Let f : R (0, ) be a twice differentiable function such that f(3) = 18, f'(3) = 0 and f''(3) = 4. Then x 1 e ( f(2 + x)f(3) ) 18(x - 1)2 is equal to :
- 9
- 18
- 1
- 2
Answer
(4) 2
Solution
e mn = n e m
L = x 1 18(x - 1)2 e f(2 + x)f(3)
Put x - 1 = h, h 0:
L = h 0 18 [ e f(3 + h) - e f(3) ]h2 ( 00 )
By L'Hôpital's rule:
L = h 0 18 f'(3 + h)f(3 + h)2h = h 0 9f'(3 + h)h f(3 + h) ( 00 ), since f'(3) = 0
Again by L'Hôpital's rule:
L = h 0 9f''(3 + h)f(3 + h) + h f'(3 + h)
= 9f''(3)f(3) = 9 418 = 2
Answer: (4)