[HINTS & SOLUTIONS] Practice Test Paper-1 Q.1 [Sol.
Given (2 + sin ) (3 + sin ) (4 + sin ) = 6. As, L.H.S 6. Equality can hold only if sin = – 1 3 7 3 7 , sum = + = 5 = k (Given) 2 2 2 2 So, k = 5 Ans.]
=
Q.2 [Sol.
M1 M2 M3 M4 M5 M6 | G1 G2 G3 | H1 H2 H3 H4 H5 H6 H7 Number of ways of selecting 4 books = 6C4 + 6C3 · 10C1 + 6C2 · 10C2 + 6C1 · 10C3 = 1610. Ans. Alternatively: Total number of ways of selection of four books – total number of ways of selecting four non maths books. 16 10 = C4 – C4 = 1610 Ans.] Q.3 [Sol.
Let
f '(x) = (x – 1) (x + 1) = (x2 – 1)
x3 Integrating on both sides, we get f(x) = x + c 3 2 Now, f(1) = – 1 – 1 = +c ......(i) 3
y 3
2 Similarly, f (– 1) = 3 3 = +c ......(ii) 3 From (1) and (2), we get 2c = 2 c = 1 and = 3
1 –1 O
1
x
–1
x3 x + 1 = x3 – 3x + 1 f(x) = 3 3 So, f(2) = 8 – 6 + 1 = 3 Ans.]
Paragraph for Question 4 to 6 [Sol.
Let P ' (x) = A(x – 1)(x – 2)(x – 3) or P ' (x) = A(x3 – 6x2 + 11x – 6) Integrating both sides with respect to x, we get A P(x) = (x4 – 8x3 + 22x2 – 24x) + B .....(1) 4 Given, P(0) = 0 B = 0 ......(2) and P(– 1) = 55 A = 4 .....(3) So, from (1), we get P(x) = x4 – 8x3 + 22x2 – 24x = x ( x 3 – 8 x 2 22 x – 24) = x(x – 4) x 2 4x 6
non real roots
(i)
Clearly, area of triangle formed by extremum points of P(x) =
1 2 1 1 2 PAGE # 1
(ii)
4 x2 As, P(x) + P(–x) = 2x 44 1
So,
even function 1
P( x ) P(x )dx
1
= 2 0
1
1
x 5 22 x 3 2 x 44 x dx = 4 x 22 x dx = 4 3 5 0 0 4
2
4
2
4 113 452 1 22 3 110 = 4 = 4 = = Ans. 15 15 5 3 15
(iii)
Q.7 [Sol.
From above graph, P(x) has two relative minimum points and one relative maximum point (A) is incorrect Clearly, from above graph, range of P(x) = [– 9, ) (B) is correct As, real roots of equation P(x) = 0 are 0 and 4. So sum of real roots = 4 (C) is incorrect As, P'(x) = 0 has 3 distinct real roots, so P''(x) = 0 has two distinct real roots. P(x) has two inflection points (D) is incorrect.] Statement-1: Let det.(B) 0, then B–1 exists. Now, AB = O ABB–1 = O A = O so det.(B) = 0. Similarly, suppose det.(A) 0, then A–1exists. Now, AB = O A–1AB = O B = O But B is not a null matrix, so det.(A) = O Statement-1 is true. Statement-2: Obviously Statement-2 is true. For example : 0 1 5 5 Let M = and N = then, MN = O, but neither M = O nor N = O. Ans.] 0 2 0 0
Q.8 [Sol.
Q.9 [Sol.
Statement-1: The equation of plane through A (1, 1, 1), B (1, – 1, 1) and C (–1, – 3, – 5) is 3x – z – 2 = 0, which clearly, passes through (2, k, 4) k R. Obviously, Statement-2 is true and not explaining Statement-1. ] We have 2 2 x sin x x 1 sin x 1 , x 0, 1 g(x) = if x 0, 1 0, Clearly, g(x) is differentiable x R. (As sum and product of differentiable functions is also differentiable function.)
cos x 2x sin x cos x 1 2( x 1) sin x 1 , Now, g '(x) = 0,
x 0, 1 x 0, 1
Clearly, g'(x) is discontinuous at both x = 0 and x = 1. Also, g(0) = 0 = g(1) (Given) So, Rolle's theorem is applicable for g(x) in [0, 1].
PAGE # 2
Q.10 [Sol.
A vector coplanar with a 2ˆi ˆj kˆ , b ˆi ˆj kˆ and perpendicular to c 5ˆi 2ˆj 6 kˆ will be along a b c = a · c b b · c a = 18 ˆi ˆj kˆ 9 2ˆi ˆj kˆ = 27ˆj 9kˆ 9 3ˆj kˆ
Vector will be along 3ˆj kˆ .
This vector will lie in the plane which will be parallel to it. Normal of plane will be perpendicular to vector. Ans.]
PART-B Q.1 [Sol. (A)
Obviously each digit must be 0 or 1
(B)
Hence number o five digit numbers are = 1 × 2 × 2 × 2 × 2 = 16 Ans. Let u = x2 du = 2x dx
1 Limit b 2
b2
1
u2 1 1
du = Lim b
1
1 1 tan 1 b 2 tan 1 1 = = =L 22 4 8 2
120L 120 = = 15 Ans. 8 ax + 3y – z = a .....(1) 2ax – y + z = 2 .....(2) bx – 2y + z = 1 – a .....(3) For no solution, D = 0
(C)
a 3 1 2 a 1 1 =0 D= b 2 1
[Note: D1 = – a]
D = 2b – a = 0 a 3 1 2 1 1 = a(1) – 3(2 – 1 + a) – 1(– 3 – a) Also Dx = 1 a 2 1 Dx = a – 3 – 3a + 3 + a = – a Dx 0 (Given a [1, 8] ) Hence for no solution a , aI 2
b=
sum =
1 36 (1 + 2 + ......... + 8) = = 18 Ans. 2 2
Note: when a = 0, b = 0 system will have infinite solution.]
PAGE # 3
PART-C Q.1 [Sol.
We have, f(x) = x3 – 3ax2 + 3(a2 – 1) x + 1 f '(x) = 3 x 2 2ax a 2 1 = 3 x (a 1) x (a 1)
=a+1 =a–1
M a–1
m a+1
a – 1 > – 2 and a + 1 < 4 a > – 1 and a < 3 So, a (– 1, 3) a1 = – 1 and a2 = 3. Hence, (a12 + a22) = (–1)2 + (3)2 = 10. Ans.] Q.2 [Sol.
Given, 2xy + 6 – 4x – 3y = 0 2x (y – 2) – 3 (y – 2) = 0 (2x – 3) (y – 2) = 0
Y 3 , 4 (0, 4) 2
Clearly, image of R (3, 4) in y = 2 is P(3, 0) and in x =
(3, 4) R
3 is Q (0, 4). 2
y=2 (3, 2)
1 43 2 Inradius of the PQR is, r = = 3 4 5 = 1. Ans.] s 2
O
x
3 2
(3, 0)
X
Q.3 x
[Sol.
Given,
f(x) = ex
2 x y · e · f ' ( y) dy ( x x 1)e
..........(1)
0
differentiating both the sides x
f '(x) =
ex
·
e–x
· f '(x) + e
x
e
y
· f ' ( y) dy – x 2 x 1 e x e x 2x 1
0
x
f '(x) = f '(x) + e x y ·f ' ( y) dy e x x 2 x
0
x
0 = f(x) + (x2 – x + 1) ex – ex (x2 + x) [Substituting
e
x y
f ' ( y) dy = f (x) + (x2 – x + 1) ex from (1)]
0
f(x) = ex (x2 + x – x2 + x – 1) f(x) = ex · (2x – 1) f(1) = e. Also, f '(x) = ex · 2 + (2x – 1) ex f '(1) = 2e + e = 3e and f '' (x) = 2ex + (2x – 1)ex + 2ex f '' (1) = 2e + e + 2e = 5e. Hence f(1) + f '(1) + f '' (1) = 9e k = 9. Ans.]
PAGE # 4
Practice Test Paper-2 Q.1 [Sol.
Given, h(x) = h'(x) =
f (x ) g( x ) g(x ) f ' (x ) f (x ) g' ( x) g2 (x)
Clearly, h'(c) = 0 So,
h'(c–)
(As, f ' (c) = 0 and g ' (c) = 0)
g ( c ) f ' (c ) f (c ) · g ' (c ) = g 2 (c )
[f '(c–) > 0; g'(c–) < 0 ; f (c–) > 0 ; g(c–) > 0]
h'(c–) > 0 |||ly h'(c+) < 0 So, h(x) has a local maximum at x = c. Ans. r Aliter-1: N i.e. f (x) is maximum at x = c and Dr i.e. g (x) is minimum at same x i.e. x = c Hence
f (x ) is maximum at x = c g( x )
Aliter-2: For example, let f (x) =
1 and g (x) = x2 and c = 0. 1 x2
2 x (1 2 x 2 ) 1 f (x ) So, h (x) = 2 = h ' (x) = 2 2 x (1 x 2 ) g ( x ) x (1 x 2 )
Clearly, for x < 0, h ' (x) > 0 and for x > 0, h ' (x) < 0 h(x) has a local maximum at x = c] Q.2 [Sol.
Given that a k1b k 2 c 0 P(a )
1 b a c a Area PQR 2 4 = 1 Area OQR bc 2
Now,
c–a
b–a O(0)
Q(b)
1 k1 b k 2 c k1b 1 k 2 c
= 4 (1 + k1) (1 + k2) – k1k2 = 4
b c
Hence, k1 + k2 = 3. Ans. ] Q.3 [Sol.
|A| = x + y+ z 2
det.(adj.(adj. A)) = det .A ( n 1) = (det. A)4 = 28 · 34 = 124 | A | = 12 x + y + z = 12 Using beggar method x + y + z = 9. Number of matrix = 11C2 = 55. Ans.]
PAGE # 5
Paragraph for Question 4 to 6 [Sol. (i)
The given lines (L1 and L2) are parallel and distance between them (BC or AD) is =
15 5 = 2 units. 5
Let BAC = AB = BC cosec = 2 cosec and AA1 = AD sec = 2 sec Clearly, area (||gm AA1BB1) = (AB)(AA1) = 4 sec cosec
L B D B1
A1
L2
C A P(4, 3)
L1
8 , which is least for = sin 2 4 Let slope of line L be m.
=
m So, 1 =
3 4
(4m + 3) = ± (4 – 3m) m =
3 1 m 4
1 or – 7 7
But m > 0 (Given) 1 (x – 4) x – 7y + 17 = 0 Ans.(i) 7 If line L : x – 7y + 17 = 0 is orthogonal to circle x2 + y2 – 6x + 4y – 9 = 0, then line L must be normal to the given circle. So, centre of circle (3, – 2) must satisfy the line L. Hence, 3 – 7(– 2) + 17 = 0 17 + 17 = 0 = – 1 Ans.(ii) Equation of circle S having the ends of diameter at (0, – 1) and (–2, 0) is x(x + 2) + y(y + 1) = 0 i.e. x2 + y2 + 2x + y = 0 So, equation of circle S', is (x2 + y2 + 2x + y) + (x – 7y + 17) = 0
The equation of line L, is (y – 3) = (ii)
(iii)
As it passes through (1, 2), so (1) 2 ( 2) 2 2(1) 2 1 14 17 = 0 9 9 + 4 = 0 = 4 Hence, the equation of circle S', is
(x2 + y2 + 2x + y) – Q.7 [Sol.
9 (x – 7y + 17) = 0 4x2 + 4y2 – x + 67y – 153 = 0 Ans.(iii)] 4
Equating the coefficient of x20, we get 20 220 – a20 = 1 a 2 20 1 (C) 1 put, x = , we get 2 a 0 – b 2
a b 2
20
20
10
1 p q 4 2 10
1 p q 0 4 2
a 1 p b and q 0 2 4 2 PAGE # 6
a + 2b = 0 (B) 20 2 20 1 . 2 x = 0 we get 1 – b20 = q10 b= put,
20
20 2 20 1 = q10; 1– 2 1 p q 0 Using, 4 2 1 p 1 0 p=–1 4 2 4 Hence B, C, D. Ans.]
1
2
20
q
1
2 20
10 ;
1 1 q10 q = 20 4 2
Q.8 y
x= – 4
y=3 3/2
[Sol. –4
x.
4
O
Now, verify alternatives. ]
PART-B Q.1 [Sol.
(A)
IN =
x
9
·e
x 2
dx =
9
x 2
dx xI · e II 0
0
I.B.P.
2 x 8 · e x + 8 IN = 2 2 0
x
7
2
· e x dx
0
IN = 0 + 4ID IN = 4ID IN I D = 4 Ans.
