If f(x) = sin x - frac{x}{2} is an increasing | vedclass

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(C) For a function $f(x)$ to be increasing,its derivative $f'(x)$ must be greater than $0$.Given $f(x) = \sin x - \frac{x}{2}$.Differentiating with re

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$-\frac{\pi}{3} < x < \frac{\pi}{3}$

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(C) For a function $f(x)$ to be increasing,its derivative $f'(x)$ must be greater than $0$.Given $f(x) = \sin x - \frac{x}{2}$.Differentiating with respect to $x$,we get $f'(x) = \cos x - \frac{1}{2}$.For the function to be increasing,set $f'(x) > 0$:$\cos x - \frac{1}{2} > 0$$\cos x > \frac{1}{2}$We know that $\cos x = \frac{1}{2}$ at $x = \pm \frac{\pi}{3}$.Since the cosine function is positive and decreasing in the interval $(0, \frac{\pi}{2})$ and symmetric,the inequality $\cos x > \frac{1}{2}$ holds for $x \in (-\frac{\pi}{3}, \frac{\pi}{3})$.Thus,the correct option is $C$.

If $f(x) = \sin x - \frac{x}{2}$ is an increasing function,then:

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