solutions to application of derivatives

7/26/2019 Solutions to Application of Derivatives

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Section-A : JEE Advanced/ IIT-JEE

A 1. )cos(ln q 2.1 1 1 1

, 0 , ; , 0,2 2 2 2

x - - - 3. 0x 4. abe

5. { })1,1(,fB 1. F 2. F

C 1. (a) 2. (a) 3. (c) 4. (b) 5. (d) 6. (d)

7. (b) 8. (b) 9. (a) 10. (c) 11. (b) 12. (d)

13. (c) 14. (d) 15. (d) 16. (b) 17. (a) 18. (c)

19. (d) 20. (a) 21. (d) 22. (a) 23. (c) 24. (a)

25. (d) 26. (b) 27. (a) 28. (b) 29. (c) 30. (c)

D 1. (c) 2. (b, c) 3. (c) 4. (a) 5. (a, b, c) 6. (a, c)

7. (d) 8. (b) 9. (b, d) 10. (b, c) 11. (a, b) 12. (b, c, d)

13. (a, b, c, d) 14. (a, c)

E 2. 2 4. ep 6.

1

4c- 9. 2 2

ud

v u- 10. (0, 0)

11. 2 4 0x y+ - p = 12.

-l

2

3,00,

2

314. )2,0(

2x+ 4y+ 3p= 0

15. fis min at7

5x= 16. units.sq

3

1018.

23 3

4r

and max. atx = 1

19. 6:6 p+ 20. 3 2 2x x x+ - + 21. 2 2x y+ = or 2 2x y- = -

22. 1x y+ = 23. ( 2, 1) (1, )b - - 24.1 3

, , 32 4

-= - = =a b c

25. units.sq9

3426. 2 kh 27. ( 1) ; 1 sq. unita xy e -=

28. min at 21

( 1)4

x b b= + - , max at 21

( 1)4

x b b= - - 29.2

,3

a

a

-

31.21 5 1 5

, , 2 , ( ) 24 4 4 4

a b c f x x x-

= = = = - + 32. 1xy=

35. (2, 1) 41. y= 2 42. 654 43. 6

F 1. (A) p (B) r

G 1. (c) 2. (a) 3. (a) 4. (c) 5. (b) 6. (d) 7. (c)

I 1. (7) 2. (0) 3. (9) 4. (1) 5. (5) 6. (9) 7. (9)

Section-B : JEE Main/ AIEEE

1. (b) 2. (a) 3. (d) 4. (a) 5. (b) 6. (d) 7. (d) 8. (a) 9. (d) 10. (b) 11. (b) 12. (a)

13. (c) 14. (c) 15. (d) 16. (c) 17. (a) 18. (b) 19. (b) 20. (a) 21. (c) 22. (c) 23. (d)24. (a) 25. (c) 26. (b) 27. (b) 28. (c) 29. (b) 30. (c) 31. (a)

Solutions Explanations


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FILLINTHEBLANKS:

1. We have/ 2

/ 2e-p < q < p

ln ln / 22

p- < q < p

cos (p/2) < cos (ln q) < cos (ln p/2)[Q cos is increasing in IV quad]

cos (ln q) > 0 ....... (1)

Also 1 cos 1- q " q

\ 1 ln(cos ) 0 0 cos 1- q " < q

ln (cos q) 0 ....... (2)From (1) and (2) we get, cos (ln q) > ln (cos q)\ cos (ln q) is larger.

2. 22 ln | |y x x= -

1 (2 1)(2 1)4

dy x xx

dx x x

+ -= - =

Critical points are 0, 1/2, 1/2

Clearlyf(x) is increasing on1 1

,0 ,2 2

-

and

f(x) is decreasing on1 1

, 0,2 2

- -

.

3. Letf(x) = log (1 +x) x forx > 1

1( ) 1

1 1

xf x

x x

-= - =

+ +We observe that,

( ) 0f x > if 1


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x = a which is not possible.

\ min1

2y gy

+ > cannot be 2.

\ minf cannot be 2.

\ Statement is false.

MCQ' WITHONE CORRECTANSWER:s

1. (a) Consider the functionf(x) = ax3+ bx2+ cxon [0, 1]then being a polynomial. It is continuous on [0, 1],differentiable on (0, 1) andf(0) =f(1) = 0 [as given a + b + c = 0]

\ By Rolle's theorem (0,1)x$ such that

2( ) 0 3 2 0f x ax bx c= + + =Thus equation 3ax2+ 2bx + c = 0 has at least one rootin [0, 1].

2. (a) Area of DABC, 2 212

A x d x= -

y

xa

dA

B

C

For max/min, 0dAdx

=

2 22 2

1 1 20

2 2 2

xd x x

d x

-- + =

-

2 2 2

2 20 / 2

2

d x xx d

d x

- -= =

-

2 2 2 2

2 2 2

2 2 2

1( 4 ).2 2 ( 2 ).( 2 )

2

4( )

x d x x d xd A d x

dx d x

- - - - --=

-

At / 2,x d= 2

2 2 2

8. 0

2 2

4( / 2)

d d

d Ave

dx d d

-+

= = --

\ Ais max at / 2x d= then / 2y d=\ Dis isosceles for max. area.

3. (c) ( sin sin cos ) cosdx

a ad

= - q + q + q q = q qq

(1)

(cos cos sin )dy

ad

= q - q + q qq

a sin= q q (2)

Dividing (2) by (1), we get

tandy

dx= q (slope of tangent)

\ Slope of normal = cot q\ Equation of normal is

y a (sin q qcos q) cos ( (cos sin ))sin

x aq

= - - q + q qq

ysin qasin2q+ asin qcos q= x cos q+ acos2q+ a qsin qcos q

xcos q+y sin q=aAs qvaries inclination is not constant.

\ (a) is not correct.Clearly does not pass through (0, 0).

It's distance from origin =2 2cos sin

aa=

q + qwhich is constant

4. (b) y = alnx+ bx2+xhas its extremum values atx= 1 and 2

\ 0dy

dx = atx= 1 and 2

2 1 0a

bxx

+ + = or 2bx2+x+ a = 0

has 1 and 2 as its roots.\ 2b 1 + a= 0 (1)

8b+ 2 + a= 0 (2)Solving (1) and (2) we get a= 2, b= 1/2.

5. (d) Fory2= 4ax,y-axis is tangent at (0, 0), while forx2= 4ay, x-axis is tangent at (0, 0). Thus the two curvescut each other at right angles.

6. (d) ( ) ( 2) ( 1) 0x xf x x e x e- -= - + = - + =

x = 1For ( , 1), ( ) 0x f x - - > and for

( 1, ), ( ) 0x f x -


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Also atx= 0,y = 0, atx= 1,y= 0, and

atx= 1/4,y> 0

\ Max. value ofyoccurs atx = 1/49. (a) Slope of tangent at (x,f(x)) is 2x+ 1

f ' (x) = 2x+ 1 f(x) =x2+x+ cAlso the curve passes through (1, 2)

\ f (1) = 2 2 = 1 + 1 + c c= 0, \f(x) =x2+x

\ Required area1

2

0( )x x dx= +

13 2

03 2

x x = +

1 1 5

3 2 6= + =

10. (c) We havef(x) = ,0 1sin

xx

x<

2

sin cos( )

sin

x x xf x

x

-=

where sin2xis always +ve, when 0 1x< . But tocheck Nr., we again let

h(x) = sinxxcosx

( )h x =xsinx> 0 for 0 1x<

h(x) is increasing h(0) < h(x), when 0 1x< 0 < sinxxcosx, when 0 1x< sinxxcosx> 0, when 0 1x<

f' (x) > 0, (0,1]x f (x) is increasing on (0, 1]

Again ( )tan

xg x

x=

2

2

tan sec( )

tan

x x xg x

x

-= , when 0 1x<

Here tan2x> 0 But to check Nr. we consider

p(x) = tanxxsec2x

2 2( ) sec sec .2sec .sec tanx x x x x x x= - -

2

( ) 2 sec tan 0p x x x x= - p (x) 0 > tanxxsec2x

\ ( ) 0g x - >

It will increasing on ( / 4,3 /8)p p .12. (d) From graph it is clear that both sin xand cosxin the

interval ( / 2, )p p are decreasing function.

