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: Recall the fundamental trigonometric inequality for acute angles.
For 0<x<π/2, sin x<x<tan x.

Step 2: Derive bounds for (sin x)/x.
From sin x<x: (sin x)/x<1. From x<tan x = sin x/cos x: x cos x<sin x → cos x<(sin x)/x.

Step 3: Apply Jordan's inequality for a sharper lower bound.
For 0<x<π/2, 2/π<(sin x)/x<1. Thus (sin x)/x>2/π.

Step 4: Verify each option.
Option (2) states 2/π<(sin x)/x, which is true. Options (1), (3), and (4) contradict the established bounds.

Step 5: Final conclusion.
The correct inequality is 2/π<(sin x)/x.