EXERCISE 1(A)

1. 3 2y 2x 13x 5x 9

2dy6x 26x 5

dxLet point be (h,k)

3 2k 2h 13h 5h 9

3 22

y 2h 13h 5h 96h 26h 5

x h

substituting 0,0

3 2 3 22h 13h 5h 9 6h 26h 5h

3 24h 13h 9 0 h 1,k 15.

2. x a t sin t cos t

2y a 1 sin t

dx

a 1 cos 2tdt

dy

2a 1 sin t cos tdt

dy 2cos t sin 2tdx 1 cos 2t

2

22 2

t tcos sin

2cos t 1 sin t 2 2t t2cos t cos sin2 2

t1 tan2t1 tan2

ttan

4 2

3.

n nx y

2a b

n 1 n 1

n n

nx ny dy0

a b dx

n n 1

n n 1

dy b xdx a y

At a,b

dy bdx a

b x a a y b 0

bx ay 2ab

x y2.

a b

4. x / ay be

x 0 y b

x / aby ' e

a

0,b

dy bdx a

a y b bx

bx y ab

x y

1.a b

5. 2y 3x bx 2

x 0 y 2

dy

6x bdx

x 0

dyb 4

dx .... Given

b 4

6.

28 x

y2

dy

x 2dx

[Given]

x 2 y 2

y 2 2 x 2 0

2x y 6 0.

7. Points are 2p,ap bp c & 2q,aq bq c

2 2 2y ap bp c a q p b q p

x p q p

2y ap bp caq ap b

x p

y aq ap b x apq c

m aq ap b

& dy

m 2ax bdx

p qx

2

8. x y a

1 1 dy

0 x x xydx2 x 2 y

dy ydx x

Y y yX x x

X 0 Y y xy

Y 0 X x xy

2

x y 2 xy OA OB x y a

9. x2/3+y2/3= a2/3

1 / 3 1 / 3

2 2 dy0

3x 3y dx

1 / 3dy ydx x

1 / 3Y y yX x x

2 / 3 1/ 3Y y x y When X 0

1/ 3 2 / 3X x x y When Y 02 2 2 4 / 3 2 / 3 2 / 3 4 / 3 2 4 / 3 2 / 3 2 / 3 4 / 3Y X y x y 2x y x 2x y x y

2 2 4 / 3 2 / 3 2 / 3 4 / 3x y 3x y 3x y

2 4 / 3 2 / 3 2 2 / 3 4 / 3 4 / 3 2 / 3 2 / 3 4 / 3x x y y x y 2 x y x y

/ 3 4 / 3 2 / 3 2 / 3 2 / 3 2 / 3x y a 2x y a

2 / 3 2 / 3 2 / 3a x y

2a

10. n n 1xy a

n n 1 dyy nxy 0 , n 1

dx

dy ydx nx

Y y yX x nx

yX 0 Y y

n

Y 0 X x nx

1 1 1XY xy 1 n 1

2 2 n , Here n is a constant.

is constant only when xy is constant but nxy is constant

n 1.

11. 3 2f x 2x 9x 12x 3

2f ' x 6x 18x 12 0 2x 3x 2 0

x ,1 2,

12. 3 2f x x ax 48x 19

2f ' x 3x 2ax 48 0 x

22a 4 3 48 0

2a 144 0

a 12,12

13. 3 2f x 2x 9x 60x 81

2f ' x 6x 18x 60 0 2x 3x 10 0

x 2,5

14. 2x

f xx 2

22x x 2 xf ' x

x 2

2x 4x0

x 2

x x 40 x , 4 2,0

x 2

15. 2f x x

xf ' x x 1 nx 0

1 n x 0

1x

e

For 1x , f ' x 0

e

Function decreases in 1

0,e

.

16. log xf x

x

2

1 log xf ' x 0

x

log x 1 x e

x e,

17. f x 2 x 2 x 3

For x 2

f x 2 2 x 3 x

7 3x is a decreasing function.For 2 x 3

f x 2 x 2 3 x

x 1 is increasing function.For x 3

f x 3x 7 is an increasing function.

x 2,

18. f x cos x sinx

f ' x sin x cos x 0 x 0,2

g x cos x sinx

g ' x cos x sin x 0 x 0,4

0 x ,4 2

sin xh x

x

2

x cos x sin xh ' x 0

x

at x 0,

2

2

x x sinx 2x x cos x sinxh '' x 0 x 0,

x 2

h' x 0 x 0,2

xsin x

being reciprocal of sinx

x is an increasing function.

