JEE Main 2026 — 24 January, Evening Shift — Question 2

$f(t)=\frac{-3}{4}+2 t-t^{2}$ $\begin{gathered} \left.f(t)\right|_{\text {maximum }}=\frac{1}{4}=e \Rightarrow e^{2}=\frac{1}{16} \Rightarrow \frac{a^{2}-b^{2}}{a^{2}}=\frac{1}{16} \end{gathered}$

$$\begin{gathered} \because \frac{2 \mathrm{~b}^{2}}{\mathrm{a}}=30 \Rightarrow \mathrm{~b}^{2}=15 \mathrm{a} \end{gathered}$$

By & (2) $16\left(a^{2}-15 a\right)=a^{2} \Rightarrow 15 a^{2}-16 \times 15 a=0$

$\mathrm{a}=16$ $\mathrm{b}^{2}=240$

$\mathrm{a}^{2}+\mathrm{b}^{2}=256+240$

$=496$