APPLICATION OF DERIVATIVES — MATHS STD 12 Science — Question

4. Neither maximum nor minimum.

Solution:

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.