19. a sin x bcos xf x

c sin x d cos x

2

a cos x bsinx c sinx d cos x a sinx bcos x c cos x d sinxf ' x

c sinx d cosx

2

ad bc0 x

c sin x d cos x

iff ad bc 0.

20. sinx bx c f x

f ' x cos x b 0 b 1

21. 3 2y 2x 3x 36x 10 f x

2f ' x 6x 6x 36 0 2x x 6 0

x 3 or x 2

f 3 2 27 3 9 36 3 10

71

f 2 2 8 3 4 36 2 10

28 82 54

22. 2f x x 3x 3

f ' x 2x 3 0

3x f '' x 2 0

2

f x has minima at 3

x2

3 9 9 3f 3

2 4 2 4

23. 3 2f x 2x 3x 12x 8

2f ' x 6x 6x 12 0

x 2 or x 1

f '' x 12x 6 0 for x 1

and 0 for x 2 x 2 is minima

24. 2 2 2 2a sec x b cosec x f x

2 2 2 2f ' x 2a sec x tanx 2b cos ec x cot x 0

f x can have minima only as maxima is .

2 2 2 2a sec x tan x b cosec x cot x

2 23 3

sin x cos xa b

cos x sin x

24

2

btan x

a

2 2b atan x , cot x

a b

For a,b 0

2 2a b a bsec x , cosec x

a b

2 2 2 2a sec x b cos ec x

22 2a ab ab b a b

25. f x sinx sinx cosx

2 2f ' x cos x cos x sin x 0 cos x cos 2x 2cosx 2cos x 1 0

cosx 1 x or1

cos x x or2 3 3

f '' x sin x 2sin 2x 0 for x3

0 for x3

x3

26. f x x sinx

f ' x 1 cos x f '' x sinx

When f ' x 0 , f '' x 0

f x has neither minimum nor maximum.

27.

a 0 11

a cos bsin 12

a cos bsin 1

12absin 2abcos sin

2

ab sin sin cos

dab cos cos 2 0

d

22cos cos 1 0

1cos 1 or cos

2

Now for a triangle,as it will be a st.line, 0 , cos 1

1cos

2

3

3sin

2

3 3 3 3abab

2 4 4

28. 2f ' x 3x 6x 6 0 for all x

Neither minimum nor maximum.

29. 4 3f x x 6 8 x

Let x 6 t 8 x 14 t

34f t t 14 t & f ' x f ' t

3 23 4f ' t 4t 14 t 3t 14 t 0

23t 14 t 4 14 t 3t 0

23t 4 t 56 7t 0

t 0 or t 14 or t 8Now t 0 & t 14 as product will become zero

t 8 & 4 3f t 8 .6

30. 2x a 2 x a 1 0

a 2, a 1

2 2 2a 4a 4 2a 2

2f a a 2a 6

f ' a 2a 2 0 f '' a 0

a 1

2 2f a 5 min

31. Function must be continuous and differentiable to apply Rolle’s theorem.

32. Function must be continuous and differentiable to apply Rolle’s theorem.

33. f x log sinx in 5

,6 6

5f f

6 6

f ' x cot x 0 at x2

c2

34. 2x ab

f x logx a b

in a,b

2

a b2xf ' x 0

x ab x a b

2 22x x ab x = GM of a & b.

35. 3 2f x x bx ax satisfies Rolle’s theorem on 1,3

1c 2

3

f 1 f 3

1 a b 27 9b 3a and

21 1

3 1 2b 2 a 03 3

Solving , we get a,b 11, 6

36. f x log x in 1,e

1f ' x

x 1

f ' cc

1 log e log1 1c e 1 e 1

c e 1.

37. Here, f 0 f 2 0 in 0,2

f ' c 0

2c 2 2c c 2 0

c 2 3c 2 0

2c

3 as c 0,2

38. 2f x x mx n in

2 2b a m b af ' c 2 x m

b a

2 x m b a m

a bx

2

39. Function should be diffrentiable in domain.

40.1

x 1 xx x 1

Now , 2x N ,2x 1 N

,x N x 1 N

x x 1 2N

1 12Nx x 1

EXERCISE 1(B)

1. y = x1/3(x – 1)

dxdy

= 323231

x31

x1·

31x

34

[4x – 1]

hence f is for x > 41

and f for x < 41

2 3 1 4minima

/x is always positive and at x /the curves has a local

now f ' (x) = 3231 x·31x

34 (non existent at x = 0, vertical tangent)

f '' (x) = 3532 x1·

32·

31

x1·

94

=

x12

x92

32 =

x1x2

x92

32

f '' (x) = 0 at x = – 21

(inflection point)

graph of f (x) is as

A = 1

0

3134 dxxx = 1

0

3473 x43x

73

= 43

73 = 28

743 = 28

9 (D) ]

2.dydx

slope fo tangent

2

1 bt a

2at 0

b a and b are of same sign.

