Question

If \(0<x<\dfrac{\pi}{2}\), then

• A. \(\dfrac{2}{\pi}>\dfrac{\sin x}{x}\)
• B. \(\dfrac{2}{\pi}<\dfrac{\sin x}{x}\)
• C. \(\dfrac{\sin x}{x}>1\)
• D. \(2<\dfrac{\sin x}{x}\)

Hint:

For \(0<x<\dfrac{\pi}{2}\), remember Jordan's inequality: \[ \frac{2}{\pi} < \frac{\sin x}{x} < 1. \] This is frequently used in limits, inequalities, and applications of Rolle's and Mean Value Theorems.

Answer 1

The Correct Option is B

Step 1: Use the standard trigonometric inequality.
For \[ 0<x<\frac{\pi}{2}, \] the well-known inequality is \[ \sin x < x < \tan x. \] Dividing the first inequality by \(x>0\), \[ \frac{\sin x}{x}<1. \] Thus, option (3) is false.

Step 2: Obtain a lower bound for \(\dfrac{\sin x}{x}\).
From \[ x<\tan x=\frac{\sin x}{\cos x}, \] we get \[ x\cos x<\sin x. \] Dividing by \(x>0\), \[ \cos x<\frac{\sin x}{x}. \] Hence, \[ \frac{\sin x}{x}>\cos x. \]

Step 3: Compare with \(\dfrac{2}{\pi}\).
For \[ 0<x<\frac{\pi}{2}, \] Jordan's inequality states that \[ \frac{2}{\pi} < \frac{\sin x}{x} < 1. \] Therefore, \[ \frac{\sin x}{x} > \frac{2}{\pi}. \]

Step 4: Verify the options.
Option (1): \[ \frac{2}{\pi}>\frac{\sin x}{x} \] is false.
Option (2): \[ \frac{2}{\pi}<\frac{\sin x}{x} \] is true.
Option (3): \[ \frac{\sin x}{x}>1 \] is false because \[ \frac{\sin x}{x}<1. \] Option (4): \[ 2<\frac{\sin x}{x} \] is impossible since \[ \frac{\sin x}{x}<1. \]

Step 5: Final conclusion.
Hence, the correct relation is \[ \boxed{\frac{2}{\pi}<\frac{\sin x}{x}}. \]