If the solve the problem


At $x=\frac{5 \pi}{6}, f(x)=2 \sin 3 x+3 \cos 3 x$ is

(a) 0

(b) maximum

(c) minimum

(d) none of these


(d) none of these

Given: $f(x)=2 \sin 3 x+3 \cos 3 x$

$\Rightarrow f^{\prime}(x)=6 \cos 3 x-9 \sin 3 x$

For a local minima or a local maxima, we must have


$\Rightarrow 6 \cos 3 x-9 \sin 3 x=0$

$\Rightarrow 6 \cos 3 x=9 \sin 3 x$'

$\Rightarrow \frac{\sin 3 x}{\cos 3 x}=\frac{2}{3}$

$\Rightarrow \tan 3 x=\frac{2}{3}$                   ......(1)

At $x=\frac{5 \pi}{6}:$

$\tan 3 x=\tan \frac{5 \pi}{2}$

$\Rightarrow \tan 3 x=\tan \frac{\pi}{2}$

So, $\tan 3 x$ is not defined.       $\left[\tan 3 \mathrm{x} \neq \frac{2}{3}\right.$ is not satisfying eq. (1) $]$

Thus, $x=\frac{5 \pi}{6}$ is not a critical point.

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