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: Analyze f(x)=sin x - x.
\(f'(x)=\cos x-1\leq0\) for all \(x\), with equality only at \(x=0\). So \(f\) is strictly decreasing except at isolated points. At \(x=0\), \(f(0)=0\).

Step 2: Determine sign of f(x).
For \(x>0\): \(f\) has decreased from \(0\), so \(f(x)<0\Rightarrow\sin x<x\). For \(x<0\): \(f(x)>0\Rightarrow\sin x>x\). So \(\sin x\leq x\) holds for \(x\geq0\). Matching the given options: \[\boxed{\left(0,\dfrac{\pi}{2}\right)}\]