Question

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

Answer 1

To get the value of $gof\left( x \right)$ we need to simplify the function $f\left( x \right)$ . Then apply the formulas of $\sin \left( {A + B} \right){\text{ = }}\sin A\cos B + \cos A\sin B$ and $\cos \left( {A + B} \right){\text{ = }}\cos A\cos B - \sin A\sin B$ in the given equation of function. Then substitute the values of trigonometric functions when needed and solve it further. After getting the value of $f\left( x \right)$ , find the value of $gof\left( x \right)$ by using the given value of $g\left( {\dfrac{5}{4}} \right)$ .

Complete step-by-step solution:
First we have to simplify the function $f\left( x \right)$ . So, it is given that
$f\left( x \right) = {\sin ^2}x{\text{ }} + {\text{ }}{\sin ^2}\left( {x + \dfrac{\pi }{3}} \right){\text{ }} + {\text{ }}\cos x\cos \left( {x + \dfrac{\pi }{3}} \right)$ ------ $(i)$
By applying the formula of $\sin \left( {A + B} \right){\text{ = }}\sin A\cos B + \cos A\sin B$ in the equation $(i)$ we get
$f\left( x \right) = {\sin ^2}x{\text{ }} + {\text{ }}{\left[ {\sin x\cos \dfrac{\pi }{3} + \cos x\sin \dfrac{\pi }{3}} \right]^2}{\text{ }} + {\text{ }}\cos x\cos \left( {x + \dfrac{\pi }{3}} \right)$
Value of $\cos \dfrac{\pi }{3}{\text{ = }}\dfrac{1}{2}$ and the value of $\sin \dfrac{\pi }{3}{\text{ = }}\dfrac{{\sqrt 3 }}{2}$ . Now by substituting these values in the above equation we get
$f\left( x \right) = {\sin ^2}x{\text{ }} + {\text{ }}{\left[ {\dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x} \right]^2}{\text{ }} + {\text{ }}\cos x\cos \left( {x + \dfrac{\pi }{3}} \right)$ -------- $(ii)$
Again by applying the formula, $\cos \left( {A + B} \right){\text{ = }}\cos A\cos B - \sin A\sin B$ in the equation $(ii)$ we get
$f\left( x \right) = {\sin ^2}x{\text{ }} + {\text{ }}{\left[ {\dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x} \right]^2}{\text{ }} + {\text{ }}\cos x\left[ {\cos x\cos \dfrac{\pi }{3} - \sin x\sin \dfrac{\pi }{3}} \right]$
Value of $\cos \dfrac{\pi }{3}{\text{ = }}\dfrac{1}{2}$ and the value of $\sin \dfrac{\pi }{3}{\text{ = }}\dfrac{{\sqrt 3 }}{2}$ . Now by substituting these values in the above equation we get
$f\left( x \right) = {\sin ^2}x{\text{ }} + {\text{ }}{\left[ {\dfrac{1}{2}\sin x + \dfrac{{\sqrt 3 }}{2}\cos x} \right]^2}{\text{ }} + {\text{ }}\cos x\left[ {\dfrac{1}{2}\cos x - \dfrac{{\sqrt 3 }}{2}\sin x} \right]$
By taking L.C.M inside the brackets we have
$f\left( x \right) = {\sin ^2}x{\text{ }} + {\text{ }}{\left[ {\dfrac{{\sin x + \sqrt 3 \cos x}}{2}} \right]^2}{\text{ }} + {\text{ }}\cos x\left[ {\dfrac{{\cos x - \sqrt 3 \sin x}}{2}} \right]$
Take $\dfrac{1}{4}$ and $\dfrac{1}{2}$ common from the brackets,
$f\left( x \right) = {\sin ^2}x{\text{ }} + {\text{ }}\dfrac{1}{4}{\left( {\sin x + \sqrt 3 \cos x} \right)^2}{\text{ }} + {\text{ }}\dfrac{1}{2}\left( {{{\cos }^2}x - \sqrt 3 \sin x\cos x} \right)$ ------ $(iii)$
By applying the formula, ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ in the equation $(iii)$ we get
$f\left( x \right) = {\sin ^2}x{\text{ }} + {\text{ }}\dfrac{1}{4}\left( {{{\sin }^2}x + 3{{\cos }^2}x + 2\sqrt 3 \sin x\cos x} \right){\text{ }} + {\text{ }}\dfrac{1}{2}\left( {{{\cos }^2}x - \sqrt 3 \sin x\cos x} \right)$
Now multiply $\dfrac{1}{4}$ and $\dfrac{1}{2}$ with the terms inside the bracket.
$f\left( x \right) = {\sin ^2}x{\text{ }} + {\text{ }}\dfrac{1}{4}{\sin ^2}x + \dfrac{3}{4}{\cos ^2}x + \dfrac{{\sqrt 3 }}{2}\sin x\cos x{\text{ }} + {\text{ }}\dfrac{1}{2}{\cos ^2}x - \dfrac{{\sqrt 3 }}{2}\sin x\cos x$
By adding ${\sin ^2}x$ terms and ${\cos ^2}x$ terms we get
$f\left( x \right) = \dfrac{{4{{\sin }^2}x + {{\sin }^2}x}}{4} + \dfrac{{3{{\cos }^2}x + 2{{\cos }^2}x}}{4}$
$f\left( x \right) = \dfrac{{5{{\sin }^2}x}}{4} + \dfrac{{5{{\cos }^2}x}}{4}$
Taking $\dfrac{5}{4}$ common we get
$f\left( x \right) = \dfrac{5}{4}\left( {{{\sin }^2}x + {{\cos }^2}x} \right)$
And we all know that ${\sin ^2}x + {\cos ^2}x = 1$ . Therefore,
$f\left( x \right) = \dfrac{5}{4}\left( 1 \right)$
$\Rightarrow f\left( x \right) = \dfrac{5}{4}$
According to the question we have to find the value of $gof\left( x \right)$ .
And here the required value of $f\left( x \right)$ is $\dfrac{5}{4}$ . So,
$gof\left( x \right){\text{ }} = {\text{ }}g\left( {f\left( x \right)} \right){\text{ = }}g\left( {\dfrac{5}{4}} \right)$
And it is given in the question that $g\left( {\dfrac{5}{4}} \right){\text{ = }}1$ . Therefore, the value of $gof\left( x \right){\text{ }} = {\text{ 1}}$
Hence, the correct option is $\left( 1 \right){\text{ }}1$

Note: $gof\left( x \right)$ means $f\left( x \right)$ function is in $g\left( x \right)$ function. The notation $gof$ is read as “ g of f ”. $gof\left( x \right)$ is a composite function. To solve $gof\left( x \right)$ always solve $f\left( x \right)$ first. $gof$ is formed by the composition of g and f.