Question

The set of all \(x\) for which \[ \sin x\leq x \] is

• A. \(\left(0,\dfrac{\pi}{2}\right)\)
• B. \(\left(-\dfrac{\pi}{2},\pi\right)\)
• C. \(\left(-\dfrac{\pi}{2},0\right)\)
• D. \(\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)\)

Hint:

For \(x\gt 0\), remember the standard inequality \(\sin x\lt x\). For \(x\lt 0\), the inequality reverses near zero.

Answer 1

The Correct Option is A

Step 1: Understand the inequality.
We need the values of \(x\) for which \[ \sin x\leq x. \] This is a standard inequality.
For positive \(x\), \[ \sin x\lt x \] for \[ x\gt 0. \] Thus, in the interval \[ \left(0,\frac{\pi}{2}\right), \] we have \[ \sin x\leq x. \]

Step 2: Check the options.
Option (1) is \[ \left(0,\frac{\pi}{2}\right). \] For every \(x\) in this interval, \[ x\gt 0 \] and \[ \sin x\lt x. \] Hence, \[ \sin x\leq x. \]

Step 3: Eliminate intervals containing negative values.
For negative values near \(0\), such as \[ x\lt 0, \] we generally have \[ \sin x\gt x. \] For example, \[ x=-0.1 \] gives \[ \sin(-0.1)\approx -0.0998\gt -0.1. \] So, \[ \sin x\leq x \] is not true for negative values near \(0\).
Therefore, intervals containing negative values cannot be the required answer.

Step 4: Final conclusion.
Hence, the correct set from the given options is \[ \boxed{\left(0,\frac{\pi}{2}\right)} \]