1

1

O p/2 px

1

1

O p/2 px

Y x= cosY x= sin

\ Sis correct.To disproveRlet us consider the counter example :

f (x) = sinxon (0, / 2)pso that ( ) cosf x x=Again from graph it is clear thatf(x) is increasing on

(0, / 2)p butf' (x) is decreasing on (0, / 2)p\ Ris wrong.

13. (c) f (x) = ( 1) ( 2)xe x x dx- -

For decreasing function,f' (x) < 0

ex(x 1) (x 2) < 0 (x 1) (x 2) < 0

1 " Q

14. (d) Slope of tangenty=f(x) is (3,4)( )dy

f x

dx

=

Therefore, slope of normal

(3,4)

1 1

( ) (3)f x f= - = -

but1 3

tan(3) 4f

p - = (given)

or1

tan 1(3) 2 4f

p p - = + = -

(3) 1f =15. (d) It is clear from figure that at x = 0, f (x) is not

differentiable. y

+2

y = xy = x

2 O x

f(x) has neither maximum nor minimum atx = 0.

16. (b) Let ( ) 1xf x e x= - - then ( ) 1 0xf x e= - > for

(0,1)x\ f (x) is an increasing function.

\ ( ) (0), (0,1)f x f x> " 1 0 1x xe x e x- - > > +\ (a) does not hold.


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(b) Letg(x) = log (1 +x) x

then1

( ) 1 0, (0,1)1 1

xg x x

x x= - = - < "

+ +\ g(x) is decreasing on (0, 1) \x> 0

g(x)


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24. (a) f(x) =x3+ bx2+cx+ d, 0 < b2 "

f(x) is strictly increasing x R" 25. (d) For Rolle's theorem in [a, b]

f(a) =f(b), ln [0, 1] f(0) =f(1) = 0Q The function has to be continuous in [0, 1]

00

(0) lim ( ) 0 lim log 0xx

f f x x x+

a

= = =

0

loglim 0

x

x

x-a=

Applying L' Hospital's Rule

10 0

1/lim 0 lim 0 0

x x

x x

ax

a

-a-

-= = a >

a-

26. (b) Let the polynominal beP(x) = ax2

+ bx + cGivenP(0) = 0 andP(1) = 1 c= 0 and a+ b= 1 a= 1 b

\ P(x) = (1 b)x2+ bx P'(x) = 2 (1 b)x+b

Given ( ) 0, [0,1]P x x> " 2 (1 b)x+ b> 0 Whenx= 0, b> 0 and whenx= 1, b< 2 0 < b< 2

\ { }2(1 ) , (0,2)S a x ax a= - + 27. (a) The equation of tangent to the curvey= exat (c, ec) is

( )c c

y e e x c- = - (1)and equation of line joining (c1, ec1) and (c+1, ec+1)is

1 11 [ ( 1)]

( 1) ( 1)

c cc e ey e x c

c c

+ -- -- = - -

+ - -

1

1 ( ) [ 1]2

cc e e ey e x c

-- -- = - + (2)

Subtracting equation (1) from (2), we get

1 11 2( )

2 2

c c c ce e e ee e e x c e- -

- - - -- = - +

11

1

1 1

12 2

2 2

2

e ee

e ex c

e e e e

---

- -

-- - - - - = =

- - - -

1

1

2.

2 ( )

e e

e e

-

-+ -

=- -

1

1

12

1

2

e e

veve

vee e

-

-

+- +

= = = ---

-

x c< 0 x< c\ The two lines meet on the left of linex= c.

28. (b) The given curves, are

C1:y2= 4x ...(1) and C2:x

2 +y2 6x+ 1 = 0...(2)

Solving (1) and (2) we get

x2 + 4x 6x+ 1 = 0 x= 1 and y= 2 or 2\ Points of intersection of the two curves are (1, 2)and (1, 2).

For C1,2dy

dx y=

\(1,2)

dy

dx

= 1= m1and (1, 2)

dy

dx -

= 1 = 1 'm

For C2,3dy x

dx y

-= \ 2

(1,2)

1dy

mdx

= =

and2

(1, 2)

1 'dy

m

dx -

= - =

Q m1= m2and 1 2' 'm m=

\C1and C2touch each other at two points.

29. (c) y

xO

The given function is

3

2 3

(2 ) , 3 1( )

, 1 2

x xf x

x x

+ - < -=

- < -

The graph ofy=f(x) is as shown in the figure. From

graph, clearly, there is one local maximum (at x= 1)

and one local minima (atx= 0)

\total number of local maxima or minima = 2.

30. (c) Given thatg(u) =1

2 tan ( )

2

ue

- p-

\ g(u) = ( )12tan2

ue- -

p- = 1

12tan

2ue

- p -

= ( )12cot2

ue- p- = ( )12 tan

22ue

- pp --

=1

2 tan ( )2

ue

- pp - - = 12 tan ( )2

ue

-p -

= g(u) \gis an odd function.

Also 2

2

'( ) 01

u

u

e

g u e= >+ , ( , )u"

\gis strictly increasing on ( , ) .


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1. (c) The given polynomial is

2 4 20 1 2( ) ,

nnp x a a x a x a x x R= + + ++

and 0 1 20 na a a a< < < < 0 and decreases for all

x< 0.\ P'(x) has no. max. value and min. value atx= 0.ALTERNATE SOLUTION2

We haveP' (x)= 2a1x + 4a

2x3+ + 2na

nx2n1

( ) 0 0P x x= =

2 2n 21 2 nP (x) 2a 12a x 2n(2n 1)a x

- = + ++ -

0( ) |xP x ve= = + as a1> 0

\ P(x) has only one minimum atx= 0.2. (b, c) Let the line ax + by + c= 0 be normal to the curve

xy= 1 at the point ( ,x y ), then

1 (1) [ ( , )x y pt x y= lies on the curve]Also differentiating the curvexy= 1 with respect tox

we get

0dy dy y

y xdx dx x

+ = = -

( , )x y

dy y

dx x

- =

\ Slope of normal'

'

x

y=

Also equation of normal suggests, slope of normal

a

b

-=

\ We must have,

x a

y b

= -

(2)

Now from eq. (1), 0 ,x y x y> are of same sign

x a a

ve ve vey b b

= + - = + = -

aand bare of opposite sign. either a< 0 and b> 0 or a> 0 and b< 0.

3. (c)

O x

y'

x'

y

x=p

x=

p/

2c=3p /2P

y=x

It is clear from the graph that the curvesy= tanxand

y=xintersect atPin ( ,3 / 2)p p .Thus the smallest +ve root of tan x x= 0 lies in

( ,3 / 2)p p .4. (a) Sincegis decreasing in [0, )

\ For , ( ) ( )x y g x g y ........ (1)Alsog(x),g(y) [0, ) and f is increasing from [0, )to [0, ).

\ Forg(x),g(y) [0, )

such that ( ) ( )g x g y

( ( )) ( ( ))f g x f g y , where x y

( ) ( )h x h y

his decreasing function from [0, ) to [0, )

\ ( ) (0)h x h , 0x" But h(0) = 0 (given)

\ ( ) 0 0h x x " (2)

Also ( ) 0 0h x x " (3)

[as h(x) [0, )]

From (2) and (3), we get h(x) = 0, 0x"

Hence, h(x) h(1) = 0 0 = 0 0x" 5. (a, b, c) We are given that

23 2 1, 1 2( )

37 , 2 3

x x xf x

x x

+ - - =

- <

Then on [ 1, 2], ( ) 6 12f x x= +For 1 2, 6 6 12x x- -

6 6 12 24x + ( ) 0, [ 1, 2]f x x> " -\ fis increasing on [ 1, 2]

Alsof(x) being polynomial for [ 1,2) (2,3]x - Uf(x) is cont. on [ 1, 3] except possibly atAt x= 2,

LHL2

0 0lim (2 ) lim 3(2 ) 12(2 ) 1h h

f h h h

= - = - + - -

= 35

RHL0 0

lim (2 ) lim 37 (2 ) 35h h

f h h

= + = - + =

and f(2) = 3.22+ 12.2 1 = 35LHL = RHL =f(2)


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f(x) is continuous atx= 2Hencef(x) is continuous on [ 1, 3]Again atx= 2

RD

0 0

(2 ) (2) 37 (2 ) 35lim lim 1

h h

f h f h

h h

+ - - + -= = =

LD0

(2) (2 )limh

f f h

h

- -=

2

0

35 3(2 ) 12(2 ) 1limh

h h

h

- - - - +=

2

0

3 24lim 24h

h h

h

- += =

AsLD RD\ f '(2) does not exist. Hencef(x) can not have max. value

atx= 2.

