JEE Main 2026 — 28 January, Morning Shift — Question 1

$g(2)=13$

$\mathrm{f}(\mathrm{g}(2))=\mathrm{f}(13)$

Now $4 g(f(x))=3 x^{2}-32 x+72$

$4\left[3 \mathrm{f}^{2}(\mathrm{x})+2 \mathrm{f}(\mathrm{x})-3\right]=3 \mathrm{x}^{2}-32 \mathrm{x}+72$

Let $\mathrm{f}(\mathrm{x})=\mathrm{t}$

$12 \mathrm{t}^{2}+8 \mathrm{t}-\left(3 \mathrm{x}^{2}-32 \mathrm{x}+84\right)=0$

$\mathrm{f}(\mathrm{x})=\frac{-8 \pm \sqrt{64+48\left(3 \mathrm{x}^{2}-32 \mathrm{x}+84\right)}}{24}$

$f(x)=\frac{-8 \pm 4(3 x-16)}{24}$

$\because f(0)=-3 \quad$

we take +ve sign

$\therefore \mathrm{f}(\mathrm{x})=\frac{-8+4(3 \mathrm{x}-16)}{24}$

$\therefore f(13)=\frac{-8+4 \times 23}{24}=\frac{84}{24}=\frac{7}{2}$