Exams › JEE Main › Maths

Let f:(0,∞)→ R and define F(x)=∫₀^x f(t) dt. If F(x²)=x²(1+x), then the value of f(4) is

1. 5/4
2. 7
3. 4
4. 2

Correct answer: 4

Solution

From F(x^2)=x^2+x^3, differentiate: F'(x^2)*2x = 2x+3x^2, and F'=f so f(x^2)=1+3x/2. Putting x^2=4 (x=2) gives f(4)=1+3=4.

Related JEE Main Maths questions

• For the function y = a²/x + b²/(a - x), where a > 0 and b > 0, the least value of y for x in the interval (0, a) is:
• Let f(x)=∫ from 2x to sin x of cos(t³) dt. Then the value of f''(x) is
• Evaluate the integral ∫_(√(ln 3))^(√(ln 2)) (xsin(x²))/(sin(x²)+sin(ln 6-x²)) dx.
• Evaluate the integral ∫ (e^x (1+xⁿ⁻¹-xⁿ))/((1-xⁿ)√(1-x²ⁿ)) dx.
• Let J = ∫ from 0 to 1 of (cos x)/√x dx and K = ∫ from 0 to 1 of (sin x)/√x dx. Which of the following statements is correct?
• A curve passes through the point (0,1) and satisfies y = ∫ x² e^(x³) dx. If this curve is expressed as x = f(y), then f(y) equals

⚔️ Practice JEE Main Maths free + battle 1v1 →