Question

Consider the real-valued function $f$ satisfying $2f\left(\sin x\right)+f\left(\cos x\right)=x$. Then,

• A. Domain of $f$ is $R$
• B. Domain of $f$ is $[-1,1]$
• C. Range of $f$ is $[-\frac{2\pi }{3},\frac{\pi }{3}]$
• D. Range of $f$ is $R$

Answer 1

Answer: (B), (C)

Range of $f$ is $[-\frac{2\pi }{3},\frac{\pi }{3}]$

Given $2f\left(\sin x\right)+f\left(\cos x\right)=x\dots \left[1\right]$
Replacing $x$ by $\frac{\pi }{2}-x$, we get
$2f\left(\cos x\right)+f\left(\sin x\right)=\frac{\pi }{2}-x‘\dots \left[2\right]$
Eliminating $f\left(\cos x\right)$ from $\left[1\right]$ and $\left[2\right]$, we get
$3f\left(\sin x\right)=3x-\frac{\pi }{2}$
or, $f\left(\sin x\right)=x-\frac{\pi }{6}$
or, $f\left(x\right)={\sin }^{-1}x-\frac{\pi }{6}$
$f\left(x\right)$ has the domain $[-1,1]$.
Also, ${\sin }^{-1}x\in [-\frac{\pi }{2},\frac{\pi }{2}]$
or, ${\sin }^{-1}x-\frac{\pi }{6}\in [-\frac{2\pi }{3},\frac{\pi }{3}]$