Question

If $f\left(x\right)={\sin }^{2}x+{\sin }^2(x+\frac{\pi }{3})+\cos x\cos (x+\frac{\pi }{3})$ and $g\left(\frac{5}{4}\right)=1$, then find $\left(gof\right)\left(x\right).$

Answer 1

We have,

$f\left(x\right)={\sin }^{2}x+{\sin }^2(x+\frac{\pi }{3})+\cos x\cos (x+\frac{\pi }{3})$ and $g\left(\frac{5}{4}\right)=1$

The composite function $g\left(f\right(x\left)\right)$ is defined if and only if Range of $f\left(x\right)$ $\cap$ Domain of $g\left(x\right)$ is not an empty set, $\varphi$.

Domain of g(x) is $\frac{5}{4}$, as given in the question.

Now, lets find the range of the $f\left(x\right)$ by simplifying the function.

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

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

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

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

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

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

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

$\Rightarrow f\left(x\right)=\frac{1}{2}\times \frac{5}{2}$

$\Rightarrow f\left(x\right)=\frac{5}{4}$

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

Hence, range fo $f\left(x\right)$ is $\frac{5}{4}$.

We can see that Range of $f\left(x\right)$ $\cap$ Domain of $g\left(x\right)$ is $\frac{5}{4}$ and not an empty set or $\varphi$. So, the composite function $g\left(f\right(x\left)\right)$ exists.

Therefore, $gof\left(x\right)=g\left(f\right(x\left)\right)=g\left(\frac{5}{4}\right)=1$ for all $x\in R$ and it is a constant function.