Explanation

(d)
$f(x) = cosx + \frac{1}{2}cos2x - \frac{1}{3}cos3x$
$\Rightarrow \ \ \ \ \ f^{'}(x)$=$-$sin$x$-sin2$x$+sin3$x$
Put $f^{'}(x) = 0$
$\Rightarrow \ 2\sin{\frac{3x}{2}\cos{\frac{x}{2} = 2\sin\frac{3x}{2}}}\cos\frac{3x}{2}$
$\Rightarrow \ \sin{\frac{3x}{2} = 0,}\cos{\frac{3x}{2} = \cos\frac{x}{2}}$
$\Rightarrow$ $x = \frac{2n\pi}{3},\frac{3x}{2} = 2n\pi \pm \frac{x}{2}$
$\Rightarrow \ \$At $x = 0$,$\frac{2\pi}{3},$
At $x = 0$
$f(x) = 1 + \frac{1}{2} - \frac{1}{3} = \frac{7}{6}$
At $x = \frac{2\pi}{3}$,
$f(x) = - \frac{1}{2} - \frac{1}{4} - \frac{1}{3} = - \frac{13}{12}$
$\therefore\$difference$= \frac{7}{6} + \frac{13}{12}$=$\frac{27}{12} = \frac{9}{4}$