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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$[0, \infty)$

Vedclass · Answer

(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

Similar Questions

In which interval does the function $f(x) = 2x^2 - \log |x|$ $(x \neq 0)$ monotonically increase?

The function $f(x) = \frac{4x^3 - 3x^2}{6} - 2 \sin x + (2x - 1) \cos x$:

The function $f(x) = \cos x - 2px$ is monotonically decreasing for

Given the function $f(x) = \left( \frac{e^{2x} - 1}{e^{2x} + 1} \right)$,the function is:

Find the intervals in which the function given by $f(x) = \sin 3x, x \in \left[0, \frac{\pi}{2}\right]$ is:
$(a)$ increasing
$(b)$ decreasing.

Vedclass Test Series

Exam Paper Generator

Online Exam Module