For ` a in [pi , 2 pi]` and `n in Z` the critical points of g
`f(x) = 1/3 sin a tan ^3 "" x + (sin a-1)tan x + sqrt(a-2)/(8-a)` are

Options

A. `x = n pi`
B. `x=2n pi`
C. `x= (2n +1)pi`
D. none of these

Answer

D

Explanation

We have
`f(x)=sin a+ tan^2 x sec^2 x + (sin a-1)sec^2`
`rArr f(x)=sin a tan ^2 x +sin a-1)sec^2 x`
At critical points ,we must have f(x)=0
`rArr sina tan^2 x+ sin a-1 =0 " " [thereforesec^2xne0 " for any "x in R]`
`tan^2x=(1-sina)/sina`
Now
` a in [pi,2 pi]`
`rArr (1-sina)/sinalt0`
`rArr tan^2x=(1-sina)/sina` does not have solution in R
Hence , f(x) has not critical points