Application of DerivativesmediumPYQ · JEE Main · 2 Apr 2026 · Shift 1 (Morning)Free

Critical Points of |sin x / x| on (−2π, 2π): 5 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
The number of critical points of the function f(x)={∣sin⁡xx∣,x≠01,x=0f(x)=\begin{cases}\left|\dfrac{\sin x}{x}\right|, & x\ne0\\[4pt] 1, & x=0\end{cases} in the interval (−2π,2π)(-2\pi,2\pi) is equal to
A11
B33
C55correct
D77
Solution
Step 1: lim⁡x→0∣sin⁡xx∣=1=f(0)\displaystyle\lim_{x\to0}\left|\frac{\sin x}{x}\right|=1=f(0), so ff is continuous everywhere. Step 2: For x≠0x\ne0, consider g(x)=sin⁡xxg(x)=\dfrac{\sin x}{x}. Then
g′(x)=xcos⁡x−sin⁡xx2=0 ⇒ xcos⁡x=sin⁡x ⇒ tan⁡x=x.g'(x)=\frac{x\cos x-\sin x}{x^2}=0\ \Rightarrow\ x\cos x=\sin x\ \Rightarrow\ \tan x=x.
 In (−2π,2π)(-2\pi,2\pi), the equation tan⁡x=x\tan x=x has 33 solutions (including xx near 00 region and one in each of two other branches) — these are stationary points. Step 3: Because of the modulus, f=∣g∣f=|g| has sharp corners where g(x)=0g(x)=0 with a sign change, i.e. at x=±πx=\pm\pi (where sin⁡x=0\sin x=0). At these points f′f' does not exist — 2 additional critical points. Step 4: Total number of critical points:
3+2=5.3+2=5.
 Correct answer: (3)
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