JEE Mathematics 2025 Question with Solution

Given: $\lim_{x \to 0} \frac{x^2 \sin \alpha x + (\gamma - 1)e^{x^2}}{\sin 2x - \beta x} = 3$ for $\alpha, \beta, \gamma \in \mathbb{R}$.

Find: $\beta + \gamma - \alpha$.

Use Maclaurin expansions near $x=0$.

As $x \to 0$, the denominator tends to $0$, so for the limit to be finite the numerator must also tend to $0$:

Hence,

Now the numerator becomes

Using

we get

For the denominator,

and

So,

Since the numerator starts with the term of order $x^3$, the coefficient of $x$ in the denominator must vanish for the limit to be finite and non-zero:

Then

Comparing coefficients of the lowest power term,

Thus,

Now,

Therefore, the correct option is A.

The solution concludes that $\alpha=-4$, $\beta=2$, and $\gamma=1$, so the required value is $7$.

Given: the limit equals $3$.

Find: $\beta+\gamma-\alpha$.

First set $x=0$ in the numerator condition for finiteness:

Then the numerator starts as

The denominator is

For the limit to be finite and non-zero, the $x$-term must vanish:

Now compare the leading $x^3$ coefficients:

Hence,

Therefore, the correct option is A.