



dz b 2  z  c1    a2 dx z  c2

. . . (iii)

In the differential equation (iii), the variables x and z are separted.

zc

Integrating (iii), we get



  b2  a 2  z  b2 2c1  a 2c2 dz   1.dx  C

z  c2 dz  x  C , where z = a2x + b2y.  b2  a 2  z  b2c1  a 2c2

This represents the general solution of the differential equation (i) Remark. In differential equation (ii),

If we take f(a2x + b2y) =

  a 2 x  b 2 y   c1 a 2x  b2 y  c2

  a 2 x  b 2 y   c1

is a function of a2x + b2y..

, then (ii) becomes

a 2x  b2 y  c2

dy = f(a2x + b2y) and we dx

have already learnt the method of solving this type of differential equation by substituting z = a2x + b2y

HOMOGENEOUS DIFFERENTIAL EQUATION A function f(x, y) is called homogeneous function of degree n if

161

IIT - MATHS - SET - III f  x, y    n f (x, y) For example: (a) f(x, y) = x2y2 – xy3 is a homogeneous function of degree four, since f(  x,  y) = (  2x2)(  2y2) – (  x)(  3y3) =  4(x2y2 – xy3) =  4 f(x, y). x3 y (b) f(x, y) = x e + + y2 ln   is a homogeneous function of degree two, since y x 2

x/y

3x 3  y  x y    2 y2 ln   f(  x,  y) = (  x ) e / + y  x 



2 2



 2 x / y x 3  y     y 2 ln      2 f (x, y) =  x e y  x    2

dy = f(x, y), where f(x, y) is a homogenous polynomial of dx degree zero is called a homogenous differential equation. Such equations are solved by sub-

A differential equation of the form

stituting v = y/x (or x/y) and then seperating the variables.

DIFFERENTIAL EQUATION REDUCIBLE TO HOMOGENOUS FORMS Equation of the form

ax  by  c dy = (aB  Ab) can be reduced to a homogenous form dx Ax  By  C

by changing the variables x, y to X, Y by equations x = X + h, y = Y + k, where h, k are constants to be chosen so as to make the given equation homogenous, we have

dy d  Y  k  dY dY dX dY    .  dx dx dx dX dx dX So equation becomes aX  bY   ah  bk  c  dY  dX AX  BY   Ah  Bk  C 

Let h and k be chosen so as to satisfy the equations ah + bk + c = 0 and Ah + Bk + C = 0 This gives h =

bC  Bc Ac  aC , k , which are meaningful except when aB = Ab aB  Ab aB  Ab

Thus the reduced equation is

dY aX  bY  , which can now be solved by means of the dX AX  BY

substitution Y = VX .

LINEAR DIFFERENTIAL EQUATION 162

DIFFERENTIAL EQUATION Form :

dy + Py = Q, where P, Q are functions of x alone dx

Integrating factor = e 

Pdx

Pdx dy Pdx Multiplying the form by e  Pdx on both sides, we get e    Py   Q.e   dx 

Or,

d dy  Pdx  Pdx = Qe  Pdx .e + y.. e dx dx

Or,

d dx

Or, y .e 

Pdx  y.e  Pdx    = Q. e  or,  

Pdx



 Q .e

d   Pdx    Pdx    P.e  since dx .  e   

d 

 dx .  y.e

 Pdx  dx  Q.e  Pdx dx   

 Pdx + C

Which is the required solution of the given differential equation.

DIFFERENTIAL EQUATION REDUCIBLE TO THE LINEAR FORM f (y)

dy  f (y)P(x)  Q(x) dx

The transformation f(y) = u

 f (y)dy  du

The equation (i) reduces to du  uP(x)  Q(x) dx

which is of the linear differential equation form The given equation can be written as

d tan  r tan 2    dr sec 2  sec2 

 sec 2   d 1 i.e.  tan 2   dr  tan   r   d  cot  = r i.e. cosec2  dr

Using the transformation cot  = u, we get – cosec2  d  du

163

. . . (i)

IIT - MATHS - SET - III the equation reduces to du  u  r dr

whose

I.F. = e  1dr  e  r .

