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

Hint: 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.