MCQ
At $x=\frac{5 \pi}{6}, f(x)=2 \sin 3 x+3 \cos 3 x$ is
  • A
    maximum
  • B
    minimum
  • C
    zero
  • D
    neither maximum nor minimum

Answer

$
\begin{array}{l}
\text { (d) : } f(x)=2 \sin 3 x+3 \cos 3 x \\
f^{\prime}(x)=2 \cos 3 x \cdot 3+3(-\sin 3 x) \cdot 3 \\
f^{\prime}(x)=6 \cos 3 x-9 \sin 3 x
\end{array}
$
For maxima or minima, $f^{\prime}(x)=0$
$
\Rightarrow 6 \cos 3 x-9 \sin 3 x=0 \Rightarrow \tan 3 x=\frac{6}{9}=\frac{2}{3}
$
$\therefore f(x)$ is neither max. nor min. at $x=\frac{5 \pi}{6}$

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