Hence, the solution of the given differential equation is given by ue–r = –  re  r dr  re  r   e  r dr  re  r  e  r  C i.e. u = r – 1 + Cer i.e. cot  = r – 1 + Cer.

GENERAL FORM OF VARIABLE SEPRATION If we can write the differential equation in the form f(f1(x, y)) d(f1(x, y)) +  (f2(x, y)) d(f2(x, y)) + ... = 0, then each term can be easily integrated separately. For this the following results must be memorized. (i)

d(x + y) = dx + dy

(iii)

 x  ydx  xdy d  y2 y

(v)

d(log xy) 

ydx  xdy xy

 1 y  xdy  ydx  (vii) d  tan x  x 2  y 2 

(ii) d(xy) = ydx + x dy  y  xdy  ydx (iv) d    x x2 y  (xdy  ydx)  (vi) d  log   x xy 

(viii) d





x 2  y2 

xdx  ydy x 2  y2

DIFFERENTIAL EQUATIONS OF FIRST ORDER AND SECOND DEGREE The differential equation of first order and second degree will be a quadratic in dy/dx. Find two values of dy/dx i.e, two differential equations of first degree and first order and solve them.

ORTHOGONAL TRAJECTORY Any curve which cuts evey member of a given family of curves at ritght angle, is called an orthogonal trajectory of the family. For example, each straight line passing through the origin, y = mx, is an orthogonal trajectory of the family of the circles x2 + y2 = a2

164

DIFFERENTIAL EQUATION Method of finding orthogonal trajectory of a given family of curves: (i)

Let f(x, y, c) = 0 be the equation, where c is an orbitary parameter, of given family.

(ii) Differentiate the given equation with respect to x and then eliminate c. (iii) Replace

dx dy by in the equation obtained in (ii) dy dx

(iv) Solve the differential equation in (iii)

165

IIT-MATHS-SET-III

7-A

DIFFERENTIAL EQUATION

ASSIGNMENT

166


WORKEDOUT ILLUATRATION ILLUSTRATION : 01 1.

The differential equation of all non-horizontal lines in a plane is d2x dy d2y 0 A) B) 2  0 C) 2 =0 dy dx dx

D)

dx 0 dy

Solution : The general equation of all non-horizontal lines in xy–plane is ax  by  1, where a  0. dx a b0 [Diff. w.r.to y] dy2 d x a 2 0 [Diff. w.r.to y] dy 2 d x  2 0 dy d2x Hence, the required differential equations is 2  0 dy

ILLUSTRATION : 02 The degree of the differential equation satisfying 1  x 2  1  y 2  a x  y  is A) 1

b) 2

c) 3

4) None of these.

Solution : We have, a(x-y) = 1  y 2  1  x 2 Putting x=sinA, y=sinB, we get Cos A+CosB = a(sinA-sinB) AB  a Þ A  B  2 cot 1 a Þ cot 2 Þ sin 1 x  sin 1 y  2 cot 1 a. Differentiating w.r. to x, we get 2 1 1 dy   0 Þ dy  1  y 1 x2 1  y 2 dx dx 1 x2 Clearly, it is differential equation of first order and first degree.

ILLUSTRATION : 03 The order of the differential equation whose general solution is given by y  C1  C 2  sin x  C 3   C 4e x C5 is A) 5

B) 4

c) 2

Solution : We have, y   C1  C2  sin  x  C3   C4 e x C5  y  C6 sin  x  C3   C4 eC5 .e x , where C6  C1  C2 167

d) 3

IIT-MATHS-SET-III  y  C6 sin  x  C3   C7 e x , where C4 eC5  C7 Clearly, the above relation contains three arbitrary constants. So, the order of the differential equation satisfying it is 3.

ILLUSTRATION : 04 A solution of the differential equation is A) y=2

B) y=2x

c) y=2x-4

d)

Solution : Clearly, y=2x–4 satisfies the given differential equation.

