JEE Advance - Mathematics (2024 - Paper 2 Online - No. 10)
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 _________.
Answer
1
Explanation
$$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\frac{\left(\mathrm{x}^{2023}+2024 \mathrm{x}+2025\right)}{\mathrm{e}^{\pi \mathrm{x}}\left(\mathrm{x}^2-\mathrm{x}+3\right)}(\sin \mathrm{x}+2) \\
& \because(\sin \mathrm{x}+2) \text { is never zero } \\
& \therefore \text { for } \mathrm{x}^{2223}+2024 \mathrm{x}+2025=0 \\
& \text { let } \phi(\mathrm{x})=\mathrm{x}^{2023}+2024 \mathrm{x}+2025 \\
& \phi^{\prime}(\mathrm{x})=2023 \mathrm{x}^{2022}+2024>0 \forall \mathrm{x} \in \mathrm{R} \\
& \therefore \phi(\mathrm{x}) \text { is an Strictly Increasing function } \\
& \therefore \phi(\mathrm{x})=0 \text { for exactly one value of } \mathrm{x} \\
& \therefore \mathrm{f}(\mathrm{x})=0 \text { has one solution }
\end{aligned}$$
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