(B)MB x4 – 13x 2 + 36 0 (x2 – 9) (x2 – 4) 0 x [– 3, – 2] [2, 3] Now, let f(x) = x3 – 3x f '(x) = 3(x2 – 1) > 0 x [– 3, – 2] [2, 3] fmax. (x = 3) = (3)3 – 3(3) = 27 – 9 = 18. Ans. (C)MB Any circle through (2, 2) and (9, 9) is (x – 2) (x –9) + (y – 2) (y – 9) + (y – x) = 0 .........(1) For the point of intersection with x-axis, we put y = 0 in (1), we get (x – 2) (x – 9) + 18 – x = 0
PAGE # 7
disc. = 0 (11 + )2 – 4 + 36 = 0 = – 23, 1 1 x= =±6 2 So, the absolute value of the difference of x-coordinate of the point of contact = | 6 – (– 6) | = 12 Ans. Put
(D)
– sin–1(3x – 4x3) = – 3 sin 1 x 2 2 1 because sin–1(3x – 4x3) = – 3 sin–1x if x , 1 2
y = cos–1(3x – 4x3) =
dy = dx
Hence dy dx x
3 2
=
3 3 1 4
3 1 x2
= 6 Ans.
dy 1 (3 12x 2 ) dy Alternatively (D): Clearly, = dx x 1 (3x 4 x 3 ) 2 dx
3 2
3(4 x 2 1) 1 x (3 4 x 2 ) = 6 Ans.]
PART-C Q.1 x
[Sol.
x
f ( x ) f ( t ) tan t dt tan t x dt = 0 0
0 Using King
x
x
f ( x ) f ( t ) tan t dt tan t dt = 0 0
0
Differentiate w.r.t. x f '(x) + f ( x ) 1 tan x = 0 f ' (x) + tan x = 0 f (x) 1 integrate w.r.t. x ln f ( x ) 1 + ln (sec x) = C f(0) = 0 C = 0. Hence
f ( x ) 1 1 cos x
f(x) = cos x – 1 f(x) = 0 cos x = 1
Hence only solution in , is x = 0. 2 2 Hence number of solution is 1. Note that domain of f(x) is , Ans.] 2 2 PAGE # 8
Q.2 [Sol.
We have, f '' (x) = 12x – 4 f ' (x) = 6x2 – 4x + c As, f ' (1) = 0 c=–2 2 f ' (x) = 6x – 4x – 2 f (x) = 2x3 – 2x2 – 2x + As, f (1) = 0 =2 Hence, f (x) = 2(x3 – x2 – x + 1) or f (x) = 2(x – 1)2 (x + 1). Now, M (x = 2, y = 6) and f ' (2) = 14. So, the equation of normal at M is (y – 6) =
1 (x – 2) 14
For x-intercept, put y = 0 we get x = 86 Ans.] Q.3 [Sol.
Let y = f 2(x) – g2(x) dy = 2 f ( x ) f ' ( x ) g ( x )g ' ( x ) = 2 f ( x ) g ( x ) g ( x ) f ( x ) = 0 dx f 2(x) – g2(x) = k (constant) So, f 2(3) – g2(3) = k (5)2 – (4)2 = k k = 9, x R Hence, f 2() – g2() = 9. Ans.]
Now,
Practice Test Paper-3 Q.1 [Sol.
We have (1 + x)10 = 10C0 + 10C1 x + 10C2 x2 + 10C3 x3 +........ + 10C9 x9 + 10C10 x10 Also (x – 1)10 = 10C0 x10 – 10C1 x9 + 10C2 x8 +........ – 10C9 x + 10C10 Multiplying (1) and (2), we get 10
10
x 1 = 2
C0 10C1x ....... 10C9 x 9 10C10 x10 10
.......(1) .......(2)
C0 x10 10C1x 9 ....... 10C9 x 10C10
.......(3)
Comparing the coefficients of x10 in (3) , we get 10C
5
(–1)5 =
10 C0 2 10 C1 2 10 C2 2 ....... 10 C9 2 10 C10 2
Ans.]
Q.2 [Sol.
f ' (x ) 1 ln (f(x)) = x + c f (x ) x = 0, f(0) = 1 c = 0 f(x) = ex g(x) = ex (x + 1)2 – ex = ex (x2 + 2x) f '(x) = f(x) =
1
1
x e f (x ) g( x ) dx = e x x 2 2x dx = e x · x 2
0
0
1
0
= e Ans.]
PAGE # 9
Q.3 x2
[Sol.
Let, I =
1 sin x 1 sin 2 x King & Add
x 2 2 2 1 sin 2 x
2I =
dx
2
2 2 1 1 sin x sin x
So, I =
dx
x 2 1 1 sin 2 x dx 1 = 2 2 2 1 1 sin x
3 x dx = 3 . Ans.]
2
Paragraph for question nos. 4 to 6 [Sol.(i) angle between a and b is obtuse hence a · b 0 (log3x)2 + 2log3x – 1 0 x (0, ) 42 + 4 0 [–1, 0]
2 1 1 1 1 1 = |2 – 3 | Volume = 6 1 1 1 6
Since, [–1, 0]
(ii)
1 1 Volume , 3 2 a c = b c
( log
3 x)
(a b ) c 0
ˆi 2ˆj kˆ
= (log
ˆ
a b c
3 x ) i ( log3 x )
ˆj kˆ + ˆi ˆj kˆ
log 3 x log 3 x ; 2 = log3x + and 1 = – 1
=2 log3x = 0 and log3x = – 2 =0 = 2, log3x = –2, = 0 a 2ˆj kˆ b 2ˆi kˆ c ˆi ˆj kˆ | b c | | 3ˆi ˆj 2kˆ | 7 |ac| | ˆi ˆj |
PAGE # 10
(iii)
( a b ) c = a 2b ( b · c ) a (a · c ) b a 2 b log3x + log3x – 1 = –1 ... (1) log3 x + 2 + 1 = 2 log3x = –1 ... (2) log3x = 1, = –1 a ˆi 2ˆj kˆ b ˆi ˆj kˆ c ˆi ˆj kˆ 2 a b b c c a = a b c = 4. Ans.]
b · c 1 , a · c 2
Q.7 ex
[Sol.
dt f(x) = 1 t2 1
e x
1
dt –1 e x –1 e – x = tan –1 (e x ) tan –1 (e – x ) – 2 tan t ]–1 tan t ]1 1 t 4 4
f(x) = tan –1 (e x ) + cot –1 (e x ) =
2
Now verify alternatives. Ans.] Q.8 [Sol.
cos C =
Also,
1 16 64 c 2 = 2 64
a c = sin A sin C
As C = 60°, A = 30°
c2 = 48
4 3 4 = sin A 3 2
c= 4 3
A = 30°
B = 90°
1 ac 44 3 16 3 4 3 ac Now, r = = 2 = = = = abc 4 4 3 8 12 4 3 s abc 3 3 2
Now verify alternatives. 4 3 3 3
(C)
r=
(D)
the length of internal angle bisector of angle C is
8 3
Hence A, B are correct.]
PAGE # 11
PART-B Q.1 [Sol. (A)
Given, C : y2 = px3 + q
.......(1)
3px 2 dy dy = 3px2 = 2y dx dx Put x = 2 and y = 3 in (1), we get 9 = 8p + q 2y
Also, from (2), we get
dy 3p ( 4) = = 2p = 4 dx M ( 2,3) 6
.......(2) .......(3) .......(4)
From (2) and (3), we get p = 2, q = – 7. Hence, (p – q) = 2 – (– 7) = 2 + 7 = 9. Ans. (B)
1 We have V 1 0
1 2 cosec 1
0 = 4 cosec2 – 1 – 2cosec 1 2 cosec
2 1 5 2 = 4 cos ec cos ec 1 = 4 cos ec 2 4 4
9 5 4 Vleast = 4 1 ] 2 16 4 4
(C)
We have
sin x 1
x 1
cos x x
x =0 1
(sin x – x) ( cos x – x2) = 0 Either sin x = x or cos x = x2 Hence, number of real roots are 3. Ans. (D)
Given, tan A =
1 2 1 , tan B = , tan C = and tan D = d 4 3 5
Now, A + B + C + D = 2 tan (A + B) = – tan (C + D)
ab cd 1 2 1 = d 1 = 1 1 ab cd 1 4 3 5
1 1 d 6 5
Solving 39 d = – 65 d=
5 5 3 | tan D | = 3 = 5. Ans.] 3 3 PAGE # 12
PART-C Q.1 [Sol.
Let O be the centre of polygon Area of rectangle = 4 OA1A2 = 6 and Area OA1A2 =
1 area of polygon n
.........(1) A1
.........(2)
Ak + 1 O
A2
6 1 (1) and (2) = 60 4 n n = 40. Ans.]
Ak
Q.2 [Sol.
Using tan–1 + tan–1 + tan–1 = tan–1 1 In L.H.S. we get
c ax bx c · ax · bx tan–1 1 c(ax) ax(bx) (bx)(c) 2 1 x where c = . x 8 1 x Now, 1 = ax + abx2 + bx x 8
1 x a b 1 = a x 2 abx 2 b x 2 x R x 8 8 8
a b 1 = (a + b) + x2 ab 8 On comparing, we get a+b=1 ..........(1) and ab =
ab 1 = 8 8
.......(2)
Now, a2 + b2 + 2ab = 1 a2 + b2 +
(Using (1)) 1 =1 4
Hence, 4(a2 + b2) = 3. Ans.] Q.3 [Sol.
Using cosine law in BKC, 18 4 1 B 4 cos = 2 3 2 2· ·2 2
9 2 = 15 = 5 = 12 2 4 2 6 2 3
B
c= 2m x 3 2 2
A
a=2
1
m K
C
PAGE # 13
Now, x=
( 2 · 2 · 2 m) 5 2ac B cos = · ac 2 2m 2 4 2
5 4m 5m 3 2 = · 3= 3m + 3 = 5m m 1 4 2 m 1 2
m=
3 2
AB = 2m c = 3 . Ans.]
Practice Test Paper-4 Q.1 [Sol.
S=
1 1 tan 3 0
Note that
1 1 tan 3 10
1 3
1 tan
.......
1 1 tan 3 80
1 3
1 tan 90
=1
1 1 1 tan 3 1 tan 3 = = = . 1 tan 3 1 cot 3 1 tan 3 1 tan 3 1 tan 3 Hence, S = 1 + (1 + 1 + 1 + 1) = 5 A. Ans.] Q.2 [Sol.
Given, f '(x) = 2f ( x ) 5 or
dy = 2dx. On integrating, we get y5
ln ( y 5) = 2x + c, if x = 0 ; y = 0 c = ln 5 y5 = 2x y + 5 = 5e2x y = 5e2x – 5 f(x) = 5(e2x – 1) ln 5 Now, f(x) + 5 sec2x = 5(e2x + tan2x) = 0 So, no solution exist (0, 2). Ans.] Alternative : f '(x) – 2 f(x) = 10
dy – 2y = 10 ; Integrating factor = e–2x dx y · e–2x = – 5e2x – 5 f(0) = 0 ; c = 5 y = 5e2x – 5 f(x) = 5 (e2x – 1) Hence, f(x) + 5 sec2x = 0 e2x + tan2x = 0 No solution. Ans.]
PAGE # 14
Q.3 2
[Sol.
I=
1
tan 1 x dx x2 x 1
.......(1)
put x =
1 t
2
2
I=
1
1 t dt 2 t t 1 tan 1
.........(2)
2
Now, (1) + (2) gives 2I = 2
2
1 2
dx 2 3 1 x 2 2
2
2
2 2x 1 tan 1 I= 4 3 3 1
2
2 0 = = A. Ans. ] 2 3 3 6 3
Paragraph for question nos. 4 and 5 [Sol.
r a · n 1 0
n1 = ^i + ^j
r · n1 a · n 1
xˆi yˆj zkˆ · ˆi ˆj = 2
L A (1,1,1)
x + y = 2 ............ plane 1 Now,
AB ˆi kˆ ; A(1, 1, 1); B(0, 1, 0)
Now,
ˆ ˆ kˆ i j AB V n 2 1 0 1 1 1 1
r = (1, 1, 1)+(1, – 1, – 1)
n2 = AB × V
^
^
^
V=i–j–k
^
B( j ) 2
n 2 ˆi 0 1 ˆj 1 1 kˆ 1 = n 2 ˆi 2ˆj kˆ Hence, equation of plane 2 is r ˆj · ˆi 2ˆj kˆ = 0 or x + 2y – z = 2 Vector along the line of intersection of planes 1 and 2 is n1 n 2 which is also perpendicular
(i)
to the normal vector of plane 1. n n = – ˆi ˆj kˆ 1
2
So, required vector = ±
ˆi ˆj kˆ =± 6 3
2 ˆi ˆj kˆ or ± 2 1, 1, 1
PAGE # 15
(ii)
If is the acute angle between 1 and 2 then ˆi ˆj · ˆi 2ˆj 2kˆ n1 · n 2 cos = = = n1 n 2 2· 6
Q.6 [Sol.
