Maths Q17 – JEE-MAIN 2026

Home Maths Limits and DerivativesQuestion #264

Let

f (x) = lim_{y \to 0} \frac{(1 - \cos (x y)) \tan (x y)}{y^{3}}

. Then the number of solutions of the equation

f (x) = \sin x

x \in R

is:

Clue 1 of 3: Limit Evaluation

First, evaluate the given limit to find the explicit form of the function $f (x)$ .

🧠 Full Solution Path

Step 1: Evaluate the limit for

f (x)

✦ Active

The function is defined as $f (x) = lim_{y \to 0} \frac{(1 - \cos (x y)) \tan (x y)}{y^{3}}$ . To evaluate this limit, we can rearrange the terms to use standard trigonometric limits. Let $z = x y$ . As $y \to 0$ , $z \to 0$ .

f (x) = lim_{y \to 0} \frac{1 - \cos (x y)}{(x y)^{2}} \cdot x^{2} \cdot \frac{\tan (x y)}{x y} \cdot x

Using the standard limits $lim_{z \to 0} \frac{1 - \cos z}{z^{2}} = \frac{1}{2}$ and $lim_{z \to 0} \frac{\tan z}{z} = 1$ , we get:

f (x) = (\frac{1}{2}) \cdot x^{2} \cdot (1) \cdot x = \frac{1}{2} x^{3}

This formula also holds for $x = 0$ , as $f (0) = lim_{y \to 0} \frac{(1 - \cos (0)) \tan (0)}{y^{3}} = 0$ , and $\frac{1}{2} (0)^{3} = 0$ .

Step 2: Formulate the equation and analyze its symmetry○ Expand

The equation to solve is $f (x) = \sin x$ , which becomes $\frac{1}{2} x^{3} = \sin x$ . Let $g (x) = \frac{1}{2} x^{3} - \sin x$ . We need to find the number of roots of $g (x) = 0$ . First, check for symmetry:

g (- x) = \frac{1}{2} (- x)^{3} - \sin (- x) = - \frac{1}{2} x^{3} + \sin x = - (\frac{1}{2} x^{3} - \sin x) = - g (x)

Since $g (- x) = - g (x)$ , $g (x)$ is an odd function. This means if $x_{0}$ is a solution, then $- x_{0}$ is also a solution. Also, $g (0) = \frac{1}{2} (0)^{3} - \sin (0) = 0$ , so $x = 0$ is one solution.

💡 Teacher's Secret Hint

Remember that for odd functions, if $x = 0$ is a root, it's counted once. If $x_{0} \neq 0$ is a root, then $- x_{0}$ is a distinct root.

Step 3: Analyze for

x > 0

using calculus○ Expand

Consider $g (x)$ for $x > 0$ . Calculate the first and second derivatives:

g^{'} (x) = \frac{3}{2} x^{2} - \cos x

g^{″} (x) = 3 x + \sin x

For $x > 0$ , $3 x > 0$ . Also, $\sin x \geq - 1$ . Thus, $g^{″} (x) = 3 x + \sin x > 0$ for all $x > 0$ . This implies that $g^{'} (x)$ is strictly increasing for $x > 0$ . Evaluate $g^{'} (0)$ :

g^{'} (0) = \frac{3}{2} (0)^{2} - \cos (0) = - 1

Since $g^{'} (x)$ is strictly increasing for $x > 0$ , $g^{'} (0) = - 1$ , and $lim_{x \to \infty} g^{'} (x) = \infty$ , there is exactly one $x_{0} > 0$ such that $g^{'} (x_{0}) = 0$ . This means $g (x)$ has a unique local minimum at $x_{0}$ for $x > 0$ . Since $g (0) = 0$ and $g^{'} (0) = - 1$ , $g (x)$ decreases from $0$ for $x \in (0, x_{0})$ , so $g (x_{0}) < 0$ . As $x \to \infty$ , $g (x) = \frac{1}{2} x^{3} - \sin x \to \infty$ . Since $g (x_{0}) < 0$ and $g (x) \to \infty$ as $x \to \infty$ , there must be exactly one root $x_{1} > x_{0}$ for $g (x) = 0$ . Therefore, for $x > 0$ , there is exactly one solution. Combining with the solutions from symmetry:

1. $x = 0$ (from $g (0) = 0$ )

2. $x = x_{1}$ (the unique positive root)

3. $x = - x_{1}$ (the unique negative root due to odd symmetry)

Thus, there are a total of 3 solutions.

💡 Teacher's Secret Hint

The behavior of $g^{'} (x)$ (increasing from negative to positive) guarantees a single minimum for $g (x)$ . If this minimum is negative, and $g (x)$ starts at zero and goes to infinity, there must be exactly one positive root.

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Maths Question 17 – JEE-MAIN 2026

Clue 1 of 3: Limit Evaluation

🧠 Full Solution Path