Question

The number of values of $x$ where the function $f\left(x\right)=\cos \left(x\right)+\cos \left(\sqrt{2x}\right)$ attains its maximum is

• A. $0$
• B. $1$
• C. $2$
• D. Infinite

Answer 1

The correct option is B

$1$

Explanation for correct option

The maximum value of $f\left(x\right)=\cos \left(x\right)+\cos \left(\sqrt{2x}\right)$

We know that the range of $\cos \left(x\right)$ is $\left[-1,1\right]$

$\cos \left(x\right)$ attains its maximum at $0,2\mathrm{\pi },4\mathrm{\pi },.....$

That is $x=\left(2n-2\right)\pi$

Clearly at $x=0$

$$\begin{gathered} f\left(x\right)=\cos \left(x\right)+\cos \left(\sqrt{2x}\right) \\ =\cos \left(0\right)+\cos \left(0\right) \\ =2\left[\because \cos \left(0\right)=1\right] \end{gathered}$$

Since $\pi$ is irrational number so for any other value of $x$, $\sqrt{2x}$ cannot become $\left(2n-2\right)\pi$

So $\cos \left(\sqrt{2x}\right)$ cannot attain its maximum at any point other than $x=0$

So $2$ which occurs at $x=0$ .

Also there is no other value of $x$ for which it will attain maximum again

Hence, option (B) is correct