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(A) Given $f(x) = \sin(2x) - x$. To find the critical points,we find $f'(x) = 2\cos(2x) - 1$. Setting $f'(x) = 0$,we get $2\cos(2x) = 1$,so $\cos(2x)

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$\pi$

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(A) Given $f(x) = \sin(2x) - x$. To find the critical points,we find $f'(x) = 2\cos(2x) - 1$. Setting $f'(x) = 0$,we get $2\cos(2x) = 1$,so $\cos(2x) = \frac{1}{2}$. In the interval $\left[ -\frac{\pi}{2}, \frac{\pi}{2} \right]$,$2x \in [-\pi, \pi]$,so $2x = \pm \frac{\pi}{3}$,which gives $x = \pm \frac{\pi}{6}$. Now,we evaluate $f(x)$ at the critical points and endpoints: $f\left(-\frac{\pi}{2}\right) = \sin(-\pi) - (-\frac{\pi}{2}) = 0 + \frac{\pi}{2} = \frac{\pi}{2} \approx 1.57$. $f\left(-\frac{\pi}{6}\right) = \sin(-\frac{\pi}{3}) - (-\frac{\pi}{6}) = -\frac{\sqrt{3}}{2} + \frac{\pi}{6} \approx -0.866 + 0.523 = -0.343$. $f\left(\frac{\pi}{6}\right) = \sin(\frac{\pi}{3}) - \frac{\pi}{6} = \frac{\sqrt{3}}{2} - \frac{\pi}{6} \approx 0.866 - 0.523 = 0.343$. $f\left(\frac{\pi}{2}\right) = \sin(\pi) - \frac{\pi}{2} = 0 - \frac{\pi}{2} = -\frac{\pi}{2} \approx -1.57$. The greatest value is $\frac{\pi}{2}$ and the least value is $-\frac{\pi}{2}$. The difference is $\frac{\pi}{2} - (-\frac{\pi}{2}) = \pi$.

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

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