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Mathematics- SolutionsPAPER - 2

SECTION - I1. (b)

1 2

A B C A B C4R sin cos cos 4R cos sin cosr r 2 2 2 2 2 2 2R 1 cosC R 1 cosC

_______________________________________________________________2. (a)

Aplying Rotation formula at P and B we get2

2p 4 2iω ω 1

2iω 4p ω

2 2 2p 2iω ω 2iω 4 ω 2iω 4 4 2iω ω

22

2 2

2 i 4iω 4ω2i 8iω 8ωp iω 2iω4iω ω 4 4ω i 4iω

so,p bz 0

2

2z 0

_______________________________________________________________3. (b)

2 2θ 0

2 n sinθ nθ 1L lim by putting xθ θ

θ 0

1cosθθlim L'H rule

θ

3

2θ 0 θ 0 3

θθ θ ......3θ tanθ 1lim limtanθ 3θ tanθ θθ

_______________________________________________________________

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4. (a)

Let sin x t and evaluate 2

lim 2 2t 1 2

t 2t 3t 4 t 6t 21 t

by rationalization

_______________________________________________________________5. (b)

a

03x 0

n 1 tan a.tan x dxlim

a

a

T

0

N : I n 1 tan a.tan x dx

king and add

a

2 2

0

2I n 1 tan a dx a n 1 tan a

2aI n 1 tan a2

Hence 2

3x 0

a n 1 tan a 12lima 2

6. (c)

x

4 2 3

0

f x t t t dt

differentiating both sides wrt x,4f3 (x) f’(x) = x2 f3 (x)

2xf ' x

4

Integrate both sides :

3xf x C

12 , since f(0) = 0

C = 0

3xf x

12

0 43 3

1 0

x xA dx dx12 12

Page : 370 44 4

–1 0

x xA48 48

41 4 257A48 48 48

Ans.

7. (b)

n

n m 1mx 0 x 0m

1x cos1xlim lim x cosxtanx . x

x

n – m > 0 so that f(x) is continuous at x = 0But f(x) is non-derivable at x = 0, so

n

n m 1mx 0 x 0

m

1x cos1xf ' 0 lim lim x cosxtan x . x . x

x

n m 1 0 for f ' 0 not existing

n m 1 Ans.8. (c)

Given expression = 1 1 1 1

2

x2x x x3sin 2 tan tan 2 tan2 2 2x1

2

when x1 12

–2 x 2

Number of integral values = 5. Ans]9. (a)

l = max (sin10x + cos5 x)sin10 x + cos5 x sin2 x + cos2 x = 1 sin A cos B + cos A B sin C = 1 sin A cos B A. sin B.

A B 1

A B C2

10. (c)y = mx + c

2 2 2y mx a m b

Page : 382 2 2 2

2 2x y x y1; 19 16 3 4

29m 16 2 5

9m2 – 16 = 20, 9m2 = 36, m 2

sum = a + b = 0, Product = k = – 4a + b + k = – 4

_______________________________________________________________

SECTION - II1. (a,b,c)

Equation of plane is

x 3 y b z 10 0 3 02 6 8

3x y b 9 0 _______________________________________________________________

2. (b, d)

1

2 2

Scot / 2

(g f c)

2 2

1

(g f c)tan / 2

S

2 2

1

1

(g f c)2 tan

S

_______________________________________________________________3. (a, d)

x I x 3

f x f xLim and lim

g x g x both exists there fore x = 1 and x = 3 must be roots of f(x)

f(x) = (x – 1) (x – 3) ( f(x) is a polynomial of least possible degree)

x l

f xLim 4

g x 4

f x4

g x .

f(x) = 4(x – 1) (x – 3)

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f(2) = 4(2 – 1) (2 – 3) = – 4f(4) = 4(3) (1) = 12 Ans.

_______________________________________________________________4. (c, d)

f(x) = ax2 + bx + cf(0) = cf(– 2) = 4a – 2b + cf(0) . f(–2) = c2 + 4ac – 2bc < 0 one root lies between – 2 and 0._______________________________________________________________

5. (a, c)

consider , 2cos sin 1 ;6cos 3sin 3

consider , 7cos 6sin x 6cos 3sin cos 3sin x cos 3sin x 3 2cos 6sin 2x 6

2x 75sin 2x 7 ;sin5

cos x 3 3sin 2x 7x 3 35

5x 15 6x 21 6 x5 5

2 2(2x 7) (6 x) 25 25x 40x 60 0

2x 8x 12 0 x 2 or 6

________________________________________________________________

SECTION - V1. (2)

________________________________________________________________2. (2)

Differentiate w.r.t y keeping x constant

x

1

xf xy f t dt xf y

Put y = 1

x

1

xf x f t dt 3x

xf ' x f x f x 3

3f ' xx

Page : 40f(1) = 3 c = 3 f(x) = 3 ln ex

3f ' xx

x

3f ' x

3

From the curve

3 = 3 ln e

1

, 1,3

The line y = kx intersects the curve two distinct points then 0 < k < 3 Number of interval values of k = 2.________________________________________________________________

3. (4)________________________________________________________________

4. (1)

5. (4)f(x) f’(–x) = f(–x) f’(x)f’(x) f(–x) – f(x) f’(– x) = 0

d f x f x 0dx

f(x) f(–x) = kput x = 0f(0) f(0) = k k = 100f(5) f(– 5) = 100f(5) = 20. Ans._______________________________________________________________