Question

The difference between the greatest and least values of the function \(f(x)=\sin 2x-x\) on \(\left[-\frac{\pi}{2},\frac{\pi}{2}\right]\) is

• A. \(\frac{\pi}{2}\)
• B. \(\frac{-\sqrt{3}}{2}+\frac{\pi}{6}\)
• C. \(\pi\)
• D. \(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\)

Hint:

Check \(f\) at \(x=\pm\frac{\pi}{6}\) (where \(f'=0\)) and at both end points, then subtract the smallest from the largest.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
On a closed interval, the greatest and least values of a continuous function occur either at critical points inside the interval or at the end points. We compare all these values.

Step 2: Find critical points:
\(f'(x)=2\cos 2x-1\). Setting it to zero gives \(\cos 2x=\frac12\). For \(x\in[-\frac{\pi}{2},\frac{\pi}{2}]\) we have \(2x\in[-\pi,\pi]\), so \(2x=\pm\frac{\pi}{3}\) and \(x=\pm\frac{\pi}{6}\).

Step 3: Value at the critical points:
\[ f\left(\frac{\pi}{6}\right)=\sin\frac{\pi}{3}-\frac{\pi}{6}=\frac{\sqrt3}{2}-\frac{\pi}{6}\approx 0.342 \] \[ f\left(-\frac{\pi}{6}\right)=-\frac{\sqrt3}{2}+\frac{\pi}{6}\approx -0.342 \]

Step 4: Value at the end points:
\[ f\left(\frac{\pi}{2}\right)=\sin\pi-\frac{\pi}{2}=-\frac{\pi}{2}\approx -1.571 \] \[ f\left(-\frac{\pi}{2}\right)=\sin(-\pi)+\frac{\pi}{2}=\frac{\pi}{2}\approx 1.571 \]

Step 5: Greatest and least:
The greatest value is \(\frac{\pi}{2}\) (at \(x=-\frac{\pi}{2}\)). The least value is \(-\frac{\pi}{2}\) (at \(x=\frac{\pi}{2}\)).

Step 6: Difference:
\[ \frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi \] Options 2 and 4 are only the values at the critical points, which are not the extremes. Option 1 is just the maximum. So option 3 is correct.

Final Answer:
The difference between the greatest and least values is \(\pi\), option 3. \[ \boxed{\pi} \]