Question

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

• A. $2$
• B. $1$
• C. $5$
• D. None of these

Answer 1

The correct option is B $1$

We have, $f\left(x\right)=\cos x+\cos \left(\sqrt{2}x\right)$

$\Rightarrow \left|f\left(x\right)\right|=|\cos x+\cos \sqrt{2x}|\le \left|\cos x\right|+\left|\cos \sqrt{2}x\right|$

$=1+1=2,\forall x\in R$

$\therefore$ Maximum value of $f\left(x\right)=2.$

This requires $\cos x=1$ and $\cos \sqrt{2}x=1$

$\Rightarrow x=2n\pi \sqrt{2}x=2m\pi ,n,m\in I.$

$\Rightarrow 2n\pi =\frac{2m\pi }{\sqrt{2}}\Rightarrow n=\frac{m}{\sqrt{2}}.$

This is possible only when $n=m=0.$

$\therefore$ There is only one value $(x=0)$ at which $f\left(x\right)$ attains its maximum value .