The maximum and minimum values of `f(x)=secx+logcos^(2)x,0 lt x lt 2pi` are respectively

Options

A. `(1,-1)` and `2(1-log2),2(1+log2)`
B. `(1,-1)` and `{2(1-log2),2(1-log2)}`
C. `(1,-1)` and `(2,-3)`
D. None of the above

Answer

B

Explanation

Given `f(x)=secx+log cos^(2)x`
`impliesf(x)=secx+2log(cosx)`
Therefore `f'(x)=sec x tan x-2 tanx=tanx (secx-2)`
Put `f'(x)=0impliestanx=0` or `secx=2`
`impliestanx=0` or `cosx=1/2`
Therefore, possible values of `x` are `x=0, x=pi` and `x=(pi)/3`
or `x=(5pi)/3`.
Again`f''(x)=sec^(2)x(secx-2)+tanx(secx tan x)`
`=sec^(3)x+sec x tan^(2)x -2 sec^(2) x`
`=secx (sec^(2)x + tan^(2) x-2secx)`
At `x=0, f''(0)=1(1+0-2)=-1lt0`
Therefore `x=0` is a point of maxima.
At `x=pi, f''(pi)=-1(1+0+2)=-3lt0`
Therefore `x=pi` is a point of maxima.
At `x=(pi)/3, f''((pi)/3)=2(4+3-4)=6 lt 0`
Therfore `x=(pi)/3` is a p[oint of minima.
at `x=(5pi)/3, f''((5pi)/3)=2(4+3-4)=6gt0`
Therefore `x=(5pi)/3` is a point of minima.
Maximum value of `y` at `x=0` is `1+0=1`.
Maximum value of `y` at `x=pi` is `-1+0=-1`
Minimum value of `y` at `x=(pi)/3` is `2+2"log"1/2=2(1-log2)`
Minimum value of `y` at `x=(5pi)/3` is `2+2 "log"1/2=2(1-log2)`.