#### At $x=\frac{5\pi}{6}$, $f (x) = 2 sin3x + 3 cos3x$ is: A. maximum B. minimum C. zero D. neither maximum nor minimum

Given  $f (x) = 2 sin3x + 3 cos3x$

Apply first derivative and get
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{\mathrm{d}(2 \sin 3 \mathrm{x}+3 \cos 3 \mathrm{x})}{\mathrm{dx}}$
Apply sum rule and take the constant terms out and get
$\Rightarrow f^{\prime}(x)=2 \frac{d(\sin 3 x)}{d x}+3 \frac{d(\cos 3 x)}{d x}$
Apply derivative,
$\Rightarrow f^{\prime}(x)=2 . \cos 3 x \cdot \frac{d(3 x)}{d x}+3 \cdot(-\sin 3 x) \cdot \frac{d(3 x)}{d x}$

\begin{aligned} &\Rightarrow f^{\prime}(x)=2 \cdot \cos 3 x \cdot 3-3 \cdot \sin 3 x \cdot 3\\ &\Rightarrow f'(x)=6 \cos 3 x-9 \sin 3 x \ldots \ldots(i)\\ &x=\frac{5 \pi}{6}\\ &\text { Now we find the value of } f^{\prime}(x) \text { at } \frac{5\pi}{6}\text { and get }\\ &\Rightarrow(f'(x))_{x=\frac{5 \pi}{6}}=6 \cos 3 x-9 \sin 3 x\\ &\Rightarrow(f'(x))_{x=\frac{5 \pi}{6}}=6 \cos \left(\frac{5 \pi}{2}\right)-9 \sin \left(\frac{5 \pi}{2}\right)\\ &\text { Splitting } \frac{5 \pi}{2} \text { as } 2 \pi+\frac{\pi}{2}\\ &\Rightarrow(\mathrm{f'}(\mathrm{x}))_{\mathrm{x}=\frac{5 \pi}{6}}=6 \cos \left(2 \pi+\frac{\pi}{2}\right)-9 \sin \left(2 \pi+\frac{\pi}{2}\right)\\ &\text { Also, } \cos (2 \pi+\theta)=\cos \theta \text { and } \sin (2 \pi+\theta)=\sin \theta\\ &\Rightarrow(\mathrm{f'}(\mathrm{x}))_{\mathrm{x}=\frac{5 \pi}{6}}=6 \cos \left(\frac{\pi}{2}\right)-9 \sin \left(\frac{\pi}{2}\right)\\ \end{aligned}

$\\\cos \left(\frac{\pi}{2}\right)=0 \quad \sin \left(\frac{\pi}{2}\right)=1\\ \\\Rightarrow(f'(x))_{x=\frac{5 \pi}{6}}=6(0)-9(1)=-9$

And we found f’(x) at $x=\frac{5\pi}{6}$ not equal to 0.

So $x=\frac{5\pi}{6}$ cannot be point of minima or maxima.

Hence, $f (x) = 2 sin3x + 3 cos3x$ at $x=\frac{5\pi}{6}$ is not minima nor maxima.

So, the correct answer is option D.