Question

If probability that $ax^2 + 2\sqrt{2}bx + c > 0 \, \forall x \in R$ where $a, b, c \in {1, 2, 3, 4}$ is $\frac{m}{n}$ where $m$ & $n$ are coprime, then $(m + n)$ is:

• A. 81
• B. 18
• C. 78
• D. 17

Answer 1

The Correct Option is A

Step 1: Determine the condition for a positive quadratic.
For $ax^2 + 2\sqrt{2}bx + c > 0$ to be true for all real $x$, the leading coefficient must be positive ($a > 0$) and the discriminant must be negative ($D < 0$).
The condition $a > 0$ is satisfied since $a ∈ \{1, 2, 3, 4\}$.
Discriminant $D = (2\sqrt{2}b)^2 - 4ac < 0$
$8b^2 - 4ac < 0 → 2b^2 < ac$.


Step 2: Count valid triplets $(a, b, c)$.
Total outcomes = $4 × 4 × 4 = 64$.
We test values of $b$:
If $b = 1$, then $2(1)^2 < ac → ac > 2$.
Pairs $(a, c)$ that satisfy $ac > 2$ are:
(1,3), (1,4), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4). (Total 13)

If $b = 2$, then $2(2)^2 < ac → ac > 8$.
Pairs $(a, c)$ are:
(3,3), (3,4), (4,3), (4,4). (Total 4)

If $b = 3$, then $2(3)^2 < ac → ac > 18$.
Since max $ac = 4 × 4 = 16$, no pairs exist.

If $b = 4$, no pairs exist.


Step 3: Calculate probability and result.
Total valid cases $m = 13 + 4 = 17$.
Probability $P = \frac{17}{64}$. Here $m = 17, n = 64$.
They are coprime since 17 is a prime number and doesn't divide 64.
$m + n = 17 + 64 = 81$.
The answer is 81 (Option 1).