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Maxima, Minima and Optimization Problems - Class XI
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The value of $a$ for which the function $f(x)=a\ \sin x+\dfrac{1}{3}\sin 3x$ has an extremum at $x=\dfrac{\pi}{3}$ is
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A
$2$
💡 Explanation:
$f\left( x \right) = a\sin x + \dfrac{1}{3}\sin 3x$
$ \Rightarrow f'\left( x \right) = a\cos x + \dfrac{1}{3}\cos 3x \times 3$
$ \Rightarrow f'\left( x \right) = a\cos x + \cos 3x$
For extremum at ${\dfrac{\pi }{3}}$
$f'\left( {\dfrac{\pi }{3}} \right) = 0$
$ \Rightarrow a\cos \left( {\dfrac{\pi }{3}} \right) + \cos 3\left( {\dfrac{\pi }{3}} \right) = 0$
$ \Rightarrow \dfrac{a}{2} - 1 = 0$
$\Rightarrow a = 2$