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MathsJEE Main 2 Apr 2026, Shift 1·Q18·MCQ·Tough

Applications of Derivatives · Critical points of a function

The number of critical points of the function f(x)={∣sin⁡xx∣,x≠01,x=0f(x) = \begin{cases} \left\lvert\dfrac{\sin x}{x}\right\rvert, & x \neq 0\\[6pt] 1, & x = 0\end{cases} in the interval (−2π,2π)(-2\pi, 2\pi) is equal to:
  1. (1)1
  2. (2)3
  3. (3)5
  4. (4)7

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