Question

The difference between the greatest and least values of the functions $f\left(x\right)$ = $cosx$ + $\frac{1}{2}$ $cos2x$ - $\frac{1}{3}$ $cos3x$ is

• A. $\frac{4}{3}$
• B. $1$
• C. $\frac{9}{4}$
• D. $\frac{1}{6}$

Answer 1

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

$f\left(x\right)=\cos x+\frac{1}{2}\cos 2x-\frac{1}{3}\cos 3x$

$f^{\prime }\left(x\right)=-\sin x-\sin 2x-\sin 3x=0⟹\sin 2x(2\cos x+1)=0$

So, either $\sin 2x=0⟹x=n\pi /2$

or $\cos x=-1/2⟹x\pm \frac{2\pi }{3}+2n\pi$

Put $x=0$

$f\left(0\right)=1+1/2-1/3=7/6$

put $x=\frac{2\pi }{3}$

$f\left(\frac{2\pi }{3}\right)=-1/2+1/2(-1/2)-1/3=\frac{-13}{12}$

$⟹f\left(0\right)-f\left(\frac{2\pi }{3}\right)=\frac{7}{6}+\frac{13}{12}=\frac{9}{4}$