JEE Advanced 2024 · Paper 2, Mathematics question 10

Factorise:
$$f(x)=\frac{(\sin x+2)\left(x^{2023}+2024x+2025\right)}{e^{\pi x}\left(x^2-x+3\right)}.$$

Three of the factors never vanish:

• $\sin x+2\ge1$
• $e^{\pi x}>0$
• $x^2-x+3>0$ (its discriminant is $1-12<0$)

So the zeros of $f$ are the zeros of $p(x)=x^{2023}+2024x+2025$.

$p'(x)=2023x^{2022}+2024>0$, so $p$ is strictly increasing. Being of odd degree, it goes from $-\infty$ to $\infty$, so it has exactly one real root.

The number of solutions is $1$.