Question

If $2f\left(sinx\right)+f\left(cosx\right)=x\forall xϵ$ R then range of f(x) is

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

Answer 1

The correct option is B $[\frac{-2\pi }{3},\frac{\pi }{3}]$

Put $x=si{n}^{-1}x$
$2f\left(x\right)+f\left(\sqrt{1-x^2}\right)=si{n}^{-1}x\to \left(1\right)$
On Putting $x=co{s}^{-1}x$
$\Rightarrow 2f\left(\sqrt{1-x^2}\right)+f\left(x\right)=co{s}^{-1}x\to \left(2\right)$
Eq.$\left(1\right)\times 2\Rightarrow 4f\left(x\right)+2f\left(\sqrt{1-x^2}\right)=2si{n}^{-1}x\to \left(3\right)$
On subtracting Eq. 2 from Eq. 3 we get -
$3f\left(x\right)=2si{n}^{-1}x-co{s}^{-1}x$
$f\left(x\right)=\frac{2}{3}si{n}^{-1}x-\frac{1}{3}(\frac{\pi }{2}-si{n}^{-1}x)$
$=si{n}^{-1}x-\frac{\pi }{6}$
$f_{max}=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3},f_{min}=-\frac{\pi }{2}-\frac{\pi }{6}=\frac{-4\pi }{6}=\frac{-2\pi }{3}$
$=[\frac{-2\pi }{3},\frac{\pi }{3}]$