JEE Advanced Model Paper 10 with solutions for online practice

lim

x→−1

f(x)

(x+1)3

= 1
⇒ f(1) = f'(-1) = f''(-1) = 0 and f'''(-1) = 6
∴ f(x) = A (x+1)4 + (x+1)3
∵ f''(0) = 0 ⇒ A = −

1

4

∴ f(x) = (x+1)3 -

(x+1)4

4

∴ f'(x) = (x+1)2 (2 - x)
⇒ f'(x) = 0
⇒ x = -1 , 2
⇒ f''(2) = -9
∴ The maximum value of f(x) = f(2) = 27 -

81

4

=

27

4

∵ f(0) = 1 -

1

4

=

3

4

, f'(0) = 2
Subnormal at x = 0 is f(0) , f'(0) =

3

4

× 2 =

3

2