ILLUSTRATION : 05 The differential equation representing the family of curves , where c is apositive parameter, is of (A) order 1

B)order 2

C)degree 3

D) degree 4

Solution :



2 We have, y  2c x  c



2 yy1  2c



yy1  c



c



y y1

Eliminating c from (i) and (ii) , we get. y2 

y 2y   x   y1  y1 



yy1  2x  2



 yy1  2x 

2

y y1

y1  4 y

Clearly, it is a differential equation of order one and degree 3.

ILLUSTRATION : 06 dx  x  1 (x is the distance described) . dt The time taken by a particle to traverse a distance of 99 metres is

A particle moves in a straight line with a velocity given by

168

DIFFERENTIAL EQUATION (A) log10 e

B) 2log e 10

1 D) log10 e 2

C) 2log10 e

Solution : dx  x 1 dt



log  x  1  t  C.

Putting t=0, x=0, we get log1=C 

C=0.

So, t = log (x+1) For x=99 ; t = log e 100=2 log e 10.

ILLUSTRATION : 07 g  x If f  x  , be twice different ial functions on 0,2 satisfying " " ' ' f  x   g  x  , f 1  2g  1  4 and f  2   3g  2   9, then f  x   g  x  at x  4 equals. (A) 0

B) 10

C) 8

D) 2

Solution : We have, f "  x   g"  x  . On integration, we get f '  x   g '  x   C ……. (i) Putting x=1, we get f '  1  g  1  C 

4  2C



C2

 f '  x   g'  x   2 Integrating w.r.t.x, we get f(x) =g(x) +2x+ Putting x = 2, we get.

f  2   g  2   4  c1 

169

9  3  4  c1

……. (ii)

IIT-MATHS-SET-III 

c1  2

 f  x   g  x   2x  2 Putting x=4, we get f(4) -g(4) =10.

ILLUSTRATION : 08 d The function f(q) = d



dx

 1  cos  cos x satisfies the differential equation. 0

(A)

df  2 f   cot   0 d

B)

df  2 f   cot   0 d

C)

df  2 f    0 d

D)

df  2 f    0 d

Solution : 

d dx 1 2 We have, f    d  1  cos  cos x  1  cos 2   cos ec  0



df    2cos ec 2 cot  d



df    2 f ( ).cot   0 d

  f    cos ec 2 

ILLUSTRATION : 09 The equation of the curve satisfying the differential equation y2  x 2  1  2 xy1 passing through the point (0,1) and having slope of tangent at x=0 as 3 is (A) y  x 2  3x  2

B) y 2  x 2  3 x  1

C) y  x3  3 x  1

D) None of these

Solution : The given differential equation is y 2  x 2  1  2 xy1



y2 2x  2 y1 x  1

Integrating both sides, we get. log y1  log  x 2  1  log C

 y1  C  x 2  1

……. (i)

It is given that y1  3 at x  0 170

DIFFERENTIAL EQUATION Putting x=0, y1  3 in (i) , we get C = 3 Substituting the value of C in (i), we obtain y1  3  x 2  1 Integrating both sides w.r.t to x, we get y  x 3  3x  C2 This passes through the point (0,1). Therefore, =1 Hence, the required equation of the curve is y  x3  3 x  1

ILLUSTRATION : 10 The form of the differential equation of the central conics, is dy (A) x  y dx

dy 0 B) x  y dx

Solution : Central conics are given by ax 2  by 2  1 

2ax  2byy1  0



ax  byy1  0



 a yy1  b x



a  by12  byy2  0



a  y12  yy2 ……. (ii) b

From (i) and (ii) , we get yy1  y12  yy2 x



171

yy1  xy12  xyy2

……. (i)

2

d2y dy  dy  x  xy y C)   D) None of these 2 dx dx  dx 

IIT-MATHS-SET-III

SECTION A - SINGLE ANSWER TYPE QUESTIONS 1.

