JEE Main 2026 — Mathematics Calculus

The critical points of a function are the points in its domain where the derivative is zero or does not exist.

First, consider $x = 0$x=0. The function is continuous at $x = 0$x=0 since $\displaystyle \lim_{x \to 0} \left|\dfrac{\sin x}{x}\right| = 1 = f(0)$x→0lim​​xsinx​​=1=f(0).

For $x \in (-\pi, \pi)$x∈(−π,π), $\dfrac{\sin x}{x} > 0$xsinx​>0, so $f(x) = \dfrac{\sin x}{x}$f(x)=xsinx​.

The derivative at $x = 0$x=0 is given by $\displaystyle f'(0) = \lim_{x \to 0} \dfrac{\dfrac{\sin x}{x} - 1}{x} = \lim_{x \to 0} \dfrac{\sin x - x}{x^2} = 0$f′(0)=x→0lim​xxsinx​−1​=x→0lim​x2sinx−x​=0.

Since $f'(0) = 0$f′(0)=0, $x = 0$x=0 is a critical point.

Next, consider the points where $f(x) = 0$f(x)=0, which occurs at $x = \pi$x=π and $x = -\pi$x=−π in the interval $(-2\pi, 2\pi)$(−2π,2π).

At $x = \pi$x=π, the inner function $g(x) = \dfrac{\sin x}{x}$g(x)=xsinx​ has $g(\pi) = 0$g(π)=0 and $g'(\pi) = -\dfrac{1}{\pi} \neq 0$g′(π)=−π1​=0. Since the derivative of the inner function is non-zero, the absolute value function $f(x) = |g(x)|$f(x)=∣g(x)∣ is not differentiable at $x = \pi$x=π. By symmetry, $f(x)$f(x) is also not differentiable at $x = -\pi$x=−π.

Thus, $x = \pi$x=π and $x = -\pi$x=−π are critical points.

For $x \neq 0, \pi, -\pi$x=0,π,−π, the function is differentiable and $f'(x) = \pm \dfrac{x \cos x - \sin x}{x^2}$f′(x)=±x2xcosx−sinx​.

Setting $f'(x) = 0$f′(x)=0 gives $x \cos x - \sin x = 0 \Rightarrow \tan x = x$xcosx−sinx=0⇒tanx=x.

We analyze the roots of $\tan x = x$tanx=x in $(-2\pi, 2\pi)$(−2π,2π):

In $(0, \pi)$(0,π), $\tan x > x$tanx>x for $x \in (0, \pi/2)$x∈(0,π/2) and $\tan x < 0 < x$tanx<0<x for $x \in (\pi/2, \pi)$x∈(π/2,π). There are no roots in this interval.

In $(\pi, 2\pi)$(π,2π), $\tan x$tanx increases from $0$0 to $\infty$∞ on $(\pi, 3\pi/2)$(π,3π/2). Since $\tan \pi = 0 < \pi$tanπ=0<π and $\tan x \to \infty$tanx→∞ as $x \to 3\pi/2^{-}$x→3π/2−, there is exactly one root in $(\pi, 3\pi/2)$(π,3π/2). In $(3\pi/2, 2\pi)$(3π/2,2π), $\tan x < 0 < x$tanx<0<x, yielding no roots.

Since $\tan x = x$tanx=x is an odd equation, there is exactly one symmetric root in $(-2\pi, -\pi)$(−2π,−π).

This gives $2$2 additional critical points where $f'(x) = 0$f′(x)=0.

The total number of critical points in $(-2\pi, 2\pi)$(−2π,2π) is $1$1 (at $x=0$x=0) $+ 2$+2 (at $x=\pm \pi$x=±π) $+ 2$+2 (roots of $\tan x = x$tanx=x) $= 5$=5.

Answer: $5$5