Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade None
Question:
In a certain problem the differentiation of product $(f(x).g(x))$ appears. One student commits mistake and differentiates as $\left(\frac{d(f(x))}{dx} - \frac{d(g(x))}{dx}\right)$ but he gets correct result if $f(x) = x^3$ & $g(4) = 9, g(2) = -9$ & $g(0) = -\frac{1}{3}$ then:
g(x) = \frac{9}{(x-3)^4}
\left(\frac{d}{dx}(f(x-3).g(x))\right)_{at\,x=100} = 0
\lim_{x\to 0} \frac{f(x).g(x)}{x(1+g(x))} = 0
None of these
Step-by-Step Solution
Key Concept: Identify the mistake in differentiation and solve the resulting separable differential equation using logarithmic integration.
Given $h(x) = f(x) \cdot g(x)$ with a mistake where $h'(x) = f'(x) \cdot g'(x)$ instead of the correct product rule. This error leads to $3x^2 g'(x) = 3x^2 g(x) + x^3 g'(x)$, simplifying to $(3x^2 - x^3)g'(x) = 3x^2 g(x)$. Solving the differential equation $\frac{g'(x)}{g(x)} = \frac{3x^2}{x^2(3-x)} = \frac{-3}{x-3}$ gives $g(x) = \frac{-9}{(x-3)^3}$ (up to constants).
Correct Answer: 1,2,3