3. f (x) = 1 4 x > 0 in (–1, 1) f is

Now f (x) + f (– x) = dtt1dtt1x

0

4x

0

4

dyy1dtt1

y

0

4x

0

4 ( t = – y)

= 0 f (x) is odd

again f (x) =

4

2 1

3

4

x

x which vanished at x = 0 and changes sign (0, 0) is inflection since f is

well defined in [–1, 1] A, B, C, D ]

4. Since intercepts are equal in magnitude but opposite in sign Pdx

dy= 1

now dxdy

= x2 – 5x + 7 = 1 x2 – 5x + 6 = 0 x = 2 or 3 ]

5. h (x) =

n f x g x

n a

( ) . ( ) =

an

an xsgn.axsgn.a |x||x|

= a xx| | sgn + a xx| | sgn = ax sgn x ( {y} + [y] = y)

= a for x

for x

a for x

x

x

0

0 0

0

h (x) is an odd function ]

6. f (x) = 100 x99 + cosx

for )1,0(x and

2,0 , cosx and x are both +ve

for

,2

x , x > 1 hence 100 x99 obviously > cosx ]

7. Note that f (x) is continuous at x= 2 and f is decreasing for (2, 3) and increasing for [–1, 2] .At x = 2 f has a maxima hence (A) is not correct. ]

8. Graph of y = f (x) (A) and (C) ]

9. If f and g are inverse then (fog)(x) = xf ' [g(x)] g ' (x) = 1if f is increasing f ' > 0 sign of g ' is also + ve (A) is correctIf f is decreasing f ' < 0 sign of g ' is – ve (B) is falsesince f has an inverse f is bijective f is injective (C) is correctinverse of a bijective mapping is bijective g is also bijective g is onto (D) is correct ]

10. f (x) = ln (1 – ln x)domain (0, e)

f ' (x) = – )xn1(

1l

·x1

< 0 decreasing x in its domain (A) & (B) are incorrect

f ' (1) = – 1 (C) is also incorrect

also f (1) = 0;

)x(fLim1ex

;

)x(fLim0x

f '' (x) = 22 )xn1(xxn

ll

f '' (1) = 0 which is a point of inflectiongraph is as showny axis and x = e are two asynmptotes ]

11. f is obvious continuous x R and not derivable at – 1 and 1f ' (x) changes sign 4 times at – 1, 0, 1, 2local maxima at 1 and – 1local minima at x = 0 and 2 ]

12. Domain is x R

Also f (x) = 21 sintancos where cot = x

=

2

21

x1

1tancos

= (cos )2 where tan = 2x1

1

=

2

2

2

x2

x1

g (x) = 2

2

x2x1

= 1 – 2x21

range is

1,21

; f ' (x) = 22 )x2(x2

hence f ' (0) = 0

also 1)x(fLimx

hence (B), (C), (D) ]

13. Let the tangent line be y = ax + bThe equation for its intersection with the upper parabola is

x2 + 1 = ax + bx2 – ax + (1 – b) = 0

This has a double root when a2 – 4(1 – b) = 0 or a2 + 4b = 4For the lower parabola

ax + b = – x2

x2 + ax + b = 0This has a double root when a2 – 4b = 0subtract these two equations to get 8b = 4 or b = 1/4add them to get 2a2 = 4 or a = ± 2

The tangent lines are y = 21x2 and y = –

21x2

14. f (x) =

0

dt)txcos(tcos ....(1)

=

0

dt)txcos(·tcos

f (x) =

0

dt)txcos(·tcos ....(2)

(1) + (2) gives

2 f (x) =

0

dt)tcos·xcos2(tcos

f (x) = cos x

0

2 dttcos = 2 cos x 2

0

2 dttcos

f (x) = 2

xcos Now verify.. Only (A) & (B) are correct.

