MCQ
Choose the correct answer from the given four options : At $\text{x}=\frac{5\pi}{6},\text{ f(x)}=2\sin3\text{x}+3\cos\text{x}$ is $6 :$
  • A
    Maximum.
  • B
    Minimum.
  • C
    Zero.
  • Neither maximum nor minimum.

Answer

Correct option: D.
Neither maximum nor minimum.
We have, $\text{f(x)}=2\sin3\text{x}+3\cos3\text{x}$
$\therefore\ \text{f}'(\text{x})=2\cdot\cos3\text{x}3+3(-\sin3\text{x})\cdot3$
$\Rightarrow\ \text{f}'(\text{x})=6\cos3\text{x}-9\sin3\text{x}\ \ \dots(\text{i})$
Now, $\text{f}''(\text{x})=-18\sin3\text{x}-27\cos3\text{x}$
$=-9(2\sin3\text{x}+3\cos3\text{x})$
$\therefore\ \text{f}'\Big(\frac{5\pi}{6}\Big)=6\cos\Big(3\cdot\frac{5\pi}{6}\Big)-9\sin\Big(3\cdot\frac{5\pi}{6}\Big)$
$=6\cos\frac{5\pi}{2}-9\sin\frac{5\pi}{2}$
$=6\cos\Big(2\pi+\frac{\pi}{2}\Big)-9\sin\Big(2\pi+\frac{\pi}{2}\Big)$
$=-9\neq0$
So, $\text{x}=\frac{5\pi}{6}$ cannot be point of maxima or minima.
Hence, $f(x)$ at $\text{x}=\frac{5\pi}{6}$ is neither maximum nor minimum.

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