Let `f(x)={{:(xe^(ax) "," x le 0 ),(x+ax^2-x^3 "," x gt 0 ):}`,where is a
positive constant .Then the interval in which f' (x ) is increasing is

Options

A. (0,a/3)
B. (-2 /a,0)
C. (-2/a,a/3)
D. non of these

Answer

C

Explanation

We have
`f(x)={{:(xe^(ax) "," x le 0 ),(x+ax^2-x^3 "," x gt 0 ):}`
Clearly f(x) is continuous and differentiable at x=0
`therefore f'(x)={:{((1+ax)e^ax "," x le 0 ),(1+2ax -3x^2 ","x gt 0 ):}`
f(x) is also continous and differentiable at x=0
`therefore f''(x) ={:{((2+ax)ae^(ax)","x ge 0 ),(2a-6x ","x gt0 ):}`
For f(x) to be increasing ,we must have `f'(x) gt 0 ` now two cases arise .
CASE I When `x le 0`
In this case m we have
`f''(x) gt 0 `
` rArr 2a-6x gt 0 rArr x lt a/3`
`therefore x in (0,a//3)`
From (i) and (ii) we find that f(x) is increasing for
`x in (-2 //a, a//3)`