Question:
Let the function \( f : \mathbb{R} \to \mathbb{R} \) be defined by
\[
f(x) = \frac{\sin x}{e^{\pi x}} \cdot \frac{x^{2023} + 2024x + 2025}{(x^2 - x + 3)} + \frac{2}{e^{\pi x}} \cdot \frac{x^{2023} + 2024x + 2025}{(x^2 - x + 3)}.
\]
Then the number of solutions of \( f(x) = 0 \) in \( \mathbb{R} \) is \_\_\_\_\_.
Let the function \( f : \mathbb{R} \to \mathbb{R} \) be defined by
\[
f(x) = \frac{\sin x}{e^{\pi x}} \cdot \frac{x^{2023} + 2024x + 2025}{(x^2 - x + 3)} + \frac{2}{e^{\pi x}} \cdot \frac{x^{2023} + 2024x + 2025}{(x^2 - x + 3)}.
\]
Then the number of solutions of \( f(x) = 0 \) in \( \mathbb{R} \) is \_\_\_\_\_.
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Check monotonicity of functions to determine the number of roots for polynomials.
Updated On: Jan 20, 2025
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Solution and Explanation
Simplifying \( f(x) \):
\[
f(x) = \frac{x^{2023} + 2024x + 2025}{(x^2 - x + 3)e^{\pi x}} \cdot (\sin x + 2).
\]
For \( f(x) = 0 \), we solve:
\[
x^{2023} + 2024x + 2025 = 0 \quad \text{(since \( \sin x + 2 \neq 0 \) and \( x^2 - x + 3>0 \))}.
\]
Let:
\[
g(x) = x^{2023} + 2024x + 2025.
\]
Derivative:
\[
g'(x) = 2023x^{2022} + 2024>0 \quad \text{(strictly increasing)}.
\]
Thus, \( g(x) \) cuts the \( x \)-axis only once, so \( f(x) = 0 \) has exactly one solution.
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