Question

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

• A. domain of f(x) is R
• B. domain of f(x) is[-1, 1]
• C. range of f(x) is $\left(-\frac{2\pi }{3},\frac{\pi }{3}\right]$
• D. range of f(x) is R

Answer 1

Answer: (B), (C)

range of f(x) is $\left(-\frac{2\pi }{3},\frac{\pi }{3}\right]$


$\mathrm{Given}2f\left(\sin x\right)+f\left(\cos x\right)=x.....\left[1\right]\mathrm{Replacing}x\mathrm{by}\frac{\pi }{2}-x,\mathrm{we}\mathrm{get},2f\left(\cos x\right)+f\left(\sin x\right)=\frac{\pi }{2}-x.....\left[2\right]\mathrm{Solving}\left[1\right]\mathrm{and}\left[2\right]\mathrm{by}\mathrm{elimination}\mathrm{method},\mathrm{we}\mathrm{get},3f\left(\sin x\right)=3x-\frac{\pi }{2}\mathrm{or},f\left(\sin x\right)=x-\frac{\pi }{6}\mathrm{or},f\left(x\right)={\sin }^{-1}x-\frac{\pi }{6}\mathrm{Clearly},f\left(x\right)\mathrm{has}\mathrm{domain}[-1,1]\mathrm{Also},{\sin }^{-1}x\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right]\mathrm{or},{\sin }^{-1}x-\frac{\pi }{6}\in \left(-\frac{2\pi }{3},\frac{\pi }{3}\right]\Rightarrow \mathrm{Range}\mathrm{of}f\mathrm{is}\left(-\frac{2\pi }{3},\frac{\pi }{3}\right]$