Integral Calculus – Differential Equations – JEE Main 22 Jan 2026 Shift 1

Solution:

Differentiating the given equation with respect to $x$:
$$ \frac{d}{dx} \left( 6\int_{1}^{x}f(t)dt \right) = \frac{d}{dx} (3xf(x) + x^3 – 4) $$
$$ 6f(x) = 3(f(x) + xf'(x)) + 3x^2 $$
$$ 6f(x) = 3f(x) + 3xf'(x) + 3x^2 $$

Rearranging the terms:
$$ 3f(x) = 3xf'(x) + 3x^2 $$
$$ f(x) = xf'(x) + x^2 $$
$$ x\frac{dy}{dx} – y = -x^2 \quad (\text{where } y = f(x)) $$
$$ \frac{dy}{dx} – \frac{1}{x}y = -x $$

This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$.
Integrating Factor (I.F.):
$$ I.F. = e^{\int -\frac{1}{x} dx} = e^{-\ln x} = \frac{1}{x} $$

The general solution is:
$$ y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C $$
$$ y \left( \frac{1}{x} \right) = \int (-x) \left( \frac{1}{x} \right) dx + C $$
$$ \frac{y}{x} = \int -1 dx + C $$
$$ \frac{y}{x} = -x + C \Rightarrow f(x) = -x^2 + Cx $$

To find $C$, put $x=1$ in the original integral equation:
$$ 6\int_{1}^{1}f(t)dt = 3(1)f(1) + 1^3 – 4 $$
$$ 0 = 3f(1) – 3 \Rightarrow f(1) = 1 $$

Substitute $f(1)=1$ in $f(x) = -x^2 + Cx$:
$$ 1 = -(1)^2 + C(1) \Rightarrow C = 2 $$
So, $f(x) = 2x – x^2$.

Now, find $f(2) – f(3)$:
$$ f(2) = 2(2) – 2^2 = 4 – 4 = 0 $$
$$ f(3) = 2(3) – 3^2 = 6 – 9 = -3 $$
$$ f(2) – f(3) = 0 – (-3) = 3 $$

Ans. (3)