Exercise 2-jeemain.guru ex2 ans PDF

Preview

Summary

test...

Description

MAXIMA & MINIMA

Page # 9

HINTS & SOLUTIONS

EXERCISE – II Sol.1

If n = 3 f(x) = (x2 – 1)3 (x2 + x + 1)

D x2  1

f(x) =

x2  1

–

+ 0

–1

0

1

4x

f ’(x) =

x = 0 is minima

( x 2  1) 2

Sol.5

f(0) = –1 Sol.2

–1

A,C,D

A,C

f ' (x) =

 3(sin 1 x)2

1  x2

f(x) = 40 (3x4 + 8x3 – 18x2 + 60)–1

–

3(cos 1 x)2 1  x2

=0

f ’(x) = – 40 (3x4 + 8x3 – 18x2 + 60)–2 (12x3 + 24x2 – 36x)

sin–1 x = cos–1 x  x =

f ’(x) = 0

 4

12x(x2 + 2x – 3) = 0 critical points are x = 1, –1,

12x (x + 3) (x – 1) = 0 –

+

x = –3, 1 x=0

–

+ 0

–3

  3 f  4 = ;   32

1

Local Maxima

Local minima f(–1) = –

Sol.3

 4

B

7 3 3 + 3 = 8 8

f(x) = a n | x | + bx2 + x Sol.6 f ’(x) =

a x

B,D

+ 2bx + 1 y = f(x) =

f ’(–1) = 0  –a –2b + 1 = 0

.....(1)

a + 4b + 1 = 0 f ’(2) = 0  2

.....(2)

x 1 x tan x

ymax when is reciprocal take min. value 1 1 1 x tan x + tan x = = y x x

Solving (1) & (2) we get a = 2, b = –1/2 Sol.4

A,C,D

Let z =

1 + tan x x

1 dz = – 2 + sec2 x = 0  x = cos x dx x

If n = 2 f(x) = (x2 – 1)2 (x2 + x + 1)

d2 z dx

–1

0

1

2

=

2 x3

+ 2 sec2 x tan x > 0

min. at x = cos x y take max. value at some x0 where x0 = cos x

394 - Rajeev Gandhi Nagar Kota, Ph. No. 0744-2209671, 93141-87482, 93527-21564 IVRS No. 0744-2439051, 0744-2439052, 0744-2439053, www.motioniitjee.com, [email protected]

MAXIMA & MINIMA

Page # 10 Sol.7

A,C 0x1

f(X) = – 1 x 2 ,

x>1

–x

d 2y

at t = 3/2 t = –1

dx

2

>0

ymax = 14

–1

f (x)

t = 3/2 ymin = – 1

–1

69 4

f(x)

Sol.9

A,C f(x) = x2/3

Sol.8

B,D x = (t) = t 5 – 5t3 – 20t + 7

f’(x) =

y = (t) = 4t3 + 4t2 – 18t + 3

2 3.x 1 / 3

f’(x) changes its sign near about origin and

 = ’(t) = 5t4 – 15t2 – 20 x

at origin does not exist

= 5 (t2 – 4) (t2 + 1)   0 when –2 < t < 2 x

Sol.10 A,D

y = ’(t) = 12t2 + 8t – 18 = 6 (2t2 – t – 3) = 6 (2t – 3) (t + 1) y = 0  t = 3/2, t = –1 –2 < t < 2

satisfied

y dy =  =0 x dx

(x  0)

Sol.11 A,C

Sol.12 A,C

d  y  dt d  y  = =    . dt  x  dx dx  x  dx 2

a+ b=9 

dx

=

a

1  r2 h v= 3

x y  y x . 1 ( x ) ( x ) 2

b

= 2

x=a

check.

d2 y

d2 y

–2

x=a

In all three definition separately differentiate &

y = 0  t = –1, 3/2

=

2

x y 1 × x x 2

1 1  b2 (9 – b)  b2 a = 3 3

dv =0 b=6a=3 db

d2 y

y 6 (4 t  1) = 2 = 2  x x 2 dx

=

a2  b2 = 3 5

Surface area =  r  at t = –1

d2 y

0 min. y=

x = – f ’’(–) =  > 0 min.

2 ax  2bx  c

Ax2  2Bx  C

x = –2 f ’’(–2) = – 2 < 0 max. Cross multiply ax2 + 2bx + c – y (A x2 + 2Bx + C) = 0

Sol.14 A,B,D

D  0 x  R x 1 f(x) = 2 x 1

Let y  [, ] Equality holds when D = 0  ,  are the extremum values

(x2 + 1) f ’(x) + 2x f(x) = 1

f ’(x) =

2 3x  2x  1 2

2

( x  1)

Sol.18 A,C

0

From options we can check by putting the value in function instead of differentiatig the function.

x 3  3x 2  3x  1

f ’’(x) =

( x 2  1)4

=0

x3 + 3x2 – 3x – 1 = 0

Sol.19 A,C,D f(x) = log (x – 2) –

1 x

(x – 1) (x2 + 4x + 1) = 0 x = 1, (x2 + 4x+1)=0 x = –2 ± 3

f ’(x) =

2 1 x  x 2 1 =0 + 2 = 2 x ( x  2) x 1 x

x2 + x – 2 = 0

Sol.15 B,C

x = 1, –2

odd cubic polynomial

–

+

f(x) = ax3 + cx

+

–2

f’(x) = 3ax2 + c Two points possible

2

 (2, )

only when a, c have opposite sign.

1

f ’’(x) = –

or

1

( x  2 )2

–

2 x3...