What is the value of p for which the function \(\rm f(x)= p \cos x - \dfrac{\cos 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{6} \ ?\)

Concept:

If function f(x) has an extreme at x = a then f'(a) = 0

Calculations:

Consider, the function \(\rm f(x)= p \cos x - \dfrac{\cos 3x}{3}\) 

Taking derivative w.r.to x , we get

⇒ \(\rm f'(x)= - \;p \sin x + \dfrac{3\sin 3x}{3}\)

⇒ \(\rm f'(x)= - \;p \sin x + \sin 3x\)

⇒ \(\rm f'(​​\dfrac {\pi}{6})= - \;p \sin ​​\dfrac {\pi}{6} + \sin 3​​\dfrac {\pi}{6}\)

⇒\(\rm f'(​​\dfrac {\pi}{6})= - ​​\dfrac p2 + 1\)

The function \(\rm f(x)= p \cos x - \dfrac{\cos 3x}{3}\) has an extremum at  \(\rm x=\dfrac{\pi}{6} \)

⇒f'(\(\dfrac{\pi}{6} \)) = 0

\(⇒ \rm - ​​\dfrac p2 + 1 = 0\)

⇒ p = 2