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QJEEAdv24Ma52ca: JEE Advanced 2024 Mathematics - Class 11 JEEAdv24Ma52ca Solution: Let the function be defined by Then the number of solutions of…

Let the function f:R→Rf: \mathbb{R} \rightarrow \mathbb{R} be defined by

f(x)=sin⁡xeπx(x2023+2024x+2025)(x2−x+3)+2eπx(x2023+2024x+2025)(x2−x+3).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)=0f(x)=0 in R\mathbb{R} is _________.

Step-by-Step Verified Solution

amp;f(x)=(x2023+2024x+2025)eπx(x2−x+3)(sin⁡x+2)amp;∵(sin⁡x+2) is never zero amp;∴ for x2223+2024x+2025=0amp; let ϕ(x)=x2023+2024x+2025amp;ϕ′(x)=2023x2022+2024gt;0∀x∈Ramp;∴ϕ(x) is an Strictly Increasing function amp;∴ϕ(x)=0 for exactly one value of xamp;∴f(x)=0 has one solution \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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