Question

An exhaustive set of values for $x$ for which $g\left(f\right(x\left)\right)$ is invertible where $f\left(x\right)=\cos x$ and $g\left(x\right)=2x^2-1$ is

• A. $[0,\pi ]$
• B. $[0,\frac{\pi }{2}]$
• C. $[-\frac{\pi }{4},\frac{\pi }{4}]$
• D. $R$

Answer 1

The correct option is B $[0,\frac{\pi }{2}]$

Given $f\left(x\right)=\cos x,g\left(x\right)=2x^2-1$
$g\left(f\right(x\left)\right)=2\left(f\right(x){)}^2-1=2{\cos }^{2}x-1=\cos 2x$

$g\left(f\right(x\left)\right)$ is invertible iff $\cos 2x$ is bijective
i.e., $g\left(f\right(x\left)\right)$ is one-one and onto.
If $\cos 2x$ is one-one, then exhaustive set of values for $2x$ is $[n\pi ,(n+1\left)\pi \right]$ where $n\in Z$
$\Rightarrow x\in [\frac{n\pi }{2},\frac{(n+1)\pi }{2}]$
For $n=0,x\in [0,\frac{\pi }{2}]$