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APPLICATION OF DERIVATIVES — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsAPPLICATION OF DERIVATIVES5 Marks
Question
Find the absolute maximum and minimum values of the function of given by
$\text{f}(\text{x})=\cos^{2}\text{x}+\sin\text{x}, \text{x}\in[0,\pi]$
✓

Answer

Given, $\text{f}(\text{x})=\cos^{2}\text{x}+\sin\text{x}$
$\Rightarrow\text{f}'(\text{x})=2\cos\text{x}(-\sin\text{x})+\cos\text{x}$
$=-2\sin\text{x}\cos\text{x}+\cos\text{x}$
For a local maximum or a local minimum, We must have f'(x) = 0
$\Rightarrow-2\sin\text{x}\cos\text{x}+\cos\text{x}=0$
$\Rightarrow\cos\text{x}(2\sin\text{x}-1)=0$
$\Rightarrow\sin\text{x}=\frac{1}{2}\ \text{or}\ \cos\text{x}=0$
$\Rightarrow\text{x}=\frac{\pi}{6}\ \text{or}\ \frac{\pi}{2}\ [\therefore \text{x}\in(0,\pi)]$
Thus, the critical points of f are $0,\frac{\pi}{6},\frac{\pi}{2}\ \text{and}\ \pi$.
Now, $\text{f}(0)=\cos^{2}(0)+\sin(0)=1$
$\text{f}\Big(\frac{\pi}{6}\Big)=\cos^{2}\Big(\frac{\pi}{6}\Big)+\sin\Big(\frac{\pi}{6})=\frac{5}{4}$
$\text{f}\Big(\frac{\pi}{2}\Big)=\cos^{2}\Big(\frac{\pi}{2}\Big)+\sin\Big(\frac{\pi}{2}\Big)=1$
$\text{f}(\pi)=\cos^{2}(\pi)+\sin(\pi)=1$
Hence, the absolute maximum value when $\text{x}=\frac{\pi}{6}$ is $\frac{5}{4}$ and the absolute minimum value when $\text{x}=0,\frac{\pi}{2},\pi$ is 1.

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