Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f(xy) = f(x) \cdot f(y)$, $\forall x > 0, y > 0$ and $f(1 + x) = 1 + x[1 + g(x)]$, where $\lim_{x \to 0} g(x) = 0$, then $$\int \frac{f(x)}{f'(x)} dx$$ is:
$$\frac{x^2}{2} + c$$
$$\frac{x^3}{3} + c$$
$$\frac{x^2}{3} + c$$
None of these
Step-by-Step Solution
Key Concept: Functional equations combined with differentiation reveal the exponential nature of $f(x)$.
From the functional equation $f(xy) = f(x)f(y)$, setting $x = y = 1$ gives $f(1) = f(1)^2$, so $f(1) = 1$. Differentiating the functional equation with respect to $x$ yields $yf'(xy) = f'(x)f(y)$. At $x = 1$, this gives $f'(y) = f'(1)f(y)$. Integrating $\frac{f'(x)}{f(x)} = f'(1)$ yields $\ln f(x) = f'(1) \cdot x + C$, which gives $f(x) = e^{f'(1) \cdot x}$.
Correct Answer: 1