Differential equation of the family of curves v = A/4 + B, where A and B are arbitrary constants, is A)

2.

d2v

1 dv  . 0 2 r dr dr

B)

d2v

2 dv  . 0 2 r dr dr

C)

d2v

2 dv  . 0 r dr dr 2

The differential equation of the family of curves y = Ae 3 x  Be 5 x , where A, B are arbitrary constants, is d2y

dy d2y dy 8  15 y  0 B) 8  15 y  0 A) 2 2 dx dx dx dx 3.

If (1 - y) x

If y = y(x) and

7.

B) log xy + x - y = c

B) 2/3

1  y  log y  c x

dy  y  0 D) None of these dx

C) log xy - x - y = c

C) -1/3

1 1 B) x  y  y 2  c

Solution of differential equation

D) None of these

D) 1

1 1 C) x  y  log y  x  c D) none of these

dy  sin( x  y)  cos( x  y ) is equal to dx

xy   xc A) log  2  sec 2  

B) log { 1 + tan (x + y)} = y + c

x  y   yc C) log 1  tan 2  

x  y   xc D) log 1  tan 2  

 dy  The solution of the equation log    ax  by is  dx 

A) 8.



The solution of the equation (x2 - yx2) dy + (y2 + x2y2) dx = 0 A)

6.

dx

2

2  sin x  dy    = - cos x, y (0) = 1, then y (p/2) equals y  1  dx 

A) 1/3 5.

C)

d2y

dy  (1  x) y = 0 then solution of above equation is dx

A) log xy + x + y = c

4.

D) none of these

e by e ax  c b a

B)

e  by e ax  c b a

C)

e  by e ax  c a b

D) None of these

Solution of the differential equation x2 dy + y (x + y) dx = 0 is A) y + 2x + c2x2y

B) y - 2x = c2x2/y

C) y + 2x = c2x2/y

D) none of these

172

DIFFERENTIAL EQUATION 9.

Solution of the differential equation x dy - ydx = ( x 2  y 2 ) dx is A) [ y +

( x 2  y 2 ) ] = ky2

B) [y +

C) [ y + ] = k (x2 + y2) 10.

12.

D) None of these

The solution of the equation (x2 - xy) dy = (xy + y2) dx is A) xy = ce  y / x

11.

C) yx2 = ce1/ x

B) xy = ce  x / y

dy The solution of the equation x dx = x2 + xy + y2 is 1  y  A) tan    log x  c x

1  x  B) tan    log x  c  y

1  x  C) tan    log x  c  y

1  y  D) tan    log y  c x

The solution of the equation

dy 3 x  2 y  5   0 is dx 2 x  3 y  5

1  y  1  B) sin  a  x 

 1 y  C) sin  a  1 x 

B) x - y = ke x  y

C) x + y = ke x  y

2

2

D) none of these

dx  x  1 (x is the distance described). The dt time taken by the particle to traverse a distance of 99 metres is

A particle moves in a straight line with a velocity given by

A) log10e

173

D) x - y = ke x  y

The differential equation of all parabolas having their axis of symmetry coinciding with teh axis of x is d 2 x  dx  d 2 y  dy  d 2 y dy 0 A) y 2     0 B) x 2     0 C) y 2  dy  dy  dx  dx  dx dx

16.

 y  D) sin  a  x 1 

Solution of the differential equation (x + y) (dx - dy) = dx + dy is A) x + y = ke x  y

15.

B) x2 + 4xy - y2 - 4x + 6y = k D) None of these

 dy  Solution of differential equation sin    a when y (0) = 1  dx  1  y  1  A) sin  a  x 

14.

D) none of these

2

A) 3x2 + 4xy + 3y2 - 10x - 10y = k C) (x + 2y)2 + 3y = k

13.

( x 2  y 2 ] = kx2

B) 2 log e 10

C) 2 log10e

D)

1 log10e 2

IIT-MATHS-SET-III 2

17.

dy  dy  A solution of the differential equation    x  y  0 is dx  dx 

A) y = 2 18.

