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What is the value of p for which the function \(\rm f(x)= p \cos x - \dfrac{\cos 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{6} \ ?\)1. 12. 23. -14. None of these |
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Answer» Correct Answer - Option 2 : 2 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 Hence, the value of p for which the function \(\rm f(x)= p \cos x - \dfrac{\cos 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{6}\) is 2.
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