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Maths Question 20 – JEE-MAIN 2025

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Let f:[1,∞)→[2,∞) be a differentiable function. If 10∫1xf(t)dt=5xf(x)−x5−9 for all x≥1, then the value of f(3) is:

The given equation involves a definite integral with a variable upper limit and the function itself, suggesting differentiation.

Step 1: Differentiate the Integral Equation✦ Active

The given equation is 10∫1xf(t)dt=5xf(x)−x5−9. Differentiate both sides with respect to x using the Leibniz Integral Rule on the LHS and the product rule on the RHS.

ddx(10∫1xf(t)dt)=ddx(5xf(x)−x5−9) 10f(x)=5(1⋅f(x)+x⋅f′(x))−5x4−0 10f(x)=5f(x)+5xf′(x)−5x4
Step 2: Solve the Differential Equation○ Expand

Rearrange the equation to form a first-order linear differential equation and solve it. First, simplify the equation:

5f(x)=5xf′(x)−5x4 f(x)=xf′(x)−x4 xf′(x)−f(x)=x4 f′(x)−1xf(x)=x3 This is a linear differential equation of the form y′+P(x)y=Q(x). The integrating factor (IF) is e∫P(x)dx=e∫−1xdx=e−ln⁡x=eln⁡(x−1)=1x. Multiply the equation by the IF:
1xf′(x)−1x2f(x)=x3⋅1x ddx(1xf(x))=x2 Integrate both sides with respect to x:
1xf(x)=∫x2dx 1xf(x)=x33+C f(x)=x43+Cx
💡 Teacher's Secret Hint

Remember to correctly identify P(x) and Q(x) for the integrating factor method.

Step 3: Find the Constant and Evaluate f(3)○ Expand

Substitute x=1 into the original integral equation to find an initial condition for f(x).

10∫11f(t)dt=5(1)f(1)−(1)5−9 0=5f(1)−1−9 0=5f(1)−10⟹5f(1)=10⟹f(1)=2 Now, use f(1)=2 in the general solution f(x)=x43+Cx to find C:
2=143+C(1) 2=13+C C=2−13=53 So, the function is f(x)=x43+53x. Finally, calculate f(3):
f(3)=343+53(3) f(3)=813+5 f(3)=27+5 f(3)=32
💡 Teacher's Secret Hint

Don't forget to use the initial condition derived from the original integral equation to determine the constant of integration.

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