Let the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x)=\frac{\sin x}{e^{\pi x}}…
Let the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by
$f(x)=\frac{\sin x}{e^{\pi x}} \frac{\left(x^{2023}+2024 x+2025\right)}{\left(x^2-x+3\right)}+\frac{2}{e^{\pi x}} \frac{\left(x^{2023}+2024 x+2025\right)}{\left(x^2-x+3\right)}$.
Then the number of solutions of $f(x)=0$ in $\mathbb{R}$ is
Solution
$f(x)=\frac{\left(x^{2023}+2024 x+2025\right)}{e^{\pi x}\left(x^2-x+3\right)}(\sin x+2)$
$\because(\sin x+2)$ is never zero
$\therefore$ for $x^{2023}+2024 x+2025=0$
$\begin{aligned} & \text { let } \phi(\mathrm{x})=\mathrm{x}^{2023}+2024 \mathrm{x}+2025 \\ & \phi^{\prime}(\mathrm{x})=2023 \mathrm{x}^{2022}+2024>0 \forall \mathrm{x} \in \mathrm{R}\end{aligned}$
$\therefore \phi(\mathrm{x})$ is an Strictly Increasing function
$\therefore \phi(\mathrm{x})=0$ for exactly one value of $\mathrm{x}$
$\therefore \mathrm{f}(\mathrm{x})=0$ has one solution