B) y = 2x

D) y = 2x2 - 4

C) y = 2x - 4

Solution of the differential equation where y (0) = 1/8, y1 (0)  0, y2 (0)  1 is equal to

1  e8 x 7 1  e8 x 7 1  e8 x 7  x   x   x   D) none of these  B) y =   C) y =  A) y = 8  8 8 8 8 8 8 8 8 

19.

dy yf 1 ( x )  y 2  Solution of differential equation is equal to dx f ( x) A) xy = f(x) + c

B) xy = f(x) + cx

C) y(x+c) = f(x)

D) y = f(x) + x + c

dy 2 2 2 dx  x cos ( x  y ) Solution of differential equation is equal to dy y3 yx dx x y

20.

x2 A) tan( x  y )  2  c y 2

2 2 C) tan( x  y ) 

x 21.

The solution of

1 A) sin

22.

x

2

A) 23.

2

y2 c x2

2

2 2 D) cot( x  y ) 

y2 c x2

dy y dx = m x2 is given by 2 2 x y

y mx 2  c x 2

1 B) sin y 

mx 2 y mx 2  cx C)  c 2 x 2

D) None

 y 2  dy  xy dx. If y (x0) = e, y(1)= 1, then value of x0 =

3e

B)

e2 

1 2

C)

e2  1 2

D)

e2  1 2

If x dy = y(dx + ydy), y > 0 and y(1) = 1, then y(-3) is equal to A) 1

24.

x2 B) cot  x  y   2  c y

2

B) 3

C) 5

The degree of the differential equation whose solution is A) 6

B) 9

C) 12

D) –1



cx  y

1/3

  y c 3

2

is

D) 15

174

DIFFERENTIAL EQUATION 25.

The differential equation y

dy  x  a represents dx

A) a set of circles whose centres are on the x-axis B) a set of circles whose centres are on the y-axis C) a set of parabolas D) a set of ellipses 26.

A particle moves in a line with velocity given by

ds  s  1 . The time taken by the particle to cover a dt

distance of 9 meters is

27.

A) 1

B) loge 10

General solution of tan y

dy  sin( x  y )  sin( x  y ) is dx

A) sec y = 2 cosx + c 28.

3 x  ex  c y

The solution of

B)

The solution of

C) 4 y  e 2 x  cx 2  d D) 4 y  e 2 x  cx 2  dx

B) 2e 2 y 2  e 4 x  e 4

C) 2e 2 y  2  e4 x  e4

D) 3e 2 y  2  e3 x  e3

 y  xdy  2  1 dx is 2 2 x y x y  B) y = x cot(c – x)

C) y = xtan(c – x)

2 d 2 y   dy   The order and degree of the differential equation 2  1     dx   dx  

B) 2, 1

C) 1, 2

D) y = xcos(c – x) 3/ 2

are D) 2, not defined

The D.E. which represents the family of plane curve y = ecx where c is parameter, is

Integrating factor of A) xex

175

3

D) xye x  c

2

A) y1 = cy 34.

3

C) xy  e x  c

 dy  The solution of log    4 x  2 y  2 given that y= 1 when x = 1 is  dx 

A) 2, 2 33.

3 y  ex  c x

B) 4 y  e 2 x  cx  d

A) y2 = x3 tan(c – x)

32.

4) secy + cosx = c

d2y  e 2 x is dx 2

A) 2e 2 y  2  e4 x  e2

31.

B) secy + 2cosx = c C) sec y = cosx + c x3

A) 4y = e 2x

30.

D) none

The general solution of ydx – xdy – 3x 2y2 e dx = 0 is

A)

29.

C) 2 loge 10

B) xy1 =lny = 0

C) xlny = yy1

D) ylny = xy1

C) (1+x)ex

D) log (1+x)

dy (1  x)  xy  1  x is dx

B) (1+x)ex

35.

IIT-MATHS-SET-III The slope of a curve at any point on it is the reciprocal of twice the ordinate at that point. If the curve passes through (4, 3) then its equation is A) y2 = x

36.

