Question

If $f\left(x\right)={\sin }^{2}x+{\sin }^2(x+\frac{\pi }{3})+\cos x\cos (x+\frac{\pi }{3})$ and $g\left(x\right)$ is a one-one function defined in $R\to R$, then $\left(gof\right)\left(x\right)$ is

• A. One-one
• B. Onto
• C. Constant function
• D. Periodic with fundamental period $\pi$

Answer 1

The correct option is C Constant function

We have,

$f\left(x\right)={\sin }^{2}x+{\sin }^2(x+\frac{\pi }{3})+\cos x\cos (x+\frac{\pi }{3})$

$\Rightarrow f\left(x\right)=\frac{1}{2}[2{\sin }^{2}x+2{\sin }^2(x+\frac{\pi }{3})+2\cos x\cos (x+\frac{\pi }{3})]$

$\Rightarrow f\left(x\right)=\frac{1}{2}[1-\cos 2x+1-\cos (2x+\frac{2\pi }{3})+\cos (2x+\frac{2\pi }{3})+\cos \frac{\pi }{3}]$

$\Rightarrow f\left(x\right)=\frac{1}{2}[\frac{5}{2}-\cos 2x-\cos (2x+\frac{\pi }{3})+\cos (2x+\frac{\pi }{3})]$

$\Rightarrow f\left(x\right)=\frac{1}{2}[\frac{5}{2}-2\cos (2x+\frac{\pi }{3})\cos \frac{\pi }{3}+\cos (2x+\frac{\pi }{3})]$

$\Rightarrow f\left(x\right)=\frac{1}{2}[\frac{5}{2}-\cos (2x+\frac{\pi }{3})+\cos (2x+\frac{\pi }{3})]$

$\Rightarrow f\left(x\right)=\frac{5}{4}$ for all $x\in R$

$\therefore ,$ for any $x\in R,$ we have

$g.f\left(x\right)g\left(f\left(x\right)\right)=g\left(\frac{5}{4}\right)=1$

Thus, $g.f\left(x\right)=1$ for all $x\in R$

Hence, $g.f:R\to R$ is a constant function.