Differential Equations
Differential Equations
star_batch_jee_advanced_2025
Grade 12
Question:
If $f : R - \{-1\} \to R$ and $f$ is differentiable function which satisfies: $f(x + f(y)) + xf(y) = y + f(x) + yf(x)\forall x, y \in R - \{-1\}, f(1) \neq 1$ then find the value of $2019\left[1 + f(2018)\right]$.
Step-by-Step Solution
Key Concept: Differentiating a functional equation with respect to different variables isolates $f'(x)$ and $f'(y)$, revealing that their ratio must be constant.
Given $f(x+f(y)+yf(y)) = y + f(x) + yf(x)$, differentiate with respect to $x$ treating $y$ as constant to get $f'(x+f(y)+yf(y))(1+f'(y)) = f'(x) + yf'(x)$. Differentiate the original equation with respect to $y$ to obtain $f'(x+f(y)+yf(y))(1+x)f'(y) = 1 + f(x)$. From these two equations, derive that $\frac{(1-x)f'(y)}{1+f(y)} = \frac{1+f(x)}{(1+x)f'(x)} = \lambda$ (constant).
Correct Answer: 1