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: Approach:
Since $f$ is an odd function, its values at $x$ and $-x$ are negatives of each other. So the least value is minus the greatest value, and the difference is twice the greatest value.

Step 2: Check the oddness:
$f(-x)=\sin(-2x)-(-x)=-\sin 2x+x=-f(x)$. So $f$ is odd on the symmetric interval.

Step 3: List the candidates:
Critical points come from $\cos 2x=\frac12$, giving $x=\pm\frac{\pi}{6}$. The candidate values are $\pm 0.342$ at those points and $\pm\frac{\pi}{2}$ at the end points.

Step 4: Compare:
Since $\frac{\pi}{2}\approx1.571$ is much larger than $0.342$, the maximum is $\frac{\pi}{2}$ and the minimum is $-\frac{\pi}{2}$.

Step 5: Compute:
Difference $=2\times\frac{\pi}{2}=\pi$, which is option 3.

Final Answer:
\[ \boxed{\pi} \]