Question

The number of values of $x$ where the function $f\left(x\right)=2(\cos 3x+\cos \sqrt{3}x)$ attains its maximum, is-

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

Answer 1

The correct option is A $1$

We know that
$cosA+cosB=2cos\left(\frac{A+B}{2}\right)\times cos\left(\frac{A-B}{2}\right)$
Given $f\left(x\right)=2(\cos 3x+\cos \sqrt{3}x)$ $=2(2\cos \left(\frac{3+\sqrt{3}}{2}\right)x.\cos \left(\frac{3-\sqrt{3}}{2}\right)x)$
Maximum value of sine or cosine function is $1$.So
$f\left(x\right)=4(\cos \left(\frac{3+\sqrt{3}}{2}\right)x.\cos \left(\frac{3-\sqrt{3}}{2}\right)x)\le 4$
It is equal to 4 when both
$\cos \left(\frac{3+\sqrt{3}}{2}\right)x$ and $\cos \left(\frac{3-\sqrt{3}}{2}\right)x$are equal to$1$
This is possible only when $x=0$