BANSAL CLASSES TARGET IIT JEE 2007
M A T H E M A T IC IC S XII & XIII
FINA FINAL PRA PRACTICE CTICE PROB PROBL LEM EMS S FOR FOR IIT JEE-20 E-2007 07 (With Hints Hints and Solutions at the End)
Q.1 to Q.29 are of 6 Marks Problems Q.30 to Q.66 are of 8 Marks Problems Q.67 to Q.82 are of 10 Marks Problems Q.83 to Q.100 Q.1 00 are Objective type problems.
Advise : Do not spend more than 10 minutes for each problem and then read
the solution and then do it. ALL
THE
BEST
FOR
JEE
-2007
SUBJECTIVE:
Q.1
333 x 2
If the sum sum of the the roots roots of the the equa equati tion on 2
111x 1
2
2
222 x 2
S1 1 is expressed in the form S find 2
S1
S1 + S2, where S is in its its lowest form. 2
[6]
Q.2
Let K is is a positiv positivee integer integer such such that 36 + K, K, 300 + K, K, 596 + K are are the squar squares es of three three consecu consecutiv tivee terms ter ms of an arithmetic progression. pro gression. Find K. [6]
Q.3
Find Find the the numb number er of 4 digi digitt numbers numbers startin startingg with with 1 aand nd havi having ng exactl exactlyy two two identi identical cal digits digits..
Q.4
A chord of of the the pa parabola y2 = 4ax touches the parabola parabo la y2 = 4bx. Show that the tangents at the t he extremities extremities 2 2 of the chord meet on the parabola parabo la by = 4a x. [6]
Q.5
Consi Consider a circl circlee S with with centre centre at the origin origin and radiu radiuss 4. Four Four circle circless A, A, B, C and and D each each with with radi radius unity and centres (–3, 0), (–1, 0), 0) , (1, 0) 0 ) and ((3, 3, 0) respectively are drawn. A chord chord PQ of the circle circle S touches touc hes the circle B and passes through the t he centre of the circle C. If the length of this chord can be expressed as
Q. 6
Integrate
2
[6]
x , find x.
[6]
x7 dx (1 x 2 ) 5
[6]
1 sin sin 2 x
a dx = where a, b are ar e relatively relat ively prime find a + b + ab. 0 (1 sin sin 2 x ) 2 b
Q. 7
If
Q.8
A bus contractor contractor agrees to run speci special al buses buses for for the emp employ loyees ees of ABC ABC Co. Ltd. Ltd. He agrees agrees to run the the buses if atleast atleast 200 persons travel by his his buses. buses. The fare per person is to be Rs. 10 per per day if 200 travel and will be decreased for everybody ev erybody by 2 paise per person perso n over 200 that travels. How many passengers will give the contract cont ractor or maximum daily daily revenue? [6]
Q.9
If the the poi point nt P(a, P(a, b) lies on the the cur curve ve 9y2 = x3 such that the normal to the curve at P makes makes equal intercepts with the axes. Find the value of (a + 3b). 3 b). [6]
Q.10
Let x(t) x(t) be the concentrati concentration on of glu glucose cose per uni unitt volum volumee of blood blood at time time t, being the amount of glucose being injected injected per unit unit volume volume per unit unit time. If the glucose is is disappearing disappearing from from the blood at a rate rate proportion proportional al to the con concen centrati tration on of glucos glucosee (K bein beingg the con constan stantt of proportion proportional ality ity), ), find find x(t) x(t).. Also find ind the ultimate ultimate concentration co ncentration of glucose glucose as t . [6]
Q.11 .11
Find th the va value(s) ue (s) of the parameter 'a' (a > 0) for each of which the area of the figure figure bounded by the a2 a x x2 2 a x 3a 2 straight stra ight line, y = & the parabola y = is the greatest. 1 a4 1 a4
B ansal C lasses
Problems for for JEE-2007 JEE-2007
[6]
[6]
[2]
SUBJECTIVE:
Q.1
333 x 2
If the sum sum of the the roots roots of the the equa equati tion on 2
111x 1
2
2
222 x 2
S1 1 is expressed in the form S find 2
S1
S1 + S2, where S is in its its lowest form. 2
[6]
Q.2
Let K is is a positiv positivee integer integer such such that 36 + K, K, 300 + K, K, 596 + K are are the squar squares es of three three consecu consecutiv tivee terms ter ms of an arithmetic progression. pro gression. Find K. [6]
Q.3
Find Find the the numb number er of 4 digi digitt numbers numbers startin startingg with with 1 aand nd havi having ng exactl exactlyy two two identi identical cal digits digits..
Q.4
A chord of of the the pa parabola y2 = 4ax touches the parabola parabo la y2 = 4bx. Show that the tangents at the t he extremities extremities 2 2 of the chord meet on the parabola parabo la by = 4a x. [6]
Q.5
Consi Consider a circl circlee S with with centre centre at the origin origin and radiu radiuss 4. Four Four circle circless A, A, B, C and and D each each with with radi radius unity and centres (–3, 0), (–1, 0), 0) , (1, 0) 0 ) and ((3, 3, 0) respectively are drawn. A chord chord PQ of the circle circle S touches touc hes the circle B and passes through the t he centre of the circle C. If the length of this chord can be expressed as
Q. 6
Integrate
2
[6]
x , find x.
[6]
x7 dx (1 x 2 ) 5
[6]
1 sin sin 2 x
a dx = where a, b are ar e relatively relat ively prime find a + b + ab. 0 (1 sin sin 2 x ) 2 b
Q. 7
If
Q.8
A bus contractor contractor agrees to run speci special al buses buses for for the emp employ loyees ees of ABC ABC Co. Ltd. Ltd. He agrees agrees to run the the buses if atleast atleast 200 persons travel by his his buses. buses. The fare per person is to be Rs. 10 per per day if 200 travel and will be decreased for everybody ev erybody by 2 paise per person perso n over 200 that travels. How many passengers will give the contract cont ractor or maximum daily daily revenue? [6]
Q.9
If the the poi point nt P(a, P(a, b) lies on the the cur curve ve 9y2 = x3 such that the normal to the curve at P makes makes equal intercepts with the axes. Find the value of (a + 3b). 3 b). [6]
Q.10
Let x(t) x(t) be the concentrati concentration on of glu glucose cose per uni unitt volum volumee of blood blood at time time t, being the amount of glucose being injected injected per unit unit volume volume per unit unit time. If the glucose is is disappearing disappearing from from the blood at a rate rate proportion proportional al to the con concen centrati tration on of glucos glucosee (K bein beingg the con constan stantt of proportion proportional ality ity), ), find find x(t) x(t).. Also find ind the ultimate ultimate concentration co ncentration of glucose glucose as t . [6]
Q.11 .11
Find th the va value(s) ue (s) of the parameter 'a' (a > 0) for each of which the area of the figure figure bounded by the a2 a x x2 2 a x 3a 2 straight stra ight line, y = & the parabola y = is the greatest. 1 a4 1 a4
B ansal C lasses
Problems for for JEE-2007 JEE-2007
[6]
[6]
[2]
Q.12 Mr. Mr. A is a compul compulsi sive ve lia liarr. He lie liess 2 5 of the time. However a clue to his validity validity is is that his ears droop dro op t he time when he is telling telling a lie. They only only droop droo p 1 10 of the time t ime when he he is telling telling the truth. t ruth. 2 3 of the Mr. A tells his his friend Mr. Mr. B that "certain " certain event has occured" and his his ears were we re dropping dropp ing as noticed by Mr. B. Find the probability that Mr. A was telling the truth. [6]
Q.13
Five Five persons entered the lift lift cabin on the ground floor of an eight eight floor floor house. Suppose that each of them , independently independent ly & with equal pro probabili bability ty can leave the cabin at any floo floorr beginning with with the first, find out the t he probability of all 5 persons perso ns leaving leaving at different different floors. floor s. [6]
Q. 14 Let u and v be non zero vectors on a plane or in 3-space. Show that the vector w | u | v | v | u bisects bisects the angle angle between u and v . [6]
Q.15 Find Find the distan distance ce from from the line ine x = 2 + t , y = 1 + t , z =
1 1 t he plane x + 2y + 6z = 10. t to the 2 2 [6]
Q. 16
If is the angle between the lines in in which the planes planes 3x – 7y – 5z = 1 and 5x – 13y + 3z + 2 = 0 cuts the plane 8x 8 x – 11y + 2z = 0, find sin. [6]
Q.17
Suppose u, v and w are twice twice dif differen ferentiab tiable le fun function ctionss of x that satisf satisfyy the relations relations au + bv + cw = 0
u v w where a, b and c are constants , not all zero. Show that t hat u ' v' w ' = 0. u ' ' v' ' w ' '
[6]
A B C 3 + cos B · sin sin2 + cos C · sin sin2 . 2 2 2 8
Q.18 In any any triang triangle le ABC, ABC, prove prove that, cos A · sin2
[6]
3 Q.19 If the the norm normal alss to the the cur curve ve y = x2 at the t he points P, P, Q and R pass through the point 0, , find the radius 2 of the circle circumscribing circumscribing the triangl tr ianglee PQR. [6] Q.20 .20 Let A = {a R | the equation (1 + 2i)x3 – 2(3 + i)x2 + (5 – 4i)x + 2a2 = 0} has at least one real root. roo t. Find Find the t he value of
Q.21
a2 . aA
[6]
Find Find the equation equation of a line line passin passing g through (– 4, –2) having having equal equal intercepts intercepts on the coordinate coordinate axes axes.. [6]
Q.22 Q.22 Let Let S be be the set set of all all x suc suchh that that x4 – 10x2 + 9 0. Find the t he maximum value of f (x) (x) = x3 – 3x on S. [6]
Q.23 Solve Solve the diff differe erenti ntial al equ equati ation, on, (x4y2 – y)dx + (x2y4 – x)dy = 0
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Problems for for JEE-2007 JEE-2007
(y(1) = 1)
[6]
[3]
Q.24 All All the face face cards from a pack of 52 playin playing g cards are removed. removed. From the remain remainin ing g pack half half of the cards are randomly removed removed without withou t looking look ing at them and then randomly drawn two cards ca rds simultaneously p( 38C 20 )
from the remaining. remaining. If the probabil pro bability ity that two cards drawn dra wn are both bot h aces is 40
C 20 · 20 C2
, find p. [6]
x 2 y2 Q.25 A circl circlee inter interse sects cts an an ell ellipse ipse 2 2 = 1 precisel pr ecisely y at three t hree points A, a b B, C as shown in the figure. AB is a diameter of the circle and is perpendic perpendicula ularr to the major major axis axis of the ellips ellipse. e. If the eccentrici eccentricity ty of the ellipse is 4/5, find the length of the diameter diamete r AB in in terms of o f a. [6] Q.26 Suppose R is is set of reals reals and C is is the set of complex complex numb numbers ers and a functi function on is defin defined ed as f : R C, f (t) =
Q.27
1 ti where t R, prove that f is is injective. 1 ti
[6]
Circless A and B are external Circle externally ly tangent tangent to each other and to lin lineet . The sum of the radii of the two t wo circles is 12 and the t he radius of circle A is is 3 tim t imes es that of circle B. The area in between the two circles and and its b external tangent is a 3 – then find the value of a + b. 2
[6]
0 0 1 Q.28 Q.28 Def Define a mat matri rixx A = 3 0 . Find a vertical vector V such that (A8 + A6 + A4 + A2 + I) V = 11 where I is a unit matrix of order 2. [6] Q.29
A circle circle is inscrib inscribed ed in a triangle triangle with with sides of lengths lengths 3, 4 and 5. A second circle, circle, interior interior to the triangle, triangle, is tangent to the first circle circle and to both sides of the larger acute acut e angle of the triangle. If the radius radius of teh second circle can be expressed expre ssed in the form
sin k where k and w are ar e in degrees and lie in in the interval cos w
(0, 90°), 9 0°), find the value of o f k + w. w.