15. (A) f (x) = x – tan–1x

f ' (x) = 1 – 2

2

2 x1x

x11

> 0 f is increasing in (0, 1)

f (x) > f (0) but f (0) = 0f (x) > 0 x > tan–1x in (0, 1)

(B) f (x) = cos x – 1 + 2

x 2

f ' (x) = – sin x + x = x – sin x > 0 in (0, 1) (B) is not correct

(C) f (x) = 1 + x ln

2x1x – 2x1

f ' (x) = 2

22

2

x1

xx1xnx1xx1

x2·211

x

l

= 22

2 x1

xx1xnx1

x

l > 0 x R

(C) is true

(D) f (x) = x – )x1(n2

x 2 l

f ' (x) = (1 – x) – x11

= x1

1)x1( 2

= – x1

x2

< 0 (D) is correct

hence f (x) is decreasing in (0, 1) f (x) < f (0)

f (x) < 0 x – 2

x 2 < ln(1 + x) ]

16. f ' (x) = 3xx2

and f '' (x) = 4x3x

. Now interpret

]

17. (A) f (x) has no relative minimum on (–3, 4)(B) f (x) is continuous function on [–3, 4]

f (x) has min. and max. on [–3, 4] by IVT(C) f '' (x) > 0 f (x) is concave upwards on [–3, 4]

(D) f (3) = f (4)By Rolle's theorem

c (3, 4), where f ' (c) = 0 critical point on [– 3, 4]

18. (A) False, e.g. f (x) = sin x(B) True, from IVT

(C) True as

x11sinLim 1

x = sin–1(a quantity greater than one) not defined

(D) True, as the line passes through the centre of the circle.

19.

(A) Let =

0

0)1xe(

dtexLim x

x

0

t

0x

2

=

2

x2

x

0

t

0x

x1xex

dtexLim

2

=

0

0x

dteLim2

x

0

t

0x

2

= 1

eLim22x

0x = –2

(B) 14x2 – 7xy + y2 = 2

y2x7y7x28

dxdy

....(1)

if x = 1 then 14 – 7y + y2 = 2 y2 – 7y + 12 = 0 y = 3 or 4hence L (1, 3) and M (1, 4)

slope of tangent at L = 672128

= 7 ; slope of tangent at M = 872828

= 0

equation of tangent at L and M arey – 3 = 7(x – 1) y = 7x – 4

and y – 4 = 0(x – 1) y = 4

hence N =

4,

78

(C)

(C) If n is odd then graph of f (x) is

a3 is the only point where

a1 a2 a3 a4 a5f (x) has its minimum value

If n is even then graph of f (x) is

From a2 to a3 at all values of x, f (x) is minimum. a1 a2 a3 a4

(D) 2lc + m = (lb2 + mb) ab

)mya( 2

l

= l(b2 – a2) + m(b – a) = l(b + a) + m; c = 2ba

20. We have f ' (x) = 5 sin4x cos x – 5 cos4x sin x = 5 sin x cos x(sin x – cos x)(1 + sin x cos x)

f ' (x) = 0 at x = 4

. Also f ' (0) =

2'f = 0

Hence some c

2,0 for which f ' (c) = 0 (By Rolle's Theorem) (C) is correct.

Also in

4,0 f is decreasing and in

2,

4 f is increasing minimum at x = 4

As f (0) =

2f = 0 2 roots (D) is correct.]

21. f (x) = tan–1(x) is defined on R and is strictly increasing but do not have its range R]

22. f (0) = 1; f (2) = 2f (1–) =f (1+) = f (1) = 2 ]

23. f (x) = ln(2 + x) – 3x2x2

is continuous in (–2, )

f ' (x) = 2)3x(4

2x1

= 2

2

)3x)(2x()2x(4)3x(

= 2

2

)3x)(2x(1x2x

= 2

2

)3x)(2x()1x(

> 0 (f ' (x) = 0 at x = – 1)

f is increasing in (– 2, )

also

)x(fLim2x

and

)x(fLimx unique root]

24. Let f (x) = 0 has two roots say x = r1 and x = r2 where r1, r2 [a, b] f (r1) = f (r2)hence there must exist some c (r1, r2) where f ' (c) = 0but f ' (x) = x6 – x5 + x4 – x3 + x2 – x + 1for |x| 1, f ' (x) = (x6 – x5) + (x4 – x3) + (x2 – x) + 1 > 0for |x| 1, f ' (x) = (1 – x) + (x2 – x3) +(x4 – x5) + x6 > 0hence f ' (x) > 0 for all x Rolles theorem fails f (x) = 0 can not have two or more roots.]

25. Consider the example of f (x) = ex and f ' (x) = ex both increasing]

Paragraph for question nos. 26 to 27(i) We have f(x) = x 2–x

So, f '(x) = 2–x (1 – x ln2)and f "(x) = 2–x ln2 (x ln 2 – 2)

Clearly, f(x) is increasing in

2n1,

l and decreasing in

,

2n1

l.

