If the function $f\left( x \right)={{\sin }^{2}}x+{{\sin }^{2}}\left( x+\dfrac{\pi }{3} \right)+\cos x\cos \left( x+\dfrac{\pi }{3} \right)$ and $g\left( x \right)$ is a one-one function defined in $R\to R$, then $\left( gof \right)\left( x \right)$ is a
A. One-one
B. Onto
C. Constant function
D. Periodic with fundamental period $\pi $

Answer Verified Verified
Hint: To solve this problem, we should know the formulae related to the trigonometric ratios. We know that for the given angles A and B, $\cos A\cos B=\dfrac{\cos \left( A+B \right)+\cos \left( A-B \right)}{2}$ and ${{\sin }^{2}}A=\dfrac{1-\cos 2A}{2}$. Using these two formulae, we rewrite the function $f\left( x \right)$ and we get a relation that $f\left( x \right)=c$ which is a constant function. We can infer from this that $\left( gof \right)\left( x \right)$ is a constant function because $g\left( x \right)$ is a one-one function.

Complete step by step answer:
Let us consider $f\left( x \right)={{\sin }^{2}}x+{{\sin }^{2}}\left( x+\dfrac{\pi }{3} \right)+\cos x\cos \left( x+\dfrac{\pi }{3} \right)$ which is given in the question. We know the formulae
We know that for the given angles A and B,
$\cos A\cos B=\dfrac{\cos \left( A+B \right)+\cos \left( A-B \right)}{2}\to \left( 1 \right)$
${{\sin }^{2}}A=\dfrac{1-\cos 2A}{2}\to \left( 2 \right)$.
Using equation-1 , we can write $\cos x\cos \left( x+\dfrac{\pi }{3} \right)$ as
A = $x$, B = $x+\dfrac{\pi }{3}$
  & \cos x\cos \left( x+\dfrac{\pi }{3} \right)=\dfrac{\cos \left( x+x+\dfrac{\pi }{3} \right)+\cos \left( x-x-\dfrac{\pi }{3} \right)}{2} \\
 & \cos x\cos \left( x+\dfrac{\pi }{3} \right)=\dfrac{\cos \left( 2x+\dfrac{\pi }{3} \right)+\cos \left( -\dfrac{\pi }{3} \right)}{2} \\
We know that $\cos \left( -\theta \right)=\cos \theta $, the above equation can be written as
  & \cos x\cos \left( x+\dfrac{\pi }{3} \right)=\dfrac{\cos \left( 2x+\dfrac{\pi }{3} \right)+\cos \left( \dfrac{\pi }{3} \right)}{2} \\
 & \cos x\cos \left( x+\dfrac{\pi }{3} \right)=\dfrac{\cos \left( 2x+\dfrac{\pi }{3} \right)}{2}+\dfrac{1}{4}\to \left( 3 \right) \\
Using equation-2, we can write ${{\sin }^{2}}x$and ${{\sin }^{2}}\left( x+\dfrac{\pi }{3} \right)$ as
  & {{\sin }^{2}}x=\dfrac{1-\cos 2x}{2} \\
 & {{\sin }^{2}}\left( x+\dfrac{\pi }{3} \right)=\dfrac{1-\cos 2\left( x+\dfrac{\pi }{3} \right)}{2} \\
Using these relations, f(x) becomes
  & f\left( x \right)=\dfrac{1-\cos 2x}{2}+\dfrac{1-\cos \left( 2x+\dfrac{2\pi }{3} \right)}{2}+\dfrac{\cos \left( 2x+\dfrac{\pi }{3} \right)}{2}+\dfrac{1}{4} \\
 & f\left( x \right)=\dfrac{5}{4}+\dfrac{\cos \left( 2x+\dfrac{\pi }{3} \right)}{2}-\left( \dfrac{\cos 2x+\cos \left( 2x+\dfrac{2\pi }{3} \right)}{2} \right) \\
We know that $\cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)$
Let us consider $\cos 2x+\cos \left( 2x+\dfrac{2\pi }{3} \right)$, we get
  & \cos 2x+\cos \left( 2x+\dfrac{2\pi }{3} \right)=2\cos \left( \dfrac{2x+2x+\dfrac{2\pi }{3}}{2} \right)\cos \left( \dfrac{2x-2x-\dfrac{2\pi }{3}}{2} \right) \\
 & \cos 2x+\cos \left( 2x+\dfrac{2\pi }{3} \right)=2\cos \left( 2x+\dfrac{\pi }{3} \right)\cos \left( \dfrac{\pi }{3} \right)=2\cos \left( 2x+\dfrac{\pi }{3} \right)\times \dfrac{1}{2}=\cos \left( 2x+\dfrac{\pi }{3} \right) \\
 & \cos 2x+\cos \left( 2x+\dfrac{2\pi }{3} \right)=\cos \left( 2x+\dfrac{\pi }{3} \right) \\
Using this result in f(x), we get
  & f\left( x \right)=\dfrac{5}{4}+\dfrac{\cos \left( 2x+\dfrac{\pi }{3} \right)}{2}-\left( \dfrac{\cos 2x+\cos \left( 2x+\dfrac{2\pi }{3} \right)}{2} \right)=\dfrac{5}{4}+\dfrac{\cos \left( 2x+\dfrac{\pi }{3} \right)}{2}-\dfrac{\cos \left( 2x+\dfrac{\pi }{3} \right)}{2}=\dfrac{5}{4} \\
 & f\left( x \right)=\dfrac{5}{4} \\
We can infer that the function f(x) is a constant function.
In the question it is given that g(x) is a one-one function which means that for a given value of x, there is only one functional value corresponding to the function g(x).
In the question, we are asked about the nature of $gof(x)$, which means that we are asked about $g\left( f\left( x \right) \right)$.
We know that $f\left( x \right)=\dfrac{5}{4}$, we can write $g\left( f\left( x \right) \right)$ as
$g\left( f\left( x \right) \right)=g\left( \dfrac{5}{4} \right)$.
As g(x) is a one-one function, for the value of $\dfrac{5}{4}$, we get a unique functional value which is also a constant. So, we can infer that $g\left( f\left( x \right) \right)$ is a constant function.
$\therefore $ $g\left( f\left( x \right) \right)$ is a constant function.

So, the correct answer is “Option C”.

Note: Students can make a mistake by thinking that gof(x) is also a one-one function because the function g(x) is a one-one function without doing any calculation. The value of gof(x) also depends on the nature of f(x) and not just g(x). This question is an example of this scenario in which f(x) became a constant function and the function gof(x) became a constant function.
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