[6]
ax 2 24x b Q.3 .300 If the the equ equat atiion = x, has exactly two distinct real solutions solut ions and their sum is 12 then find x2 1 the value value of (a – b). [8] Q.31
If a, b, c and and d are positive integers integers and a < b < c < d such that a, b, c are in A.P. A.P. and b, c, d are are in G.P. .P. and d – a = 30. Find the four numbers. [8]
Q.32
Let the set A = {a, b, c, d, e} and P and Q are two non empty empty subsets of A. Find Find the number number of ways ways in which which P and Q can be selected so that P Q has at least one o ne common element. element. [8]
Q.33 If the nor norm mals als draw drawn n to the the curve curve y = x2 x + 1 at the t he points A, B & C on the t he curve are concurrent at the point P(7/2, P (7/2, 9/2) 9 /2) then compute the sum of the t he slopes of the three normals. Also Also find find their equations and the co-ordin co-ord inates ates of o f the feet feet of the normals onto the curve. [8]
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Problems for for JEE-2007 JEE-2007
[4]
Q.34 A conic passing through the point A(1, 4) is such that the segment joining a point P(x, y) on the conic and the point of intersection of the normal at P with the abscissa axis is bisected by the y- axis. Find the equation of the conic and also the equation of a circle touching the conic at A(1, 4) and passing through its focus. [8] Q.35
A hyperbola has one focus at the origin and its eccentricity = 2 and one of its directrix is x + y + 1 = 0. Find the equation to its asymptotes. [8]
Q.36 Let A, B, C be real numbers such that (i) (sin A, cos B) lies on a unit circle centred at origin. (ii) tan C and cot C are defined. If the minimum value of (tan C – sin A)2 – (cot C – cos B)2 is a + b 2 where a, b N, find the value of a3 + b3. [8] Q.37
For a 2, if the value of the definite integral
0
dx
1 a2 x x
2
equals
5050
. Find the value of a. [8]
4
Q.38
2 ( 4) tan If , find the value of (kw), where k, w N. d = ln k – w 1 tan 4
[8]
1
Q.39 Given a function g, continuous everywhere such that g(1) = 5 and g (t) dt = 2. 0
1 If f(x) = 2
x
(x t)2 g(t) dt , then compute the value of f (1) f (1).
[8]
0
1
Q.40
Let f : [0, 1] R is a continuous function such that
0 f (x )dx = 0. Prove that there is c (0, 1) such
c
0
that f ( x )dx = f (c).
[8]
Q.41 Consider the equation in x, x3 – ax + b = 0 in which a and b are constants. Show that the equation has only one solution for x if a 0, for a = 3, find the values of b for which the equation has three solutions. [8]
Q.42
A tank consists of 50 litres of fresh water. Two litres of brine each litre containing 5 gms of dissolved salt are run into tank per minute; the mixture is kept uniform by stirring, and runs out at the rate of one litre per minute. If 'm' grams of salt are present in the tank after t minute, express 'm' in terms of t and find the amount of salt present after 10 minutes. [8]
Q.43 Urn-I contains 3 red balls and 9 black balls. Urn-II contains 8 red balls and 4 black balls. Urn-III contains 10 red balls and 2 black balls. A card is drawn from a well shuffled back of 52 playing cards. If a face card is drawn, a ball is selected from Urn-I. If an ace is drawn, a ball is selected from Urn-II. If any other card is drawn, a ball is selected from Urn-III. Find (a) the probability that a red ball is selected. (b) the conditional probability that Urn-I was one from which a ball was selected, given that the ball selected was red. [8]
B ansal C lasses
Problems for JEE-2007
[5]
Q.44 The digits of a number are 1, 2 , 3, 4 , 5, 6 , 7 , 8 & 9 written at random in any order. Find the probability that the order is divisible by 11. [8] Q.45 A number is chosen randomly from one of the two sets, A = {1801, 1802,.....,1899, 1900} & B = {1901, 1902,.....,1999, 2000}. If the number chosen represents a calender year. Find the probability that it has 53 Sundays. [8]
Q.46
A box contains 2 fifty paise coins, 5 twenty five paise coins & a certain fixed number N ( 2) of ten & five paise coins. Five coins are taken out of the box at random. Find the probability that the total value of these five coins is less than Re. 1 & 50 paise. [8]
Q.47
A hunter knows that a deer is hidden in one of the two near by bushes, the probability of its being hidden in bushI being 4/5. The hunter having a rifle containing 10 bullets decides to fire them all at bushI or II . It is known that each shot may hit one of the two bushes , independently of the other with probability 1/2. How many bullets must he fire on each of the two bushes to hit the animal with maximum probability. (Assume that the bullet hitting the bush also hits the animal). [8]
Q.48
ABCD is a tetrahedron with A( 5, 22, 5); B(1, 2, 3); C(4, 3, 2); D( 1, 2, 3). Find
AB
(BC BD) . What can you say about the values of (AB BC) BD and (AB BD) BC .
Calculate the volume of the tetrahedron ABCD and the vector area of the triangle AEF where the quadrilateral ABDE and quadrilateral ABCF are parallelograms. [8] Q.49
Find the equation of the line passing through the point (1, 4, 3) which is perpendicular to both of the lines x 1 y3 z2 x2 y4 z 1 = = and = = 2 2 1 4 3 2 Also find all points on this line the square of whose distance from (1, 4, 3) is 357.
[8]
Q.50
Find the parametric equation for the line which passes through the point (0, 1, 2) and is perpendicular to the line x = 1 + t, y = 1 – t and z = 2t and also intersects this line. [8]
Q.51
Suppose that r 1 r 2 and r 1r 2 = 2 (r 1 , r 2 need not be real). If r 1 and r 2 are the roots of the biquadratic x4 – x3 + ax2 – 8x – 8 = 0 find r 1, r 2 and a. [8]
Q.52
Express
x 2 y2 a 2 2ax xy 2ay x 2
2ax xy 2ay x 2 a 2 2x 2 2ax xy 2ax xy x 2 y 2 a 2
as a product of two polynomial.
[8]
1 2 2 2 1 1 Q.53 Given the matrices A = 2 2 3 ; C = 2 2 1 and D = 1 1 3 1 1 1 Solve the matrix equation Ax = b. Q.54
Prove that
a b c
+
10 13 and that Cb = D. 9 [8]
b c 3 for a, b, c > 0. + ca a b 2
[8]
x 2 y2 Q.55 Given x, y R, x2 + y2 > 0. If the maximum and minimum value of the expression 2 are x xy 4 y 2 M and m, and A denotes the average value of M and m, compute (2007)A. [8] Q.56
Prove that the triangle ABC will be a right angled triangle if cos
A B C A B C 1 cos cos – sin sin sin = 2 2 2 2 2 2 2
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Problems for JEE-2007
[8]
[6]
Q.57
A point P is situated inside an angle of measure 60° at a distance x and y from its sides. Find the distance of the point P from the vertex of the given angle in terms of x and y. [8]
Q.58
In ABC, a = 4 ; b = 3 ; medians AD and BE are mutually perpendicular. Find ‘c’ and ‘’.
[8]
Q.59 The lengths of the sides of a triangle are log1012, log1075 and log10n, where n N. Find the number of possible values of n. [8] Q.60 A flight of stairs has 10 steps. A person can go up the steps one at a time, two at a time, or any combination of 1's and 2's. Find the total number of ways in which the person can go up the stairs. [8] b
Q.61
e x a e b x dx = 0 Let a and b be two positive real numbers. Prove that x a
[8]
Q.62
Let f (x) = 2kx + 9 where k is a real number. If 3 f (3) = f (6), then the value of f (9) – f (3) is equal to N, where N is a natural number. Find all the composite divisors of N. [8]
Q.63
Line l is a tangent to a unit circle S at a point P. Point A and the circle S are on the same side of l, and the distance from A to l is 3. Two tangents intersect line l at the point B and C respectively. Find the value of (PB)(PC). [8]
Q.64
A triangle has one side equal to 8 cm the other two sides are in the ratio 5 : 3. What is the largest possible area of the triangle. [8] R = 3 + 1, where R is the radius of the r circumcircle and r is the radius of the incircle. Find C in degrees. [8]
Q.65 In triangle ABC, max {A, B} = C + 30° and
Q.66 The parabola P : y = ax2 where 'a' is a positive real constant, is touched by the line L: y = mx – b (where m is a positive constant and b is real) at the point T. Let Q be the point of intersection of the line L and the y-axis is such that TQ = 1. If A denotes the maximum value of the region surrounded by P, L and the y-axis, find the value of
1 . A
[8]
Q.67 A point moving around circle (x + 4)2 + (y + 2)2 = 25 with centre C broke away from it either at the point A or point B on the circle and moved along a tangent to the circle passing through the point D (3, – 3). Find the following. (i) Equation of the tangents at A and B. (ii) Coordinates of the points A and B. (iii) Angle ADB and the maximum and minimum distances of the point D from the circle. (iv) Area of quadrilateral ADBC and the DAB. (v) Equation of the circle circumscribing the DAB and also the intercepts made by this circle on the coordinate axes. [10] 7
Q.68
If
1 i x 2
i
1 and
i
7
1 (i 1)
then find the value of
2
x i 12 and
i
1 (i 2)2 x
i
123 ,
i
7
1 (i 3)2 x
7
i
.
[10]
i
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Problems for JEE-2007
[7]
Q.69
The normals to the parabola y2 = 4x at the points P, Q & R are concurrent at the point (15, 12). Find (a) the equation of the circle circumscribing the triangle PQR (b) the co-ordinates of the centroid of the triangle PQR. [10]
Q.70
The triangle ABC, right angled at C, has median AD, BE and CF. AD lies along the line y = x + 3, BE lies along the line y = 2x + 4. If the length of the hypotenuse is 60, find the area of the triangle ABC. [10]
Q.71
Let W1 and W2 denote the circles x2 + y2 + 10x – 24y – 87 = 0 and x2 + y2 – 10x – 24y + 153 = 0 respectively. Let m be the smallest positive value of 'a' for which the line y = ax contains the centre of a p circle that is externally tangent to W2 and internally tangent to W1. Given that m2 = where p and q are q relatively prime integers, find ( p + q). [10]
Q.72
If
3
dx = a – 2 ( 1 sin x ) 5 6
b 3 where a, b, c N and b, c are relatively prime, find the value of c
a + b + c + abc. 1
Q.73
If
0
Q.74
[10]
dx = 1 x 1 x 2
a b
c
where a,b,c N, find the value a2 + b2 + c2.
[10]
Suppose f (x) and g (x) are differentiable functions such that x g f ( x) f ' g (x )g ' (x) = f g (x )g ' f (x ) f '( x) a
for all real x. Moreover, f (x) is nonnegative and g (x) is positive. Furthermore,
0
e 2a f g( x ) dx 1 2
for all reals a. Given that g f (0) = 1. If the value of g f (4) = e –k where k N, find k.
[10]
Q.75 Let f (x) be a differentiable function such that f ' (x) + f (x) = 4xe –x · sin 2x and f (0) = 0. Find the value n
of Lim n
Q.76
f ( k ) . k 1
[10]
x f (x) Let f be a differentiable function satisfying the condition f = y f (y) (y 0, f(y) 0) V x, y R and f (1) = 2, then find the area enclosed by y = f(x), x2 + y2 = 2 and x – axis.
