JEE Main 2026 — Application of Derivatives Question with Solution

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

First, consider $x=0$. The function is continuous at $x=0$ since ${\lim }_{x\to 0}\left|\frac{\sin x}{x}\right|=1=f(0)$.

For $x\in (-\pi ,\pi )$, $\frac{\sin x}{x}>0$, so $f(x)=\frac{\sin x}{x}$.

The derivative at $x=0$ is given by $f^{\prime }(0)={\lim }_{x\to 0}\frac{\frac{\sin x}{x}-1}{x}={\lim }_{x\to 0}\frac{\sin x-x}{x^2}=0$.

Since $f^{\prime }(0)=0$, $x=0$ is a critical point.

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

At $x=\pi$, the inner function $g(x)=\frac{\sin x}{x}$ has $g(\pi )=0$ and $g^{\prime }(\pi )=-\frac{1}{\pi }\ne 0$. Since the derivative of the inner function is non-zero, the absolute value function $f(x)=|g(x)|$ is not differentiable at $x=\pi$. By symmetry, $f(x)$ is also not differentiable at $x=-\pi$.

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

For $x\ne 0,\pi ,-\pi$, the function is differentiable and $f^{\prime }(x)=\pm \frac{x\cos x-\sin x}{x^2}$.

Setting $f^{\prime }(x)=0$ gives $x\cos x-\sin x=0\Rightarrow \tan x=x$.

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

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

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

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

This gives $2$ additional critical points where $f^{\prime }(x)=0$.

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

Answer: $5$