B) 4y2 = 9x

B) 1, 2

d 2x 2 B) 2  p x dt

The general solution of

The solution of

43.

The solution of y - x

D) y

dy x dx

D) y2 – 2xy – x2 = c

C) sin(y – x) = ce x2

D) sin(y-x) = cex

dy dy    a  y 2   is dx dx  

B) (x+a)(1-ay) = cy

C) (x+a)(1-ay) = c

D) (x+a)(1+ay) = cy

Let P be the amount of petrol left at time t. If the rate of evaporation of petrol is proportional to the amount remaining, then B) P = ce-kt

C) P = ct

D) Pt = c(k>0)

The differential equation obtained from y = Ae3x + Be5x where A, B are parameters is

d2y dy A) 2  9  15 y  0 dx dx

d2y dy  9  15 y  0 B) 2 dx dx

d2y dy C) 2  9  15 y  0 dx dx

d2y dy  9  15 y  0 D) 2 dx dx

The general solution of xdx + ydy =

xdy  ydx is x2  y2

A) x2 + y2 = 2tan-1 x/y + c C) x2 + y2 + 2 tan-1x/y + c 44.

C) y2 + 2xy – x2 = c

B) sin(y-x) = ce x2 / 2

A) P = cekt 42.

dy y dx

dy  x tan( y  x)  1 is dx

A) (x+a) (x+ay) = cy 41.

C) x

B) (x-y)2 = c

A) 2sin(y-x) = x2 + c

40.

D) 2, 2

dy x  y  is dx x  y

A) (x+y)2 = c 39.

C) 2, 1

By eliminating the parameters A,  in x = A cos(pt + ), the differential equation is

d 2x A) 2  x dt

38.

D) y2 = x+5

The degree and order of the D.E. of all parabolas with x-axis as axis is A) 1, 1

37.

C) y2 = 2x + 1

The general solution of y

B) x2 + y2 = 2 tan-1y/x+c D) x2 + y2 + 2tan-1y/x + c

2 2 dy  xe x  y is dx

A) 1 + exp[-(x2 + y2)] = exp[-(x2 + c)]

B) c + exp(-x2) = exp(-y2)

C) e x2 - e y 2 = c

D) e x2  e  y 2 = c 176

DIFFERENTIAL EQUATION 45. The x-intercept of the tangent at any point of a curve y = f(x) is equal to twice the abscisa. Then the curve is A) x2 + y2 = c

B) x2 – y2 = c

C) xy = c

D) y = cx

46.

The slope of the tangent at (x, y) to a curve y = f(x) passing through (1, /4) is given by y/x – cos2(y/x). Equation of the curve is A) tan (y/x) = log(x/e) B) tan(y/x) = log(ex) C) tan (y/x) = log(e/x) D) log (y/x) = tanx

47

The different equation of all conics whose centre lie at the origin is of order A) 2

48

d2x B) 2  0 dy

 d3y  The degree of the differential equation  3   dx 

A) 1 50

C) 4

D) none of these

The differential equation of all non-vertical lines in a plane is

d2y A) 2  0 dx

49

B) 3

B) 2

C)

dy 0 dx

2/3

 43

D)

dx 0 dy

d2y dy  5  0 is 2 dx dx

C) 3

D) none of these

The differential equation of family of curves whose tangent form an angle of

 with the hyperbola 4

xy  C 2 is

dy x 2  C 2  A) dx x 2  C 2 51

dy C2  2 C) dx x

D) None of these

The slope of a curve at any point is the reciprocal of twice the ordinate at the point and it passes through the point (4, 3). The equation of the curve is A) x 2  y  5

52

dy x 2  C 2 B)  2 dx x  C 2

B) y 2  x  5

C) y 2  x  5

D) x 2  y  5

The order of the different equation associated with the primitive y1  c1  c2 e x  c3e 2 x  c4 , where c1 , c2 , c3 , c4 are arbitrary constants, is A) 3