Q.77
[10]
The equation Z10 + (13 Z – 1) 10 = 0 has 5 pairs of complex roots a1, b1, a2, b2, a3, b3, a4, b4, a5, b5. Each pair ai, bi are complex conjugate. Find
1 . a b i i
[10]
Q.78(i) Let Cr's denotes the combinatorial coefficients in the expansion of (1 + x)n, n N. If the integers an = C0 + C3 + C6 + C9 + ........ bn = C1 + C 4 + C7 + C 10 + ........ and cn = C2 + C5 + C8 + C11 + ........, then prove that (a) a 3n b3n c3n – 3an bncn = 2n, (ii) Prove the identity:
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(b) (an – bn)2 + (bn – cn)2 + (cn – an)2 = 2.
[10]
(C0 – C2 + C4 – C6 + .....) 2 + (C1 – C3 + C5 – C7 + .......) 2 = 2n Problems for JEE-2007
[8]
1 3 5 Q.79 Given the matrix A = 1 3 5 and X be the solution set of the equation Ax = A, 1 3 5 x 3 1 where x N – {1}. Evaluate 3 where the continued product extends x X. x 1
[10]
c Q.80 If a, b, c are the sides of triangle ABC satisfying log 1 + log a – log b = log 2. Also a a(1 – x2) + 2bx + c(1 + x2) = 0 has two equal roots. Find the value of sin A + sin B + sin C. [10] Q.81
1 For x (0, /2) and sin x = , if 3
sin( nx) a b b = then find the value of (a + b + c), n 3 c n 0
where a, b, c are positive integers. eix e ix (You may Use the fact that sin x = ) 2i
[10]
Q.82 Two distinct numbers a and b are chosen randomly from the set {2, 22, 23, 24, ......, 2 25}. Find the probability that loga b is an integer. [10] OBJECTIVE
Select the correct alternative. (Only one is correct):
Q.83
A child has a set of 96 distinct blocks. Each block is one of two material (plastic, wood), 3 sizes (small, medium, large), 4 colours (blue, green, red, yellow), and 4 shapes (circle, hexagon, square, triangle). How many blocks in the set are different from "Plastic medium red circle" in exactly two ways? ("The wood medium red square" is such a block) (A) 29 (B) 39 (C) 48 (D) 56 49
Q.84
The sum
n! n 99 (1) k 2k where r r !(n r )! equals k 0
(A) – 298 Q.85
(B) 298
(C) – 249
(D) 249
If A > 0, c, d , u, v are non-zero constants, and the graphs of f (x) = | Ax + c | + d and g (x) = – | Ax + u | + v intersect exactly at 2 points (1, 4) and (3, 1) then the value of
(A) 4
(B) – 4
(C) 2
uc equals A
(D) – 2
Q.86 Consider the polynomial equation x4 – 2x3 + 3x2 – 4x + 1 = 0. Which one of the following statements describes correctly the solution set of this equation? (A) four non real complex zeroes. (B) four positive zeroes (C) two positive and two negative zeroes. (D) two real and two non real complex zeroes. Q.87 The units digit of 31001 · 71002 · 131003 is (A) 1 (B) 3
B ansal C lasses
(C) 7
Problems for JEE-2007
(D) 9
[9]
Q.88
The polynomial f (x) = x4 + ax3 + bx2 + cx + d has real coefficients and f (2i) = f (z + i) = 0. The value of (a + b + c + d) equals (A) 1 (B) 4 (C) 9 (D) 10
Q.89
If the sum
k 1 ( k 2)
equals (A) 6
1 = k k k 2 (B) 8
a b where a, b, c N and lie in [1, 15] then a + b + c c (C) 10
(D) 11
Q.90
Triangle ABC is isosceles with AB = AC and BC = 65 cm. P is a point on BC such that the perpendicular distances from P and AB and AC are 24 cm and 36 cm respectively. The area of triangle ABC in sq. cm is (A) 1254 (B) 1950 (C) 2535 (D) 5070
Q.91
The polynomial function f (x) satisfies the equation f (x) – f (x – 2) = (2x – 1)2 for all x. If p and q are the coefficient of x2 and x respectively in f (x), then p + q is equal to (A0 0 (B) 5/6 (C) 4/3 (D) 1
Q.92
Three bxes are labelled A, B and C and each box contains four balls numbered 1, 2, 3 and 4. The balls in each box are well mixed. A child chooses one ball at random from each of the three boxes. If a, b, and c are the numbers on the balls chosen from the boxes A, B and C respectively, the child wins a toy helicopter when a = b + c. The odds in favour of the child to receive the toy helicopter are (A) 3 : 32 (B) 3 : 29 (C) 1 : 15 (D) 5 : 59
Q.93
4 5 The value of tan arc sin arc cos is equal to 5 13 (A)
25 63
(B) –
3 7
(C) –
33 56
(D)
16 63
Select the correct alternatives. (More than one are correct):
Q.94
Three positive integers form the first three terms of an A.P. If the smallest number is increased byone the A.P. becomes a G.P. In original A.P. if the largest number is increased by two, the A.P. also becomes a G.P. The statements which does not hold good? (A) first term of A.P. is equal to 3 times its common difference. (B) Sn = n(n + 11) (C) Smallest term of the A.P. is 8 (D) The sum of the first three terms of an A.P. is 36.
Q.95 If the line 2x + 9y + k = 0 is normal to the hyperbola 3x2 – y2 = 23 then the value of k is (A) 31 (B) 24 (C) – 31 (D) – 24 Q.96 The line 2x – y = 1 intersect the parabola y2 = 4x at the points A and B and the normals at A and B intersect each other at the point G. If a third normal to the parabola through G meets the parabola at C then which of the following statement(s) is/are correct. (A) sum of the abscissa and ordinate of the point C is – 1. (B) the normal at C passes through the lower end of the latus rectum of the parabola. (C) centroid of the triangle ABC lies at the focus of the parabola. (D) normal at C has the gradient – 1.
B ansal C lasses
Problems for JEE-2007
[10]
Q.97
If (x) = f (x2) + f (1 – x2) and f '' (x) > 0 for x R then which of the following are correct? (A) (x) attains its extrema at 0, ±
1 2
(C) (x) attains its local maxima at 0.
(B) (x) increases in 1 (D) (x) decreases in 1
2,0 1
2,
2,0 1
2,
2 sin x 2 x 3 = If tan where 0 < x < , then the value of x is 3 cos 2 cos x 3 sin
Q.98
(A)
(B)
12
5 12
(C)
7 12
(D)
MATCH THE COLUMN: Q.99 Column-I
(A) (B)
(C)
(D)
Q.100 (A)
Column-II
The smallest positive integeral value of n for which the complex
11 12
(P)
4
(Q)
6
(R)
8
(S)
9
number 1 3 i n 2 is real, is Let z be a complex number of constant non zero modulus such that z2 is purely imaginary, then the number of possible values of z is 3 whole numbers are randomly selected. Two events A and B are defined as A : units place in their product is 5. B : their product is divisible by 5. If p1 and p2 are the probabilities of the events A and B such that p2 = kp1 then 'k' equals For positive integers x and k, let the gradient of the line connecting (1, 1) and (x, x3) be k. Number of values of k less than 31, is
Column-I For real a and b if the solutions to the equation Z9 – 1 = 0
are written in the form of a + ib then the number of distinct ordered pairs (a, b) such that a and b are positive, is
1 e x 1 x
(B)
Lim e x
(C) (D)
Let A, B be two events with P(B) > 0. If B Athen P(A/B) equals A real number x is chosen at random such that 0 x 100.
x
Column-II
(P)
0
(Q)
1
(R)
2
(S)
e
1 a is , where a and b are relatively 3 b primes and [x] denotes the greatest integer then (b – a) equals The probability that x – [x]
B ansal C lasses
Problems for JEE-2007
[11]
HINTS AND SOLUTIONS
1.
2111x =
Let y so that log2y = 111 x equation becomes
x=
log 2 y 111
y3 + 2y = 4y2 + 1 4 y3 – 16y2 + 8y – 4 = 0 sum of the roots of the given equation is x1 + x2 + x3 = 2.
3.
log 2 y1 log2 y 2 log 2 y3 log 2 ( y1y 2 y3 ) log 2 4 2 = = = S1 + S2 = 113 Ans.] 111 111 111 111
Let the 3 consecutive terms are a – d, a, a + d d>0 hence a2 – 2ad + d 2 = 36 + K ....(1) 2 a = 300 + K ....(2) a2 + 2ad + d 2 = 596 + K ....(3) now (2) – (1) gives d(2a – d) = 264 ....(4) (3) – (2) gives d(2a + d) = 296 ....(5) (5) – (4) gives 2d 2 = 32 d 2 = 16 Hence from (4) 4(2a – 4) = 264 2a – 4 = 66 2 K = 35 – 300 = 1225 – 300 = 925 Ans.]
d = 4 (d = – 4 rejected)
2a = 70
a = 35
Case-I : When the two identical digits are both unity as shown.
any one place out of 3 block for unity can be taken in 3 ways and the remaining two blocks can be filled in 9 · 8 ways. Total ways in this case = 3 · 9 · 8 = 216 Case-II : When the two identical digit are other than unity. ; ; two x's can be taken in 9 ways and filled in three ways and y can be taken in 8 ways. Total ways in this case = 9 · 3 · 8 = 216 Total of both case = 432 Ans. ] 4.
h = a(t1 t2) k = a(t1 + t2) Equation to the variable chord 2x – (t1 + t2)y + 2at1 t2 = 0 2
y=t t 1 2
x +
2at1t 2 t1 t 2
2a 2h x + a ....(1) k k Since (1) touches y2 = 4bx , using the condition of tangency y=
2ah bk k 2a Locus is by2 = 4a2x ]
B ansal C lasses
Problems for JEE-2007
[12]
5.
Note that triangles BCM and OCN are similar now let ON = p. N will be mid point of chord PQ
p 1 = 1 2
now
R = 2 r 2 p 2 for large circle
p =
= 2 16 (1 4) = 63 Equation of large circle as x2 + y2 = 16
Alternatively:
1 C = (1, 0) with slope PQ = – (think !) 3
now
equation of PQ :
3y+x=1
P (from origin) = 6.
x7 dx = (1 x 2 ) 5
=
x 3
x
1 = – 2
7.
1
2
5
2
0
dx
1 2
x7
1 1 x 2
5
result
dx
]
Taking x2 out of the bracket
x 10
Put x –2 – 1 = t =
2 x3
dx = dt
4 1 1 dt 1 t 1 1 x8 = – = + C = + C ] 4 = 8 1 2 4 2 4 8 t4 t5 8 1 x 2 1 x
Using sin 2x =
I=
1 2
2 tan x 1 tan 2 x
2 tan x 2 2 2 2 (1 tan x ) 2 (1 tan x ) 1 tan x dx = 2 · sec 2 x dx = · ( 1 tan x ) dx 4 2 tan x (1 tan x ) (1 tan x ) 4 0 1 0 1 tan 2 x 1
put y = tan x
dy = sec2x dx
(1 y) 2 dy I= 4 0 (1 y)
now put 1 + y = z
dy = dz
3z 2 6 z 4 (2 z ) 2 dz = I= 4 3z 3 z 1
2
Alternatively:
I=
0
B ansal C lasses
1
1 = 3
a = 1, b = 3 1 + 3 + 3 = 7 Ans. ]
(cos x sin x ) 2 dx (cos x sin x ) 4 Problems for JEE-2007
[13]
1 I=– 3
2
0
d 1 dx (cos x sin x ) · dx (cos x sin x )3 I II
integrating by parts 1 (cos x sin x ) =– 3 (cos x sin x )3
using sin 2x = 2 1 = – 3 3
2
0
2
0
0
dx = – (cos x sin x ) 3 (sin x cos x )
1 (1) (1) 3
2
0
dx 1 sin 2 x
2 tan x 1 tan 2 x
sec 2 x 2 1 dx = – 3 3 (1 tan x ) 2
1 2 1 = – = 3 3 3 8.
