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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the absolute maximum value and the absolute minimum value of the following functions in the given intervals: $ f (x) = \sin x+\cos x, x \in [0,\pi] $

This is second part of multipart Q5, which appeared in model paper 2012.

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Toolbox:
  • $\large\frac{d}{dx}$$(\sin x)=\cos x$
  • $\large\frac{d}{dx}$$(\cos x)=-\sin x$
Step 1:
$f(x)=\sin x+\cos x$
Differentiate with respect to x,
$f'(x)=\cos x-\sin x$
Given interval $[0,\pi]$
For extreme values $f'(x)=0$
$\cos x-\sin x=0$
On simplifying we get
$\Rightarrow \cos x\big[1-\large\frac{\sin x}{\cos x}\big]$$=0$
But $\sin x/\cos x =\ tan x$
$\Rightarrow \cos x[1-\tan x]=0$
$1-\tan x=0$
$\tan x=1$
$x=\tan^{-1}1$
$x=\large\frac{\pi}{4}$
Step 2:
Now we find $f(x)$ at $x=0,\large\frac{\pi}{4}$$,\pi$
$f(0)=\sin 0+\cos 0$
But $\sin 0= 0$ and $\cos 0 = 1$
$\qquad=0+1$
$\qquad=1$
Step 3:
$f\big(\large\frac{\pi}{4}\big)$$=\sin\large\frac{\pi}{4}+$$\cos\large\frac{\pi}{4}$
But $\sin\large\frac{\pi}{4}=\cos\large\frac{\pi}{4}= \frac{1}{\sqrt2}$
$\qquad\;\;=\large\frac{1}{\sqrt 2}+\frac{1}{\sqrt 2}$
$\qquad\;\;=\large\frac{2}{\sqrt 2}$
$\qquad\;\;=\sqrt 2$
Step 4:
$f(\pi)=\sin \pi+\cos\pi$
$\qquad=0-1=-1$
Step 5:
Absolute maximum value $=\sqrt 2$ at $x=\large\frac{\pi}{4}$ and absolute minimum value =-1 at $x=\pi$
answered Aug 7, 2013 by sreemathi.v
edited Aug 30, 2013 by sharmaaparna1
 

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