6. (a, c) We have2( ) ( )[1 2 ( ) 3 ( )) ]h x f x f x f x= - +

=2 2 13 ( ) ( ( )) ( )

3 3f x f x f x

- +

23 ( )[{ ( ) 1/ 3} 2/ 9]f x f x= - +

Note that ( ) 0h x , thus, h(x) increases (decreases)wheneverf(x) increases (decreases).

7. (d) We have,

2

2

1

( ) 1

x

f x x

-

= +

2 2 2

4( )

( 1)( 1)

xf x

x x=

+ +

For max/min ( ) 0 0.f x x= =

2

2 3

4(1 3 )( )

( 1)

xf x

x

-=

+, 0( ) |xf x ve= = +

fis min atx= 0

\ Min value off(x) is 0 1 10 1

-= -

+ALTERNATE SOLUTION

2 2

2 2 2

1 ( 1) 2 2( ) 1

1 1 1

x xf x

x x x

- + -= = = -

+ + +

Forf(x) to be min2

2

1x +should be max, which is so

ifx2+ 1 is min.Andx2+ 1 is min atx= 0.

\ min0 1

10 1

f -= = -

+

8. (b) The maximum value off(x) = cosx+ cos ( 2 )x is 2which occurs atx= 0. Also, there is no value of xforwhich this value will be attained again.

9. (b, d)3 5( ) ( 1) ( 1)( 2) ( 3) 0x

dyf x x e x x x

dx= - - - - =

Critical points are 0, 1, 2, 3. Consider change of sign of

dy

dx atx= 3.

x< 3,dy

dx= ve andx> 3,

dy

dx= +ve

Change is from ve to +ve, hence minimum atx= 3.

Again minimum and maximum occur alternately.

\ 2nd minimum is atx= 110. (b, c) Letf(x) = ax3+ bx2+ cx + d

Then,f(2) = 18 8a+ 4b+ 2c+ d= 18 (1)f(1) = 1 a + b + c + d= 1 (2)f(x) has local max. atx= 1

3a 2b + c= 0 (3)

( )f x has local min. atx= 0 b= 0 (4)

Solving (1), (2), (3) and (4), we get

31 17( ) (19 57 34) (0)

4 2f x x x f= - + =

Also 257

( ) ( 1) 0, 14

f x x x= - > " >

Also ( )f x = 0 x= 1, 1

( 1) 0, (1) 0 1f f x- < > = -

is a point of local max.

andx= 1 is a point of local min. Distance between ( 1,

2) and (1,f(1)), i.e. (1, 1) is 13 2 5

11. (a, b) Q g(x) =0

( )x

f t dt

1

, 0 1

( ) ( ) 2 , 1 2

, 2 3

x

x

e x

g x f x e x

x e x

-

= = - < - <

\ 1( ) 0 2xg x e -= = or 0x e- =

1 log 2x - = or x= e

x= 1 + ln 2 or e

1

, 0 1

( ) , 1 2

1, 2 3

x

x

e x

g x e x

x

-

= - < <

NOTE THIS STEP

\ (1 ln 2) 2g + = - and ( ) 1 ( )g e g x= has local max.atx= 1 + ln 2 and local min. atx = e.


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1 2 3

e(1+ ln 2)

3

2

1

X

Graph of '( )g x

Also graph of ( )g x suggests,g(x) has local max. atx= 1 and local min. atx= 2

12. (b, c, d) We have,

1

( ) cos , 1f x x xx=

\1 1 1

'( ) cos sinf xx x x

= +

lim '( ) cos0 (0)

= +x

f x (some finite value)

lim '( ) 1x

f x

=

Also2 2 3

1 1 1 1 1 1"( ) sin sin cos= - -f x

x x xx x x

3

1 1

"( ) cos 0, [1, )

-

= < " f x xxx

'( )f x is strictly decreasing in [1, )

\ '( ) lim '( )

>x

f x f x

( 2) ( )

1( 2)

+ ->

+ -

f x f x

x x

( 2) ( ) 2+ - >f x f x13. (a, b, c, d)

We have2

0

( ) ( 2)( 3)x

tf x e t t dt= - -

f (x) =2

.( 2)( 3)xe x x- -

f (x) = 0 x= 2, 3

f (x) =2 22.2 ( 5 6) (2 5)x xe x x x e x- + + -

f (2) = ve and f (3) = + ve\x= 2 is a point of local maximaandx = 3 is a point of local minima

Also orx(2, 3)f (x)< 0 f is decreasing on (2, 3)Also we observe

f (0) < 0 andf (1) > 0\There exists some C(0, 1) such thatf (C) = 0\All the options are correct.

14. (a, c) LetL= 8x,B= 15xandybe the length of square cut offfrom each corner. Then volume of box= (8x 2y) (15x 2y)yV= 120x2y 46xy2+ 4y3

dV

dy = 120x2

92xy+ 12y2

Now 0dV

dy= aty= 5 for maximum value of V.

[30x2 23xy+ 3y2]y= 5

= 0 6x2 23x+ 15 = 0

x= 3,5

6

Forx= 3, sides are 45 and 24.

SUBJECTIVEPROBLEMS:

1. ( )( )( ) , , ,( )

a x b xf x a b c x cc x

+ += > > -+

( ) ( )a c x c b c x c

x c

- + + - + +=

+

( )( )( ) 2

a c b cx c a b c

x c

- -= + + + + -

+

f '2

( )( )( ) 1

( )

a c b cx

x c

- - -= +

+

\ ( ) 0 ( ) ( )f x x c a c b c= = - - -

( ) ( )x c a c b c= - + - - [+ve sign is taken x c> -Q ]

Also3

2( )( )( ) 0

( )

a c b cf x

x c

- -= >

+for a, b> candx> c

\ f(x) is least atx= c ( )( )a c b c+ - -

\ min( )( )

( )( )( )( )

a c b cf a c b c

a c b c

- -= + - -

- -

( ) ( )a c b c+ - + -

= ( ) ( ) 2 ( )( )a c b c a c b c- + - + - -

=2

( )a c b c- + -2. Given thatxandyare two real variables such thatx> 0

andxy= 1.To find the minimum value ofx+y.Let S = x + y

1

S xx

= + (usingxy= 1)

\ 21

1= -dS

dx x

For minimum value of , 0dS

Sdx

=

2

11 0

x- = 1x=


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10

Butx > 0, \x= 1

Now2

2 3

2d S

dx x=

2

21

2

x

d S vedx =

= = +

\ Sis minimum whenx= 1

\ min1

1 21

S = + =

3. We are given that

[0,1],| ( ) | 1x f x


11/31

11


6. (0, c),y=x2, 0 5c .Any point on parabola is (x,x2)Distance between (x,x2) and (0, 1) is

2 2 2( )D x x c= + -

To minimumDwe considerD2=x4 (2c 1)x2+ c2

22 2 1 1

2 4

cx c

- = - + -

which is minimum when 22 1

02

cx

-- = 2

2 1

2

cx

-=

min1

4D c= -

7. Given 2b

ax cx

+ (1)

0, 0, 0" > > >x a b

To show that 272 3

4ab c .Let us consider the functionf (x) = ax2+ b/x

then2

( ) 2 0b

f x axx

= - =

x3= b/2a x= (b/2a)1/3

\ 32

( ) 2b

f x ax

= +

1/ 3

22 2 6 0

2

b bf a a a

a b

= + = >

\ fis minimum at1/ 3

2

bx

a

=

As (1) is true " x

\ so is for1/ 3

2

bx

a

=

2 / 3

1/ 32 ( / 2 )

b ba c

a b a

+

1/ 3

2

( / 2 )

ba b

ac

b a

+

1/ 33 2

2

b ac

b

As a, bare +ve, cubing both sides we get

3327 2.

8

b ac

b 2 327 4ab c Hence proved.

8. To show

1 +xln2 2( 1) 1x x x+ + + for 0x

Considerf(x) = 1 +xln 2 2( 1) 1 )x x x+ + - +

Here, 2

2( ) ln( 1)

1

xf x x x

x x

= + + ++ +

2 21

1 1

x x

x x

+ - + +

2ln ( 1)x x= + +

As 2 1 1x x+ + for 1x

\2

ln ( 1) 0x x+ + \ ( ) 0, 0f x x " Hencef(x) is increasing function.

