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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2024 · Paper 2 · Mathematics · Application of Derivatives

Let the function f: ℝ arrow ℝ be defined by f(x)= x e^ π x (x^ 2023...

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

Correct answer

1

Step-by-step solution

f(x)=(x2023+2024x+2025)eπx(x2−x+3)(sin⁡x+2)∵(sin⁡x+2)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 x2023+2024x+2025=0x^{2023}+2024 x+2025=0 let ϕ(x)=x2023+2024x+2025ϕ′(x)=2023x2022+2024>0∀x∈R\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}∴ϕ(x)\therefore \phi(\mathrm{x}) is an Strictly Increasing function ∴ϕ(x)=0\therefore \phi(\mathrm{x})=0 for exactly one value of x∴f(x)=0\mathrm{x} \therefore \mathrm{f}(\mathrm{x})=0 has one solution
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