2
dt
1 t 2
=
2 1 t 2 1 + t 0 = + [(0) – (1) 3 3 3 3
a = 1, b = 3 1 + 3 + 3 = 7 Ans. ]
Let the number of passengers be x ( x > 200) Fair changed per person = 10 – (x – 100)
2 100 2
2x 2x 2 10 x 4x 10 x ( x 200 ) Total revenue = x . 10 ( x 200) = = 100 100 100 2x 2 f (x) = 14x – 100 f (x) = 14 – f (x) < 0 9.
4x = 0 x = 350 100 x = 350 gives maxima]
Given 9y2 = x3 Let the point on the curve be x = t2 and dx = 2t ; dt
t3 y= 3
dy = t2 dt
dy dy dt t 2 t2 = × = = slope of the normal = – dx dt dx 2 t 2t normal makes equal intercept
hence –
2 = – 1 t
t=2
8 ) 3
a + 3b = 4 + 3 ·
Hence P = (4,
B ansal C lasses
8 = 4 + 8 = 12 Ans. ] 3
Problems for JEE-2007
[14]
10.
Amount of glucose in blood at time t is x (t)
hence
dx dt K x
dx = – K x dt
1 ln ( – K x) = t + C K ln ( – K x) = – Kt + C – K x = e – K t + C e K t C x= K Lim x (t ) ] t K
–
x2
11.
A=
(a 2 a x) (x 2 2 a x 3 a 2 ) 1 a4
x1
dx
where x1 & x2 are the roots of , x2 + 2 a x + 3 a2 = a2 a x x = a or x = 2 a dA a3 A= = 0 gives a = 31/4 Ans. ] 4 da 6(1 a )
12.
A : ears of Mr A formed to be drooping B1 : Mr A was telling a truth P(B1) = 3/5 B2 : Mr B was telling a false P(B2) = 2/5 P(A/B1) = 1/10 P(A/B2) = 2/3 3 1 · 9 3 5 10 P(B1/A) = = = Ans. ] 40 49 3 1 2 2 3 · · 3 5 10 5 3
13.
E : all the 5 persons leave at different floors n(S) = 85 n(A) = 8C5 · 5!
105 C 5 5! = ans. ] 512 85 u·w u · (| u | v | v | u ) (u · v) | u | | v | | u |2 = cos = | w = || u | | u || w | | u || w | ( u · v ) | v | | u | cos = ....(1) |w| v·w v · (| u | v | v | u ) | v |2 | u | ( v · u ) | v | = cos = | w = || v | | w || v | | w || v| 8
14.
P(E) =
B ansal C lasses
Problems for JEE-2007
[15]
| v | | u | ( v · u ) cos = ....(2) |w| from (1) and (2) cos = cos = ]
1 x 2 y 1 2 t The line is 1 1 1 2 z
15.
....(1)
1ˆ 1ˆ ˆ ˆ 2 i j k V ˆi jˆ k line passes through and is parallel to the vector 2 2 ˆ vector normal to the plane x + 2y + 6z = 10, is n ˆi 2 jˆ 6 k line (1) is | | to the plane V . n = 1 + 2 – 3
16.
d=
2 2 3 10 = 1 4 36
9 Ans 41
]
Vector v1 along the line of intersection of 3x – 7y – 5z = 1 and 8x – 11y + 2z = 0 is given by
ˆi jˆ ˆ k ˆ) v1 n1 n 2 = 3 7 5 = – 23( 3ˆi 2 jˆ k 8 11 2 |||ly vector v 2 along the line of intersection of the planes 5x – 13y + 3z = 0 and 8x – 11y + 2z = 0 is
ˆi ˆ jˆ k ˆ) v 2 n 3 n 4 = 5 13 3 = 7 ( ˆi 2 jˆ 7 k 8 11 2 now v1 · v 2 = 0 angle is 90° sin90° = 1
17.
Given and
]
au + bv + cw = 0 ....(1) au + bv + cw = 0 ....(2) au + bv + cw = 0 ....(3)
u v w For non trivial solution (non zero) solution of a, b and c . We must have u ' v' w ' = 0 ] u ' ' v' ' w' ' 18.
Let
A B C + cos B · sin2 + cos C · sin2 2 2 2
y = cos A · sin2 =
1 [cosA (1 – cosA) + cosB (1 – cosB) + cos C (1 – cos C)] 2
=
1 [(cosA – cos2A) + (cosB – cos2B) + (cosC – cos2C)] 2
2 2 2 1 1 1 1 1 1 1 cos A cos B cos C = 2 2 4 2 4 2 4
B ansal C lasses
Problems for JEE-2007
[16]
1 y= 2
2 2 2 3 1 1 1 cos A cos B cos C 2 2 2 4
now y will be maximum if cosA = cos B = cos C =
1 2
hence ymax = 3/8 ] 19.
y = x2; x = t; y = t2 dy = 2x = 2t dx
slope of normal m = –
1 2t
equation of normal y – t2 = – if
1 (x – t) 2t
x = 0; y =
2t(y – t2) = – x + t
or
3 2
3 2 2t t = t t = 0 2 or 3 – 2t2 = 1 t = 1 or – 1 hence one of the point is origin and the other two are (–1, 1) and (1, 1) PQR is a right triangle radius of the circle is 1 its equation is x2 + (y – 1)2 = 1 x2 + y2 – 2y = 0 ] 20.
Let x be a real root. Equating real and imaginary part x3 – 6x2 + 5x + 2a2 = 0 .....(1) and 2x3 – 2x2 – 4x = 0 .....(2) 2 2x(x – x – 2) = 0 2x(x – 2)(x + 1) = 0 the given x = 0, 2 or – 1 if x=0 a=0 x= –1
a2 = 6
a=±
6
a2 = 3
a=±
3
x=2
a { 0, 6 , 6 , 3 , 3 } S = 0 + 6 + 6 + 3 + 3 = 18 Ans. ] 21.
For non zero intercepts slope = – 1 y=–x+c point (– 4, – 2) – 2 = 4 + c
B ansal C lasses
c=–6 Problems for JEE-2007
[17]
lines is y = – x – 6 x+y+6=0 for zero intercept line is y = mx – 2 = m(– 4) m = 1/2 2y = x lines are 2y = x and x + y + 6 = 0 ] 22.
x4 – 10x2 + 9 0 (x2 – 9)(x2 – 1) 0 hence – 3 x – 1 or 1 x 3 now f (x) = x3 – 3x f ' (x) = 3x2 – 3 = 0 x=±1 maximum occurs when x = 3 f (3) = 18 ]
23.
x4y2dx + x2y4dy = xdy + ydx x2y2(x2dx + y2dy) = xdy + ydx x2dx + y2dy =
d (xy) ( xy) 2
x
Integrating,
2
y
dx +
2
dy =
d (xy) ( xy) 2
1 x3 y3 + = – + C xy 3 3 3 = C; now if x = 1; y = 1 C = 5, xy hence x3 + y3 + 3(xy) –1 = 5 Ans. ] (x3 + y3) +
24.
face card removed 40 20 drawn randomly 52 Let E0 : 20 cards randomly removed has no aces. E1 : 20 cards randomly removed has exactly one ace. E2 : 20 cards randomly removed has exactly 2 aces. E: event that 2 drawn from the remaining 20 cards has both the aces. P(E) = P(E E0) + P(E E1) + P(E E2) = P(E0) · P(E / E0) + P(E1) · P(E / E1) + P(E2) · P(E / E2)
= 40 /\ 4
=
4 aces 36 other
C0 · 36 C 20 40
36
=
B ansal C lasses
C 20
4
4
C2
· 20
C2
+
C1 · 36 C19 40
C 20
3
C2
· 20
C2
4
+
C2 · 36 C18 40
C 20
2
C · 20 2 C2
C 20 · 4 C 2 4C1 · 36 C19 · 3C 2 4C 2 · 36 C18 · 2C 2 40
C 20 · 20 C 2 Problems for JEE-2007
[18]
=
= 25.
e=
6 · 36C 20 12 · 36 C19 6 · 36C18 40
C 20 · 20C 2
6( 37 C20 37 C19 ) 40
C 20 · 20 C2
=
=
6 [ 36 C20 36C19 36C19 36C18 ]
6( 38C 20 ) 40
C 20 · 20 C 2
40
C 20 · 20 C2
p = 6 Ans. ]
4 5
16 9 b 3 b 2 = 1 – = ; = ....(1) 25 25 a 5 a2 now radius of the circle r = a – (where , 0 is the centre of the circle) also r = AC = b sin a – = b sin where = a cos a(1 – cos ) = b sin a2(1 – cos )2 = b2(1 – cos )(1 + cos ) a2(1 – cos ) = b2(1 + cos ) 1 cos 9 = 1 cos 25 25 – 25 cos = 9 + 9 cos 16 = 34 cos cos =
26.
sin =
AB = 2b sin = 2 ·
Let
a, b R, such that f (a) = f (b)
27.
8 ; 17
15 17
3a 15 18 · = a Ans. ] 5 17 17
1 ai 1 bi = 1 ai 1 bi 1 – bi + ai + ba = 1 + bi – ai + ab 2ai = 2bi a=b f is injective. ]
Let r be the radius of circle A and R be the radius of circle B r + R = 12 and r = 3R 4R = 12; R = 3 and r = 9 Area of trapezium ABCD =
1 (3 + 9) (12) 2 62 2
= 6 108 = 36 3 Area of arc ADC =
B ansal C lasses
27 1 81 = 2 3 2 Problems for JEE-2007
[19]
Area of arc BCE =
28.
1 2 9 = 3 2 3
27 3 33 = 36 3 – required area = 36 3 – 2 2
a = 36, b = 33 a + b = 69 Ans. ]
0 1 0 1 3 0 A2 = 3 0 3 0 = 0 3 = 3I A4 = 9I; A6 = 27; A8 = 81I (A8 + A6 + A4 + A2 + I) = 121 I 1 0 0 hence 121 0 1 V = 11 ;
121 0 a 0 0 121 b = 11
1 121 a 0 = a = 0, b = ; 121 b 11 11 29.
Radius of the first circle = sin
S
=
0 V1 ] 11
6 =1 6
C 1 r = ....(1) (r < 1) 2 1 r
also
sin C =
now
2sin2 sin2
4 5
C 3 2 = 1 – cos C = 1 – = 2 5 5
C 1 = 2 5 2
1 1 r = 5 1 r
30.
5 – 1 = ( 5 1 )r
5(1 – r)2 = (1 + r) 2
r=
5 (1 r ) = 1 + r
5 1 sin 18 = 5 1 cos 36
k + w = 54° Ans. ]
Cross multiplication and rearranging gives the cubic. x3 – ax2 + 23x – b = 0 2 + = a 2 + 2 = 23 and 2 = Also given + = 12 from (2) and (4) 2 + 2(12 – ) = 23 2 + 24 – 22 = 23
B ansal C lasses
....(1) ....(2) ....(3) ....(4)
Problems for JEE-2007
[20]
31.
