JEE Main 2021 · 27 August · morning · Single correct · +4 / −1
Let us consider a curve, y = f(x) passing through the point (-2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x². Then :
- (a) x² + 2xf(x) - 12 = 0
- (b) x³ + xf(x) + 12 = 0
- (c) x³ - 3xf(x) - 4 = 0
- (d) x² + 2xf(x) + 4 = 0
Answer and solution
Answer (c) x³ - 3xf(x) - 4 = 0
Solution
Maths sir · 4 steps
Spot that f + xf' is the derivative of x f(x), and the equation integrates in one line.
- Spot the product rule. By the product rule, d/dx(x f(x)) = f(x) + x f'(x). So the given condition is really d/dx(x f(x)) = x².
x f(x) = ∫ x² dx = x³/3 + C
- Use the point. At (-2, 2) the left side is (-2)(2) = -4 and the right side is -8/3 + C. So C = -4 + 8/3 = -4/3.
- Clear fractions. x f(x) = x³/3 - 4/3. Multiply by 3 to get 3x f(x) = x³ - 4.
- So. The curve satisfies x³ - 3x f(x) - 4 = 0, which is option (c).
The idea to keep. Before using an integrating factor, check whether the left side is already a product-rule derivative. Here f + xf' is (xf)', so you can integrate both sides directly.
Answer from the NTA final answer key. Solution by PrepWiser.