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Mathematics · Application of Derivatives

JEE Main 2026 — 2 April, Morning Shift — Question 35

The number of critical points of the function f(x)={sin⁡xx,x≠01,x=0f(x) = \begin{cases} \frac{\sin x}{x}, & x \neq 0 \\ 1, & x=0 \end{cases} in the interval (−2π,2π)(-2\pi, 2\pi) is equal to:

  1. Option A:

    1

  2. Option B:

    3

  3. Option C:

    5

    Correct
  4. Option D:

    7

Answer: C

Step-by-step solution

lim⁡x→0∣sin⁡xx∣=1=f(0)→f(x)\lim _{x \rightarrow 0}\left|\frac{\sin x}{x}\right|=1=f(0) \rightarrow f(x) is continuous Now ddx(sin⁡xx)=xcos⁡x−sin⁡xx2\frac{d}{d x}\left(\frac{\sin x}{x}\right)=\frac{x \cos x-\sin x}{x^{2}} ⇒f′(x)=0⇒tan⁡x=x\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=0 \Rightarrow \tan \mathrm{x}=\mathrm{x} 3 solution in (−2π,2π)(-2 \pi, 2 \pi) also f′(0)\mathrm{f}^{\prime}(0) does not exist at x=−π,π\mathrm{x}=-\pi, \pi total 5 points

Solution figure

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Exam
JEE Main 2026
Paper
2 April, Morning Shift
Subject
Mathematics
Chapter
Application of Derivatives
Topic
Monotonicity
The number of critical points of the function f(x) = begin cases sin… | JEE Main 2026 PYQ with Solution · DhiX AI