Question
Download Solution PDFWhat is the value of p for which the function \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{3} \ ?\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
If the function f(x) has an extremum at x = a then f'(a) = 0
Calculations:
Given, the function is \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\)
⇒ f'(x) = \(\rm p\;cos \; x + \dfrac {3\;cos \;3x}{3}\)
⇒ f'(x) = \(\rm p\;cos \; x + cos \;3x\)
⇒ f'(\(\rm \dfrac {\pi}{3}\)) = \(\rm p\;cos \; (\dfrac {\pi}{3}) + cos \;3(\dfrac {\pi}{3})\)
⇒ f'(\(\rm \dfrac {\pi}{3}\)) = \(\rm p\;cos \; (\dfrac {\pi}{3}) + cos \; \pi\)
The function \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{3}\)
Therefore, \(\rm f'(\dfrac {\pi}{3}) = 0\)
⇒ \(\rm p\;cos \; (\dfrac {\pi}{3}) + cos \; \pi\) = 0
⇒ \(\rm \dfrac p 2 -1 = 0\)
⇒ \(\rm \dfrac p 2 =1\)
⇒ \(\rm p = 2\)
Hence, the value of p for which the function \(\rm f(x)= p \sin x + \dfrac{\sin 3x}{3}\) has an extremum at \(\rm x=\dfrac{\pi}{3}\) is 2.
Last updated on Jun 11, 2025
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