Let f:[1, ∞) arrow ℝ be a differentiable function. If 6 ∫_ 1 ^ x f(t) d t=3 x f(x)+x^ 3 -4 for all x ≥ 1 , then the value of f(2)-f(3) is

Correct answer: B. 3 Differentiating 6∫_1^x f(t)dt = 3xf(x) + x^3 - 4 : 6f(x) = 3f(x) + 3xf'(x) + 3x^2 f(x) - xf'(x) = x^2 . (d)/(dx)((f(x))/(x)) = (xf'(x)-f(x))/(x^2) = -1 . Integrating: (f(x))/(x) = -x + C f(x) = -x^2 + Cx . At x = 1 : 0 = 3f(1) - 3 f(1) = 1 . So -1 + C = 1 C = 2 . f(x) = -x^2 + 2x . f(2) = 0 , f(3) = -3 . f(2) - f(3) = 3 .

JEE Main 202622 JanuaryMorning Shift

Exam: JEE Main
Year: 2026
Date: 22 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Differential Equations

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Differential Equations

Let f:[1, ∞) arrow ℝ be a differentiable function. If 6 ∫_ 1 ^ x...

Let

f:[1, \infty) \rightarrow \mathbb{R}

be a differentiable function. If

6 \int_{1}^{x} f(t) d t=3 x f(x)+x^{3}-4

for all

x \geq 1

, then the value of

f(2)-f(3)

Options

Correct answer

B. 3

Step-by-step solution

Differentiating

6\int_1^x f(t)dt = 3xf(x) + x^3 - 4

6f(x) = 3f(x) + 3xf'(x) + 3x^2 \Rightarrow f(x) - xf'(x) = x^2

\frac{d}{dx}\left(\frac{f(x)}{x}\right) = \frac{xf'(x)-f(x)}{x^2} = -1

.
Integrating:

\frac{f(x)}{x} = -x + C \Rightarrow f(x) = -x^2 + Cx

.
At

x = 1

0 = 3f(1) - 3 \Rightarrow f(1) = 1

. So

-1 + C = 1 \Rightarrow C = 2

f(x) = -x^2 + 2x

f(2) = 0

f(3) = -3

f(2) - f(3) = 3

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