Question

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

• A. $2$
• B. $1$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$1$

Step 1: Simplify the given function

The given function is $f\left(x\right)={\sin }^2\left(x\right)+{\sin }^2\left(x+\frac{\mathrm{\pi }}{3}\right)+\cos \left(x\right)·\cos \left(x+\frac{\mathrm{\pi }}{3}\right)$ and $g\left(\frac{5}{4}\right)=1$.

It is known that $\mathbf{sin}\left(a+b\right)\mathbf{=}\mathbf{sin}\left(a\right)\mathbf{cos}\left(b\right)\mathbf{+}\mathbf{sin}\left(b\right)\mathbf{cos}\left(a\right)$ and $\mathbf{cos}\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}\mathbf{=}\mathbf{cos}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{-}\mathbf{sin}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\mathbf{b}\mathbf{\right)}$.

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

Step 2: Determine the value of $\left(g\circ f\right)\left(x\right)$

It is also given that, $g\left(\frac{5}{4}\right)=1$.

So, $\left(g\circ f\right)\left(x\right)=g\left(f\left(x\right)\right)$

$$\begin{gathered} \Rightarrow \left(g\circ f\right)\left(x\right)=g\left(\frac{5}{4}\right) \\ \Rightarrow \left(g\circ f\right)\left(x\right)=1 \end{gathered}$$

Therefore, $\left(g\circ f\right)\left(x\right)$ is equal to $1$.

Hence option(B) is the correct option.