2 – 24 + 23 = 0 = 1 (rejected) = 23;
since x ± 1 = – 11
a = 35 from (4)
and
b = 2 = 529 × – 11
b = – 5819
a – b = 35 – (–5819) = 5854 Ans. ]
Let the numbers be G.P. ( A D) 2 A D, A, A D, A ( a ) ( ) ( c ) b ( d ) A .P.
Given d – a = 30
(A D) 2 – (A – D) = 30 A D2 + 3AD = 30 A D2 = 3A(10 – D)
(A + D)2 – A(A – D) = 30A
D2 A= ....(1) 3(10 D) since 'A' is a + ve integer 0 < D < 10 ....(2) Also since '3' is prime and A is an integer D2 must be divisible 3 D must be of the form of 3K possible values of D are 3, 6, 9 3 (rejected) 7 D=6 A = 3 (rejected) D=9 A = 27 Numbers are 18, 27, 36, 48 Ans. D=3
32.
A=
]
Total number of ways in which P and Q can be chosen simultaneously = (25 – 1)(25 – 1) = 45 – 26 + 1 number of ways when P and Q have no common element = 5C1(24 – 1) + 5C2 (23 – 1) + 5C3(22 – 1) + 5C4(21 – 1) + 5C5(20 – 1) = 5C1 · 24 + 5C2 · 23 + 5C3 · 22 + 5C4 · 2 + 5C5 – ( 5C1 + 5C2 + 5C3 + 5C4 + 5C5) = (5C0 · 25 + 5C1 · 24 + 5C2 · 23 + 5C3 · 22 + 5C4 · 2 + 5C5 – 25) – (25 – 1) = (35 – 25) – (25 – 1) = 35 – 26 + 1 Hence P and Q have atleast one common element = (45 – 26 + 1) – (35 – 26 + 1) = 45 – 35 Ans. ]
B ansal C lasses
Problems for JEE-2007
[21]
33.
Slope of the normal m =
1 2 x1 1
m1 x1 = 2m
3 m2 1 y1 = ; 4 m2
equation of the normal in terms of slope of the normal is y = mx +
5 m2 2 m 3 1 . 4 m2
It passes through (7/2, 9/2) 12 m3 13 m2 + 1 = 0 sum = 13/12. Also (m 1) (3m 1) (4m + 1) = 0 m1 = 1 ; m2 = 1/3 ; m3 = 1/4 the normals are x y + 1 = 0 ; x 3y + 10 = 0 & 2x + 8y 43 = 0 Point A (0, 1) ; B ( 1, 3) ; C (5/2, 19/4) ] 34.
Equation of normal, 1 (X x) m
Y y =
X = 0 gives Y = x2 +
Y = 0 gives X = x + my and
x my m
Hence
x x my = 0 2
y2 = C ; passes through (1, 4) 2
1 x2 y2 conic is = 1 with e = 9 18 2
2x + y
dy = 0 dx
C=9
focii are (0, 3) & (0, 3)
Equation of the circles are ; (x 1)2 + (y 4)2 + (x + 2 y 9) = 0 where x + 2y 9 = 0 is the tangent to the ellipse at (1, 4)] 35.
Equation to the hyperbola where S = (0, 0) ; directrix is x + y + 1 = 0 and e = 2 is
x y 1 2
x 2 y2 2
x2 + y2 = (x + y + 1) 2 2xy + 2x + 2y + 1 = 0 Let the combined equation of the asymptotes is 2xy + 2x + 2y + c = 0 put D = 0 to get c = 2 hence combined equation of the asymptotes are xy + x + y + 1 = 0 (x + 1)(y + 1) = 0 x + 1 = 0 and y + 1 = 0 36.
]
Note that (tan C – sin A)2 + (cot C – cos B)2 denotes the square of the distance PQ now d2PQ = (Q – OP) 2 2
d 2
(tan 2 C cot 2 C) 1 PQ =
d 2
(tan C cot C) 2 2 1 PQ =
d 2min =
2
2 2 1 = 3 – 2 2 a = 3; b = 2 a3 + b3 = 27 – 8 = 19 Ans. ]
B ansal C lasses
Problems for JEE-2007
[22]
37.
I=
0
x 2 dx = 4 2 2 0 x (a 2) x 1
dx
1 x 2 2 (a 2 2) x
(a2 – 2 = k 0)
x 2 dx 1 ( x 2 1) ( x 2 1) dx = = 2 4 2 4 2 x kx 1 0 x kx 1 0
1 1 (1 x 2 ) 1 1 (1 x 2 ) dx dx + = 2 0 x 2 (1 x 2 ) k 2 0 x 2 (1 x 2 ) k
I2
I1
now proceed, I1 =
38.
=
Let
0
=
I
2
4
;
2a
and I2 = 0
2a
2a
x
=
5050
a = 2525 Ans. ]
d = dx
or
4 = + 4x
x x (1 tan x ) 0 4 dx = – 4 1 tan x dx = – 4 1 tan x 2 1 1 tan x 1 tan x 4
( 4x ) tan
0
– 4 = – 4x
0
x (1 tan x ) (1 tan x ) · dx 1 tan x ( 2 ) tan x 2
0
x (1 tan x ) x x dx dx = 2 =2 tan x tan x 2 2
I= x
I=–
0
2 +
2 0
2 4
2
+ 2
0
x dx tan x 2
t dt tan t
x=–t
2
now
2
0 t cotII t dt = t ln sin t 0
I1 =
I
I1 = 0 + Hence 2 ·
2
B ansal C lasses
2
2
–
0 ln sin t dt
ln 2
ln 2 –
2 4
= ln 2 –
2 4
k = 2, w = 4
Problems for JEE-2007
kw = 8 Ans. ]
[23]
1
39.
g(1) = 5 and
0
g (t) dt = 2
x
2f (x) =
x
x
x
0 (x2 – 2xt + t2) g(t) dt = x g( t) dt 2x t g( t) dt t 2g( t) dt 0 0 0 2
Differentiating
2 x g ( t ) dt · 2 x – 2x g ( x ) t g( t ) dt + x2g(x) 0 0 x
2 f '(x) = x2 g(x) + x
x
0
0
2 f '(x) = 2x g( t )dt – 2 t g( t ) dt x
x
0
0
f " (x) = x g (x) + g( t )dt – x g (x) = g ( t )dt 1
0
hence f " (1) = g ( t ) dt = 2 also
f ''' (x) = g (x) f ''' (1) = g (1) = 5 f ''' (1) – f ''(1) = 5 – 2 = 3 Ans. ] x
40.
Consider a function
0
g (x) = e –x f ( t )dt in [0, 1]
obvious continuous and derivable g (0) = 0 and g (1) = 0 (given) hence some c (0, 1) such that g ' (c) = 0 x
now
g ' (x) =
e –x f
(x) –
e –x
0 f ( t)dt c
41.
0
x
g ' (c) = e –c f (c) – e –c f ( t )dt = 0
0 f (t )dt = f (c)]
Consider f (x) = x3 – ax + b f '(x) = 3x2 – a if a 0 then f ' (a) 0 for all x hence f is strictly increasing hence f (x) = 0 has exactly one root for a = 3 f ' (x) = 3x2 – 3 = 0 x = 1 or – 1 in order that f (x) may have 3 roots f (x1) · f (x2) 0 where x1 and x2 and the roots of f ' (x) = 0 hence (1 – a + b)(– 1 + a + b) 0 put a = 3 (b – 2)(b + 2) 0 or – 2 b 2 ]
B ansal C lasses
Problems for JEE-2007
[24]
42.
Let m gms of salt is present at time t differential equation of the process is dm m(1) = 10 – dt 50 t dm + dt
1 m = 10; 50 t dt
(50 t ) 2 I.F = e = 50 + t; m(50 + t) = (50 t )dt = 10 + C 2 m(50 + t) = 5(50 + t)2 + C; t = 0; m = 0, C = – 5.(50)2 m(50 + t) = 5(50 + t) 2 – 5 (50)2
50 t
m = 5(50 +
t)2 –
5(50) 2 50 t
5(50) 2 m(t = 10) = 5 · 60 – 60 m= 43.
250 25 11 2 11 = 91 = 50 6 = 50 · 60 3 3 6
]
A : red ball is selected B1 : Face card is drawn B2 : ace card is drawn B3 : neither face nor ace is drawn P(A) =
12 3 4 8 36 10 107 · + · + · = Ans. 52 12 52 12 52 12 156
12 · 3 156 9 · P(B1/A) = = Ans. ] 52 12 107 107 44.
1, 2, 3, 4, 5, 6, 7, 8, 9
x + y = 45 ; x y = 11 x = 28 ; y = 17 Now to realise a sum 17 using 4 digits we can have different cases ,
7 6 3 1 8 6 2 1 ; ( 9 cases ) 7 5 4 1 ; 6 5 4 2 8 5 3 1 7 5 3 2 8 4 3 2 If we use five digits then 7, 1 , 2 , 3 , 4 ( 2 cases ) 6, 5, 3, 2, 1 4 ! 5 ! 9 5! 4 ! 2 11 5 ! 4 ! 11 Hence p = = = 9 4 3 1 9 5 2 1 ;
9!
9!
[ odd in favour 11 : 115 ] 45.
126
A = {1801, 1802,.....,1899, 1900} B = {1901, 1902,.....,1999, 2000}
B ansal C lasses
Problems for JEE-2007
[25]
E : randomly chosen year has 53 sundays P (E) = P (E L) + P (E O) = P (L). P(E/L) + P (O). P(E/O) 1 24 2 76 1 1 25 2 75 1 . . + . . = 2 100 7 100 7 2 100 7 100 7 =
249 Ans.] 1400
46.
P(E) = 1 P (value of 5 coins is more than or equal to Rs. 1.50) = 1 P(A A B B B or A A B B C or A B B B B) ]
47
6 on bush-I & 4 on bush-II
48.
AB
(BC BD) = 0 ; (AB BC) BD = 0 ; (AB BD) BC = 0 ;
Note that AB ; BC ; BD are mutually perpendicular Þ BC × BD is collinear with AB and so on Volume =
1 6
220 , BC , BD ] = cu. units [ AB 3
Vector area of triangle AEF = 49.
1 AF 2
1
] AE = BC BD = 3 i 10 j k 2
Equation of the line passing through (1, 4, 3) x 1 y 4 z 3 ....(1) a b c since (1) is perpendicular to
x 1 y 3 z 2 x 2 y 4 z 1 = = and = = 2 2 1 4 3 2
hence 2a + b + 4c = 0 and 3a + 2b – 2c = 0
a b c 2 8 12 4 4 3
a b c 10 16 1
x 1 y 4 z 3 ....(2) Ans. 10 16 1 now any point P on (2) can be taken as 1 – 10 ; 16 + 4 ; + 3 distance of P from Q (1, 4, 3) (10)2 + (16)2 + 2 = 357 (100 + 256 + 1)2 = 357 Hence Q is (–9, 20, 4) or (11, – 12, 2) = 1 or – 1 hence the equation of the lines is
50.
Ans.]
Equation of the line through (0, 1, 2) x 0 y 1 z 2 ....(1) a b c now given line
B ansal C lasses
x 1 y 1 z 0 = t 1 1 2
....(2)
Problems for JEE-2007
[26]
ˆ (2) is along the vector V ˆi jˆ 2k a – b + 2c = 0 ....(3) since (1) and (2) intersect; hence must be coplanar
hence
1 0 2 1 1 2 = 0 a b c
2a + 4b + c = 0 solving (3) and (4),
x y 1 z 2 = t Ans. ] 1 2 3
required equation is 51.