3 3 3 = . Ans.] = = cos–1 6 2· 6 2 2
A is true by the Extreme Value Theorem, B is true by the Intermediate Value Theorem and C is not always true, D is true because g is continuous.]
Q.7 [Sol.mb As, cos–1x 0 x [– 1, 1] and tan–1x > 0, x in (0, ) So, a solution must be a positive number only. Now, cos–1x = tan–1x 1 cos–1x = cos–1 2 1 x
x=
1 x2 = 1 x2
x4 + x2 – 1 = 0 x2 =
1 5 2
1 5 (0, 1) 2
Now verify alternatives.] Q.8 [Sol.
Q.9 [Sol.
Suppose , be two distinct roots in (0, 1) of the equation x3 – 3x + p = 0 Let f(x) = x3 – 3x + p. As f() = 0 = f() f(x) satisfies hypothesis of Rolle's theorem in [, ] (0, 1). So f '(c) = 0 3c2 = 3 c = ±1 But c must lies between and . Hence p . So, S1 is false. Also, S2 is obviously true. Ans.]
We have 3
f(x) = tr.(A ) =
a ii i 1
x As,
2
9 x
= x+
9 +2 x
9 9 x x + 6 (as x > 0 is given) x x
so, f(x) 8 fMin. (x = 3) = 8 Also, S2 is true and explaining S1 also. Ans.] PAGE # 16
PART-B Q.1 [Sol. (A)
Let the point of intersection of curves y2 = 2ax (a > 0) and xy = 4 2 be P(x1, y1). Now, y2 = 2ax Also,
xy = 4 2
dy dx P ( x
y1
= 1 , y1 )
a = m1 y1
= m2
... (2)
As,
m1 × m2 = –1 x1 = a
...(3)
As,
x1y1 = 4 2 y1 =
So,
4 2 P x1 a , y1 a
1 , y1 )
Also,
(B)
dy dx P ( x
x1
... (1)
4 2 a
Point P must satisfy y12 = 2ax1 32 = 2a2 a4 = 16 a = 2. Ans. a2 f (x) is differentiable in R
Hence Lim f ( x ) must exist and is finite
y=f(x)
(0,4)
x
y = f (x) must have a horizontal asymptote as x then only Lim f ( x ) will exist
y=1 O
x
x
Lim f ' ( x ) = 0 x
(C)
Lim f ( x ) f ' ( x ) = 3 x
Lim f ( x ) = 3 Ans. x
y = mx + (n + 2 – x) or y + x – (n + 2) – mx = 0 y (n 2) + (1 – m) (x – 0) = 0 m R Fixed point is the point of intersection of lines y = n + 2 and x = 0. Family of lines which passes through (0, n + 2) Hence n + 2 = 3 n = 1. Ans.]
PAGE # 17
PART-C Q.1 [Sol.
Q.2 [Sol.
Let the 3 consecutive terms are a – d, a, a + d ; d > 0 hence a2 – 2ad + d2 = 36 + K ....(1) 2 a = 300 + K ....(2) 2 2 a + 2ad + d = 596 + K ....(3) now (2) – (1) gives d(2a – d) = 264 ....(4) (3) – (2) gives d(2a + d) = 296 ....(5) (5) – (4) gives 2d2 = 32 d2 = 16 d = 4 or Hence from (4) 4(2a – 4) = 264 2a – 4 = 66 2 K = 35 – 300 = 1225 – 300 = 925 Ans.]
d = – 4 (rejected as increasing A.P.)
2a = 70
a = 35
Given, f (y) = f (x) + y y2 + ay + b = x2 + ax + b + y y2 + y(a – 1) – x2 – ax = 0 As, y R, so D 0 x R (a – 1)2 + 4(x2 + ax) 0 x R 4x2 + 4ax + a2 – 2a + 1 0 x R Now, D 0
16a2 – 16(a2 – 2a + 1) 0 2a – 1 0 a
1 2
1 2 Hence, maximum value of 100a = 50 Ans.]
So, amax. =
Q.3 [Sol.
For number of solutions of given equation to be 9, so n must be 5. 2
Now (3 + cosec2x) + 2sin y = 5 possible only if cosec2x = 1 and sin2y = 0 3 5 7 , , , 2 2 2 2 possible values of y = 0, , 2, 3, 4 Hence number of ordered pairs (x, y) = 4 × 5 = 20 Ans.]
possible values of x =
Q.4 [Sol.
Put f (x) = t f (t) = 4 t = – 2, 0, 4 Hence f (x) = – 2 2 values of x. or f (x) = 0 2 values of x. or f (x) = 4 3 values of x. Hence, the number of real solutions of the equation f (f(x)) = 4 are 7. Ans.] PAGE # 18
Practice Test Paper-5 Q.1 [Sol.
Lim f ( x ) = Lim x
So,
h0
2
Lim f ( x ) = Lim
and
x
2
h0
sin (1 ) 1 . h
Lim f ( x ) does not exist. Ans.] x
Q.2 [Sol.
sin (sinh) =–1 h
2
1, x, y G.P. Let x = k, y = k2 and x, y, 3 A.P. 2y = x + 3 2k2 = k + 3 2k2 – k – 3 = 0 k = – 1,
3 . 2
2
1 1 x + y = k2 + k = k . 2 4
Q.3 [Sol.
1 15 3 at k = . Ans.] 4 4 2
x + y|max = 4
Possible cases : (i)
000003
6! 5!
(when digit 0 comes at first place then number of arrangement =
Number of six digit numbers with 0 0 0 0 0 3 =
(ii)
6! 5! = 30 Similarly for 0 0 0 2 2 1 2! 2! 3! 2 !
(iii)
1 1 1 1 1 2
5! ) 4!
6! 5! – =1 5! 4!
6! =6 5!
Total = 1 + 30 + 6 = 37. Ans.] Paragraph for question nos. 4 to 6
[Sol.
3 4 We have, M(0) = 0 0
0 1 2 0
0 0 = diag 2 3
3 1 2 , , 4 2 3
PAGE # 19
3 2 1 2 2 2 M(0) diag , , 4 2 3 2
M(0)3 diag 3
3
1 , 4 2
3
2 , 3
3
and so on
Now, M(0) M(0)2 M(0)3 ......... M(0)n
n 2 n 2 n 3 3 2 3 1 1 1 2 2 2 diag ...... , ...... , ..... = 4 4 4 2 2 2 3 3 3
So,
2 2 3 3 2 1 1 2 2 ..... , ........ , ..... P = Lim diag 4 4 n 2 2 3 3
2 1 3 3 4 2 = diag (3, 1, 2) diag , , = 1 3 1 1 1 2 2 3 4 1 1 Also, Q = diag , 1, 3 2
1 0 0 PQ = 0 1 0 = I3 0 0 1 (PQ)m = I3 m N Rm = M(x) R100 = M(x) Sk = M(x) · ( PQ ) ( PQ ) 2 ..... ( PQ ) k S360 = M(x) · 360 · I3 = 360 M(x)
(i)
adj (adj P) = P
(ii)
n2
· P = 6 · P = 6P
Tr. adj adj P = Tr (6P) = 36.
Tr . (R100) = f(x) = Tr . M( x ) 1 1 2 2 23 f(x) = 3 x x x 2 x = 3x2 + 2x + . 4 2 3 12 2
f (sin x) f (cos x) 0
2
=
3 sin 0
2
x 2 sin x
41 dx 6
23 23 41 3 cos 2 x 2 cos x dx 12 12 6
2
= 2 sin x cos x dx = 4 0
PAGE # 20
(iii)
Tr. (S360) = g(x) = 360 Tr . M( x ) 23 23 2 2 2 = 360 3x 2 x = 360 3 x x 12 3 36 2 2 1 23 1 1 19 360 3 x 360 3 x = = 3 36 9 3 36
g(x)|min = 360 × 3 ×
19 = 570. Ans.] 36 graph of g(x) = x3 – 3x2 + 5x – 1
Q.7 [Sol.
We have f ( x )
x 12 .e x 2 2
1 x
f ' (x) =
x 1x 3 3x 2 5x 1.e x 3
2
x 1
1
–1
Let g (x) = x3 – 3x2 + 5x + 1, g ' (x) = 3x2 – 6x + 5, discriment < 0 so g (x) has only one real root.
0
Also g (–1) g (0) < 0 so one root of g(x) = 0 belong to the interval (–1, 0) say g(x)
3 2 x 1 x 3x 5x 1 .e x Now sign scheme of f '(x) = ( x 2 1)3
+ve
–ve
At x = 1
1 +ve
At x =
–ve
Clearly f (x) has two points of extremum, maxima at x (–1, 0) and minima at x = 1.] Q.8 [Sol. (A) (B) (C) (D)
a · b = a · c either b c or a b c . a b = a c either b c or a || b c True True. Ans.]
PART-B Q.1 [Sol. (A)
Given cos =
2 1 sinsin =
cos
=
2 1
2 1 cos
2 1 cos = sin 2 cos = sin + cos
cos + sin = 2 cos but given that cos + sin = k cos
PAGE # 21
Hence, K =
(B)
2
L = 11 – 2 (As, K + L = 11 (given)) 9 < L < 10 2 < log3L < 3 Sum of integers = 2 + 3 = 5. Ans. OC = 2 OC2 = 4 h2 + k2 = 4 x2 + y2 = 4. Y
2 O
Ans. X
2
2
(C)
C(h,k)
sin sin 1 1 2 x = sin where = sin–1x 3
3 1 3 cos sin = 2 2 2
1 – 2x =
1 x x2 2
3 1 x 2 x 2 3 3x 2 3x (2 – 3x)2 = 3(1 – x2) 4 – 12x + 9x2 = 3 – 3x2 Hence, (12x – 12x2) = 1. 2 – 4x =
10
(D)
g (10) =
2
f (t) d t = 5
0
0
1 1 f (t) d t = 5 f ( t )dt f (2 t )dt 0 0
[As, f is periodic with period 2.]
2a
a
a
f ( x ) dx f ( x ) dx f ( 2 a x ) dx 0 0 0
1 1 = f ( t )dt f ( t )dt (Since f(x) is periodic with period 2 hence f (2– t) = f (– t)) . 0 0 1 1 = f ( t )d t f ( t )d t 0 0
(f (– t) = – f(t), as f is odd.)
= 0. Ans.]
PART-C Q.1 [Sol.
x2 + y2 + y – 1 + k (x + y – 1) = 0 .......(1) Hence C passes through the intersection of x2 + y2 + y – 1= 0 and y = 1 – x Solving x2 + (x – 1)2 + (1 – x) – 1 = 0 2x2 – 3x + 1 = 0 2x2 – 2x – x + 1 = 0 2x (x – 1) – (x – 1) = 0 PAGE # 22
x = 1 or x =
1 2
y = 0 or y =
1 2
1 1 Hence, points are (1, 0) and , 2 2 Now minimum radius will be when the two points are the extremities of the diameter of the circle C.
diametermin = Hence, r |min =
1 1 1 = 2 4 4 1 2 2
=
1 R = 8. Ans.] 8
Q.2 [Sol.
=
1 1 1 8b = ×8×a= × 10 × c 2 2 2
a = b and c =
4a 5
.......(1)
16a 2 b 2 c2 a 2 c2 cos A = = = 4a 2bc 2bc 25 2 a 5
cos A =
2 sin A = 5
21 5
From (1) sin C = =
4 4 21 sin A = 5 25
1 1 ab sin C = × hc × c 2 2
=
1 2 4 21 1 a × = × 10 × c 2 2 25
=
25c 2 4 21 = 10 × c 16 25
p 40 40 = q ×c c= 21 21 p + q = 40 + 21 = 61 Ans.
c2 =
PAGE # 23
Alternatively: =
1 ah 2 a
2 = ha = 8 a
= 4 .....(2) and b (1) and (2) a=b
|||ly
and
a 5 b = = c 4 c
a=b=
s=
abc 7c = 2 4
= 4 .....(1) a
= 5 ......(3) c
5c 4
s–a=
7c 5c 2c c = = =s–b 4 4 4 2
s–c=
7c 3c c = 4 4
2 = s(s – a)(s – b)(s – c) =
Also
=
7c c c 3c · · · 4 2 2 4
.....(4)
1 · c · 10 = 5c 2 2 = 25c2 .....(5)
(4) and (5) 25c2 =
7c c c 3c 25 64 21c2 · · · 25 = c2 = 4 2 2 4 21 64
p 5 8 40 = = q 21 21 p + q = 40 + 21 = 61 Ans.]
c = AB = Q.3 100
[Sol.
