JEE Main 2026 — 22 January, Morning Shift — Question 9

$\quad 6 \int_{1}^{x} f(t) d t=3 x f(x)+x^{3}-4$ Diff. both side $6 \mathrm{f}(\mathrm{x})=3 \mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})+3 \mathrm{f}(\mathrm{x})+3 \mathrm{x}^{2}$

$3 \mathrm{f}(\mathrm{x})=3 \mathrm{x} \mathrm{f}^{\prime}(\mathrm{x})+3 \mathrm{x}^{2}$

$x \frac{d y}{d x}-y=-x^{2}$

$\frac{x \frac{d y}{d x}-9}{x^{2}}=-1$

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=-1$

$\frac{\mathrm{y}}{\mathrm{x}}=-\mathrm{x}+\mathrm{C}$

$\Rightarrow \mathrm{f}(\mathrm{x})=-\mathrm{x}^{2}+\mathrm{Cx}$ at $\mathrm{x}=1, \mathrm{y}=1 \Rightarrow \mathrm{C}=2$

$f(x)=-x^{2}+2 x$ $f(2)-f=3$