1n2l

1e n2l

O

y

x

Graph of f(x) = x 2– x

(ii) Clearly, f(x) = k has two distinct roots for k

2ne

1,0l

.

(iii) Given f(x) = x 2–x and g(x) = max. {f(t) : x t x + 1}

As f(x) is increasing in

2n1,

l, hence maximum value of g(x) occurs at t = x + 1

g(x) = f (x + 1) = (x + 1) 2– (x + 1)

Let I = 1

2n1

0

dx)x(gl

= dx2)1x(II

)1x(

.)P.B.I(I

= 1

2n1

0

)1x(

2n2)1x(

l

l + 2n1

l

12n

1

0

)1x( dx2l

= – 1

2n1

0

)1x(22n

)1x( l

l –

12n

1

0

)1x(2 22n

1

l

l

= – 2n

1l

21

e1

2n1

l

2n12l

21

e1

= – 2ne

12l

+ 2n2

1l

– 2ne

12l

+ 2n2

12l

= 2n2

12l

+ 2n2

1l

– 2ne

22l

Ans.]

Paragraph for question nos. 29 to 31

(1)

x11nxLim

0xl =

x1x

1xnLim

0x

l

Using L'Hospital's Rule

l = 2

0xx

x1

1x1Lim

=

2

0xx·

1x1

x1Lim

=

20x

x·)1x(x

1Lim

= )1x(xLim

0x = 0 Ans.Ans.

(2) 1)x(fLim0x

(can be verified)

e)x(fLimx

Also f is increasing for all x > 0 (D) (can be verified)

(3) l =

n1n

1k

nk

kn1

{given f (x) = (1 + 1/x)x and f (k/n) =

nk

kn1

}

taking log,

ln l =

n

1k

nk

n kn1n·

n1Lim l =

n

1kndx

nk11n

nk·

n1Lim l

=

1

0I

IIdx

x11nx

l =

1

0

21

0

2dx

2x·

1x1

x1

2x·

x11nl

=

1

0

dx1x

11x2102n

21 l = 10)1x(nx

212n

21

ll

= 0)2n1(212n

21

ll = 21

l = e

Paragraph for question nos. 32 to 34

y = 1x

x2

2

; not defined at x = ± 1

= 1 + 1x

12

; y ' = – 22 )1x(x2

dxdy

= 0 x = 0 (point of maxima)

as x 1+, y ; x 1– , y – |||ly x – 1+, y – ; x – 1– , y

The graph of y = 1x

x2

2

is as shown

verify all alternativels from the graph.

Paragraph for question nos. 35 to 37

(i) a = 1f (x) = 8x3 + 4x2 + 2bx + 1f ' (x) = 24x2 + 8x + 2b = 2(12x2 + 4x + b)for increasing function, f ' (x) 0 x R D 0 16 – 48b 0

b 31

(C)

(ii) if b = 1f (x) = 8x3 + 4ax2 + 2x + af ' (x) = 24x2 + 8ax + 2 or 2(12x2 + 4ax + 1)

for non monotonic f ' (x) = 0 must have distinct rootshence D > 0 i.e. 16a2 – 48 > 0 a2 > 3; a > 3 or a < – 3 a 2, 3, 4, .......

sum = 5050 – 1 = 5049 Ans.(iii) If x1 , x2 and x3 are the roots then log2x1 + log2x2 + log2x3 = 5

log2(x1x2x3) = 5x1x2x3 = 32

– 8a

= 32

a = – 256 Ans. ]

38. (A) R; (B) R, S, T ; (C) Q; (D) Q

(A) I =

12

1222

222dx

)1x()1x()1x(

=

12

1222

2dx

)1x()1x(1 = 2 –

1I

12

1222

2dx

)1x()1x(

I1 = a

a122

2dx

)1x()1x(

where (a = 12 ) ; put x = t1

dx = – 2t1

dt

=

a1

a22

2

2dt

t1·

1t1

1t1

= – a1

a224

42dt

)t1(tt)t1(

= – a1

a22

2dt

)t1()t1(

= a1

a22

2dt

)1t(1t

= – a

a122

2dt

)1t(1t

= – I1

2I1 = 0 I1 = 0 2 is the answer.](B) Domain of f (x) is (0, 1) (1, )

ln f (x) = 1 f (x) = e = constantf ' (x) = 0, for all in (0, ) – {1}

(C) Clearly (1, 0) is the point of intersection of given curves.

