Question

The difference between greatest and least value of
$f\left(x\right)=\cos x+\frac{1}{2}\cos 2x+\frac{1}{3}\cos 3x$, is

• A. $\frac{2}{3}$
• B. $\frac{8}{7}$
• C. $\frac{9}{4}$
• D. None of the above.

Answer 1

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

$f\left(x\right)=\cos x+\frac{1}{2}\cos 2x+\frac{1}{3}\cos 3x$
Differentiate w.r.t 'x'

$\Rightarrow f^1\left(x\right)=-\sin x-\sin 2x$

$\Rightarrow f^1\left(x\right)=\sin 3x-(\sin x+\sin 2x)$

For max|min value $f^1\left(x\right)=0$

$\Rightarrow \sin 3x=\sin x+\sin 2x\Rightarrow 3\sin x-4{\sin }^{3}x=\sin x+2\sin x\cos x$

$\Rightarrow 4{\sin }^{3}x-2\sin x+2\sin x\cos x=0\Rightarrow \sin x\left(4{\sin }^{2}x-2+2\cos x\right)=0$

$\Rightarrow \sin x[4-4{\cos }^{2}x-2+2\cos x]=0$

$\Rightarrow \sin x\left[4{\cos }^{2}x-2\cos x-2\right]=0$

$\Rightarrow -2\sin x\left[2{\cos }^{2}x-\cos x-1\right]=0$

$\Rightarrow -2\sin x\left[\left(2\cos x+1\right)\left(\cos x-1\right)\right]=0$

So either $sinx=0x=0°$ or $cosx=1x=0°$ or $\cos x=-\frac{1}{2}x=120°$

Let a be the maximum value

Let b be the minimum value

for $x=0°,f\left(x\right)=1+\frac{1}{2}-\frac{1}{3}=\frac{6+3-2}{6}=\frac{7}{6}=a$

$x=120°,f\left(x\right)=\cos 120°+\frac{1}{2}\cos 240°-\frac{1}{3}\cos 360°=-\frac{13}{12}=b$

$a-b=\frac{7}{6}-\left(\frac{-13}{12}\right)$

$=\frac{7}{6}+\frac{13}{12}$

$=\frac{84+78}{6\times 12}$

$=\frac{162}{6\times 12}$

$=\frac{9}{4}$

Hence option C is the correct answer