....(4) a:b:c=–3:1:2
Since r1 r 2 = 2,
x2 +
r 1
px + 2 = 0
r 2
r1r 2r 3r 4 = – 8
and
r 3r 4 = – 4
x4 – x3 + ax2 – 8x – 8 = (x2 + px + 2)(x2 + qx – 4) compare coefficient of x3 and x p + q = – 1 .....(1) and 2q – 4p = – 8 q – 2p = – 4 ....(2) p = 1 and q = – 2 on comparing coefficient of x2; a = – 4 p = 1 x2 + x + 2 = 0
x 52.
r 1, 2 =
y a
a x x x a y
1 i 7 2
x a
Ans. ]
x
x a y x a = y x a x y a x
x a y
2
= [x (xy – ax) – a(y2 – a2) + x (xy – ax) ]2 = [2x2 (y – a) – a (y – a) (y + a) ]2 = (y – a)2 [2x2 – a(y + a)]2 Hence D = (y2 + a2 – 2ay) (2x2 – ay – a2)2
53.
Let
]
a1 b = a 2 a 3
2 1 1 a1 2 2 1 a 2 = 1 1 1 a 3
10 13 9
2a1 a 2 a 3 10 2a 2a a 13 2 3 = 1 a a a 1 2 3 9 i.e.
a1 = 1 ; a2 = 3 ; a3 = 5
B ansal C lasses
Problems for JEE-2007
[27]
1 2 2 x1 2 2 3 x 2 = 1 1 3 x 3
1 3 5
x1 2x 2 2 x 3 1 2x 2x 3x 3 2 3 = 1 x1 x 2 3x 3 5
54.
i.e.
x1 = 1 ; x2 = – 1 ; x3 = 1
TPT
9 a b c a b c a b c + + 2 b c c a a b x1
x2
Ans.
]
....(1)
x3
Consider AM between the numbers x1, x2, x3 a b c 1 1 1 b c c a a b 3 now HM between the numbers x1, x2, x3 =
3
=
b c ca a b a b c a b c a b c AM HM
=
3(a b c) 3 = 2(a b c) 2
a b c 1 1 1 3 b c c a a b 2 3
1 1 1 9 (a + b + c) b c c a a b 2 55.
Let x = r cos and y = r sin y r 2 = x2 + y2; tan = x
Hence proved ]
(0, /2)
r 2 r 2 2 N= 2 = = 5 sin 2 3 cos 2 r [cos 2 sin cos 4 sin 2 ] (1 cos 2) sin 2 4(1 cos 2)
Nmax = Nmax = A= 56.
2 5 10 2 5 10
=
2 5 10 = M 15
=
2 5 10 = m 15
2 ·10 2 Mm = 15 · 2 = 2 3
2007 ×
2 = 1338 Ans. ] 3
Transposing 2 on RHS using 2 cos A · cos B relation, cos
A 2
B ansal C lasses
cos B C cos B C A – sin 2 2 2
cos B C cos B C = 1 2 2
Problems for JEE-2007
[28]
or cos
A BC A A A BC A A BC sin ) sin + cos cos – sin cos + sin2 – 1 = 0 ( cos 2 2 2 2 2 2 2 2 2
BC A A A A A cos cos sin + cos sin – cos2 = 0 2 2 2 2 2 2 cos
BC A A A cos sin – cos 2 2 2 2
cos A sin A 2 2 if
cos B C cos A = 0 2 2
A A – sin = 0 2 2
tan
A = 1 2
BC A = cos 2 2 B–C =A B=C +A B–C=–A B + A = C = 90° hence triangle must be right angled. ] OAMB is a cyclic quadrilateral using sine law in OBM and OAM if
57.
cos
cos A sin A = 0 2 2
cos
x d = sin( 60 ) sin 90 and
A = 90°
d y = sin 90 sin
(1) and (2)
B = 90°
.....(1) ....(2)
x y = sin( 60 ) sin
x sin( 60 ) 1 3 = = cot – y sin 2 2
2x + 1 = y
3 cot
2x y = cot 3y
(2 x y) 2 d 2 = y2 1 2 3y
from (2) d = y cosec d 2 = y2(1 + cot2
)
d 2 =
3y 2 4x 2 y 2 4xy 3
d=
2 x 2 y 2 xy Ans. ] 3
B ansal C lasses
d 2 =
d 2 =
y2 +
(2 x y ) 2 3
4x 2 4 y 2 4 xy 3
Problems for JEE-2007
[29]
58.
Let G be the centroid : AD = x ; BE = y
AG =
2x x 2y y ; GD = ; BG = ; GE = 3 3 3 3
4x 2 y 2 9 or 16x2 + 4y2 = 81 .....(1) In AGE : 9 9 4 x 2 4y 2 4 or In BGD : 9 9 (i) – (ii) , 15x2 = 45
x2 + 4y2 = 36 .....(ii) x= 3
9 4 3 5 9 16 c 2 In ADC, cosC = 2(2) (3) 6 2(4) (3)
20 = 25 – c2
or
c=
5
2 1 1 5 = ab sinC = (3) (4) 1 11 sq. units ] 2 2 6
59.
From triangle inequality log1012 + log1075 > log10n log10900 > log10n also log1012 + log10n > log1075 log1012n > log1075 12n > 75 n>
75 12
or
n>
n < 900
....(1)
25 4
Hence no. of values = 900 – 7 = 893 Ans. ] 60.
x + 2y = 10 where x is the number of times he takes single steps and y is the number of times he takes two steps Cases Total number of ways I: x = 0 and y = 5
5! = 1 (2 2 2 2 2) 5!
II: x = 2 and y = 4
6! 2!· 4! = 15 (1 1 2 2 2 2)
III: x = 4 and y = 3
7! 4!· 3! = 35 (1 1 1 1 2 2 2)
IV: x = 6 and y = 2
8! 2!· 6! = 28 (1 1 1 1 1 1 2 2)
9C = 9 (1 1 1 1 1 1 1 1 2) V: x = 8 and y = 1 1 VI: x = 10 and y = 0 1 (1 1 1 1 1 1 1 1 1 1) hence total number of ways = 1 + 15 + 35 + 28 + 9 +1 = 89 Ans. ]
B ansal C lasses
Problems for JEE-2007
[30]
b
61.
e x a e b x dx I= x a
x = at
let
b a
=a
1
dx = a dt
e t e b at dt at
t
e e t dt I= t 1
put
t=
(where b/a = )
....(1)
y
dt = –
y2
dy
( e y e y ) y · 2 dy I=– y 1
( e y e y )dy I= y
or
from (1) and (2)
2I = 0
1
62.
63.
( e t e t )dt I=– t 1
....(2)
I = 0 Ans. ]
f (3) 1 23k 9 = 6 k = ; f (9) – f (3) = (29k + 9) – (23k + 9) = 29k – 23k f (6) 2 9 3 3(23k + 9) = 26k + 9 26k – 3(23k ) – 18 = 0 23k = y y2 – 3y – 18 = 0 (y – 6)(y + 3) = 0 y = 6; y = – 3 (rejected) 23k = 6 now f (9) – f (3) = 29k – 23k { from (1) } = (23k )3 – 2 3k = 63 – 6 = 210 hence N = 210 = 2 · 3 · 5 · 7 Total number of divisor = 2 · 2 · 2 · 2 = 16 number of divisors which are composite = 16 – (1, 2, 3, 5, 7) = 11 Ans. ]
....(1)
Radius of the circle is 1
r B tan = = 2 PB s(s b) PB =
r · s(s b)
=
s · · (s b) = (s – b); s
|||ly
PC = (s – c)
(PB)(PC) = (s – b)(s – c) =
B ansal C lasses
s(s a )(s b)(s c) s(s a ) Problems for JEE-2007
[31]
=
· =r· s(s a ) (s a )
=
= (s a ) a
=
64.
r 1 s s
3 3a = = 3 Ans. ] 3 2 3a 2 a 2
5x + 3x > 8 x>1 5x + 8 > 3x x>–4 and 3x + 8 > 5x x<4 Hence, x (1, 4). Now perimeter of the triangle = 8(x + 1) s = 4x + 4 A2(x) = ( 4(x + 4)(4 – x)(4x – 4)(x + 4) ) = – 16(x2 – 1)(x2 – 16) A2(t) = – 16(t – 1)(t – 16), where x2 = t, t (1, 16) A2 (t) = – 16[t2 – 17t + 16] = f (t)
f ' (t) = 0
t=
17 2
17 1 17 16 15 15 = 16 × × = (2 × 15) 2 A (t) = – 16 max 2 2 2 2 2
(Area)max = 30 sq. units ] 65.
From the identity r = 4R · sin
A B C · sin · sin 2 2 2
or
r = 4 3 1 r · sin
let
A B A – C = 30°
then
Let
B C A · sin · sin or 2 2 2
3 1 = 4
AC B A C cos cos sin 2 2 2
3 1 = 4
6 2 B B sin sin 4 2 2
sin
B = x yields 2
x2 –
6 2 x+ 4
1 C B A = 2 sin · sin · sin 2 3 1 2 2 2
3 1 = 0, 4
B B 6 2 2 and x = . It follows that = 15° or = 45°. The second 2 2 4 2 solution is not acceptable, because A B. Hence B = 30°, A = 90° and C = 60° ] whose solutions are x =
B ansal C lasses
Problems for JEE-2007
[32]
66.
y = ax2
dy dt T = 2ax0 = m hence line is y = (2ax0)x – b
.....(1)
(x0, a x 20 ) lies on parabola and the line (1)
a x 02 = 2a x 20 – b b = a x 20 . Hence Q = (0, – b) = (0, – a x 20 ) now using (TQ)2 = 1 x 20 + 4a2 x 04 = 1 a2 =
(1 x 20 )
x0
now
A=
0
=
A2 =
let
A2 =
.....(2)
4 x 40
ax 3 mx 2 2 bx (ax mx b )dx = 3 2
ax 30 3
ax 30
a 2 x 60
ax 30 =
= max
0
3
mx 20 2
bx 0
ax 30 3 2 2 x 0 (1 x 0 ) = 36
x 02 (1 x 20 ) 36
This is maximum when x 02 = A2
=
ax 30
x 60 1 x 02
= 9 4 9 4 x 0
f (x0) =
x0
1 1 1 1 · · = ; 2 2 36 144
1 2
Amax =
1 12
1 = 12 Ans. ] A
67.
(i)
Equation of tangent from point (3, –3) to the given circle is y + 3 = m(x – 3) mx – 3m – y – 3 = 0
B ansal C lasses
Problems for JEE-2007
[33]
4m 3m 2 3 = 5 1 m2 (1 + 7m)2 = 25(1 + m2) 1 + 49m2 + 14m = 25 + 25m2 12m2 + 7m – 12 = 0
and also
(4m – 3)(3m + 4) = 0 m = 3/4 or m = – 4/3 equation of tangent at point A and B are
4 (x – 3) and 3 3y + 9 = – 4x + 12 4x + 3y = 3 Equation of normals to these 2 tangents are y+3=–
(ii)
3 (x + 4) 4 4y + 8 = 3x + 12 3(3x – 4y + 4 = 0) 9x – 12y = – 12 16x + 12y = 12 —————— x = 0; y = 1 y+2=
(iii)
4 (x + 4) 3 3y + 6 = – 4x – 16 4(4x + 3y= – 22) 16x + 12y = – 88 9x – 12y = 63 —————— 25x = – 25 x = – 1; y = – 6 y+2=–
points A and B are (0, 1) and (–1, – 6) Ans.
angle between the 2 tangents = 90° ADB = 90° | AD |max = CD + radius CD =
(iv)
and
3 (x – 3) 4 4y + 12 = 3x – 9 3x – 4y = 21 y+ 3=
50
| AD |max = 5 2 + 5 | AD |min = 5 2 – 5 Area of quadrilateral ADBC = AC × AD AD =
(iv)
7 2 12 25 = 25 = 5 area of quadrilateral ABCD = 5 × 5 = 25 sq. units. 1 area of triangle DAB = 25 = 12.5 sq. units. 2
Circle circumscribing DAB will have points A and B as its diametrical extremities x2 + y2 – x(–1) – y(–5) – 6 = 0 x2 + y2 + x + 5y – 6 = 0 Ans. x-intercept = 2 g 2 c = 2 (1 4) 6 = 5
Ans.
y-intercept = 2 f 2 c = 2 (25 4) 6 = 7 Ans. ]
B ansal C lasses
Problems for JEE-2007
[34]
68.