Let I =
1
Put, 10
f (x) dx = x
10
100
f (x) dx x 1
10
f (x ) dx = x
10
f (x) dx x 1
100 f
10
100 x dx x
dx dt 100 dx 100 dx = dt – t =t = dt 2 x x x t x 1
10
f (x ) f (t) f (x ) dx dt = 2 dx = 2 × 5 = 10. Ans.] I= x x t 1 1 10
PAGE # 24
Practice Test Paper-6 Q.1 [Sol.
A
As, ar.(ABC) = ar. (ACD) + ar. (BCD) 1 1 1 (3) (4) = (4) (CD) sin 60° + 3 CD sin 30 2 2 2
CD =
c=5 b=4
24 8 24( 4 3 3) ( 4 3 3) . Ans.] = = 4 33 13 39
D 60º
Q.2
C 100
Sol.
k0
k3 k 1! k 2 ! k 3!
100
=
k0 100
=
k0
Q.3 [Sol.
k3
k 1! k 2 6k 9 k 2 k 3 1 = k 3 ! S=
= 100
100
=
30º a=3
B
k 3
k 0 k 1! 1 k 2 k 3 k 2 k3
k 1! k 32 1
1
= k 0 k 2 ! k 3 !
100
=
1
k 0 k 3 k 1!
1 1 2 ! 103 !
1 B. Ans.] 2
Clearly, = 0
1 k 1 k 1 1 = 0 k 1 Ans.] 1 1 1 Paragraph for Question 4 to 6
[Sol. (i)
Given 1, b, c are in A.P. 2b = 1 + c ......(1) Also, 2, 5b, – 10c are in G.P. 25b2 = – 20c or 5b2 = – 4c .......(2) Now, putting c = (2b – 1) from (1) in (2), we get 5b2 = – 4(2b – 1) 5b2 + 8b – 4 = 0 b = – 2 (As b I) So, c = – 5 f (x) = x2 – 2x – 5 = (x – 1)2 – 6 Clearly in interval x [0, 4], m = minimum value of f (x) = fmin.(x = 1) = – 6 and M = maximum value of f (x) = fmax.(x = 4) = (4 – 1)2 – 6 = 9 – 6 = 3. Hence, (M + m) = 3 + (– 6) = – 3 Ans.(i)
PAGE # 25
(ii)
Given g (x) = a 2 1 x 2 a 2 4 x 3
Graph of g ( x ) is parabola opening upward
(iii)
Also, g (0) = – 3 g (x) < 0 is true, for atleast one real x provided – a2 + 4 0 –2a2 So, the number of integral values of a are 5 (i.e., a = – 2, – 1, 0, 1, 2). Ans.(ii) We have y = f (x) + h(x) = (x2 – 2x – 5) + x2 – (p – 3)x + p = x2 + (1 – p)x + (p – 5) As, range of y is [0, ), so put D = 0 6p 21 0 (1 – p)2 = 4(p – 5) 1 – 2p + p2 = 4p – 20 p2 Always So, p Ans.(iii)] positive p R
Q.7 [Sol. Q.8 [Sol.
Q.9 [Sol.
Statement-1 is true and Statement-2 is false. Statement-2 could be true only if f(x) is continuous. Rolle's Theorem can also be used for validity of Statement-1. Ans.] Statement 1: B1 + B2 + B3 + B4 = 10 B1 + B2 + B3 + B4 = 6 Using beggar total number of ways 9C3 Statemen-2: Number of ways of choosing any 3 places from 9 different places is 9C3 Hence S-1 is true, S-2 is true and statement-2 is the correct explanation for statement-1. Ans.] S-1 : Let r be the common ratio of geometric progression. So, x S = r 8 (Given) 1 r
x = 8(r – r2)
As,
1 r (– 1, 1) – {0} so r – r2 2, – {0} 4
........(1) .......(2)
From equations (1) and (2), we get x = 8 (r – r2) (– 16, 2] – {0} Hence, S - 1 is false and obviously S-2 is true. Ans.]
Q.10 y
[Sol.(A) O
(B)
(C) (D)
x=a
x
point of non-differentiability. y = x3 f '(0) = 0 But at x = 0, f(x) is increasing. f "(c) can be zero also. e.g. f(x) = – x4. false e.g. f(x) = x3 at x = 0 is increasing.]
y O
x
PAGE # 26
Q.11 [Sol.
As, a1 + a2 + a3 = 126 a1 a 3 126 = = 42. 3 2 The numbers in A.P. are 42 – d, 42, 42 + d. Let g1 = A, g2 =AR; g3 = AR2 AR = 34 and A + AR2 = 85 and a2 =
34 + 34R = 85 R 34R2 – 68R – 17R + 34 = 0 For R = 2, A = 17 g1 = 43 + d = 17 1 , A = 68 2 g1 = 43 + d = 68
for
34R2 – 85R + 34 = 0 (R – 2)(2R – 1) = 0
d = – 26 (Rejected)
R=
d = 25
Common difference of A.P. = 25 and common ratio of G.P. is
1 Ans.] 2
PART-B Q.1 [Sol. (A)
(B)
We know that | adj A | = | A |2 for a 3 × 3 matrix Given adj A = KAT |adj A| = |KAT| = K3 | A | (|AT| = | A | ) K3 | A | = | A |2 K3 = | A | ; Now det A = – 1 (1 – 4) – 2(– 2 – 4) + 2 (4 + 2) = 27 k3 = 27 K = 3. Given, f (x) = (2x + 1)50 (3x – 4)60 f ' (x) = 220(2x + 1)48 (3x – 4)58 (2x + 1)(3x – 4)(3x – 1) +
– –
(C)
1 3
–1 2
+
–
4 3
Sign scheme of f '(x) Least positive integer is k = 2 By applying continuity and differentiability at x = 1, we get a = 3, b = – 1. Hence, (2a + b) = 2 (3) – 1 = 5. Ans.]
PART-C Q.1 [Sol.mb Sn = 1 · 2 + 2 · 3 + 3 · 4 + ......... upto n terms =
n
n
n
n 1
n 1
n 1
n (n 1) = n 2 n =
Sn =
n ( n 1) ( n 2) 3
n (n 1) ( 2n 1) n ( n 1) 6 2
......(1) PAGE # 27
1 1 1 Also, n–1 = 1 · 2 · 3 · 4 2 · 3 · 4 · 5 3 · 4 · 5 · 6 ....... upto (n – 1) terms n 1
1 1 n 1 ( n 3) n = = 3 n 1 n ( n 1) ( n 2) ( n 3) n 1 n ( n 1) ( n 2) ( n 3)
1 n 1 1 1 n–1 = 3 n (n 1) (n 2) (n 1) (n 2) (n 3) n 1
n–1 =
1 1 1 1 n–1 = 18 3n ( n 1) ( n 2) 18 9Sn
n–1 =
Sn 2 18Sn n–1 – Sn + 2 = 0 18Sn
Hence, Sn 18 n 1 1
[using (1)]
= | – 2 | = 2. Ans.
For objective: Put n = 2 ] Q.2 [Sol.
f (x h) f (x) h0 h
We have f '(x) = Lim
2 x 2h 2 x 2 0 f ( 2 x ) f ( 2 h ) f ( 2 x ) f ( 0) f f 2 h 2 2 Lim f '(x) = Lim = h0 h0 h h
f (2h ) f (0) = f '(0) = k (say) h0 2h
= Lim
f '(x) = k f(x) = kx + c ; f(0) = 0 c = 0 f(x) = kx 2 2 kx sin x dx
Now, I =
0
2
x3 = k · 3 0 2
– 2k
2
=
k
2 2
x 2kx sin x sin 2 x dx
0
2
2
0
0
2 2 x sin x dx sin x dx = k ·
8 3 8k 2 3 2k 2 = 4 k . 3 3
8 3 2 k (4) k . Hence I(k) = 3
This is a quadratic in k and its minimum value occurs when k =
3 f(x) = 2 x ; 4
4 · 3 3 3 =– 16 4 2
f(– 42) = 3. Ans.]
PAGE # 28
Practice Test Paper-7 Q.1 [Sol.
m = 5! · 2 · 5! = 10 · 4! · 5! n = 4! · 5! Hence m = 10n k = 10 Ans.]
Q.2 [Sol.
Q.3 [Sol.
Let r xˆi yˆj zkˆ ; x + y + z 12, y 3 and x, z 1 (Give y = 3 and 1 to each x and z, add one extra beggar) x' + y' + z' + u' = 7 Number of possible r = 10C3 (Using beggar Method) Ans.] y
( , f( ))
f–1(b)
Let f() = b = f ( + h) = b – k f–1(b – k) = + h f ( – h) = b + k f–1(b + k) = – h –1 Now L.H.D. of f (x) at x = f () = b
x
O
y = f(x)
1 h f 1 (b k ) f 1 (b) 1 1 Lim (f –1) ' (b–) = Lim = Lim = = = h 0 f ( h ) f ( ) f ' ( ) k 0 (b k ) b h 0 f ( h ) f ( ) r h ||ly Now R.H.D. of f–1(x) at x = f () = b
( h ) f 1 (b k ) f 1 (b) 1 1 1 = Lim = = = Lim h 0 f ( h ) f ( ) k 0 (b k ) b f ' ( ) l h 0 f ( h ) f ( ) h Aliter: Verification by taking an example (f –1) ' (b+) = Lim
3x for x 0 y = f (x) = 2 x for x 0 f ' (0–) = – 3 = l f ' (0+) = – 2 = r
y
y = – 3x
y for y 0 3 –1 g (y) = x = f (y) = y for y 0 2
O
x
y = – 2x y = f(x)
1 1 = 3 l 1 1 g ' (0–) = f ' (0+) = = 2 r
g ' (0+) = f ' (0+) =
Note: If l and r are positive, then L.H.D and R.H.D of f–1 are then L.H.D. and R.H.D are
1 1 & and if l and r are negative l r
1 1 & . (Think) Ans.] r l
PAGE # 29
Q.4 Y y=x3+x2–10x y = sinx ( /2, 1)
y = 1 + cos x
[Sol. x= –1
X
(0,0)
1 2 Graph of f(x)
( ,0)
Clearly, from above graph, f(x) has local maximum at x =
and absolute maximum at x = – 1. Ans.] 2
Q.5 [Sol.
f(x) = cos 1 cos x sin 1 sin x 2 2
2 2 = 2 2
2
x ,
0x 2
x x , 2 2 3 2 2 2 x · x x x x , 2 2 2 4 2 2 3 5 (2 x ) · ( x 2 ) x x ( x 2 ) 2 , 2 2 2 4 x
2
x 2 3 x 2 3 x 2
y
1 1 f(x) = cos cos x sin sin x 2 2
2/4
2
f(x) = x x 0, , 2 2 f(x) = x x 2 2 2 f(x) = – x ; x , 2 2
f(x) = ( – x)2 –
O
/2
x 3 /2
2
– 2/4
3 2 ,
3 2 f(x) = – (2 – x)2 < x 2 2 4 At x = , f(x) is not differentiable PAGE # 30
At x = , it is local as well as global minimum. 2 2 Range : 4 , 4 , 2
2
2
I = f ( x )dx = x dx 2 0 0
Using king 2
x 3 2 3 I = x 2 dx = Ans.] = 24 3 0 0
Q.6 [Sol.
Given,
a b k = = k (say) 13 7 15 a = 13k, b = 7k, c = 15 k
Now, cos A = cos B =
49 225 169 1 105 b2 c2 a 2 = = = A= 2 ( 7 ) ( 15 ) 2 105 2 3 2bc 169 225 49 23 a 2 c2 b2 = 2(13) (15) = 26 2ac
169 49 225 1 a 2 b2 c2 = = . 2(13) (7) 26 2ab cos C < 0, C is obtuse. So ABC is obtuse.
cos C = As,
Also,
Also,
9k 9 r s a 2s 2a 35k 26k s = = 35k 35 r1 s 2s 35k sa 1 bc sin A · 4 2 2 = s (35k ) (35k )
1 3 · 4 · 7 k · 15k · 105 3 2 = = 2 = 35 · 35 3 3 : 35 35 · 35 · k 2
B C bc 7 k 15k A 8 4 = cot · 3 = 3 = 3. tan = 2 bc 7 k 15k 2 22 11 Now, verify alternatives. Ans.]