Now for 0 ( ) (0)x f x f

2 21 ln( 1) 1 0x x x x+ + + - +

2 21 ln( 1) 1x x x x+ + + +

9. Let the house of the swimmer be atB.

\ d(AB) =Lkm.Let the swimmer land at C, on the shore and let d (AC) =xkm

\ d(SC) = 2 2 and ( )x d d CB L x+ = -

S

d

A x C BL x

2 2x d+

Now using timedistance

speed=

Time from StoB= time Sto C+ time from CtoB

\2 2x d L x

Tu v

+ -= +

Hence we take, 2 21

( )L x

f x x du v v

= + + -

2 2

1 1.2 1'( ) 0

2

xf xu vx d

= + -+

For either maximum or minimum, f ' (x) = 0

2 2

1'( ) 0

xf x

vu x d

= - =+

v2x2= u2(x2+ d2) x2(v2 u2) = u2d2

2 2

2

2 2

u dx

v u=

-

2 2

ud

x v u= - (asx ve)


12/31


12

Now,

2 2

2 2

2 2 2

2.1 .

1 2"( )

( )

xx d x

x df x

u x d

+ - + =

+

2

2 2 2 2

1

.( )

d

u x d x d

= + +

2

2 2 3/ 20,

( )

dx

u x d= > "

+

\ fhas minimum at2 2

udx

v u

=-

10. Equation of the curve is given by

21

xy

x=

+...(1)

Differentiating with respect to x, we get

2 2

2 2 2 2

1 (2 ) 1

(1 ) (1 )

dy x x x x

dx x x

+ - -= =

+ +

Again let2

2 2

1( )

(1 )

x dyf x

dxx

-= =

+

Now,f '(x)2 2 2 2

2 4

(1 ) ( 2 ) (1 )2(1 )(2 )

(1 )

x x x x x

x

+ - - - +=

+

2 2

2 3

(1 )( 2 ) (1 )2.2

(1 )

x x x x

x

+ - - -=

+

2

2 3

(2 6)

(1 )

x x

x

-=

+For the greatest value of slope, we have

2

2 3

(2 6)'( ) 0

(1 )

x xf x

x

-= =

+ 0, 3x=

Again we find,

2 2 2

2 4 2 3

12 (3 ) 6(1 )"( )

(1 ) (1 )

x x xf x

x x

- -= -

+ +

\ f "(0) = 6 and3

"( 3)16

f =

Thus, second order derivative atx= 0 is negative and second

order derivative at 3x= is positive.Therefore, the tangent to the curve has maximum slope at(0, 0).

11. Equation of given curvey= cos (x+y), 2 2x- p pDifferentiating with respect to x,

sin( ). 1dy dy

x ydx dx

= - + +

[1 sin( )] sin( )dy

x y x ydx

+ + = - +

sin( )

1 sin( )

dy x y

dx x y

+= -+ +

...(1)

Since the tangent to given curve is parallel tox+ 2y= 0

\ sin( ) 11 sin( ) 2

x y

x y

- += -

+ +[For parallel line m

1= m

2]

2sin (x+y) = 1 + sin (x+y)

sin (x+y) = 1Thus, cos (x+y) = 0

Using equation of curve and above result, we get,y= 0

sinx= 1 x= np+ (1)np/2, nZ x= p/2, 3p/2which belong to the interval [2p, 2p]Thus the points on curve at which tangets are parallel togiven line are (p/2, 0) and ( 3p/2, 0)The equation of tangent at (p/2, 0) is

10 ( / 2)

2y x- = - - p

2y= x+ p/2 2x+ 4y p= 0The equation of tangent at (3p/2, 0) is

10 ( 3 / 2)

2y x- = - + p

2y = x 3p/2 2x+ 4y+ 3p= 0Thus the required equations of tangents are

2x+ 4yp = 0 and 2x+ 4y+ 3p= 0.12. The given function is,

f(x) = sin3x+ lsin2xfor p/2 < x < p/2\ f '(x) = 3 sin2xcosx+ 2lsinxcosx

1sin 2 (3sin 2 )

2x x= + l

So, fromf '(x) = 0, we getx= 0

or 3 sinx+ 2l= 0

Also,f "(x) = cos 2x(3 sinx+ 2l)3

sin2 cos2

x x+

Therefore, for3

sin2

x-

l = , we have

f "(x) = 3 sinxcos2x= 2lcos2xNow, if 0 0 so thatf(x) has a minimum.Thus, for exactly one maximum and minimum value off(x),lmust lie in the interval

3/2 < l < 0 or 0


13/31

13


C

E D

AM F

B

x

y

b y

(r , r)2

(p , p)2

(q , q)2

Then from similar D'sCEDand CAB, we have

b yor

CE CA b

ED AB x c

-= = ...(1)

LetPdenote the area of ||gmAFDE.

ThenP= 2DAEF=1

2. sin

2

xy A

. ( ) sinc

y b y Ab

= - [using eq. (1)]

2sin ( )c

P A by yb

= -

\ sin ( 2 ) 0dP c

A b ydy b

= - =

Above gives2

2and sin ( 2)

2

b d P cy A ve

bdy= = - = -

Hence area is maximum wheny=b/2 and its value is

1 1. sin sin . ( )

2 2 4 2

b c bP b A bc A ar ABC

b

= - = = D

Now,

2

2

2

11

( ) 12

1

p p

ar ABC q q

r r

-

D =

-

Operating,R2R

2R

1andR

3R

3R

1, we get

2

2 2

2 2

11

02

0

p p

q p q p

r p p r

-

= - +

- -

11( )( )

( ) 12

q pq p p r

r p

-= + -

- +

1( ) ( ) ( )

2q p p r q r = + - +

Thus max. area of ||gmAFDEis

1( )

2ar ABC = D

1( )( )( )4 q q r p r = + + -

Hence Proved.

14. The equation of given curve can be expressed as

2 2

21

4

x y

a+ = where 4 < a2< 8

Clearly it is the question of an ellipse

O

P a b( cos , sin )qq

(0, 2)

( , 0)aX

Let us consider a pointP(acos q, 2 sin q) on the ellipse.Let the distance ofP(acos q,2 sin q) from (0, 2) isL.Then,L2= (acos q 0)2+ (2 sin q+ 2)2

Differentiating with respect to q, we have

22( ) cos [ 2 sin 8sin 8]

d La

d= q - q + q +

qFor max. or min. value ofLwe should have

2( )0

d L

d=

q cosq[2a2sin q + 8 sin q + 8] = 0 Either cos q = 0

or (8 2a2) sin q+ 8 = 0

2

pq = or

2

4sin

4aq =

-Since a2< 8 a2 4 < 4

2

41

4a>

- sin q> 1 which is not possible

Also

2 22

2

( )cos [ 2 cos 8cos ]

d La

d= q - q + q

q+ ( sin q) [ 2a2 sin q+ 8 sin q + 8 ]

At 2 2 2 22

( ), 0 [16 2 ] 2( 8) 02

d L a ad

pq = = - - = -

\ f(x) is minimum atx= 7/5.

16. We havey=x(x 1)2, 0 2x

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

dyx x x x x

dx= - + - = - -

For max. or min. 0dy

dx=

(x 1) (3x 1) = 0 x= 1, 1/3

2

23 1 3( 1) 6 4

d yx x x

dx= - + - = -

Atx= 1,2

22( )

d yve

dx= + \ yis min. atx= 1

At2

21/ 3, 2( )

d yx ve

dx= = - - \yis max. atx= 1/3

\ Max value ofyis

21 1 4

13 3 27

= - =

Min value ofyis = 1 (1 1)2= 0

O

D C (2, 2)

y

x

y= 2

A1

B2

Now the curve cuts the axisxat (0, 0) and (1, 0). Whenxincreases from 1 to 2,yalso increases and is +ve.Wheny= 2,x(x 1)2= 2

x3 2x2+x 2 = 0 (x 2) (x2+ 1) = 0 x= 2Using max./min. values ofyand points of intersection withx-axis, we get the curve as in figure and shaded area is therequired area.\ The required area

=

2

0Areaof squareOBCD y dx- 2 2

02 2 ( 1)x x dx= - -

23 2

3

00

( 1) 14 ( 1) .1

3 3

xx x dx

- = - - -

243

0

( 1)4 ( 1)

3 12

x xx -= - - -

2 1 1 2 104 4 sq.units.