53

B) 4

C) x  A1 y 2  A2 y

D) None of these

The solution of ydx  xdy  3 x 2 y 2 e x dx  0 is A)

177

D) none of these

The solution of the differential equation y1 y3  3 y22 is A) x  A1 y 2  A2 y  A3 B) x  A1 y  A2

54

C) 2

3 x  ex  C y

B)

x x3 e  0 y

C) 

3 x  ex  C y

D) None of these

55

IIT-MATHS-SET-III The curve in which the slope of the tangent at any point equals the ratio of the abscissa to the ordinate of the point is A) an ellipse

56

B) a parabola

A differential equation associated to the primitive y  a  be5 x  ce7 x is A) y3  2 y2  y1  0

57

58

A) y1  y 2  x 2   2 xy  a 2  0

B) y1 y 2  xy  a 2 x 2  0

C) y1  y 2  x 2  a 2   2 xy  0

D) None of these

The solution of

dy ax  h  represents a parabola when dx by  k C) a  0, b0, or a 0, b = 0D) a = 2, b = 1

B) a  1, b  2

x2  x  1 dy dy y 2  y  1  0 The family of curves represented by = 2 and the family represented by y  y 1 dx dx x 2  x  1 A) touch each other

60

B) 4 y3  5 y2  20 y1  0 C) y3  2 y2  35 y1  0 D) none of these

The differential equation of the family of circles passing through the fixed points (a, 0) and (-a, 0) is

A) a  0, b  0

59

C) a rectangular hyperbola D) a circle

B) are orthogonal

C) are one and the same

The differential equation of all conics whose axes coincide with the axes of coordinates is der A) 2

B) 3

C) 4 2

61

D) 1

2

B) 2

C) 3

D) none of these

The degree of the differential equation corresponding to the family of curves y = a(x+a)2, where a is an arbitrary constant is A) 1

63

of or-

 d 2 y   dy   d2y    x sin The degree of the differential equation  2      is  dx   dx   dx 

A) 1 62

D) none of these

B) 2

C) 3

D) none of these

The differential equation of all parabolas whose axes are parallel to y-axis is

d3y 0 A) dx 3

d2x B) 2  C dy

d 3 y d 2x  0 C) dx 3 dy 2

D)

d2y dy 2 C 2 dx dx

178


KEY 1

2

3

4

5

6

7

8

9

10

11

c

d

b

a

c

d

b

a

b

d

a

16

17

20

21

22 23

24

25

26

b

c

a

a

c

a

31

32

35

36

39

c

a

d

b

46

47

50

51

a

b

a

b

c

61

62

63

d

c

a

179

18 19

c

c

33 34

d

b

48 49

b

a

b

37 38

c

c

52 53

a

a

12 13

14

15

c

a

27 28

29

30

b

b

b

c

40

41

42 43

44

45

b

b

b

b

a

c

54

55

56

59

60

a

c

c

b

a

a

b

a

c

57 58

c

c

IIT-MATHS-SET-III

SECTION B - MORE THAN ONE ANSWER TYPE QUESTIONS 2

1

dy x  dy  (e  e x )  1  0 are given by   Solution of the differential equation  dx  dx 

A) y + e  x  c 2

C) y + e x  c

B) order 2

C) degree 3

D) degree 4

The differential equation of the curve such that the initial ordinate of any tangent is equal to the corresponding subnormal is A) a linear equation C) an equation with seprable variable

4

D) y - e x  c

The differential equation representing the family of curves y2 = 2c (x + ), where c is a positive parameter, is of A) order 1

3

B) y - e  x  c

B) a homogeneous equation D) none of these

Solution of the differential equation xdy  y 2 e xy  e x / y   ydx(e x / y  y 2 e xy ) is equal to

A) e xy  e x / y  c

x/ y B) e 

y  xy x

C) xy  log(e x / y  c)

D) e xy  log( xy  c )

KEY

1

2

3

4

A,D A,C A,B A,C

180


181

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