Also,
PAGE # 31
PART-C Q.1 [Sol.
Equation of S is (x – 1)2 + (y – 1)2 + (x + y – 2) = 0 x2 + y2 + x( – 2) + y( – 2) + 2(1 – ) = 0 ........(1) x2 + y2 + 2x + 2y – 2 = 0 ........(2) Given (1) and (2) are orthogonal, so
S=0
P(1,1)
y=2–x
2( 2) 2 (1) 2 1 = 2 (1 – ) – 2 – 2 + – 2 = – 2 2 2 4 = 4 = 1 Hence equation of S is x2 + y2 – x – y = 0.
Now, length of tangent from (2, 2) is Q.2 [Sol.
4 4 2 2 = 2. Ans.]
Given a1 > 0. Let d = common difference of A.P. Now, 3a8 = 513 3 (a1 + 7d) = 5(a1 + 12d) 2a1 + 39d = 0 As, a1 > 0 so d < 0.
.......(1)
n n 39d n 1 d = d n 2 40n = d n 20 2 400 2a1 n 1 d = 2 2 2 2 Clearly, Sn will maximum when n = 20. (As d < 0). Ans.]
Now, Sn =
Q.3 [Sol.
Given, y = x + 2 Equation of CD y – 2 = – 1 (x – 0) y=2–x 0 1 8 Area of triangle ABC = 2 3 8
0 1 8 = 6 8 0 1 0 = 6 8 =
2 14 6
1 3 1
2 20 6
1 2 1
.......(1) y
.......(2) 2 14 3 6
1 8 · 16 64 84 20 = = . 6 6 3
64 3 21 . Ans.]
1 1 1
8 14 , 3 3
P2
C P1 (0,2)
45°
y = – 2x + 10
y=2–x 135° O
x
(0,2)
P3(8, – 6)
D m= – 1 y= –2x+10
PAGE # 32
Q.4 [Sol.
Equating the components, 3x + 2y+ 4z = x ; 2x + 2z = y & 4x + 2y+ 3z = z hence (3 – )x + 2y + 4z = 0 2x – y + 2z = 0 4x + 2y + (3 – )z = 0 for non trivial solution
3 2 4 2 2 =0 4 2 3
Use : C1 C1 – C3 and C3 C3 – 2C2 ( 1) 2 0 2(1 ) = 0 0 1 2 ( 1)
( + 1)2[– 1( – 4) – 2(– 2)] = 0 Hence = –1 or 8 sum = 7. Ans. ]
1 2 0 ( + 1)2 0 2 1 2 1
( + 1)2 ( 8) = 0
Q.5 [Sol.
Given, r a b But a b
a b sin x b c cos y 2 c a and r · a b c = 0 c (sin x + cos y + 2) = 0 c 0, so sin x + cos y = – 2, which is possible when sin x = – 1 and cos y = – 1
For (x2 + y2) to be minimum, x2 =
Hence, the minimum value of
2 and y2 = 2. 4
20 2 2 20 2 20 52 2 (x + y ) = = 2 4 = 25. Ans.] 2 2 4
Q.6
[Sol.
1 x 1 f ( x ), g ( x ), x ,1 1, f (1) g (1) , x 1 F(x) = 2 f (1) g (1) , x 1 2 continuous at x = 1
b – a = 3. continuous at x = – 1
F(1–) = F(1+) = F(1) f (1) g (1) f(1–) = g(1+) = 2 ( 4 a ) (1 b) 5a b 4+a=1+b= = . 2 2 .......(1) F(–1–) = F(–1+) = F(–1) ( 4 a ) (b 1) b–1=4–a= 2 PAGE # 33
b–1=4–a=
3 b a 2
a+b=5 .......(2) (1) and (2) a = 1, b = 4 a2 + b2 = 17. Ans.]
Practice Test Paper-8 Q.1 [Sol.
Given, z3 + iz2 = (1 + i) z1 z3 – z1 = i (z1 – z2)
A(z1)
i z3 z1 z3 z1 = – i = e6 z 2 z1 z 2 z1
| z3 – z1 | = | z2 – z1 | ABC is isosceles and A =
/6
. ] 6
B(z2)
C(z3)
Q.2 [Sol.
Urn-I
m1 W 1 ball n1 B
Urn-II
m 2 W 1 ball n 2 B item 1
Mixed lot
P(item belongs to Lot-I) =
1 = P(item belong to Lot-II) 2
P(item from the mixed lot is W) =
m2 1 m1 Ans.] 2 m1 n1 m 2 n 2
Q.3 Sol.
F(x) = P( x ) dx = x2013 – x2012 – 8x2 + 8x + C, where C is constant of integration. F(x) = x(x – 1) (x2011 – 8) + C F (0) = F(1) = F (81/2011) = C F'(x) = 0 has atleast two real roots. (Using Rolle's Theorem) 1 2011 Note that P(x) = 0 has exactly two real roots in x 0, 8
Q.4 Sol.
y
Equation of normal at P is (y – 1) = –2(x – 1) 2x + y = 3
1 , y 4
1 7 At x = ,y= 4 2
2
radius =
]
1 7 1 1 2 4
2
=
1 x 25 25 5 5 4 = Ans. 16 4 4
C P(1, 1)
1 , 0 4 PAGE # 34
[Sol.
= ax 3
+ bx2
Paragraph for Question no. 5 to 7 + cx + d
Let f(x) f(0) = 5 d = 5 So, f(x) = ax3 + bx2 + cx + 5 f '(x) = 3ax2 + 2bx + c Now, f '(–2) = 0 12a – 4b + c = 0 .....(1) and y = f(x) passes through P (–2, 0), so 0 = – 8a + 4b – 2c + 5 Also, f '(0) = 3 c = 3 .....(3) On solving, we get a =
......(2)
1 3 ,b= 2 4
1 3 3 2 x x + 3x + 5 2 4
f(x) =
y
×
5 4 3 2 1
×
(i) 5 2
–1
O1
y=3 ×
×
x
5 2
Clearly, from above graph, we get, number of solutions of equation f (| x |) = 3 are 4. Ans. (ii)
Equation of normal at Q(0, 5) is (y – 5) =
1 (x – 0) x + 3y = 15. Ans. 3 1
(iii) Q.8 Sol.
1
1
19 1 3 3 2 3 2 x x 3x 5 dx = 2 x 5 dx = Required area = f ( x ) dx = . Ans.] 2 2 4 4 0 1 1
Put x = tan 1 sin sin 2 ; f(x) = sin–1 sin = 2 sin 1 sin ; 2
0 2 0
2
1 1 2 tan x ; x 0 f(x) = 1 1 tan x ; 0 x 2 PAGE # 35
1 2(1 x 2 ) ; x 0 f '(x) = 1 ;0 x 2(1 x 2 ) 1 1 ; f ' (–1) = (A) is correct 4 4
f '(1) =
Range of f (x) = 0, . Hence (B) is incorrect. 4 Also f '(x) is an odd function. (C) is correct Lim x0
Q.9 [Sol.
f (x) 1 x 2
1 1 As, f (0 ) and f (0 ) 2 2
(D) is incorrect]
The system will always have a solution (x, y) unless the two lines are parallel. i.e. m = 2m – 1 m=1 ] PART-C
Q.1 [Sol.
A : outcome of the disease is positive B1 : He has disease d1 B2 : He has disease d2 P(Bi ) 1 3 B3 : He has disease d3 8 P(A / B1) = 10 P(A / B2) =
6 10
P(A / B3) =
4 10
1 8 8 4 3 10 P(B1/A) = = = (p + q) = 4 + 9 = 13. Ans. 1 8 6 4 18 9 3 10 10 10 P(B2/A) =
6 18
P(B3/A) =
4 ] 18
PAGE # 36
Q.2
[Sol.
ˆi ˆj kˆ 2ˆi 2ˆj kˆ As, n r2 r3 = 1 3 4 = 10ˆi 10ˆj 5kˆ nˆ 3 2 1 2
2ˆi 2ˆj kˆ d = Pr ojection of OP on n. = 3ˆi 2ˆj kˆ · 3
=
6 4 1 9 = = 3. Ans. 3 3
Aliter: Equation of plane OQR is 2x – 2y + z = 0 So, distance of P from plane OQR = Q.3 [Sol.
2(3) 2(2) 1 4 4 1
=
9 = 3. Ans.] 3
Equation of normal at P1 (4 cos 1, 3 sin 1) is 4x 3y 7 cos 1 sin 1
... (1)
Also, equation of CQ1 is
sin 1 y = cos x ... (2) 1 Solving (1) and (2), we get
4x 3 – cos 1 sin 1
Q1(4 cos 1, 4sin 1) y
sin 1 x=7 cos 1
P1(4 cos 1, 3sin 1) 1
x cos 1 = 7
x = 7 cos 1, y = 7 sin 1
So, K1 = (7 cos 1, 7 sin 1) Similarly, CK2 = CK3 = ..... = CKn = 7
CKi 175
7n = 175
x
C (0, 0)
CK1 = 7
n
K1
x2+y2=16
n=
i 1
175 = 25. Ans.] 7
Practice Test Paper-9 Q.1 [Sol.
y
Obviously for c (0, 1), f (x) lies obove the g (x) also x2 = cx3 x = 0 or x = 1c
hence
x
2
cx 3 dx =
0
or
1 2 3 = 12c 3
x2
1 c
1 12c3 O (0,0)
x3
c3 =
1 8
1 1 Hence, 2 = 2 + 4 = 6 Ans.] c c
c=
1 2
x= 1 c
x
PAGE # 37
Q.2 Sol.
Equation of chord of contact with respect to point (–4, 2) is 4x 2 y 2x y 1. 2 – 2 = 1 and with respect to point (2, 1) is a b a 2 b2 Now, according to given condition, 4 2 2 2 4 2 a a = – 1 b = 1 b = 1 4 2 1 a4 a2 2 2 2 b b
b2 1 3 Now, e = 1 2 1 2 2 a
Q.3 [Sol.
Ans.]
P(E) = P( R R W W B or R R R W W or W W W R R) 5! 5! 5! 1 1 = 2!2! · 5 + 3!2! · 5 + 3!2!35 3 3
Simplifying P(E) = Q.4 Sol.
50 35
k = 50 Ans.]
2 = 12 3 = 22 = 14 2 Hence 1
3
]
2 Paragraph for Question no. 5 to 7
[Sol.
C1 : z + z = 2 | z – 1 | 2x = 2 | x – 1 + iy | x2 = (x – 1)2 + y2 1 y2 = 2x – 1 y2 = 2 x 2
C2 : arg z (1 i) = Curve C2 is a ray emanating from (–1, –1) and making an angle from the positive real axis. C1 and C2 has exactly one common point C2 must be a tangent to C1. C2 : y + 1 = m(x + 1) Solving, C1 and C2 y 1 1 – 1 y2 = 2 m 2 my = 2(y + 1 – m) – m my2 – 2y + 3m – 2 = 0
(–1,–1)
1 , 0 2
PAGE # 38
Put
D = 0 4 – 4m (3m – 2) = 0
1 3m2 – 2m – 1 = 0 (3m + 1) (m – 1) = 0 m = , 1. 3 1 m= rejected m = 1 (As, a (0, ) ) 3 C2 : y + 1 = x + 1 y=x Putting y = x in the curve C1 x2 = 2x – 1 (x – 1)2 = 0 x = 1 P (1, 1) Complex number corresponding to P is z0 = 1 + i
(i)
| z0 | =
2
(ii)
(iii)
From above graph , (x1 + y1) = 3 + 4 = 7 Ans. Area of the shaded region (1,1)
1
=2
1 2
=
3 (2 x 1) 2 2 x 1 dx = 2 3 2 2
1
1
P(z0)
4 4
Q(z0')
2
2 2 [1 – 0] = sq. units.] 3 3
PART-B Q.1 [Sol. (A)
The equation of circle passing through A(1, 2), B(2, 3) and having least posssible perimeter, is (x – 1) (x – 2) + (y – 2) (y – 3) = 0 (Circle described on AB as diameter) x2 + y2 – 3x – 5y + 8 = 0 ..........(5) If above equation of circle intersects orthogonally the circle x2 + y2 + 2x + 2ky – 26 = 0 then, using condition of orthogonality, we get
(B)
3 5 2 1 k = 8 – 26 – 3 – 5k = – 18 5k = 15 2 2 k = 3 Ans.
