Question:
Let the function \(f:\R→\R\) 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 \(\R\) is __.
Let the function \(f:\R→\R\) 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 \(\R\) is __.
\(f(x)=\frac{\sin x}{e^{\pi x}}\frac{(x)^{2023}+2024x+2025}{(x^2-x+3)}+\frac{2}{e^{\pi x}}\frac{(x^{2023}+2024x+2025)}{(x^2-x+3)}\)
Then the number of solutions of f (x) = 0 in \(\R\) is __.
Updated On: Mar 8, 2025
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Solution and Explanation
Simplifying f(x)
We have:
\[ f(x) = x^{2023} + 2024x + 2025 \] \[ e^{\pi x} \cdot (x^{2023} + 2024x + 2025) \] \[ (x^2 - x + 3) \cdot (x^2 - x + 3) \cdot e^{\pi x} \cdot (\sin x + 2) \]
For \( f(x) = 0 \), we solve:
\[ x^{2023} + 2024x + 2025 = 0 \quad (\text{since } \sin x + 2 = 0 \text{ and } x^2 - x + 3 > 0) \]
Let:
The derivative of \( g(x) = x^{2023} + 2024x + 2025 \) is:
\[ g'(x) = 2023x^{2022} + 2024 > 0 \quad (\text{strictly increasing}) \]
Thus, \( g(x) \) cuts the x-axis only once, meaning \( f(x) = 0 \) has exactly one solution.
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