Let the function f : R → R be defined by f(x) = dfracsin xe^π x dfrac (x^2023 + 2024x + 2025 ) (x^2 - x + 3 ) + dfrac2e^π x dfrac (x^2023 + 2024x + 2025 ) (x^2 - x + 3 ). Then the number of solutions of f(x) = 0 in R is ______.

Let the function f : R → R be defined by f(x) = dfracsin… — JEE Advanced 2024 Mathematics PYQ

JEE Advanced 2024 · Paper 2 · numerical answer

Let the function

f : \mathbb{R} \to \mathbb{R}

be defined by

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

Then the number of solutions of

f(x) = 0

\mathbb{R}

is ______.

Show answer

Answer: 1

[JEE Advanced 2024 · Paper 2]

Practise Limits Continuity and Differentiability the way it is asked

A timed set of real JEE Advanced Limits Continuity and Differentiability questions, graded as the exam grades. No sign-up.