Using the limit y 0 We have, Lim x 0
1 cos y 1 = y2 2
1 cos1 cos(1 cos x · 1 cos(1 cos x)2 xa 1 cos(1 cos x )2 2
4 8 1 cos(1 cos x ) [(1 cos x )]4 1 1 1 cos x x · Lim · = · = Lim 2 x 0 (1 cos x ) 2 8 x 0 x 2 x a xa
PAGE # 39
1 Lim x 8 a 128 x 0 For finite limit 8 – a 0
l=
(C)
a 8 Ans.
If matrix A is non-singular, so det. (A) 0 1 3 p2 2 4 8 0 – 2(p + 2) + 12 0 p 4 3 5 10
So, p R – {4}. Ans. (D)
dy t 1 ·y dt t 1 t 1 –t I.F. = (t + 1) e (t + 1) e–t y = – e– t + C Put t = 0 and y = – 1 C=0 At t = 1
2e–1 y = – e–1 y =
1 ] 2
PART-C Q.1 [Sol.
Since it is isosceles. So AB = AC Now, tan Now
r = 1 r1 = a tan 2 a
r2 = 4r1 sin
A1
sin
B1
2 2 r2 = 4r1 sin 15° sin 15° sin 60°
sin
= 2 tan 15° = 2 2 3 2
90°– 90°–
A A1
r2 I
C1
B1 B
2
r1
C1
C
2a = 4
3 3 r2 = 2r1 (2sin2 15°) · sin 60° = 2r1 (1 – cos 30°) · sin 60° = 2r1 1 2 2 2 3 = 2 · 2 2 3 2
3 2
=
2
3 2 3 = 7 3 12
Now, (2r2 + 7r1) = 2 7 3 12 + 7 · 2 2 3 = – 24 + 28 = 4. Ans.] Q.2
Sol.
Given,
8g( x ) sin
2
x g ( x ) 2 dx 6
0
0
2
16 sin 4 x dx g ( x ) 4 sin 2 x .dx 6
0
PAGE # 40
2 3 32 g ( x ) 4 sin 2 x dx 6 16 0
6 g ( x ) 4 sin 2 x dx 6
2
0
g (x) =4 sin2x
Max. of g (x) = 4 sin2 x is 4. Ans.]
Practice Test Paper-10 Q.1 n
[Sol.
Given,
r2 r 1 · n Cr = r0 n
=
2n
+
r 0
n
S=
r0
r 1 1 · n C r 1
n
n
Cr = 2n + S r 1
(where S =
r 0
r
n
Cr ) r 1
n 1 C0 n C1 n C2 Cn 2n 1 1 ........ = (1 x ) n dx = 1 2 3 n 1 0 n 1
Now, 2n + S = 2n + Now, Q.2 [Sol.
n
n 32 n 1 n 1
1 1 n (2n + 1 – 1) = n 1 n 1 2 n 3 1
=
28 1 n = 5 Ans.] 6
As, 1 sin (cos–1x) = sin sin x = cos (sin–1x) 2 1
Hence,
tan
1
cot 1 d , where = cos (sin–1x) =
1
Q.3 [Sol.
Putting (, ) in y2 = 4x, we get = 0, 4 but 0 (Given) So, =4 P = (4, 4) t1 = 2 Hence t2 = – 3
Using t 2 t1 2 t1
(2) = . Ans.] 2
y2=4x
y P( , ) P(4,4)
L(1,2) m1 (0,0)V
x
S (1,0) m2
Q = (9, – 6)
4 3 Q ; m2 = (9, –6) 3 4 = 90° Hence equation of line passing through (1, 2) and inclined at an angle = 90° is x = 1. ]
Now
m1 =
PAGE # 41
Paragraph for question nos. 4 to 6 Sol. (i)
No American together A1A2A3A4 B, C, D, E For A1 we have four (B, C, D, E) favourable cases out of total cases 7. 4 (say) 7
Hence, probability =
A1 and B are paired and the remaining
are, A2A3 A4 C D E for A2, favourable cases 3 out of total cases 5 Hence, probability =
3 say A2 and C are paired and the remaining are 5
Now, A3 A4 D, E for A3, favourable cases 2 out of total 3. Hence probability =
4 3 2 8 · · (B) 7 5 3 35 P (delegates of the same country form both pairs) A1A2A3A4 E, B, C, D
(ii)
2 . 3
P (No two Americans together) =
31 ·1·1 75 [For A1 we have 3 favourable and of 7 and two A2 only 1 favourable out of 5 are for the remaining, no constraints.] EBCD
=
3 (B) 35 P (delegates of the same country not forming any pair + forming both pair + forming exactly one pair) = 1
=
(iii)
3 24 8 P (forming exactly one pair) = 1 – = (C). Ans. 35 35 35 Alternatively: A1, A2, A3, A4, B, C, D, E
n(S) =
8! = 105 ( 2!) 4 4! 4
(i)
C1 · 3C1 · 2C1 The probability that no two delegates of the same country are paired = 105
(ii)
The probability that delegates of the same country form two pairs =
(iii)
=
4 ·3· 2 105
=
8 35
3 3 3 = 105 35 (As A1, A2, A3, A4 can be paired in 3 ways and B, C, D, E can be paired in 3 ways.) The probability that exactly two delegates of the same country are paired together 4
=
C2 4 3 24 = Ans.] 35 105 PAGE # 42
PART-B Q.1 [Sol. (A)
(z + i) 2 = (z – i)2 or – (z – i)2 ; if (z + i)2 = (z – i)2 4zi = 0
z =0 (z + i)2 = – (z – i)2 = i2 (z – i)2 z + i = i (z – i) or – i (z – i) (1 – i) z = – i2 – i = 1 – i z = 1. if z + i = – iz + i2 (1 + i) z = – 1 – i = – (1 + i) z=–1 Hence, z = 0 or 1 or – 1 3 solutions Ans. if
4
zi Aliter: Given, 1 z i
(B)
zi = (1)1/4 z i
we get, z = 1, 0, – 1 ]
We have,
2 2 2 · = –1 t1 t2 = –4 ; also t2 = – t1 – ....(1) t1 t 2 t1
t1t2 = – t12 – 2 – 4 + 2 = – t12 t12 = 2 Also,
t 22 = t12 +
t 22
4 +4 t12
squaring (1)
y
P(t1) 90°
O
x
S(a,0)
x+a=0 Q(t2)
=2+2+4=8
Now, SQ = a(1 + t 22 ) = a(1 + 8) = 9a and SP = a(1 + t12 ) = a(1 + 2) = 3a SQ = 3 SQ = 3SP = 3 Ans.] SP On differentiating both sides with respect to x, we get 0 – sin2x · f(sin x) cos x = – cos x
(C)
(D)
1 sin 2 x
f(sin x) =
1 2 f ( 2) 2 2
Ans.
Given, 4y2 + 2 cos2x = 4y – sin2x 4y2 – 4y + 1 + cos2x = 0 (2y – 1)2 + cos2x = 0
y=
1 and cos x = 0 2
x=
3 , 2 2
1 3 1 So, two ordered pairs are possible i.e., , and , Ans.] 2 2 2 2 PAGE # 43
PART-C Q.1 [Sol.
Given, Re(z) – 2 = | z – 7 + 2i |
y
(x – 2)2 = (x – 7)2 + (y + 2)2
9 (y + 2)2 = 10 x 2
P(7, 3) P(z 1)
O
x (2, )
2
V 9, 2 2
S(7, – 2)
So, the given locus is that of a parabola with directrix Q(z2 ) Q(7, –7)
x = 2 and focus (7, – 2). x=2
Clearly, minimum PQ = l(L.R.) = 10 Ans.] Q.2 1/ 2
Sol.
An =
x
n
1
An =
dx
2
0
2n A n n n
Hence,
n 1
Given,
n
=
2 An n
n 1
( n 1)
2nAn =
1 2(n 1)
y
1 2(n 1)n 1 = 2
n
1
n n 1
x
O
1 1 1 = 1 n 1 n 1 2
1 1 1 1 2 1 = 1– = n 1 3 2 n 1 3
1 2
1 1 = n = 2. Ans.] 3 n 1
Q.3 [Sol.
We have, (3p2 – pq + 2q2) u v w = 0 But, u v w 0 3p2 – pq + 2q2 = 0 Now divide by q2 (assuming q 0) p 3t2 – t + 2 = 0, where t = and t R q but D < 0, hence no real values of t only solution is p = 0 and q = 0 Hence exacly one ordered pair of (p , q) i.e. (0, 0).
Aliter: We have, (3p2 – pq + 2q2) u v w = 0 2
q 7q 2 But, u v w 0 3p2 – pq + 2q2 = 0 2p2 + p2 – pq + + =0 4 2 2
q 7q 2 + p + =0 2 4 This is possible only when p = 0, q = 0 i.e. exacly one ordered pair of (p , q)]
2p2
p = 0, q = 0, p =
q 2
PAGE # 44
Practice Test Paper-11 Q.1 [Sol.
Given, f(x) = x3 – ax2 + 2x, x [0, 2] Now, f ' (x) = 3x2 – 2ax + 2 f(2) = (2)3 – 4a + 4 = 12 – 4a and f(0) = 0 Using LMVT, we have 1 f ( 2) f ( 0) 3 (12 4a ) 0 f ' a 2 20 4 2 2
Q.2 [Sol.
11 11 13 a = 6 – 2a a = 6 – a= . Ans.] 4 4 4
Equation of line through O(0, 0, 0) and perpendicular to the plane 2x – y – z = 4, is x 0 y0 z0 = t (let) 2 1 1 Any point on it is (2t, – t, – t) As above point lies on the plane 3x – 5y + 2z = 6, so
6t + 5t – 2t = 6 9t = 6 t =
2 . 3
4 2 2 , (x0, y0, z0) Co-ordinates of point of intersection are , 3 3 3 Hence, (2x0 – 3y0 + z0) = 4 Ans.]
[Given]
Q.3 [Sol.
Given, z + w = 0 ..........(1) 2 2 and z +w =1 .........(2) Putting w = – z from (1) in (2), we get
2z2 = 1 z = ±
1 . 2
1 1 1 1 ,w= and for z = ,w= . 2 2 2 2 So, from both possibility, we get
For z =
zw Q.4 [Sol.
2 . Ans.] D(z)
We must have |2 – 4| < 5 –5 < 2 – 4 < 5 2 F1 –1 < ( – 2)2 < 9 0 ( – 2)2 < 9 –3 < – 2 < 3 F2 –1 < < 5 0 < < 5 = 1, 2, 3, 4 4 values. ] [Note : = 0, 4 is not possible because, | z – 2 | + | z – 4| = 5 will represent a circle.]
PAGE # 45
Paragraph for question nos. 5 to 7 [Sol.mb Given, f(x) = x2 e–x f '(x) = x e–x (2 – x)
y 4 x 2, y 2 e
+
–
0 2 Sign scheme of f '(x)
–
y
–
O
x=2
1 4
x
Graph of f(x) = x2 e–x 2e x
As, g(x) =
0
x
2x 2e (1 e x ) f ' ( 2e x ) · 2e x 8e · e f ' (t ) dt g'(x) = = . (1 4e2 x ) (1 4e 2 x ) 1 t2
Now, verify alternatives. Q.8 [Sol. (A)
(B)
]
P(A/B) = P(B/A) P(A) = P(B) Now, P(A B) = P(A) + P(B) – P(A B) P(A B) = P(A) + P(B) – P(A B)
P(A B) = 2P(A) – 1 > 0; P(A) >
P(B) =
3 1 ; P(A/ B) = 4 2
1 2
True
P(A B) 1 1 3 3 = P(A B) = · = P(B) 2 2 4 8 Now, P(A) + P(B) – P(A B)= P(A B) 1 P(A) +
3 3 – 1 4 8
P(A)
5 8
(C)
P(A)]max. =
5 8
True
P(AC BC)C = (a 3 a 4 ) (a 4 a 4 ) C = (a4)C = a1 + a2 + a3 = P(A + B) P(AC BC) = (a1 + a3 + a4)C = a2 = P(AB) Hence P(AC BC) + P(AC BC) = P(A) + P(B)] =
(D)
1 1 3 2 5 = . 2 3 6 6
Given, A is subset of B. P(B A) P(A) Now, P(B/A) = = = 1. P(A) P(A)
B A
S
]
PAGE # 46
Q.9 [Sol.
As, r3 r1 r2 r2 = r1 · r2 r2 r2 · r2 r1 = 6( 2ˆi ˆj kˆ ) 6(ˆi ˆj 3kˆ ) = 6ˆi 12kˆ 6(ˆi 2kˆ ) · 5 = ˆi 2kˆ or ˆi 2kˆ Row 3 vector ± 6 5 1 1 3 1 1 3 2 1 1 A= or 2 1 1 Tr.(A) = 0, 4. 1 0 2 1 0 2 Also r2 r3 r2 r3 = | r2 | | r3 | | r2 r3 | = 6 · 5 · 6 · 5 = 30. Since r1 , r2 and r3 are coplanar they are linearly dependent. r1 r2 r2 r3 r3 r1 = r1 r2 r3 2 = 0]
PART-C Q.1 [Sol.
