Question

If $2f\left(\sin x\right)+f\left(\cos x\right)=x\forall x\in R$ then range of f(x) is

• A. $\left[\frac{-\pi }{6},\frac{\pi }{3}\right]$
• B. $\left[\frac{-2\pi }{3},\frac{\pi }{3}\right]$
• C. $\left[\frac{-2\pi }{3},\frac{\pi }{6}\right]$
• D. $\left[\frac{-\pi }{3},\frac{\pi }{6}\right]$

Answer 1

The correct option is A $\left[\frac{-\pi }{6},\frac{\pi }{3}\right]$

$2f\left(\sin x\right)+f\left(\cos x\right)=x\forall x\in R$

Let $y=f\left(x\right),g\left(x\right)=\sin x,h\left(x\right)=\cos x$

Range of $\sin x=[-1,1]$ and range of \cos x=[-1,1]\Rightarrow x\in[-1,1]$at$x=\dfrac{\pi}{2}$,

$2f(-1)+f\left(0\right)=\frac{\pi }{2}$

at $x=0$, $2f\left(0\right)+f\left(1\right)=0;\Rightarrow f\left(1\right)=\frac{+\pi }{3}$ and $f\left(0\right)=\frac{-\pi }{6}$

at $x=\frac{-\pi }{2}$,

$2f(-1)+f\left(1\right)=\frac{-\pi }{2}$

$\Rightarrow f(-1)=\frac{-5\pi }{12}<\frac{-\pi }{6}$

So , $f\left(x\right)<f\left(x\right)<f\left(1\right)$

$\Rightarrow f\left(x\right)\in [\frac{-\pi }{6},\frac{\pi }{3}]$