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Applications of Derivative — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplications of Derivative5 Marks
Question
In interval $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$, Find the difference in maximum value and minimum value of function $f(x)=\sin 2 x-x$.
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Answer

Given function
$f(x)=\sin 2 x-x$
So, $\quad f^{\prime}(x)=2 \cos 2 x-1$
For finding critical values :$
\begin{array}{rlrl} 
& & f^{\prime}(x) & =0 \\
\Rightarrow & 2 \cos 2 x-1 & =0 \\
\Rightarrow & & \cos 2 x & =\frac{1}{2} \\
\Rightarrow & & 2 x & =\frac{-\pi}{3}, \frac{\pi}{3} \\
& \Rightarrow & x & =\frac{-\pi}{6}, \frac{\pi}{6}
\end{array}
$
now $\begin{aligned} f\left(\frac{-\pi}{2}\right) & =\sin (-\pi)-\left(\frac{-\pi}{2}\right) \\ & =-\sin \pi+\frac{\pi}{2}=\frac{\pi}{2} \\ f\left(\frac{-\pi}{6}\right) & =\sin \frac{-\pi}{3}+\frac{\pi}{6}=\frac{-\sqrt{3}}{2}+\frac{\pi}{6} \\ f\left(\frac{\pi}{6}\right) & =\sin \frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt{3}}{2}-\frac{\pi}{6} \\ \text { and } \quad f\left(\frac{\pi}{2}\right) & =\sin \pi-\frac{\pi}{2}=\frac{-\pi}{2}\end{aligned}$
Maximum value $=\frac{\pi}{2}$ and minimum value $=\frac{-\pi}{2}$ Hence such difference $=\frac{\pi}{2}-\left(\frac{-\pi}{2}\right)=\pi \quad$

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