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JEE Advanced · Mathematics · 25. AOD

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

  1. A 5
  2. B 3
  3. C 1
  4. D 9
Verified Solution

Answer & Solution

Correct Answer

(C) 1

Step-by-step Solution

Detailed explanation

\(f(x)=\frac{\left(x^{2023}+2024 x+2025\right)}{e^{\pi x}\left(x^2-x+3\right)}(\sin x+2)\)
\(\because(\sin x+2)\) 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
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