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

Answers (1)

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.

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