Now, f '(x) = x2x

+ 2x (ln2) (lnx)

Slope of tangent to the curve f (x) at (1, 0) = m1 = 2

Similarly, g'(x) = dxd

(e2x lnx – 1) = x2x

nx2

x1x2 l

Slope of tangent to the curve g (x) at (1, 0) = m2 = 2since m1 = m2 = 2

Two curves touch each other, so angle between them is 0.Hence cos = cos 0 = 1

(D) 3y2y' – 3y – 3xy' = 0 y' = xy

y2

y' = 0 y = 0, no real x.y' = y2 = x y3 = 1, y = 1The point is (1, 1)

39. (A)R,(B)Q,(C)P,(D)S

(A)43

dy tdx

, Tangent is 4 343ty at x at

x-intercept 3

4at

y-intercept 4

3

at

the point of intersection of tangent with the axes are 3

,04

at and

4

0,3

at

4

Aat

0,3

3

Bat

,04

2 4P at ,at

P divids AB externaly in 4 : 3

m 4n 3 m 4 & n 3

as m & n are coprime to each otherm n 7

(B)sin cos :ydx e y

dy slope of normal 1

equation of normal is 1x y

Area 12

(C) 2 3

1 1: :dyyx dx x

slope of tangent 2

2 2 2 2: . 2 :x xdyy e edx

slope of tangent 2

tan 0

(D) Length of subtangent

/3

/ 331'

2

x

x

y bey b e

40 (A) (q),(B) (r),(C) (q),(D) (q)

1y f xx

so 0y

Now 1 1f y f f yy

also 1f x f y

1 1 1

1/

f x f f y f

y f x f x x

also f(x) is increasing

1 1

1/x

f x f x x

1 52

f xx

now 1 52

f xx

is decreasing so discarding it 1 5

2f x

x

.

Exercise 1(C)

1 Given S = x2 + 4xh = 1200and V = x2h

V(x) = x4

)x1200(x 22 ; V(x) = 4

1(1200x – x3)

Put V ' (x) = 0 gives x = 20If x = 20, h = 10Hence Vmax. = x2h = (400) (10) = 4000 cubic cm.

2 Note that C1 is a semicircle and C2 is a rectangular hyperbola.PQ will be minimum if the normal at P on the semicircle is also anormal at Q on xy = 9Let the normal at P be y = mx ....(1) (m > 0)solving it with xy = 9

mx2 = 9 x = m3

; y = 3m9

Q

m3,3

3

differentiating xy = 9

x dxdy

+ y = 0 dxdy

= – xy

Qdx

dy = –

3m·m3

= – m

tangent at P and Q must be parallel

– m = – m1

m2 = 1 m = 1

normal at P and Q is y = xsolvingP(1, 1) and Q(3, 3) (PQ)2 = d2 = 4 + 4 = 8 Ans. ]

3 The given expression resembles with (x1 – x2)2 + (y1 – y2)

2, where y1 = 20x2

1 and

y2 = )13x)(x17( 22 Thus, we can thing about two points P1(x1, y1) and P2(x2, y2) lying on the curves x2 = 20y and(x – 15)2 + y2 = 4 respectively.Let D be the distance between P1 and P2 then the given expression simply represents D2.Now, as per the requirements, we have to locate the point on these curves (in the first quadrant) suchthat the distance between them is minimum.Since the shortest distance between two curves always occurs along the common normal, it implies thatwe have to locate a point P(x1, y1) on the parabola x2 = 20y such that normal drawn to parabola at thispoint passes through (15, 0).

Now, equation of the normal to the parabola at (x1, y1) is

20xy

21

= )xx(x10

11

. It should pass

through (15, 0).

31x + 200x1 – 3000 = 0 x1 = 10 y1 = 5

D = 22 5)1510( – 2 = 225

The minimum value of the given expression is 2225 = 2a 2 b

a 5 & b 2

4 x = t2 ; y = t3

dtdx

= 2t ; dtdy

= 3t2

dxdy

= 2t3

y – t3 = 2t3

(x – t2) ....(1)

2k – 2t3 = 3th – 3t3

3t 3th 2k 0

1 2 3t t t 2k (put t1t2 = – 1); hence t3 = 2know t3 must satisfy the equation (1) which gives 4y2 = 3x 1.Comparing with ay2 = bx 1 , we have a = 4 and b = 3.

