The difference between the greatest and least values of the function \( f(x)=\sin 2x-x \) on \( \left[-\displaystyle\frac{\pi}{2},\frac{\pi}{2}\right] \) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \pi \)
On a closed interval, the greatest and least values occur at the critical points or at the end points.
\( \displaystyle f'(x)=2\cos 2x-1=0 \)
\( \displaystyle \cos 2x=\frac{1}{2} \)
\( \displaystyle 2x=\pm\frac{\pi}{3}\quad(2x\in[-\pi,\pi]) \)
\( \displaystyle x=\pm\frac{\pi}{6} \)
The values at the critical points and at the end points:
\( \displaystyle f\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}-\frac{\pi}{6}\approx 0.34 \)
\( \displaystyle f\left(-\frac{\pi}{6}\right)=-\frac{\sqrt{3}}{2}+\frac{\pi}{6}\approx -0.34 \)
\( \displaystyle f\left(\frac{\pi}{2}\right)=\sin\pi-\frac{\pi}{2} \)
\( \displaystyle =-\frac{\pi}{2}\approx -1.57 \)
\( \displaystyle f\left(-\frac{\pi}{2}\right)=\sin(-\pi)+\frac{\pi}{2} \)
\( \displaystyle =\frac{\pi}{2}\approx 1.57 \)
Greatest value \( \frac{\pi}{2} \) (at \( x=-\frac{\pi}{2} \)), least value \( -\frac{\pi}{2} \) (at \( x=\frac{\pi}{2} \)).
\( \displaystyle \text{Difference}=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right) \)
\( \displaystyle =\pi \)
Why the other options are incorrect
Option (1) \( \displaystyle\frac{\pi}{2} \) — Incorrect. \( \frac{\pi}{2} \) is the greatest value alone, not the difference between the greatest and least values.
Option (2) \( \displaystyle\frac{-\sqrt{3}}{2}+\frac{\pi}{6} \) — Incorrect. \( -\frac{\sqrt{3}}{2}+\frac{\pi}{6} \) is only the local minimum value at \( x=-\frac{\pi}{6} \); the least value on the interval is at the end point \( x=\frac{\pi}{2} \).
Option (4) \( \displaystyle\frac{\sqrt{3}}{2}-\frac{\pi}{6} \) — Incorrect. \( \frac{\sqrt{3}}{2}-\frac{\pi}{6} \) is only the local maximum value at \( x=\frac{\pi}{6} \); the greatest value on the interval is at the end point \( x=-\frac{\pi}{2} \).
NCERT Exemplar:
Application of Derivatives
