Differential Calculus
Differential Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f : R \to R$ be a differentiable function satisfying $f(x) = f(x)f(x-y), \forall x, y \in R$ and $f'(0) = \int_0^{3}\{2x\}dx$, where $\{.\}$ denotes the fractional part function and $f'(-3) = \alpha e^\beta$. Then, $|\alpha + \beta|$ is equal to_____.
Step-by-Step Solution
Key Concept: Recognizing the functional equation $f(x+y) = f(y)f(x)$ as Cauchy's exponential equation whose only continuous solution is $f(x) = e^{kx}$.
Given $f(x) = f(y)f(x-y)$, substituting $x$ with $x+y$ gives $f(x+y) = f(y)f(x)$. This is Cauchy's exponential functional equation with solution $f(x) = e^{kx}$, so $f'(x) = ke^{kx}$. From the integral condition $\int_0^4 (2x)dx = 2$, we get $k = 2$, giving $f'(x) = 2e^{2x}$. Therefore $f'(-3) = 2e^{-6}$ and $|\alpha + \beta| = 4$.
Correct Answer: 3