JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Definite Integrals

Let

f : [1, \infty) \to 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)

Answer: (D)

Solution

Differentiating

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

gives

6 f = 3 f + 3 x f^{'} + 3 x^{2}

, i.e.

x f^{'} - f = - x^{2}

, so

\frac{d}{d x} (\frac{f}{x}) = - 1

and

f (x) = - x^{2} + C x

. Setting

x = 1

in the original relation gives

0 = 3 f (1) - 3

, so

f (1) = 1

and

C = 2

; thus

f (x) = 2 x - x^{2}

. Then

f (2) - f (3) = 0 - (- 3) = 3

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