The difference between the greatest and least values of the function
f (x) = sin(2x) – x, on $(- \frac{π}{2}, \frac{π}{2}]$ is

\frac{π}{2}

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\frac{π}{3}

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π

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2 π

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Solution

The correct option is C $π$
$f (x) = sin (2 x) - x \Rightarrow f' (x) = 2 cos (2 x) - 1 ∴ f' (x) = 0 \Rightarrow 2 cos (2 x) - 1 = 0 \Rightarrow cos 2 x = \frac{1}{2} \Rightarrow 2 x = - \frac{π}{3} or \frac{π}{3} \Rightarrow x = - \frac{π}{6} or \frac{π}{6} f (- \frac{π}{2}) = sin (- π) + \frac{π}{2} = \frac{π}{2} f (\frac{π}{2}) = sin (π) - \frac{π}{2} = - \frac{π}{2} f (- \frac{π}{6}) = sin (- \frac{π}{3}) + \frac{π}{6} = - \frac{\sqrt{3}}{2} + \frac{π}{6} f (\frac{π}{6}) = sin (\frac{π}{3}) - \frac{π}{6} = \frac{\sqrt{3}}{2} - \frac{π}{6} Clearly, \frac{π}{2} is the greatest value and - \frac{π}{2} is the least . Therefore, difference = \frac{π}{2} + \frac{π}{2} = π .$

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