3 12 12 3 3

= - - + = - =

17. Letf(x) = 2 sinx+ tanx 3x

\ f '(x) = 2 cosx+ sec2x 3= (sec2x 1) 2 (1 cosx)

22 2 2

2

sintan 4sin / 2 4sin / 2

cos

xx x x

x= - = -

2 2 2

2

4sin / 2cos / 24sin / 2

cos

x xx

x= -

22

2

cos / 24sin / 2 1

cos

xx

x

= -

...(i)

As cosxis decreasing function

\ / 2x x cos / 2 cosx x

2 2cos / 2 cosx x (as both are +ve)

2

2

cos / 21

cos

x

x

2

2

cos / 21 0

cos

x

x

-

\2

2

2

cos / 24sin / 2 1 0

cos

xx

x

-

\ From eq. (1), we get '( ) 0f x

f (x) is an increasing function on (0, p/2) for 0x ( ) (0)f x f

2 sinx+ tan 3 0x x- 2 sinx+ tan 3 .x x Hence ProvedALTERNATE SOLUTION

Let f(x) = 2 sinx+ tanx 3xon 0 / 2x < pthenf '(x) = 2 cosx+ sec2x 3andf "(x) = 2 sinx+ 2 sec2xtanx

= 2 sinx[sec3x 1]

for 0 / 2x < p "( ) 0f x

f '(x) is an increasing function on 0 / 2.x < p\ For 0,x '( ) '(0)f x f '( ) 0 0 / 2f x or x < p

f(x) is an increasing function on 0 / 2x < p

\ For 0,x ( ) (0)f x f

2sin tan 3 0, 0 / 2x x x x+ - < p 2sin tan 3 , 0 / 2x x x x+ < p Hence proved


15/31

15


18. As QR||XYdiameter throughPis ^ QR.

Q A R

O

r r

r

PX Y

q

2q

Now area of DPQRis given by1

.2

A QR AP=

But QR= 2.QA= 2rsin 2q

andPA = OA + OP= rcos 2q+r\

1.2 sin 2 .( cos 2 )

2A r r r= q + q

= r2. 2 sin qcos q. 2 cos2q = 4 r2sin qcos3q

For max. value of area, 0dA

d=

q 4r2[cos4 q 3 sin2qcos2q] = 0 cos2q(cos2q 3 sin2 q) = 0

1

tan3

q = q= 30

Also2

2 3 3

2

4 [ 4cos sin 6sin cosd A

r

d

= - q q - q qq

36sin cos ]+ q q= 4r2[10 sin qcos3 q + 6sin3q cos q]

22

230

1 3 3 1 34 10. . 6. .

2 8 8 2

d Ar

d q=

= - +

q

2 15 3 3 3

48 8

r -

= +

2 12 3

48

r ve -

= = -

\ Ais maximum at q= 30AndA

max= 4r2sin 30 cos330

2 21 3 3 3 342 8 4

r r= =

19. LetABCEDAbe the window as shown in the figure and letAB = x mBC = y m

D

E

C

A B

Then its perimeter including the baseDCof arch

2 22

xx y m

p = + +

\ 2 22

P x yp

= + + ...(1)

Now, area of rectangleABCD = xy

and area of arch

2

2 2

xDCED

p =

Let lbe the light transmitted by coloured glass per sq. m.Then 3lwill be the light transmitted by clear glass per sq. m.Hence the area of light transmitted

2

3 ( )2 2

xxy

p = l + l

2

38

xA xy

p= l +

...... (2)

Substituting the value of y from (1) in (2), we get

21 43

2 2 8

xA x P x

+ p p = l - +

223 3(4 )

2 4 8

Px xx

+ p p= l - +

\3 3(4 )

2 2 4

dA P xx

dx

+ p p = l - +

ForAto be maximum 01dA

dx=

32

12 3 )

4 2

Px=

-p + p +

3 4

2 5 24

Px=

p +

6

5 24

Px=

p +

Also2

2

3(4 )0

2 4

d A

dx

- + p p = l +


16/31


16

3a 2b + c= 0 ... (2)and a+ 2b+ 3c= 0 ... (3)

Also,1

1

14( )

3f x

- =

14 3 2

1

14

4 3 2 3

ax bx cxdx

-

+ + + =

14

4 3 2 4 3 2 3

a b c a b cd d

+ + + - - + - =

7

3 3

bd+ =

b+ 3d = 7 ...(4)From (1), (2), (3), (4) on solving, we get

a= 1, b= 1, c= 1, d= 2\ The required cubic isx3+x2x+ 2.

21. The given curve isy=x2

...(1)Consider any pointA(t, t2) on (1) at which normal chorddrawn is shortest.Then eq. of normal to (1) atA(t, t2) is

2

2

( , )

1( )

t t

y t x tdy

dx

- = - -

[where 2dy

xdx

= from (1)]

2 1( )

2y t x t

t- = - -

x+ 2ty = t+ 2t3 ...(2)This normal meets the curve again at pointBwhich can be

obtained by solving (1) and (2) as follows :Puttingy=x2in (2), we get2t x2+x (t+ 2t3) = 0,D= 1 + 8t(t+ 2t3) = 1 + 8t2 + 16t4= (1+ 4t2)2

\2 21 1 4 1 1 4

,4 4

t tx

t t

- + + - - -=

1,

2t t

t= - -

\ 2 22

1, 1

4y t t

t= + +

Thus, 22

1 1, 1

2 4B t t

t t

- - + +

\ Length of normal chord

22

2

1 12 1

2 4AB t

t t

= + + +

Consider

222

2

1 12 1

2 4Z AB t

t t

= = + + +

24 2

1 33 4

16 4Z t

t t= + + +

For shortest chord, we have to minimize Z, and for that

0dZdt =

5 3

1 38 0

4 2t

t t- - + =

1 6t2 + 32t6= 0 32 (t2)3 6t2 1 = 0 (2t2 1) (16t4+ 8t2+ 1) = 0

2 12

t = (leaving ve values of t2)

1 1

,2 2

t= - ,

2

2 6 4

5 98

4 2

d Z

dt t t = + +

2 2

2 21 1

2 2

also

t t

d Z d Z ve ve

dt dt = =-

= + = +

\ Zis minimum at 1 1or2 2

t= -

For1

2t= normal chord is (from (2)) 2 2x y+ =

For1

2t= - normal chord is 2 2x y- = -

22. The given curve isy= (1+x)y+ sin1(sin2x)Here atx= 0,y= (1 + 0)y+ sin1(0) y= 1\ Point at which normal has been drawn is (0, 1).For slope of normal we need to find dy/dx, and for that weconsider the curve as

y= u + v dy du dv

dx dx dx= +

where u= (1 +x)y ...(i)and v= sin1(sin2x) ` ...(ii)Taking log on both sides of equation (i) we get

log u=ylog (1+x)

1

log(1 ).1

du y dyx

u dx x dx= + +

+

(1 ) log(1 )1

ydu y dyx xdx x dx

= + + + +

Also v= sin1(sin2x)

4

1.2 sin cos

1 sin

dvx x

dx x

=-

2

2sin.

1 sin

dv x

dx x

=+

Thus, we get,

2

2sin(1 ) log(1 )

1 1 sin

ydy y dy xx xdx x dx x

= + + + + + +

[1 (1 ) log(1 )]ydy

x xdx

- + +

1

2

2sin(1 )

1 sin

y xy x

x

-= + ++


17/31

17


1

2

2sin(1 )

1 sin

1 (1 ) log(1 )

y

y

xy x

dy x

dx x x

-+ ++=

- + +

(0,1)1dy

dx = , \ Slope of normal = 1

\ Equation of normal to given curve at (0, 1) isy 1 = 1 (x 0)

x + y= 1.

23. We have,

3 23

3

1, 0 1

( ) 3 2

2 3, 1 3

b b bx x

f x b b

x x

- + -- +


18/31


18

\ Max. area =4 6

1 2 2 1 2 2

4 163 3

-

= 1 64 1 5124 9 16 27 -

16 32

9 27= -

4 4 3

93 3= = sq. units.

