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← JEE Main 2026 · 2 Apr · Shift 1
MathematicsSingle correct+4 / -1Question 18 of 75

JEE Main 2026 · 2 Apr · Shift 1, Mathematics question 18

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 :

Choose one option
Answer and worked solution

Official answer: C

Smooth stationary points. sin⁡xx\frac{\sin x}x is smooth with a maximum at x=0x=0 (critical). Its derivative vanishes where tan⁡x=x\tan x=x, which in (−2π,2π)(-2\pi,2\pi) happens once in (π,3π2)(\pi,\frac{3\pi}2) and once in (−3π2,−π)(-\frac{3\pi}2,-\pi).

Corners. Where sin⁡x=0\sin x=0 (x=±πx=\pm\pi), ∣⋅∣|\cdot| makes ff non-differentiable: also critical.

Total 1+2+2=51+2+2=5.

(C)