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

The correct option is A

$1$

Explanation for the correct option.

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

$\begin{array}{rcl}f\left(x\right)& =& {\sin }^{2}x+{\sin }^2\left(x+\frac{\mathrm{\pi }}{3}\right)+\cos x·\cos \left(x+\frac{\mathrm{\pi }}{3}\right)\\ & =& \frac{1}{2}\left[2{\sin }^{2}x+2{\sin }^2\left(x+\frac{\mathrm{\pi }}{3}\right)+2·\cos x·\cos \left(x+\frac{\mathrm{\pi }}{3}\right)\right]\\ & =& \frac{1}{2}\left[1-\cos 2x+1-\cos \left(2x+\frac{2\mathrm{\pi }}{3}\right)+\cos \left(2x+\frac{\mathrm{\pi }}{3}\right)+\cos \frac{\mathrm{\pi }}{3}\right]\mathbf{\left[}\mathbf{\because }\mathbf{2}\mathbf{·}\mathbf{cos}\mathbf{a}\mathbf{·}\mathbf{cos}\mathbf{b}\mathbf{=}\mathbf{cos}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{+}\mathbf{cos}\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}\mathbf{}\mathbf{\&}\mathbf{\&}\mathbf{}\mathbf{2}{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{-}\mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{\right]}\\ & =& \frac{1}{2}\left[\frac{5}{2}-\cos 2x-\cos \left(2x+\frac{2\mathrm{\pi }}{3}\right)+\cos \left(2x+\frac{\mathrm{\pi }}{3}\right)\right]\\ & =& \frac{1}{2}\left[\frac{5}{2}-2\cos \left(2x+\frac{\mathrm{\pi }}{3}\right)\cos \frac{\mathrm{\pi }}{3}+\cos \left(2x+\frac{\mathrm{\pi }}{3}\right)\right]\\ & =& \frac{1}{2}\left[\frac{5}{2}-\cos \left(2x+\frac{\mathrm{\pi }}{3}\right)+\cos \left(2x+\frac{\mathrm{\pi }}{3}\right)\right]\\ & =& \frac{5}{4}\end{array}$

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

Now, find $gof\left(x\right)$.

$\begin{array}{rcl}gof\left(x\right)& =& g\left(f\right(x\left)\right)\\ & =& g\left(\frac{5}{4}\right)\\ & =& 1\mathbf{}\mathbf{[}\mathbf{Given}\mathbf{:}\mathbf{}\mathbf{g}\left(\frac{5}{4}\right)\mathbf{=}\mathbf{1}\mathbf{]}\end{array}$

Hence, the correct option is A.