JEE Main 2026 — 2 April, Morning Shift — Question 38

For $ax^2 + 2\sqrt{2}bx + c > 0$ for all real $x$, we need $a > 0$ and discriminant $D < 0$.

$D = (2\sqrt{2}b)^2 - 4ac = 8b^2 - 4ac$.

$D < 0$ gives $2b^2 < ac$.

Since $a,b,c \in \{1,2,3,4\}$, $a > 0$ holds automatically. Total number of triples = $4^3 = 64$. Count favorable: For each $b$, count ordered pairs $(a,c)$ with $ac > 2b^2$. $b = 1$: $2b^2 = 2$ → $ac > 2$ gives 13 pairs. $b = 2$: $2b^2 = 8$ → $ac > 8$ gives 4 pairs. $b = 3,4$: $2b^2 = 18,32$ → no $ac$ exceeding these values. Total favorable = $13 + 4 = 17$. Probability = $\frac{17}{64}$, so $m = 17, n = 64$, $m + n = 81$.