The number of critical points of the function f(x) = { |sin x|/x, x ≠ 0 ; 1, x = 0 } in the interval (-2π, 2π) is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q18 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #251

The number of critical points of the function

f (x) = {\begin{cases} \frac{| \sin x |}{x}, & x \neq 0 \\ 1, & x = 0 \end{cases}

in the interval

(- 2 π, 2 π)

is equal to :

Clue 1 of 3: Understand Critical Points

Critical points of a function $f (x)$ are points $x = c$ in the domain where either $f^{'} (c) = 0$ or $f^{'} (c)$ does not exist.

🧠 Full Solution Path

Step 1: Identify Potential Critical Points and Discontinuities✦ Active

Critical points occur where $f^{'} (x) = 0$ or $f^{'} (x)$ does not exist. The function is defined piecewise and involves an absolute value, so we must check points where the definition changes or where the function might be non-differentiable. These points are $x = 0$ and where $\sin x = 0$ , i.e., $x = \pm π$ within the interval $(- 2 π, 2 π)$ .

Step 2: Analyze Differentiability at

x = 0, \pm π

○ Expand

1. **At $x = 0$ **: We evaluate the limits from left and right. $lim_{x \to 0^{+}} f (x) = lim_{x \to 0^{+}} \frac{\sin x}{x} = 1$ . $lim_{x \to 0^{-}} f (x) = lim_{x \to 0^{-}} \frac{- \sin x}{x} = - 1$ . Since the left and right limits are not equal, $f (x)$ is discontinuous at $x = 0$ . As $f (0) = 1$ , $x = 0$ is in the domain, making it a critical point.

2. **At $x = π$ **: For $x \to π^{-}$ , $f (x) = \frac{\sin x}{x}$ , so $f^{'} (x) = \frac{x \cos x - \sin x}{x^{2}}$ . $lim_{x \to π^{-}} f^{'} (x) = \frac{π (- 1) - 0}{π^{2}} = - \frac{1}{π}$ . For $x \to π^{+}$ , $f (x) = \frac{- \sin x}{x}$ , so $f^{'} (x) = \frac{- x \cos x + \sin x}{x^{2}}$ . $lim_{x \to π^{+}} f^{'} (x) = \frac{- π (- 1) + 0}{π^{2}} = \frac{1}{π}$ . Since LHD $\neq$ RHD, $f (x)$ is not differentiable at $x = π$ , making it a critical point.

3. **At $x = - π$ **: Similarly, for $x \to - π^{-}$ , $f^{'} (x) = \frac{1}{π}$ , and for $x \to - π^{+}$ , $f^{'} (x) = - \frac{1}{π}$ . Since LHD $\neq$ RHD, $f (x)$ is not differentiable at $x = - π$ , making it a critical point.

So far, we have 3 critical points: $x = 0, x = π, x = - π$ .

💡 Teacher's Secret Hint

Remember that points of discontinuity within the domain are critical points. Also, check for sharp corners where the derivative changes sign.

Step 3: Find points where

f^{'} (x) = 0

○ Expand

For $x \neq 0, \pm π$ , we consider the derivative based on the sign of $\sin x$ :

1. If $\sin x > 0$ (i.e., $x \in (- 2 π, - π) \cup (0, π)$ ), then $f (x) = \frac{\sin x}{x}$ . $f^{'} (x) = \frac{x \cos x - \sin x}{x^{2}}$ . Setting $f^{'} (x) = 0$ implies $x \cos x = \sin x$ , or $\tan x = x$ . There are no solutions to $\tan x = x$ in $(0, π)$ or $(- 2 π, - π)$ .

2. If $\sin x < 0$ (i.e., $x \in (- π, 0) \cup (π, 2 π)$ ), then $f (x) = \frac{- \sin x}{x}$ . $f^{'} (x) = \frac{- x \cos x + \sin x}{x^{2}}$ . Setting $f^{'} (x) = 0$ implies $- x \cos x + \sin x = 0$ , or $\tan x = x$ . The equation $\tan x = x$ has two non-zero solutions in $(- 2 π, 2 π)$ : $x_{1} \approx 4.493$ and $x_{2} \approx - 4.493$ .

- $x_{1} \approx 4.493$ lies in $(π, 3 π / 2)$ , which is part of $(π, 2 π)$ . In this interval, $\sin x < 0$ . So $x_{1}$ is a critical point.

- $x_{2} \approx - 4.493$ lies in $(- 3 π / 2, - π / 2)$ , which is part of $(- π, 0)$ . In this interval, $\sin x < 0$ . So $x_{2}$ is a critical point.

Thus, we have 2 additional critical points: $x_{1} \approx 4.493$ and $x_{2} \approx - 4.493$ .

Combining all critical points: $x = 0, x = π, x = - π, x_{1} \approx 4.493, x_{2} \approx - 4.493$ . All 5 points are within the interval $(- 2 π, 2 π)$ . Therefore, the total number of critical points is 5.

💡 Teacher's Secret Hint

Graphing $y = \tan x$ and $y = x$ can help visualize the solutions to $\tan x = x$ . Ensure the solutions fall within the correct intervals where the derivative was derived.

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Maths Question 18 – JEE-MAIN 2026

Clue 1 of 3: Understand Critical Points

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