Let, f (x) = x2 x1 + (x +1)2x2 + ........ + (x + 6) 2x7 [if x = 1, we get 1st relation, and so on] note that degree of f (x) is 2 hence f (x) = ax2 + bx + c where f (1) = 1, f (2) = 12 and f (3) = 123 to find f (4) = ? hence a + b + c = 1 4a + 2b + c = 12 9a + 3b + c = 123 solving a = 50, b = – 139, c = 90 f (4) = 16a + 4b + c = 800 – 556 + 90 = 334 Ans. ]
69.
Suppose, circle x2 + y2 + 2gx + 2fy + c = 0 Solving with x = at2 , y = 2at a2t4 + 4a2t2 + 2gat2 + 4aft + c = 0 t 1 + t2 + t3 + t4 = = 0 ....(1) N : y + tx = 2at + at3 passing through (h, k) at3 + t(2a – h) – k = 0 ....(2) t 1 + t2 + t3 = 0 ....(3) from (1) and (3) t4 = 0 hence circle passes through the origin c = 0 equation of the circle after cancelling –at at3 + 4at + 2gt + 4f =0 at3 + 2(2a + g)t + 4f = 0 ....(3) Now (2) and (3) must be represent the same equation 2(2a + g) = 2a – h 2g = – (2a + h) and 4f = – k 2f = – k/2 equation of circle is x2 + y2 – (2a + h)x – (k/2)y = 0 x2 + y2 – 17x – 6y = 0 Ans. a ( t12 t 22 t 32 ) 2a ( t1 t 2 t 3 ) Centroid of PQR = , 3 3 xa =
a [(t 1 + t2 + t3)2 – 2 3
2a = – 3 =
( 2a h ) 2a 2 t1 t 2 = – . = – (2a – h) a 3 3
26 (a = 1 ; h = 15 ) 3
70.
t1 t 2 ]
26 ,0 ] 3
C :
1 ab ; also a2 + b2 = 3600 2 AD : y = x + 3 solve to get G (1, 2) BE : y = 2x + 4 acute angle between the medians is given by Area =
m1 m 2 2 1 1 1 tan = 1 m m = = tan = 1 2 3 3 1 2
B ansal C lasses
Problems for JEE-2007
[35]
In quadrilateral GDCE, we have (180 – ) + 90° + + = 360° = + – 90° cot = – tan( + )
tan tan –3= 1 tan tan 9ab = 2 × 3600 Area = 400 sq. units ]
71.
or
2 b 2a a b –3= 2 b 2a 1 · a b
2(a 2 b 2 ) 9= ab
1 ab = 400 2
W1: C1 = (–5, 12) W2: C2 = (5, 12) r 1 = 16 r2 = 4 now, CC2 = r + 4 CC1 = 16 – r let C(h, k) = c(h, ah) CC12 = (16 – r)2 (h + 5)2 + (12 – ah)2 = (16 – r)2 CC22 = (4 + r) 2 (h – 5)2 + (12 – ah)2 = (4 + r)2 By subtraction 20h = 240 – 40r h = 12 – 2r 12r = 72 – 6h ...(1) By addition 2[h2 + 25 + a2h2 – 24ah + 144] = 272 – 24r + 2r 2
12 h h2(1 + a2) – 24ah + 169 = 136 – 12r + r 2 = 136 + (6h – 72) + 2
2
[using (1)]
4[h2(1 + a2) – 24ah + 169] = 4[64 + 6h] + (12 – h)2 = 256 + 144 + h2 h2(3 + 4a2) – 96ah + 105 · 4 – 36 · 4 = 0 h2(3 + 4a2) – 96ah + 69 · 4 = 0; for 'h' to be real D 0 (96a)2 – 4 · 4 · 69 (3 + 4a2) 0 576a2 – 69.3 – 276a2 0 69 13 300a2 207 a2 ; hence m (smallest) = 100 10 69 m2 = ; p + q = 169 Ans. ] 100
So,
2 4 ( 1 sin x ) sec x dx ( 1 2 sin x sin x ) sec x dx I=3 = 3 5 6 5 6
72.
2
4
2 2 2 2 2 = 3 sec x (1 tan x )dx 2 sec x tan x sec x dx sec x (tan x )dx 5 6 5 6 5 6
2 2 2 = 3 (1 2 tan x ) sec x dx 2 (sec x tan x sec x ) dx 5 6 5 6
B ansal C lasses
Problems for JEE-2007
[36]
1 0 2 2 = 3 (1 2 t ) dt 2 t dt = 1 3 2 3
2 t 3 0 t 3 1
3
1 2 t3 1 3
3
1 2 1 2 8 (1) 3 3 3 3 3 3 3
= 3 (0)
11 6 3 16 11 2 3 3 8 = 3 9 3 3 9 3 3 3
1 2 2 8 = 3 1 = 3 9 3 3 3 3
6 3 5 5 3 b 3 = 2 – = a – 3 3 9 c 1
73.
I=
dx 1 x 1 x 2
0
put x = cos 2 4
I=2
0
4
0
I=
4
0
=
dx = – 2 sin 2 d
sin 2 d = 2 cos 2 sin 2
4
sin 2 d
= cos 1 4
0
=
2
0
4
1 d – 2 (1 cos ) d = 1 cos 0
4
sin 2 d = cos sin 2 4
0
4
0
2
0
sin 2 d
4
(1 cos ) d – 2 (1 cos ) d sin 2 0
4
cot cosec ) d – 2 (1 cos ) d = cot cosec
1 cos 1 = 0 cos 2 4
0
2
8 1
4
1 2 sin 2 d 1 cos
4
2 1 Lim
2 cos 2 4
0 (cosec
2
= 2 2 1 74.
4
cos 2 d = cos 1
4
=
a = 2, b = 5, c = 9 a + b + c + abc = 106 Ans.]
2 1
2
2 sin
4 0
0
2
a = 8, b = 1, c = 4 a2 + b2 + c2 = 81 Ans. ]
x · g f ( x) f ' g (x )g ' (x) = f g (x )g ' f (x ) f '( x) x · g f ( x )
d d f g ( x ) = f g ( x ) g f ( x ) dx dx
d d f g ( x ) g f ( x ) dx dx x· f g ( x ) g f ( x )
B ansal C lasses
Problems for JEE-2007
[37]
x·
d d ln f g ( x ) ln g f ( x ) dx dx
...(1)
e 2 a f g (x) dx 1 2 0 a
now,
differentiate w.r.t. 'a' f g (a) = e –2a from (1) and (2) we get
put
d ln g f (x) dx x = 0, C = 0
g f (x) e
75.
– 2x =
x 2 ;
ln f g (x) = – 2x
....(2)
ln g f (x) = – x2 + C
Hence g f (4) = e –16 k = 16 Ans. ]
Let f (x) = y dy + y = 4xe –x · sin 2x dx I.F. ex
I
f g (x) = e –2x
(linear differenial equation)
x dx yex = 4 x sin 2
II
cos 2 x 1 cos 2 x dx 2 2 x cos 2x sin 2x + C yex = 4 2 4 yex = 4 x
now
yex = (sin 2x – 2x cos 2x) + C f (0) = 0 C=0 –x y = e (sin 2x – 2x cos 2x) f (k ) = e –k (sin 2k – 2k · cos 2k ) = e –k (0 – 2k ) f (k ) = – 2 (k · e –k )
f (k ) = – 2 kek
k 1
e – +
2e –2 +
S –3 3e +
S =1· ......... + S e – = + e –2 + 2e –3 + ......... + —————————————————— S(1 – e – ) = e – + e –2 + e –3 + ...... 1 e – S(1 – e ) = = e 1 1 e 1 e S= = (e 1)(1 e ) (e 1) 2
2 e Ans. ] 2 (e 1)
B ansal C lasses
Problems for JEE-2007
[38]
76.
f (x) = Limit f (x h) f (x) h0 h
f ( x h ) x h 1 f 1 x f ( x ) = f(x) · Limit h0 h h
f ( x )
= Limit h 0
h f 1 1 x = f (x) Limit f 1 t 1 = f(x) · Limit x t 0 t h h0 x x Now putting x = 1, y = 1 in functional rule f(1) =
f (1) = 1 f (1)
f (x) =
f (x) · f (1) x
=
2f (x) x
f '(x) 2 = f (x) x ln (f(x)) = 2lnx + C x = 1; f(1) = 0 C = 0 Now solving y = x2 and x2 + y2 = 2 y2 + y – 2 = 0 (y + 2) (y – 1) = 0 y=1 1
A =2
;
f(x) = x2
2 y 2 y dy
0
1 1 2 2 y dy y dy = 2 0 0 1
1
= now
2 12 1 2 y dy y = 3 3 0
0
1
and
y=
2 y 2 dy
2sin
0
/ 4
/4
2 cos 2 cos d
0
/ 4
B ansal C lasses
d =
0
1 = + sin 2 2 0 Hence
2 cos
/ 4 2
1 2 A= 2 4 2 3
;
(1 cos 2) d 0
1 4
2
1 sq. units ] 2 3
A=
Problems for JEE-2007
[39]
10
77.
1 Z10 + Z10 13 = 0 Z 10
13 1 = – 1 = cos + i sin Z 13 –
1 = cos2m 1 i sin 2m 1 10 Z = e
i
( 2 m 1) 10 ( 2m 1)
i 1 = 13 – e 10 Z substituting m = 0, 1, 2,.......9 we get
i 1 e = 13 – 10 Z1
1 1 and are complex conjugate note Z Z 1 10
3
i 1 = 13 – e 10 Z2
5
i 1 = 13 – e 10 Z3
19
i 1 = 13 – e 10 Z10
Let
1 1 1 1 = and = Z1 a1 Z10 b1 and so on
i i 1 = 169 – 13 [ e 10 + e 10 ] + 1 a b i i
= 169 – 13
3 [ e 10
+ e
3 10
]+1
i 1 = 170 – 26 Re e 10 a b i i
and
3 1 i 10 etc a 2 b 2 = 170 – 26 Re e
3 5 3 9 1 = 850 – 26 cos cos cos cos cos 10 10 10 10 a b 10 i i = 850 – 26[cos18º + cos54° + cos90° + cos126° + cos162°] = 850 Ans. ]
B ansal C lasses
Problems for JEE-2007
[40]
78.(i)
|||ly
an + bn + cn = C0 + C1 + C2 + C3 + C4 + ................ an + bn + cn = 2 n ....(1) (1 + x)n = C0 + C1 x + C2 x2 + C3 x3 + ................ x = (1 + )n = C0 + C1 + C2 2 + C3 3 + C4 4 +................ = (C0 + C3 + C6 + .......) + (C1 + C4 + C7 + ........) + 2(C2 + C5 + C8 + ........) (1 + )n = an + bn + 2cn ....(2) 2 n 2 (1 + ) = an + bn + cn ....(3)
now
a 3n b3n c3n – 3an bncn = (an + bn + cn) (an + bn + 2cn) (an + 2 bn + cn)
now put
= 2n(1 + )n (1 + 2)n = 2n(– 2)n (– )n = 2n also
(a n b n ) 2 = 2(an + bn + 2cn) (an + 2 bn + cn) (a n b n ) 2 = 2 Ans.