5 We have F(x) =

3x1,1x4x3x

1x2),8k6k(logx2

23

2

21

Also F(x) is increasing on [–1, 3] because F '(x) > 0 x [–1, 3].And F '(x) = –2 x [–2, –1), so F(x) is decreasing on [–2,–1). If F(x) has smallest value at x = –1, then we must have

)1(F)h1(FLim0h

2 + log1/2 (k2 – 6k + 8) –1 log1/2(k2 – 6k + 8) –3 k2 – 6k + 8 8 k2 – 6k 0 k [0, 6] ....(1)But in order to define log1/2(k2 – 6k + 8),We must have k2 – 6k + 8 > 0 (k – 2) (k – 4) > 0 k < 2 or k > 4 ......(2) From (1) and (2), we get k [0, 2) (4, 6] Possible integer(s) in the range of k are 0, 1, 5, 6Hence the sum of all possible positive integer(s) in the range of k = 1 + 5 + 6 = 12 Ans. ]

6 We have F(x) = 3

x3 + (a – 3) x2 + x – 13

For F(x) to have negative point of local minimum, the equation F '(x) = 0 must have two distinctnegative roots.Now, F '(x) = x2 + 2(a – 3) x + 1 Following condition(s) must be satisfied simultaneously.(i) Discriminant > 0; (ii) Sum of roots < 0 ; (iii) Product of roots > 0Now, D > 0 4(a – 3)2 > 4 (a – 3)2 – 1 > 0 (a – 2) (a – 4) > 0 a (– , 2) (4, ) ........... (i)Also –2(a – 3) < 0 a – 3 > 0 a > 3 .......... (ii)And product of root(s) = 1 > 0 a R (i) (ii) (iii) a (4, ) ....(iii)Hence sum of value(s) of a = 5 + 6 + 7 + ......... + 100 = 5040 Ans. ]

7. Consider y = x + x1

– 3

2x11

dxdy

= 0

0dxdy

x = 1 or – 1

y = p – 3

1

1

–1

–5

5

A B

253Band

253AWhere

O X

y

(0, 0)

As x 0+, y and x 0–, y –

Also roots of 3x1x = 0 x2 – 3x + 1 = 0

x = 2

493 =

253

For two distinct solutions either p – 3 = 0 p = 3or 1 < p – 3 < 5

4 < p < 8Hence p {3} (4, 8)p = {3, 5, 6, 7} Sum = 21 Ans. ]

8 Let f ''(x) = 6a (x – 1) (a > 0) then f '(x) =

x

2xa6

2 + b = 3a(x2 – 2x) + b.

Now f ' (–1) = 0 9a + b = 0 b = – 9a. f '(x) = 3a (x2 – 2x – 3) = 0 x = – 1 and 3.So y = f(–1) and y = f(3) are two horizontal tangents.Hence distance between its two horizontal tangents = |f(3) – f(–1)| = |–22 – 10| = 0032. Ans. ]

9 Volume (V) = 11 hA31

h1 = 1A

V3

|||ly h2 = 2A

V3, h3 =

3AV3

and h4 = 4A

V3

So (A1 + A2 + A3 + A4)(h1 + h2 + h3 + h4) = (A1 + A2 + A3 + A4)

4321 AV3

AV3

AV3

AV3

= 3V(A1 + A2 + A3 + A4)

4321 A1

A1

A1

A1

Now using A.M.-H.M inequality in A1, A2, A3, A4, we get

4321

4321

A1

A1

A1

A1

44

AAAA

(A1+A2+A3+A4)

4321 A1

A1

A1

A1

16

Hence the minimum value of (A1+A2+A3+A4)(h1+h2+h3+h4) = 3V (16) = 48V = 48 × 5 = 240 Ans. ]

10 y = x2 and y = – x8

; q = p2 and s = – r8

....(1)

Equating dxdy

at A and B, we get

2p = 2r8

....(1) pr2 = 4 x

B(r, s)

A(p,q)f(x)=x2

O

y

Now mAB = rpsq

2p = rpr8p2

p2 = 2pr + r

8 p2 = r

16

4r16

= r16

r = 1 (r 0) p = 4

r = 1, p = 1Hence p + r = 5

11 x = 0 and x = 1 ]

12 y = x2

dtdy

= 2x . dtdx

dtdx

= 10m/sec.

tan = xx2

= x

sec2 . dtd

= dtdx

dtd

= 10 × cos2 = 10 × 101

= 1 { at x = 3m }

13 3x2 – 2y dxdy

= 0 dxdy

= y2x3 2

slope of tangent at (4a2 , 8a3) = )a8(2)a16(3

3

4

= 3a

let this tangent at this point also cuts the curve at (4b2, 8b3) and normal at this point slope of

normal at (4b2 , 8b3) = – b3

1 .

