- 0
- maximum.
- minimum.
- none of these.
Solution:
Given, $\text{f}(\text{x})=2\sin3\text{x}+3 \cos3\text{x}$
$\Rightarrow \text{f}'(\text{x})=6 \cos3\text{x}-9\cos3\text{x}$
to find maxima or minima f'(x) = 0
$6 \cos3\text{x}-9\cos3\text{x}=0$
$\Rightarrow \tan3\text{x}=\frac{2}{3}$
$\text{f}'\Big(\frac{5\pi}{6}\Big)=\tan\Big(3\times\frac{5\pi}{6}\Big)$
$\text{f}'\Big(\frac{5\pi}{6}\Big)=\tan\Big(\frac{5\pi}{2}\Big)$
$\Rightarrow\text{f}'\Big(\frac{5\pi}{6}\Big)=\tan\Big(2\pi+\frac{\pi}{2}\Big)$
$\Rightarrow\text{f}'\Big(\frac{5\pi}{6}\Big)=\tan\Big(\frac{\pi}{2}\Big)$ which is not defined.
Hence, $\text{x}=\frac{5\pi}{6}$ is not a critical point.
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$\frac{1}{2}$
$\frac{\sqrt3}{2}$
$-\frac{1}{2}$
$\text{none of these}$