Application of Derivatives — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplication of Derivatives5 Marks
Question
Find the absolute maximum and minimum values of the function f given by $f(x) = \cos^2 x + \sin x, x \in [0, \pi]$
✓
Answer
It is given that $f(x) = \cos^2 x + \sin x, x \in[0, \pi]$
$f^\prime(x) = 2 \cos x (-\sin x) + \cos x$
$= -2 \sin x \cos x + \cos x$
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,
$f\left(\frac{\pi}{6}\right)=\cos ^{2} \frac{\pi}{6}+\sin \frac{\pi}{6}=\left(\frac{\sqrt{3}}{2}\right)^{2}+\frac{1}{2}=\frac{5}{4}$
$f(\frac{5\pi}{6}) = \cos^2(\frac{5\pi}{6})+ \sin(\frac{5\pi}{6}) = \cos^2(\pi-\frac\pi6)+\sin(\pi-\frac\pi6)=\cos^2\frac\pi6-\sin\frac\pi6=\frac54$
$f(0) = \cos^2 0 + \sin 0 = 1 + 0 = 1$
$f(\pi)=\cos ^{2} \pi+\sin \pi = (-1)^2 + 0 = 1$
$f\left(\frac{\pi}{2}\right)=\cos ^{2} \frac{\pi}{2}+\sin \frac{\pi}{2}=0+1=1$
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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