Question

The difference between the greatest and least values of the function $f\left(x\right)=\cos x+\frac{1}{2}\cos 2x-\frac{1}{3}\cos 3x$ is :

• A. $\frac{2}{8}$
• B. $\frac{8}{7}$
• C. $\frac{9}{4}$
• D. $\frac{3}{8}$

Answer 1

The correct option is C $\frac{9}{4}$

$f\left(x\right)=\cos x+\frac{1}{2}\cos 2x-\frac{1}{3}\cos 3x$
Given function is periodic, with a period of $2\pi$.
$f^{\prime }\left(x\right)=-(\sin x+\sin 2x-\sin 3x)$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow 4\sin \left(\frac{x}{2}\right).\sin \left(x\right).\sin \left(\frac{3x}{2}\right)=0$
$\Rightarrow x=0,\frac{2\pi }{3},\pi ,2\pi$

$f\left(0\right)=\frac{7}{6}$

$f\left(\frac{2\pi }{3}\right)=-\frac{13}{12}$

$f\left(\pi \right)=-\frac{1}{6}$

$f\left(2\pi \right)=\frac{7}{6}$
Greatest value $=\frac{7}{6}$ and Least Value $=-\frac{13}{12}$
Required Difference $=\frac{9}{4}$
Hence, Option C.