26. Let the given line be 1x y

a b+ = , so that it makes an intercept

of a units onx-axis and bunits ony-axis. As it passes throughthe fixed point (h, k), therefore we must have

O

b

y

x

Q

Pa

( )h, k

1k h

b a= - akb

a h=

- (1)

Now Area of DOPQ=A= 12ab

\ 12

akA a

a h

= -

[using (1)]

or2

2

k aA

a h

=

-

For min. value ofA, 0dA

da=

2

2

2 ( )0

2 ( )

k a a h a

a h

- -=

-

2

2

20

2 ( )

k a ah

a h

-=

- a= 2h

Also,2 2 2

2 4

(2 2 )( ) 2( ) ( 1)( 2 )

( )

d A a h a h a h a ah

da a h

- - - - - -=

-

\2 3

2 42

(2 2 )(0) 20,[ 0]

a h

d A h hh

hda h=

+= = > >Q

\ Ais min. when a= 2h

\2

min

42

2

k hA kh

h

= =

27. The normal to the curve atPisa (y 1) + (x 1) = 0

First we consider the case when 0a

Slope of normal atP(1 , 1) is1

a

= -

Slope of the tangent at (1, 1) is = a

(1,1)

dya

dx

= (1)

But we are given that

dy dyy ky

dx dx =

dykdx

y=

log |y| = kx + C |y| =ekx+c= ec.ekx

c kxy e e= y=Aekx

WhereAis constant. As the curve passes through (1, 1)

\ 1 =Aek

A= ek

\ y= ek(x1) ( 1)k xdy kedx

-=

(1,1)

dyk

dx

=

From (1) and (2),

1,1

dya k

dx

= =

\ y= ea(x 1)which is the required curve.Now the area bounded by the curve,y-axis and normal tocurve at (1, 1) is as shown the shaded region in the fig.

O

y

x

B

C

A

P (1, 1)

a y + x =0( 1) ( 1)

\ Req. area = ar(PBC)

= ar(OAPBCO) ar(OAPCO)1 1

normal curve0 0y dx y dx= -

1 1( 1)

0 0

1( 1) 1 a xx dx e dx

a

- = - - + - 1 1

2 ( 1)

0 0

1 1( 1)

2

a xx x ea a

- = - - + -

1 1 1 1 11 1

2 2

a ae ea a a a a

- -= + - + = + -

Now we consider the case when a= 0. Then normal at (1, 1)becomesx 1 = 0 which is parallel toy-axis, therefore tangentat (1, 1) should be parallel tox-axis. Thus


19/31

19


(1,1)

0dy

dx

= (3)

Sincedy

y

dx

givesy= ek(x 1)

(as in 0a case)

( 1)k xdy

kedx

-=

(1,1)

dyk

dx

= (4)

From (3) and (4), we get k= 0 and required curve becomesy= 1

O

y

x

y = 1

x = 1

In this case the required area= shaded area in fig. = 1 sq. unit.

28.21

( ) , 0, 0

8

f x nx bx x x b= - + > l

1( ) 2

8f x b x

x= - + (1)

2( ) 0 16 8 1 0f x x bx= - + = (for max. or min.)

\ 21

14

x b b = - (2)

Above will give real values ofxif b2 1 0 i.e. 1b or

1b - . But bis given to be +ve. Hence we choose 1b

If b= 1 then1

4x= ; If b> 1 then 2

11

4x b b = -

2

2 2

1 16 1( ) 2

8 8

xf x

x x

-= - + =

Its sign will depend onNr, 16x2 1 as 8x2is +ve. We shall

consider its sign for1

4x= and 2

11

4x b b = -

( ) 0f x = atx= 1/4

\ Neither max. nor min. as "( ) 0f x =

N roff ''(x) = 16x2 1 =2 2[ 1] 1b b+ - -

= +ve for b> 1 \ Minima

or N rof 2 2( ) ( 1) 1f x b b= - - - = ve for b> 1 \ Maxima

29. Given that,2 3

, 0( )

, 0

axxe x

f xx ax x x

=

+ - >Differentiating both sides, we have

2

, 0( )

1 2 3 , 0

ax axaxe e xf x

ax x x

+ = + - >

Again differentiating both sides, we have

22 ; 0( )

2 6 ; 0

ax axae a x e xf x

a x x

+ =

- >

For critical points, we put ( ) 0f x =

2xa

= - , if 0x

3

a

= , ifx> 0 +

2/a a/3

It is clear from number line that

( )f x is +ve on2

,3

a

a

-

( )f x increases on2

,3

a

a

-

30. Let b a= t, where a + b= 4

42

ta -= and 42

tb +=

as given a< 2 and b> 2 t> 0

Now0 0

( ) ( )a bg x dx g x dx+

=

4 4

2 2

0 0( ) ( ) ( )

t t

g x dx g x dx t

- +

+ = f [say]

4 1 4 1

( )2 2 2 2

t tt g g

- + f = - +

NOTE THIS STEP

( )

( )Using ( ) [ ( )]. ( ) [ ( )]. ( )

v x

u x

df t dt f v x v x f u x u xdx

= -

1 4 4

2 2 2

t tg g + - = -

Sinceg(x) is an increasing function (given)

\ forx1>x

2 g(x1) >g(x2)

Here we have4 4

2 2

t t+ - >

4 4

2 2

t tg g

+ - >

1 (4 ) (4 )

( ) 02 2 2

t tt g g+ - f = - >

( ) 0tf >


20/31


20

Hence f (t) increase as tincreases.

0 0

( ) ( )a bg x dx g x dx+ increases as (b a) increases.

31. Applying3 3 1 2

2R R R R - - we get

2 2 2 1

( ) 1 1

0 0 1

ax ax a ax b

f x b b

- + +

= + -

2 2 1 2 1

1 1

ax ax ax

b b b

- -= =

+

[Using 2 2 1]C C C -

( ) 2f x ax b= +Integrating, we get ,f(x) = ax2+ bx + Cwhere Cis an arbitrary constant. Since f has a maximum atx

= 5/2,

(5/ 2) 0 5 0f a b= + = (1)

Alsof(0) = 2 C= 2andf(1) = 1 a + b + c= 1\ a + b= 1 (2)Solving (1) and (2) for a, bwe get,

a= 1/4, b= 5/4

Thus, 21 5

( ) 2.4 4

f x x x= - +

32. Equation of the tangent at point (x,y) on the curve is

Y( )

dyy X x

dx= -

This meets axes in

0'

dxA x y

dy

-

and 0,dy

B y xdx

-

Mid-point ofABis1 1

,2 2

dx dyx y y x

dy dx

- -

We are given

1

2

dxx y x

dy

- =

and1

2

dyy x ya

dx

- =

dy dy dxx ydx y x

= - = -

Intergrating both sides,

log logdy dy

y x cy x

= - = - + Putx= 1,y= 1,

log 1 = log 1 + c c = 0 logy+ logx= 0 logyx= 0 yx = e0= 1Which is a rectangular hyperbola.

33. Given that,p(x) = a

0+ a

1x+ a

2x2+ + a

nxn (1)

and |p(x) |

| e

x1

1|, 0x" (2)To prove that,

1 2| 2 | 1na a n a+ ++

It can be clearly seen that in order to prove the result it is

sufficient to prove that | (1) | 1p We know that,

0

(1 ) (1)| (1) | lim

h

p h pp

h

+ -=

0

| (1 ) | | (1) |lim

| |

+ +

h

p h p

h NOTE THIS STEP

[Using |xy] |x| + |y| ]

But 0| (1) | | 1|p e -[Using equation (2) forx= 1]

| (1) | 0p

But being absolute value, | (1) | 0p .

Thus we must have | (1)| 0p =

Also | (1 ) | | 1 |h

p h e+ -(Using eqn(2) forx= 1 + h)

Thus0

| 1 || (1) | lim 1

| |

h

h

ep

h

- =

or | (1) | 1p

|a1+ 2a2+ .. + nan| 1

34. Given that 1 1p- .Considerf(x) = 4x3 3xp= 0

Now,1 3

(1/ 2) 1 02 2

f p p= - - = - - as ( 1 )p-

Alsof(1) = 4 3 p= 1 0p as ( 1)p \ f(x) has at least one real root between [1/2, 1].

Also 2( ) 12 3 0f x x= - > on [1/2, 1]

fis increasing on [1/2, 1] fhas only one real root between [1/2, 1]To find the root, we observef(x) contains 4x3 3xwhich ismultipe angle formula of cos 3qif we putx= cos q.\ Let the req. root be cos qthen,

4 cos3q 3 cos qp= 0 cos 3q=p 3q= cos1p

11cos ( )3

p-q =

\ Root is 11

cos cos ( )3

-

.

