If `f''(x) gt forall in R, f(3)=0 and g(x) =f(tan^(2)x-2tanx+4y)0ltxlt(pi)/(2)`,then g(x) is increasing in

Options

A. `(0,(pi)/(4))`
B. `((pi)/(6),(pi)/(3))`
C. `(0,(pi)/(3))`
D. `((pi)/(4),(pi)/(2))`

Answer

4

Explanation

`g(x) =f((tanx-1)^(2)+3) 2 (tan x-1) sec^(2)x`
Since `f''(x) gt0 f(x)` is increasing .So
`therefore f(tanx-1)^(2)+3gtf(3)=0 forall x in (0,(pi)/(4))cup(pi)/(4),(pi)/(2)`
also `(tan x-1)gt0x in ((pi)/(4),(pi)/(2))`
so g(x) is increasing `in ((pi)/(4),(pi)/(2))`