If \( f(x)=2\sin x-x,\ -\displaystyle\frac{\pi}{2}\lt x\lt\displaystyle\frac{\pi}{2} \), then local maximum and local minimum values respectively are
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \sqrt{3}-\displaystyle\frac{\pi}{3} \) & \( \displaystyle\frac{\pi}{3}-\sqrt{3} \)
For critical points,
\( \displaystyle f'(x)=2\cos x-1=0\ \)
\( \displaystyle \Rightarrow\ \cos x=\frac{1}{2}\ \)
\( \displaystyle \Rightarrow\ x=\pm\frac{\pi}{3} \)
The second derivative is \( f''(x)=-2\sin x \).
At \( x=\frac{\pi}{3} \), \( f''\lt 0 \), so there is a local maximum, with value \( f\left(\frac{\pi}{3}\right)=\sqrt{3}-\frac{\pi}{3} \).
At \( x=-\frac{\pi}{3} \), \( f''\gt 0 \), so there is a local minimum, with value \( f\left(-\frac{\pi}{3}\right)=\frac{\pi}{3}-\sqrt{3} \).