Question

The absolute maximum and minimum value of $f\left(x\right)=\sin x+\cos x,x\in [0,\pi ]$ are respectively

• A. $\sqrt{2},-1$
• B. $\sqrt{2},1$
• C. $\sqrt{2},-\sqrt{2}$
• D. $\sqrt{3},\sqrt{2}$

Answer 1

Answer: (A)

$\sqrt{2},-1$

Differentiate the given function $f\left(x\right)=\sin x+\cos x$

$f^{\prime }\left(x\right)=\cos x-\sin x$

Put $f^{\prime }\left(x\right)=0$,

$\cos x-\sin x=0$

$\cos x=\sin x$

$\tan x=1$

$x={\tan }^{-1}\left(1\right)$

$x=\frac{\pi }{4}$

Put the value of $x$ and the end points of the given interval in the given function.

$f\left(\frac{\pi }{4}\right)=\sin \left(\frac{\pi }{4}\right)+\cos \left(\frac{\pi }{4}\right)$

$=\frac{2}{\sqrt{2}}$

$=\sqrt{2}$

$f\left(0\right)=\sin \left(0\right)+\cos \left(0\right)$

$=1$

$f\left(\pi \right)=\sin \left(\pi \right)+\cos \left(\pi \right)$

$=-1$

The maximum value of the given function is $\sqrt{2}$ and the minimum value is $-1$.