Exams › JEE Main › Maths
A function f(x) (with x > 0) satisfies integral from 0 to 1 of f(tx) dt = n f(x). Then f(x) is:
- f(x) = c * x^((1-n)/n)
- f(x) = c * x^(n/(n-1))
- f(x) = c * x^(1/n)
- f(x) = c * x^(1-n)
Correct answer: f(x) = c * x^((1-n)/n)
Solution
The substitution turns the equation into integral₀^x f(u) du = n*x*f(x); differentiating gives a separable ODE whose solution is a power law f(x) = c*x^((1-n)/n).
Related JEE Main Maths questions
- If √(1-x²ⁿ)+√(1-y²ⁿ)=a(xⁿ-yⁿ), then the value of (√(1-x²ⁿ) dy)/(√(1-y²ⁿ) dx) is
- For a curve that passes through the point (4, 0), the slope is governed by
dy/dx = y/x + 5x/((x + 2)(x − 3)).
If the point (5, a) lies on this curve, what is the value of a?
- Which differential equation represents the family of all conics whose axes are aligned with the coordinate axes?
- Find the equation of the curve that satisfies (xy - x²) (dy)/(dx) = y² and passes through the point (-1, 1).
- For the differential equation y = y/x + x/y, if its general solution is written as y = x / log|Cx|, then the function φ(x/y) is
- For the differential equation dy/dx = [y f'(x) − y²]/f(x), where f(x) is a specified function, the solution is
⚔️ Practice JEE Main Maths free + battle 1v1 →