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

Answer 1

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

The given function is periodic, with period $2\pi$. So the difference between the greatest and least values of the function is the difference between these values on the interval $[0,2\pi ]$ . We have
$f^{\prime }\left(x\right)=-(\sin x+\sin 2x-\sin 3x)$
$=-4\sin x\sin \left(3x/2\right)\sin \left(x/2\right)$
Hence $x=0$,$2\pi /3$,$\pi$ and $2\pi$ are the critical points. Also,$f\left(0\right)=1+1/2-1/3=7/6$, $f\left(2\pi /3\right)=-13/12$, $f\left(\pi \right)=-1/6$ and $f\left(2\pi \right)=7/6$. Hence the greatest value is 7/6 and the least value is -13/12.

Thus the difference is
$\frac{7}{6}-(-\frac{13}{12})=\frac{27}{12}=\frac{9}{4}$