f(x)=1+2 +3 ^ 2 x, 0<x<2 3 is : - Vidyadip

MCQ

$\text{f}(\text{x})=1+2\sin\text{x}+3\cos^{2}\text{x}, 0<\text{x}<\frac{2\pi}{3}$ is :

✓
Minimum at $\text{x}=\frac{\pi}{2}$
B
Maximum at $\text{x}=\sin^{-1}(\frac{1}{\sqrt{3}})$
C
Minimum at $\text{x}=\frac{\pi}{6}$
D
Maximum at $\sin^{-1}(\frac{1}{6})$

✓

Answer

Correct option: A.

Minimum at $\text{x}=\frac{\pi}{2}$

Given, $\text{f}(\text{x})=1+2\sin\text{x}+3\cos^{2}\text{x}$
$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}-6\cos\text{x}\sin\text{x}$
$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}(1-3\sin\text{x})$
For a local maxima or a local minima.
We must have $f'(x) = 0$
$\Rightarrow2\cos\text{x}(1-3\sin\text{x})=0$
$\Rightarrow2\cos\text{x}=0$ or $(1-3\sin\text{x})=0$
$\Rightarrow \cos \text{x}=0$ or $\sin\text{x}=\frac{1}{3}$
$\Rightarrow \text{x}=\frac{\pi}{2}$ or $\text{x}=\sin^{-1}(\frac{1}{3})$
Now, $\text{f}''(\text{x})=-2\sin\text{x}-6\cos2\text{x}$
$\Rightarrow\text{f}''(\frac{\pi}{2})=-2\sin\frac{\pi}{2}-6\cos(2\times\frac{\pi}{2})$
$=-2+6=4>0$
So, $ \text{x}=\frac{\pi}{2}$ is a local minima.
Also, $\text{f}''(\sin^{-1}\big(\frac{1}{3}\big))=-2\sin(\sin^{-1}\big(\frac{1}{3}\big))-6\cos(\sin^{-1}\big(\frac{1}{3}\big))$
$=\frac{-2}{3}-6\times\frac{2\sqrt{2}}{3}$
$=-\Big(\frac{2}{3}+4\sqrt{2}\Big)<0$
So, $\text{x}=\sin^{-1}(\frac{1}{3})$ is a local maxima.

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