78.(ii) Let and
x = C0 – C2 + C4 – C6 + ..... y = C1 – C3 + C5 – C7 + ....... (1 + i)n = C0 + C1 i + C2 i2 + C3 i3 + C4 i4 + ......... equating the real and imaginary part xn + i yn = (1 + i)n | xn + iyn | = | 1 + i |n = 2n/2
x 2n y 2n = 2 n/2
hence x 2n y 2n = 2n hence proved ] 79.
1 3 5 1 3 5 1 3 5 A2 = 1 3 5 1 3 5 = 1 3 5 = A matrix A is idempotent 1 3 5 1 3 5 1 3 5 Hence A2 = A3 = A4 = ....... = A x = 2, 3, 4, 5, ..........
x3 1 Lim 3 n x 2 x 1 n
now
n
x 1 Lim n x 1 x 2
x2 x 1 2 x 2 x x 1 n
2 3 4 5 n 1 n (n 1) 3 7 13 n · · ....... 2 Lim · · ....... n 1 2 3 ( n 1) 7 13 21 n n 1
n (n 1) 3 3 · 2 = n 1· 2 n n 1 2
Lim
80.
Ans.
]
a c a + log = log 2 a b
Given log
a c = log 2 b
log
B ansal C lasses
Problems for JEE-2007
[41]
a + c = 2b .....(1) 2 also a – ax + 2bx + c + cx2 = 0 (c – a)x2 + 2bx + (c + a) = 0 has equal roots D=0 4b2 – 4(c2 – a2) = 0 b2 = c2 – a 2 ....(2) from (1) and (2) b2 = (c – a)(c + a) b2 = (c – a) 2b 2(c – a) = b ....(3) 2 2 from (2) c = a + b2 triangle is a right at C. C = 90° A + B = 90° from (3) using sine law 2(sin C – sin A) = sin B C = 90° sin C = 1 A + B = 90° B = 90° – A 2(1 – sin A) = sin(90 – A) = cos A squaring both sides 4(1 – sin A)2 = cos2A = (1 – sin2A) 4(1 – sin A) = (1 + sin A) 3 = 5 sin A sin A = 3/5 B = 90 – A sin B = cos A = 4/5 and sin C = 1
81.
n 0
sin A + sin B + sin C =
3 4 12 + + 1 = Ans. ] 5 5 5
sin( nx ) 3n
put
e nix e nix sin (nx) = 2i
e ix ix e nix e nix e sin( nx ) 1 1 = = 3n 3n 2i n 0 2i n 0 3 n 0 3 n 0 n
n
1 1 3 1 3 = = 2i eix eix 3 eix 3 e ix 1 3 1 3 3 (3 e ix ) (3 eix ) 3 2i sin x = = 2i 9 3(e ix e ix ) 1 2i 10 6 cos x =
3 sin x 1 1 = = 2(5 3 cos x ) 2(5 2 2 ) 2 5 3 1 (1 9)
B ansal C lasses
Problems for JEE-2007
[42]
52 2 34 a = 5, b = 2 , c = 37 =
82.
a + b + c = 5 + 2 + 37 = 41 Ans. ]
log b log a (let b = 2m and a = 2n where m and n denotes the exponents on the base 2 in the given set) loga b =
m n hence loga b is an integer only if n divides m now total number of ways m and n can be chosen = 25 × 24 = 600 For favourable cases let n=1 hence m can take values 2, 3, 4, 5, 6, ......., 24 if n=2 m = 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24 n=3 m = 6, 9, 12, 15, 18, 21, 24 n=4 m = 8, 12, 16, 20, 24 n=5 m = 10, 15, 20, 25 n=6 m = 12, 18, 24 n=7 m = 14, 21 n=8 m = 16, 24 n = 9, 10, 11, 12 m = 1 for each =
Hence P =
= 24 = 11 = 7 = 5 = 4 = 3 = 2 = 2 = 4 —— 62
62 31 = Ans. ] 600 300
Select the correct alternative. (Only one is correct):
Q.83 A [Sol. There are 4C2 = 6 ways a block can differ from the given block in exactly two ways (1) material and size, (2) material and colour, (3) material and shape, (4) size and colour, (5) size and shape, and (6) colour and shape. Since there is only 1 choice for different material, 2 choices for different size, 3 choices for a different colour, and 3 choices for a different shape, it follows that the number of blocks in each of the above categories is (1 × 2), (1 × 3), (1 × 3), (2 × 3), (2 × 3) and (3 × 3), respectively. The answer is the sum of these six numbers = 29 Ans ] Q.84 C [Sol. Consider the expansion of (1 + x)99 and put x = i and equate the real part to get C] Q.85 B [Sol. f (x) = | Ax + c | + d g (x) = – | Ax + u | + v
figure is parallelogram and diagonals bisect each other
B ansal C lasses
Problems for JEE-2007
[43]
u c + = 3 + 1; A A
uc = – 4 Ans. ] A
Q.86 D [Sol. f (x) = x4 – 2x3 + 3x2 – 4x + 1 = 0 f ' (x) = 4x3 – 6x2 + 6x – 4x + 1 = 2(x – 1)(2x2 – x + 1) f ' (x) = 0 only at x = 1 also f '' (x) = 12x2 – 12x + 6 = 6(2x2 – 2x + 1) > 0 x R f is concave up f '' (1) > 0 x = 1 is minima also f (0) > 1; f (1) = – 1 two positive and two non real complex roots (D) ] Q.87 D [Sol. Notice the patterns Power of 3 go like this : 3, 9, 7, 1 Power of 7 go like this : 7, 9, 3, 1 Power of 13 go like this : 3, 9, 7, 1 so knowing that we can see that 31001 3 71002 9 131003 7 so therefore 3 · 9 · 7 = 9 ] Q.88 C [Sol. If a polynomial has real coefficients then roots occur in complex conjugate and roots are 2i, – 2i, 2 + i, 2 – i hence f (x) = (x + 2i)(x – 2i)(x – 2 – i)(x – 2 + i) f (1) = (1 + 2i)(1 – 2i)(1 – 2 – i)(1 – 2 + i) f (1) = 5 × 2 = 10 Also f (1) = 1 + a + b + c + d 1 + a + b + c + d = 10 a + b + c + d = 9 Ans. ] Q.89
D
[Sol.
1 1 1 ( k 2) k k k 2 ( k 2) k k k 2 Tk = = = 2 k k 2 2k ( k 2) k ( k 2) 2 k 2 ( k 2) T1 =
1 1 1 2 1 3
T2 =
1 1 1 2 2 4
1 1 1 T3 = and so on 2 3 5 1 1 1 2 1 2 = as k , sum = 2 1 = a + b + c = 11 Ans. ] 2 2 2 8
B ansal C lasses
Problems for JEE-2007
[44]
Q.90
C
[Sol. A =
1 2 b sin 2 = b2 sin cos ....(1) 2
now
x 65 x = 24 36 60 x = 24 · 65 x =26 sin =
12 5 and cos = 13 13
sin =
again,
b 65 = sin sin 2
from (1)
65 ·13 132 65 b = = 2 · 5 = 2 sin 2
134 12 5 A= · · = 169 · 15 = 2535 Ans. ] 13 13 4
Q.91 B [Hint: Let f (x) = ax3 + px2 + qx + r now use f (x) – f (x – 2) = x2 – 4x + 1 compare the coefficients to get a = 2/3; p = 1; q = – 1/6; hence p + q = 5/6 Ans. ] Q.92 B [Sol. n(S) = 4 × 4 × 4 = 64 n(A) = 211 or 312 or 413 or 431 or 422 6 3 3 = = 64 32 3 29 odds in favour 3 : 29 Ans. ]
P(E) =
Q.93
D
[Sol.
1 4 1 5 tan sin cos 5 13
= – tan sin 1
4 5 cos 1 5 13
= – tan( – ) where sin =
4 5 and cos = 5 13
4 12 3 5 tan tan 20 36 16 = = Ans. ] = – = – 4 12 63 63 1 tan tan 1 · 3 5
B ansal C lasses
Problems for JEE-2007
[45]
Select the correct alternatives. (More than one are correct):
Q.94 A,B [Sol. Let the numbers are a – d, a , a + d (a – d + 1), a, (a + d) in G.P. and (a – d), a, (a + d + 2) also in G.P. (a – d + 1)(a + d) = a2 hence (a2 – d 2) + (a + d) = a2 ....(1) 2 and (a – d)(a + d + 2) = a (a2 – d 2) + 2(a – d) = a2 ....(2) (2) – (1) 2a – 2d – a – d = 0 a = 3d 2nd term of A.P. = 3 times its common difference from (1) d2 – d 2 – 3d = 0 d=4 d 4 ; A.P. is 8, 12, 16 ]
a 12
Q.95 B,D [Sol. Let the line is normal to the hyperbola at P(x1, y1) hence 2x1 + 4y1 + k = 0 ....(1) 3x1 dy differentiate the curve dx P = y1
line
y1 slope of normal = – 3x ; 1
slope of line = –
2 9
y1
2 – 3x = – 2x1 = 3y1 9 1 (x1, y1) lies on the hyperbola 3x12
–
y12
= 23
3x12
–
4x12 9
= 23
23 x12 = 23 · 9
x1 = 3 or – 3 hence P is (3, 2) or (–3, –2) k = 24 or – 24 Q.96 A,B,C [Sol. Equation of AB y = 2x – 1 solving it y2 = 4x y2 = 2(y + 1) but y1 + y2 + y3 = 0 but y1 + y2 = 2 y3 = – 2 putting in y2 = 4x x3 = 1
B ansal C lasses
B, D Ans. ]
y2 – 2y – 2 = 0
Problems for JEE-2007
[46]
hence coordinates of c are (1, – 2) sum of the x and y coordinates of C are – 1 (A) is correct. obviously normal at C passes through the lower end of the latus rectum (B) is correct
x1 x 2 x 3
again centroid of ABC = now solving
3
1 with y2 = 4x
y = 2x – (2x – 1)2 = 4x 4x2 – 8x + 1 = 0
centroid of the ABC =
x1 + x2 = 2;
also x3 = 1
2 1 = (1, 0) 3
again equation of the normal at C
2 (x – 1) y+2=– 2 y+2=x–1 x–y–3=0 hence gradient of chord at C is 1
(D) is incorrect ]
Q.97 A,B,C [Sol. ' (x) = 2x[ f ' (x2) – f ' (1 – x2)] now f ' (x) is an increasing function then f ' (x2) > f ' (1 – x2)
x2 > 1 – x2
so ' > 0
x>
1 2
or
x<–
1 2
1 2 , 0 1 2 , increasing
] Q.98
B,D
[Sol.
x 2 cos x sin 2 x 5 3 3 2 = tan = – cot x = tan x = tan x 3 2 3 3 2 sin x sin x 6 3 3 2 5 2 + x = n + – x 6 3 x=
n – 2 12
n=
2x = n +
6
5 11 or Ans. ] 12 12
MATCH THE COLUMN:
Q.99 [Sol.
Ans. (A) P, (B) Q, (C) R, (D) P (A)
2 cos 3 i sin 3
n 2
is real
n i sin n 2n 2 cos is real 6 6 B ansal C lasses
Problems for JEE-2007
[47]
