Indefinite Integration
Integral Calculus-1
star_batch_jee_advanced_2025
Grade 12
Question:
Let $f$ be a function such that $f(x).f(y) + 2 = f(x) + f(y) + f(xy), \forall x, y \in \mathbb{R} - \{0\}$ and $f(0) = 1, f'(1) = 2$, then:
$3(\int f(x)dx) - x(f(x) + 2)$ is constant
$3(\int f(x)dx) - x(f(x) - 2)$ is constant
$\int f(x)dx - x(f(x) - 2)$ is constant
the only possible value of $f(1)$ is $2$
Step-by-Step Solution
Key Concept: Transform the functional equation into a multiplicative form $g(xy) = g(x)g(y)$ by substituting $g(x) = f(x) - 1$, which yields a power function solution.
Rearranging the functional equation: $f(x)f(y) - f(x) - f(y) + 2 = f(xy)$, which factors as $(f(x)-1)(f(y)-1) + 1 = f(xy)$. Let $g(x) = f(x) - 1$; then $g(x)g(y) + 1 = g(xy) + 1$, so $g(xy) = g(x)g(y)$. This means $g(x) = x^a$ for some constant $a$, giving $f(x) = x^a + 1$. Using $f(0) = 1$ confirms this works for $x \neq 0$. From $f'(1) = 2$: $f'(x) = ax^{a-1}$, so $a = 2$. Thus $f(x) = x^2 + 1$ and $f(1) = 2$. For option 1: $\int f(x)dx = \frac{x^3}{3} + x$, and $3(\frac{x^3}{3} + x) - x(x^2 + 1 + 2) = x^3 + 3x - x^3 - 3x = 0$ (constant). Option 4 is confirmed since $f(1) = 2$ is the unique value.
Correct Answer: 1,4