Let the function f: mathbb R rightarrow mathbb R be defined by f(x)=frac sin x e π x frac (x 2024 +2024 x+2025 ) (x 2 -x+3 ) +frac 2 e π x frac (x 2024 +2024 x+2025 ) (x 2 -x+3 ) . Then the number of solutions of f(x)=0 in mathbb R is

Mathematics · Functions

JEE Advanced 2024 — Paper 2 — Question 10

Let the function $f: \mathbb{R} \rightarrow \mathbb{R}$ be defined by $f(x)=\frac{\sin x}{e^{\pi x}} \frac{\left(x^{2024}+2024 x+2025\right)}{\left(x^{2}-x+3\right)}+\frac{2}{e^{\pi x}} \frac{\left(x^{2024}+2024 x+2025\right)}{\left(x^{2}-x+3\right)} .$

Then the number of solutions of $f(x)=0$ in $\mathbb{R}$ is $\qquad$

Answer: 1

Numerical answer — enter this value.

Step-by-step solution

$f(x)=\frac{(\sin x+2)\left(x^{2023}+2024 x+2025\right)}{e^{\pi x}\left(x^{2}-x+3\right)}=0$ $\because \sin x+2>0 \forall x \in R$ $\mathrm{e}^{\pi x}>0 \forall x \in R$ $x^{2}-x+3>0 \forall x \in R$ Now, let $g(x)=x^{2023}+2024 x+2025$ $\because \mathrm{g}^{\prime}(\mathrm{x})>0 \forall \mathrm{x} \in \mathrm{R}$ (strictly increasing) $\therefore$ Number of solution of $f(x)=0$ is 1 Solution is $(x=-1)$

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Exam: JEE Advanced 2024
Paper: Paper 2
Subject: Mathematics
Chapter: Functions
Topic: Functional Equations

Answer: 1

Step-by-step solution

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