A twice differentiable function f(x)is defined for all real numbers and satisfies the following conditions `f(0) = 2; f'(0)--5 and f"(0) = 3`. The function `g(x)` is defined by `g(x) = e^(ax) + f (x) AA x in R`, where 'a' is any constant If `g'(0) + g"(0)=0`. Find the value(s) of 'a'

Options

A. 1
B. `-1`
C. 2
D. `-2`

Answer

A, D

Explanation

`g(x)=e^(ax)+f(x)`
`rArr" "f'(x)=ae^(ax)+f'(x)`
`therefore" "f'(0)=a+f'(0)=a-5`
Now, `g''(x)=a^(2)e^(ax)+f''(x)`
`rArr" "g''(0)=a^(2)+f'(0)=3+a^(2)`
We have `g'(0)+g''(0)=0`
`rArr" "a-5+a^(2)+3=0`
`rArr" "a^(2)+a-2=0`
`rArr" "(a+2)(a-1)=0`
`rArr" "a=1,-2`