Question

The number of critical points of the function} \[ f(x)= \begin{cases} \dfrac{|\sin x|}{x}, & x\neq0\\ 1, & x=0 \end{cases} \] in the interval \((-2\pi,2\pi)\) is equal to:

• A. \(1\)
• B. \(3\)
• C. \(5\)
• D. \(7\)

Answer 1

The Correct Option is C

Concept: Critical points occur where

• \(f'(x)=0\), or
• derivative does not exist but the function is defined.

Step 1:Consider \(x\neq0\).} \[ f(x)=\frac{|\sin x|}{x} \] Critical points occur where \[ \sin x=0 \] or where derivative vanishes. Within \((-2\pi,2\pi)\), \[ x=-\pi,0,\pi \]
Step 2:Check derivative conditions.} Differentiating piecewise gives stationary points when \[ \tan x = x \] This equation has two solutions in the interval \((-2\pi,2\pi)\).
Step 3:Count all critical points.} Total critical points: \[ -\pi,\;0,\;\pi \] plus two stationary points. Thus total \[ 5 \] Hence \[ n=5 \]