Clearly, the given curves intersect at x = 0,
k k 1 2
k
Required area =
k 2 1
0
2 1 x kx 2 x dx = 2 k 1 6 k k
2 1 Now, for maximum area, k 1 will be minimum k + = 2 k = 1. Ans.] k k
Q.2 [Sol.
Let = 3h – 2, = 3k
2 =h 3
and
=k 3
As, (h, k) lies on given circle, so h2 + k2 – 2h – 4k – 4 = 0
( 2) 2 2 2 4 ( 2) 4 9 9 3 3
2 + 2 – 2 – 12 = 44 ( – 1)2 + ( – 6)2 = 44 + 37 Locus of (, ) is (x – 1)2 + (y – 6)2 = 81, which represents circle whose radius = 9.]
PAGE # 47
Q.3 2x
[Sol.
Let
e
fn(x) =
tn
dt
x
n
f n' ( x ) = 2 ·e ( 2 x ) e x
n
n
For maxima and minima, f n' ( x ) = 0 2e ( 2 x ) e x Taking log on both sides, we get
n
2 ·e ( 2
n n
x )
e x
n
1
ln 2 ln 2 n ln 2 = xn (2n – 1) xn = n x = n = an 2 1 2 1
ln 2 – 2nxn = – xn
Also,
1 n ln 2 fn'' x n < 0 2 1
1
ln 2 n fn(x) is maximum at x = n . 2 1
ln 2 ln n n 2 1 ln (ln 2) ln ( 2 1) Now, ln an = = n n n 1 l n 2 1 n n l n ( l n 2 ) 2 l n ( l n 2 ) l n ( 2 1 ) ln (a n ) = Lim Hence L = Lim = Lim n n n n n n 1 n · ln 2 ln 1 n 2 = – ln 2 L = Lim 0 n n
Hence, e–L = 2 Ans.] Q.4 [Sol.mb We know that | adj. A–1 | = | A–1 |2 =
det . adj. A 1
1 | A |2
1 = | A |2 = 22 = 4 Ans.] = | adj. A 1 |
1
PAGE # 48
Practice Test Paper-12 Q.1 [Sol.
x 1 y 3 1............(i) · a 2 b 2 Auxiliary circle is x2 + y2 = a2.................(ii) C is the centre. Combined equation of CL, CM is obtained by homgenising (ii) with (i), i.e.,
The equation of the tangent is
2
x 3y 0 x2 + y2 –a2 2 a 2 b Since LCM = 90° 2
y
M
P( =60º) 2
1 3a 3a 7 0 +1 – 2 2 4 4 4b 4b 2 2 2 2 2 7b = 3a 7 a (1–e ) = 3a
Hence e = Q.2 [Sol.
L
90º
1–
x
C (0, 0)
2 Ans. ] 7
Originally the number of white balls in the bag vary from 0 to 100.
initially Bag has no W ball B0 : 1 B1 : Bag has 1 W ball P(B0) = P(B1) = P(B100) = Bag has 2 W ball 101 B2 : B : Bag has 100 W ball 100
B1
B0
A B2
B100
After the white has been dropped in the bag A = A B0 + A B1 + ........+ A B100 P(A) = P(B0) · P(A/B0) + P(B1) + P(A/B2) + ...... + P(B100) · P(A/B100) = Q.3 [Sol.
2 101 1 1 101 102 51 1 1 ...... = 101 . Ans.] 101 101 101 2 101 101 101
From above figure,
S
1 1 1 P(C (A B)') = 1 – 10 15 5
= Q.4 [Sol.
30 3 2 6 19 = Ans. 30 30
B
A
C
(a + b + c2) (a + b2 + c) = | a + b + c2 |2 = (a2 + b2 + c2 – ab – bc– ca) (a – b)2 + (b – c)2 + (c – a)2 = 2 (a = b, | b – c | = 1) or (b = c, | a – c | = 1) or (c = a, | a – b | = 1) |b–c|=1 (b, c) = {(2, 1), (1, 2), (3, 2), (2, 3), (4, 3), (3, 4), (5, 4), (4, 5), (6, 5), (5, 6)} Required probability =
3 10 5 = Ans.] 6 6 6 36 PAGE # 49
Paragraph for question nos. 5 to 7 [Sol.
Let dr's of line L be
, so equation of line L is r t aˆi bˆj ckˆ ......(1)
If L intersects L1 at P, so shortest distance between them is zero. 0=
10 10 4 , , 3 3 3
2ˆi ˆj kˆ · aˆi bˆj ckˆ ˆi 2ˆj kˆ aˆi bˆj ckˆ ˆi 2ˆj kˆ
Q
L2: r = 8i – 3j + k 3 + (2i – j+ k)
(5, – 5, 2) P
ˆi ˆj kˆ 2ˆi ˆj kˆ · a b c = 0 a + 3b + 5c = 0 .......(2) 1 2 1
L1: r = (2i + j – k) + (i – 2j + k) O 0
Similarly, L intersects L2 at Q, so shortest distance between them is zero. 8ˆi ˆ ˆ 3 j k · aˆi bˆj ckˆ 2ˆi ˆj kˆ 3 0= aˆi bˆj ckˆ 2ˆi ˆj kˆ
3a + b – 5c = 0
ˆ ˆj 8ˆi ˆ ˆ i 3 j k · a b 3 2 1
kˆ c =0 1
.......(3)
On solving (2) and (3), we get
a b c . 5 5 2
So, the equation of line L is r t 5ˆi 5ˆj 2kˆ . Any point on line L is A' (5t, – 5t, 2t). If A' is the point of intersection of L and L1, so A' will also satisfy
5 t 2 5t 1 2 t 1 1 2 1 – 2 (5t – 2) = – 5t – 1 – 10t + 4 = – 5t – 1 t = 1 So, co-ordinates of P are (5, – 5, 2). Similarly, if A' (5t, – 5t, 2t) is the point of intersection of L and L2, so A' will also satisfy L2, we get
L1, we get
8 2 3 5t 3 2t 1 – 5t + 3 = 1 – 2t t = 3 2 1 1
5t
(i)
10 10 4 , So, co-ordinates of Q are , 3 3 3 Given M (1, 2, 3) and N (2 + , 1 – 2, – 1)
MN = P.v of N – P.v ot M = ( 1) ˆi (1 2 ) ˆj ( 4) kˆ Also, n = normal vector of plane r · ˆi 4ˆj 3kˆ = ˆi 4ˆj 3kˆ . Now, MN · n = 0 1 ( + 1) + 4 (1 + 2) + 3 ( – 4) = 0
12 = 7 =
7 Option (B) is correct. 12 PAGE # 50
(ii)
10 10 4 , and perpendicular to the The normal vector of the plane through P (5, – 5, 2) and Q , 3 3 3 plane r · ˆi ˆj kˆ + 1 = 0 is parallel to the vector = n PQ ˆi ˆj kˆ
ˆi ˆj kˆ ˆi ˆj kˆ 1 1 ˆ ˆ 5 5 2 5 5 2 = 3i 3 j 0kˆ = ˆi ˆj = = 3 1 1 1 3 3 3 3 1 1 1
The required equation of plane, is 1 (x – 5) + 1 (y + 5) + 0 (z – 2) = 0 x + y = 0 or r · ˆi ˆj = 0 Option (D) is correct
(iii)
Volume of tetrahedron OPAB (where O is origin) =
OP OA OB 6
5 5 2 1 51 17 1 = | 0 1 3 | = [5(– 5) + 5(–6) + 2(2)] = = . Option (C) is correct.] 6 6 2 6 2 0 5 Q.8 [Sol.
S = {HTH, THH, TTH, HHH, HTT, THT, TTT, HHT} A = {HTH, HHH, HTT, HHT} B = {HTH, TTH, HTT, TTT} C = {HTH, THH, TTH, HHH} D = {TTH, HTT, THT} E = {HHH, TTT} A B C = {HTH} As,
P(C) =
1 3 1 , P(D) = , P(E) = 2 8 4
3 = P(C) + P(E) (A) is correct 4 Also, A B C = {HTH, THH, TTH, HHH, HTT, TTT, HHT} A B C S, as THT is not included in A B C. A, B, C are not exhaustive (B) is incorrect
2 P(D) =
Also,
P(A B C) =
1 P(A) P(B) P(C) 8 A, B, C are independent events.
(C) is correct
1 = P(B) = P(C) 2 A,B,C are equally likely.
(D) is correct
Note that, P(A) =
Ans.]
Q.9 [Sol.
Given, + = – p, = q ; Now, q1 q2 =
1 1 = – p2 , + = – p1 , = q1 ; and + = q2
· = 1. and p1 + p2 = –
p ( ) ( 1) 1 = – = (q + 1) q PAGE # 51
= 2, as is the root of x2 + px + q = 0.
Also,
qq2 = ·
So,
2 + p + q = 0
q q2 p q2 q = 0
)2
)2
( 1) 2 (q 1) 2 2 = (p – 4q) ( ) 2 q
Note that
(p1 – p2 = ( –
(qp1 – qp2)2 = (p2 – 4q) (q + 1)2. Ans.] PART-C
Q.1 [Sol.
Let c = 14 – x, a = 14, b = 14 + x 5 5 (14 + x)2 = (14 – x)2 + 142 – 2(14 – x) · 14 · x = 1. 13 13 a = 14, b = 15, c = 13.
Now, cos B = So,
Hence, r = Q.2 Sol.
Sn = C0C1 + C1C2 + ... + Cn–1 Cn Sn = nC0 .nCn–1 + nC1 .nCn–2 + ... + nCn–1 .nC0 = Sn = 2nCn–1 Now,
Q.3 Sol.
84 = = 4. Ans.] s 21
Sn 1 15 Sn = 4
2n 2 2n
Cn
C n 1
=
2nC
n–1
( 2n 2)! ( n 1)!( n 1)! 15 15 n! ( n 2)! 2n! 4 4
(n 1)(2n 1) 15 = 8(2n2 + 3n + 1) = 15n2 + 30n (n 2)n 8 16n2 + 24n + 8 = 15n2 + 30n n2 – 6n + 8 = 0 (n – 4) (n – 2) = 0 n = 2 or 4 Sum of all values of n = 6 ]
Given equation of planes are P1 : x + y + 1 = 1 ... (1) P2 : x + 2ay + z = 2 ... (2) P3 : ax + a2y + z = 3 ... (3) If 3 planes intersect in a line, then 1 1 1 1 2a 1 0 a a2 1 Applying C1 C1 C2 and C2 C2 C3
0 1 2a a a2
0 1 2a 1 1 0 a 2 1 1
PAGE # 52
On expanding along C1, we get
(1 – 2a) (a2 –1) – (a –a2)(2a –1) = 0 – 1 + 3a – 2a2 = 0
a=
1 , 1. 2
1 , planes P1 and P2 are parallel. 2 Hence, we conclude that 3 planes will never intersect in a line for any real value of a.
But, for a = 1, planes P1 and P3 are parallel and for a =
Practice Test Paper-13 Q.1 Sol.
Given, tan–1 x + tan–1 y + tan–1(xy) =
11 12
...(1)
11 2 + 2 tan–1 y = 2 tan–1 y = y = 3 4 12 3 Differentiate both sides of equation (1) with respect to x, we get
At x = 1;
1 1 1 (xy' + y) = 0 2 y'+ 2 + 1 y 1 ( xy) 2 1 x
Q.2 [Sol.
1 1 1 + y' + (y' + 2 4 4
So,
y' = –1 –
3)=0
1 1 3 + + y' = 0 2 + 2 2 4
3 + 2y' = 0
3 . Ans.] 2
As, f(0+) = f(0–) = f(0) = 0, so f(x) is continuous at x = 0. Further, f(0 + h) > f(0) and f(0 – h) > f(0) where h is sufficiently small positive quantity. Hence, f(x) has local minimum at x = 0. ]
Q.3 [Sol.
We have, As,
x dy y dx dy y2
y(1) = 1 C = 2
Now, y() = – 3
x x d = – dy = – y + C y y x =–y+2 y
= 5 = –15. Ans.] 3
Q.4 [Sol.
1 tan 2 x 2(2 tan x ) tan 2 x 4 tan x 9 4 Given, f (x) = = + 2 + 5 = 2 sin 2x + 4 cos 2x + 5 1 tan 2 x 1 tan x 1 tan 2 x
Rf = 5 20 , 5 20 Hence, (M + m) = 10 Ans.
