Differentiating 6∫1xf(t)dt=3xf(x)+x3−4:
6f(x)=3f(x)+3xf′(x)+3x2⇒f(x)−xf′(x)=x2.
dxd(xf(x))=x2xf′(x)−f(x)=−1.
Integrating: xf(x)=−x+C⇒f(x)=−x2+Cx.
At x=1: 0=3f(1)−3⇒f(1)=1. So −1+C=1⇒C=2.
f(x)=−x2+2x. f(2)=0, f(3)=−3.
f(2)−f(3)=3.
JEE Main 2026 — Mathematics Calculus
Let f:[1,∞)→R be a differentiable function. If 6∫1xf(t)dt=3xf(x)+x3−4 for all x≥1, then the value of f(2)−f(3) is
Held on 22 Jan 2026 · Verified 6 Jul 2026.
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