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Application of Derivatives — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsApplication of Derivatives4 Marks
Question
Find the absolute maximum and minimum values of the function f given by f(x) = cos2 x + sin x, x $\in$ [0, $\pi$]
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Answer

It is given that f(x) = cos2 x + sin x, $x \in[0, \pi]$ 
$f^\prime$(x) = 2 cos x (-sin x) + cos x
= -2 sin x cos x + cos x
Now, if $f^\prime$(x) = 0
$\Rightarrow$ 2 sin x cos x = cos x
$\Rightarrow$ cos x (2 sin x - 1) = 0
$\Rightarrow$ sin x = $\frac{1}{2}$ or cos x = 0
$\Rightarrow$ $x=\frac{\pi}{6},\frac{5\pi}{6}$ or $\frac{\pi}{2}$ as $\mathrm{x} \in[0, \pi]$ 
Next, evaluating the value of f at critical points $x=\frac{\pi}{2}$ and $x=\frac{\pi}{6}$ and at the end points of the interval $[0, \pi]$, (i.e. at x = 0 and $x=\pi$ ), we get,
$f\left(\frac{\pi}{6}\right)=\cos ^{2} \frac{\pi}{6}+\sin \frac{\pi}{6}=\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{2}=\frac{5}{4}$ 
$f(\frac{5\pi}{6}) = cos^2(\frac{5\pi}{6})+ sin(\frac{5\pi}{6}) = cos^2(\pi-\frac\pi6)+sin(\pi-\frac\pi6)=cos^2\frac\pi6-sin\frac\pi6=\frac54$
f(0) = cos2 0 + sin 0 = 1 + 0 = 1
$f(\pi)=\cos ^{2} \pi+\sin \pi$ = (-1)2 + 0 = 1
$f\left(\frac{\pi}{2}\right)=\cos ^{2} \frac{\pi}{2}+\sin \frac{\pi}{2}=0+1=1$ 
Therefore, the absolute maximum value of f is $\frac{5}{4}$ occurring at x = $\frac{\pi}{6}$ and the absolute minimum value of f is 1 occurring at x = 1, $\frac{\pi}{2}$ and $\pi$.

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