Question

Find the range of `f(x)=sec(pi/4cos^2x),` where `-oo<x<oo`

Answer 1

Correct Answer - `[1,sqrt(2)]`
`f(x)="sec"((pi)/(4)"cos"^(2)x)`
We know that `0 le cos^(x) le 1,` i.e.,
`0 le (pi)/(4)cos^(2)x le (pi)/(4)`
For the above value of `theta =(pi)/(pi)/(4)cos^(2)x,sec x` is an increasing function.
` :. f(x) in [f(0),f((pi)/(4))] " or " f(x) in [1,sqrt(2)]`