The minimum value of x + x is - Vidyadip

MCQ

The minimum value of $\sin x + \cos x$ is

A
$-2 \sqrt{2}$
B
$\sqrt{2}$
C
$0$
✓
$-\sqrt{2}$

✓

Answer

Correct option: D.

$-\sqrt{2}$

Let $f(x)=\sin x+\cos x$
$\therefore f^{\prime}(x)=\cos x-\sin x$
$\Rightarrow f^{\prime \prime}(x)=-\sin x-\cos x$
Now, $f^{\prime}(x)=0$
$\Rightarrow \cos x-\sin x=0$
$\Rightarrow \sin x=\cos x$
$\Rightarrow \tan x=1$
$\Rightarrow x=n \pi+\frac{\pi}{4}, n \in z $
$\text { At } x=\pi+\frac{\pi}{4}$
$f^{\prime \prime}(x)=-\sin \left(\pi+\frac{\pi}{4}\right)-\cos \left(\pi+\frac{\pi}{4}\right)$
$=\sin \left(\frac{\pi}{4}\right)+\cos \left(\frac{\pi}{4}\right)$
$=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2}>0$
$\therefore x =\pi+\frac{\pi}{4}$ is point of minimum
Minimum value $=\sin \left(\pi+\frac{\pi}{4}\right)+\cos \left(\pi+\frac{\pi}{4}\right)$
$=-\sin \left(\frac{\pi}{4}\right)-\cos \left(\frac{\pi}{4}\right)$
$=-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}$
$=-\frac{2}{\sqrt{2}}$
$=-\sqrt{2}$

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