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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES3 Marks
Question
Find the absolute maximum value and the absolute minimum value of the following function in the given intervals:
$\text{f}\text{(x)} = \sin \text{x} +\cos \text{x}, \text{x}\in[0,\ \pi]$
✓

Answer

Given: $\text{f}\text{(x)} =\sin \text{x}+ \cos \text{x}, \text{x}\in[0,\pi]\ \Rightarrow\ \text{f}'\text{(x)}=\cos \text{x}-\sin\text{x}$
$\text{Now }\text{ f}'\text{(x)}=0\ \Rightarrow\ \cos \text{x}-\sin \text{x}=0\ \Rightarrow\ -\sin\text{x}=-\cos \text{x}$
$\Rightarrow\ \tan \text{x}=1\ \ [\text{Posituive}]$
$\therefore\ \text{x }\text{ is in I quadrant.}\ \ [\text{ x}\in(0,\ \pi)]$
$\therefore\ \tan \text{x} =1 = \tan \frac{\pi}{4}\ \Rightarrow\ \text{x}=\frac{\pi}{4}$
$\therefore \ \text{f} \bigg(\frac{\pi}{4}\bigg)=\sin\frac{\pi}{4}+\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}= \sqrt{2}$
$\text{f} (0)=\sin0+ \cos0=0+1=1$
$\text{f}(\pi)=\sin\pi+\cos\pi = 0-1=-1$
Therefore, absolute minimum value is -1 and absolute maximum value is 1.

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