3a = – b31

ab = – 91 ......(i)

slope of line = 22

33

b4a4b8a8

= )ba()ba(2

22

33

= ba

)abba(2 22

= 3a [it is equal to slope of target] 2a2 + 2b2 + 2ab = 3a2 + 3ab

2b2 = a2 + ab 2a812

= 9

1a2

2 = 81a4 – 9a2

81a4 – 9a2 – 2 =081a4 – 18a2 + 9a2 – 2 = 09a2 (9a2 – 2) + (9a2 – 2) = 0 (9a2 – 2) (9a2 + 1) = 09a2 = 2

14. Let x = r cos and y = r sin

r2 = x2 + y2; tan = xy

(0, /2)

N = ]sin4cossin[cosr

r222

2

=

)2cos1(42sin)2cos1(r2

=

2cos32sin52

Nmax = 1052

= 105152

= M

Nmax = 1052

= 105152

= m

A = 2mM

= 2·1510·2

= 32

2007 × 32

= 1338 Ans. ]

15. )6(f)3(f

= 9292

k6

k3

= 31

; f (9) – f (3) = (29k + 9) – (23k + 9) = 29k – 23k ....(1)

3(23k + 9) = 26k + 9 26k – 3(23k) – 18 = 0

23k = yy2 – 3y – 18 = 0(y – 6)(y + 3) = 0y = 6; y = – 3 (rejected)23k = 6

now f (9) – f (3) = 29k – 23k { from (1) }= (23k)3 – 23k

= 63 – 6 = 210hence N = 210 = 2 · 3 · 5 · 7Total number of divisor = 2 · 2 · 2 · 2 = 16number of divisors which are composite = 16 – (1, 2, 3, 5, 7) = 11 Ans. ]

16. f (–3) = f (3) = 2 [ f (x) is an even function, f (– x) = f (x) ]again f (–1) = f (1) = – 3 2 | f (–1) | = 2 | f (1) | = 2 | – 3 | = 6

from the graph, – 3 <

87f < – 2

87f = – 3

f (0) = 0 (obviously from the graph)cos–1 )2(f = cos–1 )2(f = cos–1(1) = 0f (–7) = f (– 7 + 8) = f (1) = – 3 [f (x) has period 8]f (20) = f (4 + 16) = f (4) = 3 [ f (nT + x) = f (x) ]

sum = 2 + 6 – 3 + 0 + 0 – 3 + 3 sum = 5

17 We have f (x) = (b2 – 3b + 2) (cos2x – sin2x) + (b – 1) x + sin 2 f '(x) = (b – 1) (b – 2) (– 2 sin 2x) + (b – 1)Now, f '(x) 0 for every x R,so (b – 1) x2sin)2b(21 0 x R b 1

Also,)2b(2

1

> 1 b

2,

23

25,2

Now, when b = 2, f(x) = x + sin 2 f '(x) = 1 ( 0).

Hence, b

25,

23

b1 = 23

and b2 = 25

(b1 + b2) = 25

23 = 2

8 = 4

18. Let x = r cos and y = r sin E = (x + 5)(y + 5) = (r cos + 5)(r sin + 5) = r2 sin cos + 5r(cos + sin ) + 25Now put x = r cos and y = r sin in x2 + xy + y2 = 3 r2 cos2 + r2 sin cos + r2 sin2 = 3

r2(1 + sin cos ) = 3 r2 = cossin1

3 =

2sin26

hence r2]min. = 2 + sin 2]max. = 3 occurs at sin 2 = 1 2 = 2

or 25

i.e. 4

or 45

Hence E = )2(sin2r2

+ 5r(cos + sin ) + 25

put r2 = 2 and = 4

E = 1 + 25

2

12

1 + 25 = 36

put r2 = 2 and = 45

E = 1 + 25

2

12

1 + 25 = 16

hence minimum value of E is 16

19. Using LMVT for f in [1, 2]

c (1, 2) 12)1(f)2(f

= f ' (c) 2

f (2) – f (1) 2 f (2) 4 ....(1)again using LMVT in [2, 4]

d (2, 4) 24)2(f)4(f

= f ' (d) 2

f (4) – f (2) 48 – f (2) 44 f (2) f (2) 4 ....(2)

from (1) and (2) f (2) = 4

20. Let x tree be added thenP(x) = (x + 50) (800 – 10x)now P'(x) = 0 x = 15