35. The given curve is2 2

16 3

x y+ = (an ellipse)

Any parametric point on it is ( 6 cos , 3 sin )P q q .Its distance from linex+y= 7 is given by

6 cos 3 sin 7

2D

q + q -=

For min. value ofD, 0dDd

=q


21/31

21


6 sin 3 cos 0- q + q =

tan 1/ 2q =

2

cos

3

q = and1

sin

3

q =

\ Required pointPis (2, 1)ALTERNATE SOLUTION :

Any point on given ellipse2 2

16 3

x y+ = is

( 6 cos , 3 sin )P q qDistance ofPfromx + y= 7 will be minimum if measured

along the normal of ellipse atP, and then givenline.PN^

(slope of normal atP) (slope of given line) = 1

( 6 cos , 3 sin )

( 1) 1dx

dy q q

- - = -

( 6 cos , 3sin )

2y

x q q

= 1

2 3 sin 1 21 tan cos6 cos 2 3

q= q = q =

q

1sin

3q =

\ Required point is (2, 1).36. Given that 2 (1 cosx)

2tan1 2

x

-

\

2 2

2

11 tan cos

2( )cos

x x

f xx

- ->

2

2

2

1sin 1

2cos

cos

xx

x

-

=

2

4

sin (cos 2 )0, [0, / 4)

2cos

x xx

x= > " p

\ ( ) 0 ( )f x f x> is an increasing function.

\ For [0, / 4)x p ,

0 ( ) (0)x f x f

sin (tanx) sinx (tan 0) 0

sin (tanx) 0x sin (tanx) x

Hence proved.37. Given thatf is a differentiable function on [0, 4]

\ It will be continuous on [0, 4]\ By Lagrange's mean value theorem, we getNOTE THIS STEP

(4) (0)( )

4 0

f ff a

-=

-, for (0,4)a (1)

Again sincefis continuous on [0, 4] by intermediate meanvalue theorem, we get

(4) (0)( )

2

f ff b

+= for (0,4)b (2)

[Iff(x) is continuous on [a, b] then ( )$m a,b

such that( ) ( )

( )2

f ff

a + bm = ]

Multiplying (1) and (2) we get

2 2[ (4)] [ (0)]( ) ( ); , (0, 4)

8

f ff a f b a b

-=

or [f(4)]2 [f(0)]2= 8 ( ) ( )f a f bHence Proved.

(ii) To prove

42 2

0( ) 2[ ( ) ( )] 0 , 2f t dt f f= a a + b b " < a b


22/31


22

Fis continuous on [0, 2]

2 2

2 ( )2 ( ).2( ).22

f ff

a a + b bm m =

2 2 2( ).2 ( ) ( )f f fm m = a a + b b (3)From (2) and (3) we get

22 2 2

0( )2 2[ ( ) ( )]f u u du f f= a a + b b

where 0 < a, b < 24

2 2

0( ) 2[ ( ) ( )]f t dt f f= a a + b b

where 0 < a, b < 2 (Using eqn(1))Hence Proved.

38. We are given that,

( )( ), , 1

dP xP x x

dx

> " andP(1) = 0

( )

( ) 0dP x

P xdx

- >

Multiplying by ex, we get,

( )x xdP xe edx

- -- P(x) > 0

( ) 0xd

e P xdx

- >

exP(x) is an increasing function.

\ 11, ( ) (1) 0xx e P x e P- -" > > = [UsingP(1) = 0]

( ) 0, 1x

e P x x-

> " > ( ) 0, 1P x x> " > [ 0]xe- >Q

39. We are given,P(x) = 51x101 2323x100 45x+ 1035

To show that at least one root of P(x) lies in (451/100, 46),using Rolle's theorem, we consider antiderivative ofP(x)

i.e.102 101 22323 45

( ) 10352 101 2

x x xF x x= - - +

Then being a polynominal functionF(x) is continuous anddifferentiable.

Now,

102 101

100 1001/100 (45) 2323(45)(45 )2 101

F = -

21

100100

45.(45)1035(45)

2- +

2 1

100 10045

(45) 23 45(45)2

= -

21

100100

45.(45)1035(45) 0

2- + =

And102 101 2(46) 2323(46) 45(46)

(46)2 101 2

F = - - + 1035(46)

= 23 (46)101 23 (46)101 23 45 46 + 1035 46 = 0

\1

100(45 ) (46) 0F F= =\ Rolle's theorem is applicable.

Hence, there must exist at least one root of ( ) 0F x =

i.e.P(x) = 0 in the interval

1

10045 , 46

40. Let us consider,

f(x)3 ( 1)

sin 2x x

x x +

= + -p

3( ) cos 2 (2 1)f x x x= + - +p

6

( ) sin 0, [0, / 2]f x x x= - - < " p p f '(x) is a decreasing function. .... (1)

Also3

(0) 3 0f = - >p

(2)

and3 3

( / 2) 2 ( 1) 1 0f p = - p + = - -

Letx=pbe the point at which the max. off(x) occurs [NOTEThere will be only one max. point in [0, p/2]. Sincef '(x) = 0 isonly once in the interval.]

Consider , [0, ]x p

f '(x) > 0 f(x) is an increasing function.

(0) ( ) [ 0 ]f f x as x

( ) 0f x ..... (4)

Also for [ , / 2]x p p

f '(x) < 0 f(x) is decreasing function. forxf(p/2) > 0 ..... (5)

O

y

x

Y =f x( )

Inc.

Dec.

x = p x= p /2

Hence from (4) and (5) we conclude that( ) 0, [0, / 2]f x x " p .


23/31

23


41. Given that,

|f(x1) f(x

2) | < (x

1x

2)2, 1 2,x x R

Letx1=x + handx

2=xthen we get

|f(x + h) f(x) | < h2

|f(x + h) f(x) | < | h|2

( ) ( )

| |f x h f x

hh

+ - 0 , (1, )x"

Also let h(x) =

2( 1)

1

x

x

-

+ lnx

h(x) =2

2 2

4 1 ( 1)0, (1, )

( 1) ( 1)

xx

xx x x

- -- = <

+ +

\ h(x) is decreasing function.\ Forx> 1

h(x) So g is concave upward.

Also g(0) = g(1) = 0g(x) < 0, x (0, 1)exf(x) < 0 f(x) < 0, x (0, 1)

7. (c) g(x) = exf(x)

g(x) = ex

f(x) ex

f(x)= ex(f(x) f(x))

As1

x4

= is point of local minima in [0, 1]

\ g(x) < 0 for1

x 0,4

and g(x) > 0 for1

x , 14

\ In1

0,4

, g(x) < 0

ex(f(x) f (x)) < 0 f(x) < f(x)

1. (7) The given function is 3 2( ) 2 15 36 48= - + -f x x x x

and2{ | 20 9 }A x x x= +

{ }2| 9 20 0A x x x= - + { }| ( 4)( 5) 0A x x x= - - A = [4, 5]

Also 2( ) 6 30 36= - +f x x x 26( 5 6)= - +x x

6( 2)( 3)= - -x xClearly , ( ) 0x A f x" >\ fis strictly increasing function onA.\ Maximum value off onA

= 3 2(5) 2 5 15 5 36 5 48f = - + -= 250 375 + 180 48 = 430 423 = 7.

2. (0) Let4 3 2( )= + + + +p x ax bx cx dx e

Now20

( )lim 1 2

+ =

x

p x

x

20

( )lim 1

=x

x

x

...(1)

(0) 0=p 0=e

Applying LHospitals rule toneq (1), we get

0

( )lim 1

2

=

x

x

x (0) 0p =

0=dAgain applying L Hospitals rule, we get

0

( )lim 1

2

=

x

p x (0) 2=p

2 c=2 or c=1

\ 4 3 2( )= + +p x ax bx x

3 2( ) 4 3 2p x ax bx x= + +


25/31

25


As p(x) has extremum atx =1 and 2

\ (1) 0 and (2) 0= = p p

4 3 2 0+ + =a b ...(i)

32 12 4 0a b+ + = or 8 3 1 0+ + =a b ...(ii)

Solving eqs (i) and (ii) we get1

and 14

= = -a b

\ 4 3 21

( )4

= - +x x x x

So, that16

(2) 8 4 04

= - + =p

3. (9) The equation of tangent to the curve

y f (x)= at the point P (x,y) is

( ) ( )

0

Y y dy dyor X x Y y

X x dx dx= =

dy dyX Y x ydx dx

=

Its y-intercept3

dyy x x

dx= =

2 dy y

xdx x

=

I.F.

1 1dx

xex

= =

221 1 .

