If `f(0) = f'(0) = 0` and `f''(x) = tan^(2)x` then f(x) is

Options

A. `log sec x - (x^(2))/(2)`
B. `log cos x + (x^(2))/(2)`
C. `log sec x + (x^(2))/(2)`
D. None of these

Answer

A

Explanation

Given, `f''(x) = tan^(2)x`
or `f''(x) = sec x -1`
On intergrating, we get
`f''(x) = tan x - x + c`
At `x = 0, f'(0) = 0 = tan 0 - 0 +c`
`rArr c =0`
`therefore f'(x) = tan x -x `
Again integrating, we get
`f(x) = log sec x - (x^(2))/(2) + c`
At `x = 0, f(0) = log sec 0-0 +c`
`rArr c = 0`
`therefore f(x) = log sec x - (x^(2))/(2)`