JEE Advanced 2024 Paper 2 · Q10 · Monotonicity & Differentiability
Let the function $f : \mathbb{R} \to \mathbb{R}$ be defined by $$f(x) = \frac{\sin x}{e^{\pi x}}\,\frac{\bigl(x^{2023} + 2024x + 2025\bigr)}{\bigl(x^{2} - x + 3\bigr)} + \frac{2}{e^{\pi x}}\,\frac{\bigl(x^{2023} + 2024x + 2025\bigr)}{\bigl(x^{2} - x + 3\bigr)}.$$ Then the number of solutions of $f(x) = 0$ in $\mathbb{R}$ is ______.
Reveal answer + step-by-step solution
Correct answer:1
Solution
Factor: $f(x) = \dfrac{(\sin x + 2)\,(x^{2023} + 2024x + 2025)}{e^{\pi x}\,(x^{2} - x + 3)}$. The denominator is non-zero ($e^{\pi x} > 0$ and $x^{2} - x + 3$ has discriminant $1 - 12 < 0$). The factor $\sin x + 2 \ge 1 > 0$. Hence $f(x) = 0$ iff $g(x) := x^{2023} + 2024x + 2025 = 0$. Now $g'(x) = 2023x^{2022} + 2024 > 0$, so $g$ is strictly increasing on $\mathbb{R}$ and has exactly one real root. Number of solutions $= 1$.
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