The set of all x for which sin x leq x is | vedclass

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(D) Let $f(x) = x - \sin x$. Then $f'(x) = 1 - \cos x$. Since $\cos x \leq 1$ for all $x \in \mathbb{R}$,we have $f'(x) \geq 0$ for all $x$. This mean

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$[0, \infty)$

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(D) Let $f(x) = x - \sin x$. Then $f'(x) = 1 - \cos x$. Since $\cos x \leq 1$ for all $x \in \mathbb{R}$,we have $f'(x) \geq 0$ for all $x$. This means $f(x)$ is a non-decreasing function. Since $f(0) = 0 - \sin(0) = 0$,it follows that $f(x) \geq 0$ for all $x \geq 0$. Thus,$x - \sin x \geq 0$,which implies $\sin x \leq x$ for all $x \geq 0$. For $x  0$. Then $\sin(-t) = -\sin t$. The inequality becomes $-\sin t \leq -t$,which simplifies to $\sin t \geq t$. Since $\sin t  0$,the inequality $\sin t \geq t$ is only satisfied at $t = 0$. Therefore,the set of all $x$ for which $\sin x \leq x$ is $[0, \infty)$.

The set of all $x$ for which $\sin x \leq x$ is

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