2

xy x dx C

x x\ = = +

3

2

xy Cx= +

As (1) 1 1, 1f At x y= = =

11 3 2

2C C\ = + =

33

2 2

x xy\ = +

27 9

At 3, 92 2x y= = =

(3) 9.f\ =

4. (1) We have ,

f(x) 22010 ( 2009)( 2010)x x= 3 4( 2011) ( 2012)x x

As ( ) ( )f x ln g x=

( )( ) xg x e = ( )'( ) . '( )f xg x e f x =

For max/min, '( ) 0 '( ) 0g x f x= =Out of two points one should be a point of maxima and

other that of minima.\ There is only one point of local maxima.

5. (5) We have f(x) = 2 1x x+ -

2

2

2

2

1, 1

1, 1 0

1, 0 1

1, 1

x x x

x x x

x x x

x x x

- + - < -- - + -

= - + <

Critical pts are1 1

, , 1,02 2

-- and 1

ve

1

+ve ve +ve ve +ve

0 1We observe at five pointsf (x) changes its sign

\There are 5 points of local maximum or local minimum.6. (9) Qp(x) has a local maximum atx= 1 and a local minimumatx= 3 andp(x) is a real polynomial of least degree\Letp(x) = k(x 1)(x 3 ) = k(x2 4x+ 3)

p(x) =3

22 33

xk Cx x

+- + Givenp(1) = 6 andp(3) = 2

4

63

k C+ = and 0 + C= 2 k = 3

\p(x) = 3(x 1)(x 3) p(0) = 9

7. (9) Vertical line x= h, meets the ellipse

2 2

14 3

x y+ = at

23, 4

2P h h

-

and 2

3, 4

2Q h h

--

By symmetry, tangents atPand Qwill meet each otheratx-axis.

y

0

P

Q

( , 0)h Rx


26/31


26

Tangent atPis23 4 1

4 6

xh yh+ - =

which meetsx-axis at4

, 0Rh

Area of DPQR= 2

1 43 4

2h h

h

- -

i.e., D(h) =2 3 23 (4 )

2

h

h

-

d

dh

D=

2 2

2

4 ( 2)3 0

h h

h

- + - <

\ D(h) is a decreasing function.

\ max1 1

12 2

h

D = D

and Dmin

= D(1)

\ D1=

3 2

143 454

512 8

2

- =

D2=

3 3 3 9

2 1 2=

\ 1 28

85

D - D = 45 36 = 9

1. (b) Distance of origin from (x,y) = 2 2x y+

=2 2 2 cos

ata b ab t

b

+ - - ;

2 2 2a b ab + + min

cos 1at

tb

- = -

= a+ b

\Maximum distance from origin = a+ b

2. (a) Letf (x) =3 2

3 2

ax bx cx+ + f (0) = 0 andf (1)

=2 3 6

3 2 6

a b a b cc

+ ++ + = = 0

Alsof(x) is continuous and differentiable in [0, 1] and

[0, 1[. So by Rolles theorem, f (x) = 0.i.e ax2+ bx+ c= 0 has at least one root in [0, 1].

3. (d) 3 2 2

2 2

( ) 2 9 12 1

'( ) 6 18 12 ; ''( ) 12 18

f x x ax a x

f x x ax a f x x a

= - + +

= - + = -

For max. or min.

2 2 2 26 18 12 0 3 2 0x ax a x ax a- + = - + =

2 . At max.x aor x a x a= = = and at 2 minx a=\p= aand q= 2a

2

2

As per question

2 2or 0

but 0, therefore, 2.

p q

a a a a

a a

=

\ = = =

> =

4. (a) 29

18 2 18dy dy

y x ydx dx y

= = =

Given 9 92 22

dy ydx y

= = =

Putting in 29

188

y x x= =

\Required point is

2

9,

8

9

5. (b) ( ) 6( 1).f x x= - Inegrating, we get

2( ) 3 6f x x x c= - +

Slope at (2, 1) = (2) 3f c= =

[Q slope of tangent at (2,1) is 3]2 2( ) 3 6 3 3( 1)f x x x x\ = - + = -

Inegrating again, we get 3( ) ( 1)f x x D= - +

The curve passes through (2, 1)

31 (2 1) 0D D = - + =

\f (x) = (x 1)3

6. (d) sin and cosdx dy

a ad d

= - q = qq q

cot .dy

dx

\ = - q

\The slope of the normal at q= tan q\The equation of the normal at qis

sin tan ( cos )y a x a a- q = q - - q

cos sin cos sin sin

sin cos

y a x a

a

q - q q = q - q

- q q

sin cos sin

( ) tan

x y a

y x a

q - q = q

= - q

which always passes through (a, 0)

7. (d) Let us define a function

3 2

( )3 2

ax bxf x cx= + +


27/31

27


Being polynomial, it is continuous and differentiable,also,

(0) 0 and (1)3 2

a bf f c= = + +

2 3 6(1) 0 (given)6

a b cf + + = =

(0) (1)f f\ =\f (x) satisfies all conditions of Rolles theoremthereforef (x) = 0 has a root in (0, 1)

i.e. 2 0ax bx c+ + = has at lease one root in (0, 1)

8. (a) Area of rectangle 2 cosABCD a= q(2 sin ) 2 sin 2b abq = q

Y

X

AB

C D

)sinb,cosa( qq- )sinb,cosa( qq

)sinb,cosa( q-q- )sinb,cosa( q-q

Area of greatest rectangle is equal to 2abWhen 12sin =q .

9. (d) ( )cos sinx a= q + q q

( )sin sin cosdx

ad

= - q + q + q qq

cos

dxa

d= q q

q .....(1)

( )sin cosy a= q - q q

[ ]cos cos sindy

ad

= q - q + q qq

sindy

ad

= q qq

.....(2)

From equations (1) and (2) we get

dy

dx= tan q Slope of normal = cot q

Equation of normal at ''q isy a(sin q q cosq )

= cot q (x a(cos q + q sin q ) ysin q a q2sin + a q cosq sin q

= xcosq + a q2cos + a q sin q cosq xcosq +ysin q = aClearly this is an equation of straight line which is ata constant distance a from origin.

10. (b) Given that

dv

dt= 50 cm3/min

34 503

dr

dt

p =

2

4dr

rdt

p = 50

250

4 (15)

dr

dt=

p=

p181

cm/min (here r= 10+5)

11. (b) Letf (x) = 11

n nn na x a x

--+ + ........... + 1a x = 0

The other given equation,

1nnna x

- + (n 1) 21

nna x

-- + ....+ 1a = 0 =f (x)

Given 1 0a f(0) = 0Againf(x) has root a, ( ) 0f a =\ f(0) =f(a)

\ By Rolls theoremf(x) = 0 has root between ( )0,a

Hence ( )f x has a positive root smaller than a.

12. (a)2

2

x

x+ is of the form

1y

y+ where

12y

y+ and

equality holds fory= 1

\ Min value of function occurs at 12

x= i.e.,

atx= 2

ALTERNATE SOLUTION :

f(x) =2

2

x

x+

2

1 2'( ) 0

2f x

x= - =

x2= 4 orx= 2, 2;3

4''( )f x

x=

] 2 ( )''( ) x ve f xf x = = + has local min atx= 2.

13. (c) Area =21

sin

2

x q

qx x

Maximum value of sinqis 1 at2

pq =

2max

1sin 1,

2 2A x at

p = q = q =

14. (c) Using Lagrange's Mean Value Theorem

Letf(x) be a function defined on [a, b]

then, ( ) ( )'( ) f b f af cb a

-= -....(i)

c[a, b]

\ Givenf(x) = logex \ f ' (x) =

1

x

\ equation (i) become

1 (3) (1)

3 1

f f

c

-=

-

log 3 log 11

2

e e

c

-=

log 3

2

e=

2

log 3ec= c= 2 log3e


28/31


28

15. (d) Givenf(x) = tan1(sinx+ cosx)

f '(x) =2

1.(cos sin )

1 (sin cos )x x

x x-

+ +

2

1 12. cos sin

2 2

1 (sin cos )

x x

x x

-

=+ +

2

cos .cos sin .sin4 4

1 (sin cos )

x x

x x

p p - =

+ +

\ f '(x) =2

2 cos4

1 (sin cos )

x

x x

p +

+ +

iff (x) > Othenf (x) is increasing function.

Hencef(x) is increasing, if2 4 2

xp p p

- < +

solutions to application of derivatives

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