Let the function $f : \mathbb{R} \to \mathbb{R}$ be defined by $$f(x) = \frac{\sin x}{e^{\pi x}} \frac{(x^{2023} + 2024x + 2025)}{(x^2 - x + 3)} + \frac{2}{e^{\pi x}} \frac{(x^{2023} + 2024x + 2025)}{(x^2 - x + 3)}$$ Then the number of solutions of $f(x) = 0$ in $\mathbb{R}$ is _______....

The correct numerical answer is 1. Click to view the complete step-by-step calculation.

JEE Advanced 2024 Paper 2 Mathematics Question 10 Solution

Factorize the expression: $f(x) = \frac{x^{2023} + 2024x + 2025}{e^{\pi x}(x^2 - x + 3)} (\sin x + 2)$ .
For $f(x) = 0$ , we have either $(\sin x + 2) = 0$ or $(x^{2023} + 2024x + 2025) = 0$ .
$\sin x + 2 = 0 \implies \sin x = -2$ , which has no real solutions.
Let $g(x) = x^{2023} + 2024x + 2025$ .
$g'(x) = 2023x^{2022} + 2024$ . Since $x^{2022} \ge 0$ , $g'(x) > 0$ for all $x \in \mathbb{R}$ .
Thus $g(x)$ is strictly increasing.
Since $g(x)$ is an odd-degree polynomial, $\lim_{x \to \infty} g(x) = \infty$ and $\lim_{x \to -\infty} g(x) = -\infty$ .
By the Intermediate Value Theorem, $g(x)$ has exactly one real solution.
The total number of solutions is 1.