PAGE # 53
9 tan 2 x 4 tan 2 x Aliter : f(x) = + + = sin2x + 2 sin 2x + 9 cos2x 2 2 2 1 tan x 1 tan x 1 tan x = 1 + 4(1 + cos 2x) + 2 sin 2x = 5 + 2 sin 2x + 4 cos 2x. ] Paragraph for question nos. 5 to 7
Sol.
5 16 3 x 2 y2 . So, eccentricity of hyperbola = . 1 , e 1 For the given ellipse, 3 25 5 25 16
Let the hyperbola be,
x2 y2 1 ... (1) A2 B2
25 16 2 Then, B2 = A2 1 = A . Also, foci of ellipse are (±3, 0). 9 9
As, hyperbola passes through (±3, 0). So,
9 = 1 A2 = 9, B2 = 16 A2
x 2 y2 1 9 16 Vertices of hyperbola are (±3, 0) (A) is correct. Focal length of hyperbola = 10 (B) is incorrect. 9 Equation of directrices of hyperbola are x = ± . (C) is incorrect. 5 Any point of hyperbola is P(3sec, 4tan). Equation of auxiliary circle of ellipse is x2 + y2 = 25. Equation of chord of contact to the circle x2 + y2 = 25, with respect to P(3 sec, 4 tan), is 3x sec + 4y tan = 25 ... (1) If (h, k) is the mid point of chord of contact, then its equation is hx + ky – 25 = h2 + k2 – 25 hx + ky = h2 + k2 ... (2) As, equations (1) and (2) represent the same straight line, so on comparing, we get
Equation of hyperbola is (i)
(ii)
3 sec 4 tan 25 2 h k h k2
25 h 25 k · sec = 2 2 . , tan = 2 2 h k 3 h k 4
25 Eliminating , we get, 2 2 h k
2
h2 k2 2 2 9 16 = 1. (As, sec – tan = 1) 2
(iii)
x 2 y 2 x 2 y 2 Locus of (h, k) is 9 16 25 Required area of quadrilateral = 2(A2 + B2) = 2(9 + 16) = 50 Ans.]
PAGE # 54
Q.8 Sol.
S = {1, 2, 3, ... n} E1 = No. is divisible by 2. E2 = No. is divisible by 3. If n = 6k say n = 6 S = {1, 2, 3, 4, 5, 6} 3 1 2 1 = ; P(E2) = = 6 2 6 3
P(E1) =
1 = P(E1).P(E2) (B) is correct. 6 If n = 6k + 2 say n = 8 S = {1, 2, 3, 4, 5, 6, 7, 8}
P(E1 E2) =
4 1 2 1 = ; P(E2) = = 8 2 8 4
P(E1) =
Here, P(E1 E2) =
1 = P(E1) P(E2) (C) is correct. 8
1 P(E1) · P(E2) 10 Not independent dependent. ]
Note that : P(E1 E2) =
Q.9 [Sol.
dy = x2 + y2 + 1 dx Put y2 = t
2xy
x
dt = x2 + t + 1 dx
dt t x2 1 – = dx x x
I.F. =
1 x 1 x2 1 x 2 dx = x – x + C
Hence
t = x
y2 = x2 – 1, C = 0 as y(1) = 0
Now, y (x0) =
3
3 = x 02 – 1
x 02 = 4
x0 = ± 2 Ans. Alternatively: We have, 2xydy = (x2 + 1)dx + y2dx
2 xy dy y 2dx 1 = 1 2 dx 2 x x
y2 1 d d x x x PAGE # 55
y2 1 x c x x
We get,
As,
y (1) = 0 y2 = x2 – 1
Now, y (x0) =
3
c=0
3 = x 02 – 1
x 02 = 4
x0 = ± 2 Ans.]
PART-C Q.1 [Sol. 3
3 13 1 5 1 1 13 1 5 1 1 · · = P(HHH) = p + q = 9 Ans.] 1 = 18 8 18 8 8 18 2 18 3 6
Q.2 2
[Sol.
Clearly, d =
2
3 1 1 3 = 4 2 2 4
1 1 = 16 16
1 1 = 2 2 8 y
So, | sin | = 2 2 d | sin | = 1
1 0, 2
1 3 , 2 4
O
3 = , , . 2 2 2
y
=
x
3 1 , 4 2
x
1 , 0 2
Hence, the number of values of are 3.] Q.3 [Sol.
Given,
dy 2 x dy y = x3y2. Now dividing by y2, we get – 2y = x4 y2 or dx x dx
1 dy 2 1 = x3 2 dx y x y
........(i)
This is a Bernouli's differential equation, substituting
2 dy dt . So, equation (i) becomes y 2 dx dx
1 dt t dt 2 = x3 t = 2x3 2 dx x dx x 2
IF = e
x dx
2 t , we get y
2
e 2 ln x e ln x x 2 PAGE # 56
So, general solution is given by x2t =
2 C 2x6 x2 · 2 x6 x4 + C = +C = + 2 y x 6 3 3 y
If x = 1, y = – 6 C = 0
Q.4 [Sol.
6 2 x4 y= 4 y 3 x
Now,
dy dy 24 = 24x–5 = 5 . Hence, dx x dx x
i.e. f(x) =
1 35
=
6 x4 24 = 8. Ans.] 3
Given, y = x2
y
dy = 2 x x a = 2a dx x a Equation of tangent is (y – a2) = 2a (x – a) Put y = 0, we get
A (a, a2)
Now,
O
B a , 0 2
x
a a = . 2 2 Also, equation of OA is y = ax
x=a–
a
a
ax 2 x 3 a a2 ka 3 a3 a3 ka 3 a3 ka 3 Area = (ax x ) dx = k · · = = = 2 3 2 2 4 2 3 4 6 4 0 0 2
k=
p 2 (Given) (p + q)least = 5. Ans.] q 3
Practice Test Paper-14 Q.1 [Sol.
Tangent to the parabola 1 = 4x is y = mx + ....(i) m m2x – my + 1 = 0, As, it touches the circle x2 + y2 = 1, so
y
y2
1 4
m m
2
O
S(1,0)
x
1 m4 + m2 – 1 = 0
m2 = tan2 =
1 1 4 = 2
5 1 5 1 = 2 sin 18° Ans.] 2 2 4
PAGE # 57
Q.2 3
Sol.
3
3 x ) f ' ' ( x )dx 7 (3 x ) f ' ( x )3 + f ' ( x )dx 7 Given, ( 2 2
I
II
2
0 – (f '(2)) + f(3) – f(2) + 7 f(3) = f '(2) + f(2) + 7 = 4 + (–1) + 7 = 10. Ans.
Q.3
[Sol.
a11 a12 Let A = a 21 a 22 a 31 a 32
a13 a 23 , where aij {0, 1} a 33
As, trace A = 1 So, any two elements in main diagonal are 0 and one element is 1. Also, non-diagonal elements an be 0 or 1. So, number of matrices = 3 × 23 = 3 × 8 = 24. Ans.] Q.4 Sol.
As, z lies on the curve arg(z + i) = imaginary axis making an angle
, which is a ray originating from (–i) and lying right side of 4
with the real axis in anticlockwise sense. 4 Im(z)
(– 4+3i) 45° O
Re(z)
(4–3i)
The value of | z – (–4 + 3i) | + | z – (4 – 3i) | will be minimum when z, –4 + 3i, 4 – 3i are collinear. Minimum value = distance between (–4 + 3i) and (4 – 3i) =
(4 4) 2 (3 3) 2 =
64 36 = 100 = 10. Ans.]
Q.5 [Sol.
Statement-1: We have a x a b a x b 0 x b t a , for some scalar t. x b t a = (2 t ) ˆi (1 2 t ) ˆj (1 3t ) kˆ
1 Now, a · x = 0 t = 2
3ˆi kˆ 9 1 x x = = 2 2 4 4 Obviously, Statement-2 is true. ]
5 Statement-1 is false. 2
PAGE # 58
Q.6 [Sol.
Q.7 [Sol.
Option (B) is true. S-1: As, f ' (1+) = 0 = f '(1–1) f is differentiable at x = 1. So, f is differentaible x R. S-2: As, f(1 + h) < f(1) < f (1 – h). Where h is sufficienty small positive quantity, so f(x) is decreasing at x = 1. f(x) has neither local maximum nor local minimum at x = 1. But, S-2 is not explaining S-1. Ans.] Given, = 2 tan = tan 2
y
2 tan 2 y 0 x 0 1 y0 = 2 2 1 tan x 0 2 x 0 1 y 02 2( x 0 1) 1 x 0 2 ( x 0 1) 2 y 02 3x02 – y02 = 3 Locus of M is hyperbola 3x2 – y2 = 3 Now, verify alternatives. Ans.]
M(x0, y0)
tan =
(–1,0) A (0,0)
(2,0) B
x
Q.8 [Sol.
1 x · x · 1 cos (2x ) , 0 x 1 Given, f(x) = x0 0, Note: f (x) is not defined at x = 1 for 0 ......(1) f(x) is continuous at x = 0 but for continuous at x = 1. f (1–) = f(1) 2 Lim h sin 2 h = 0 Lim h 2 = 0 h 0
h 0
+ 2 > 0 > – 2 ......(2) Hence, (1) (2) 0 Note: f (x) is differentiable in (0, 1). Ans.] Q.9 [Sol.
Now, slope of LM =
2 = – t (given) t0
– 2 = t 2 = 2 + t2 Let mid point of LM is (h, k) Now, 2h = t and 2k = + 2 = t2 + 4 On eliminating t, we get k = 2(h2 + 1)
Locus of (h, k) is y = 2(x2 + 1) x2 =
y (h, k)
M (0, )
L (t, 2) (0, 0)
x slope = m = –t (given)
1 (y – 2) 2
Now, verify alternative. Ans.] PAGE # 59
PART-B Q.1 [Sol. (A)
Clearly, ABC is equilateral. Now, ar. (ABC) =
3 z z 4
3 3 z = 4
z
(B)
(C)
2
64
Im(z)
B( z)
2
A(z)
2
2 /3 O (0,0)
= 48 3 (Given)
z 8 . Ans.]
Re(z) 2 /3
2
C( z)
x 2 y2 Equation of chord of hyperbola 1 , whose mid-point is (h, k) is 2 1 hx h2 k2 – ky = – (using T = S1) 2 2 1 As, it is tangent to the circle x2 + y2 = 4, so
h2 k2 2 2 h2 h2 2 k2 k 4 2 = 2 2 4 h k2 4 Locus of (h, k) is (x2 – 2y2)2 = 4(x2 + 4y2) = 4. We know that ac = (semi-minor axis)2 = 4 4
Now,
8
1/ 2
{2x} dx
4
= 16 {2x} dx = 0
1 2
1 2
{2x}dx
1 2
1 2
= 16 {2 x} dx = 16 2 x dx = 32 x dx
1 8 2
0
0
0
1
x2 2 1 = 32 = 32 × = 4 Ans. ] 8 2 0
PART-C Q.1 [Sol
Given, A = (I + B) (I – B)–1 Now, AT = I B (I B) 1
T
= ( I B) 1
T
(I + B)T
= ( I B) T 1 (I + BT) = (I + B)–1 (I – B) Also, AAT = (I + B) (I – B)–1 (I + B)–1 (I – B) = (I + B) ((I + B) (I – B))–1 (I – B) = (I + B) ((I – B) (I + B))–1 (I – B) = (I + B) (I + B)–1 (I – B)–1 (I – B) = I I = I2 = I. | AAT | = | I | = 1 | A |2 = 1 As, | A | > 0 | A | = 1. Hence, det.(2A) – det. (adj A) = 8 det.( A) – (det.( A))2 = 8(1) – (1)2 = 7 Ans.]
PAGE # 60
Q.2 [Sol.
2 v | c |2 v · c 2 |c|
v c p = v sin = = |c|
=
(6)(3) 4 = 3 30p2
r v=(1, – 2, – 1)
14 3
A(–2,3,1) p
B(–3,5,2) M
14 = (30) = 140 Ans.] 3
r r =v^c
Q
r c=(1, 1, 1)
Q.3 1 4
2
[Sol.
5 4
2
F( x) dx cos x dx sin x dx cos x dx 0
sin x =
=
1 4
0
1 4 0
cos x
1 2 2 2 4 2
5 4 1 4
y
2
sin x
5 4
5/4 O 1/4
2
x
1 2 2 = 2
10 10
5 4
2
F( x ) dx
10 10 2 2 F( x ) dx × = 5. Ans.] 4 2 0 4 2
PAGE # 61
