Let the function f: R rightarrow R be defined | vedclass

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(A) Given $f(x) = \frac{(x^{2023} + 2024x + 2025)}{e^{\pi x}(x^2 - x + 3)} (\sin x + 2)$.Since $e^{\pi x} > 0$ for all $x \in R$ and the discriminant

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$1$

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(A) Given $f(x) = \frac{(x^{2023} + 2024x + 2025)}{e^{\pi x}(x^2 - x + 3)} (\sin x + 2)$.Since $e^{\pi x} > 0$ for all $x \in R$ and the discriminant of $x^2 - x + 3$ is $D = 1 - 12 = -11 Also,$-1 \leq \sin x \leq 1$,so $(\sin x + 2) \geq 1$,which means $(\sin x + 2)$ is never zero.Thus,$f(x) = 0$ if and only if $x^{2023} + 2024x + 2025 = 0$.Let $\phi(x) = x^{2023} + 2024x + 2025$.Then $\phi'(x) = 2023x^{2022} + 2024$. Since $x^{2022} \geq 0$,$\phi'(x) > 0$ for all $x \in R$.Therefore,$\phi(x)$ is a strictly increasing function.Since $\phi(x)$ is continuous and strictly increasing,it takes every value in $R$ exactly once.Hence,$\phi(x) = 0$ has exactly one real solution.Therefore,the number of solutions of $f(x) = 0$ is $1$.

Let the function $f: R \rightarrow R$ 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

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