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Probability a Quadratic is Always Positive: m + n = 81 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let a,b,c∈{1,2,3,4}a,b,c\in\{1,2,3,4\}. If the probability that ax2+22 bx+c>0ax^2+2\sqrt2\,bx+c>0 for all x∈Rx\in\mathbb{R} is mn\dfrac{m}{n}, gcd⁡(m,n)=1\gcd(m,n)=1, then m+nm+n is equal to
Solution
Answer: 81 (± 0.01)
Step 1: A quadratic ax2+22bx+cax^2+2\sqrt2 bx+c is positive for all real xx iff a>0a>0 (always true here) and its discriminant is negative:
(22b)2−4ac<0 ⇒ 8b2−4ac<0 ⇒ 2b2<ac.(2\sqrt2 b)^2-4ac<0\ \Rightarrow\ 8b^2-4ac<0\ \Rightarrow\ 2b^2<ac.
 Step 2: Count favourable (a,b,c)(a,b,c) with 2b2<ac2b^2<ac, going by bb: - b=1b=1 (2b2=22b^2=2, need ac>2ac>2): a=1⇒c∈{3,4}a=1\Rightarrow c\in\{3,4\} (2); a=2⇒c∈{2,3,4}a=2\Rightarrow c\in\{2,3,4\} (3); a=3⇒c∈{1,2,3,4}a=3\Rightarrow c\in\{1,2,3,4\} (4); a=4⇒c∈{1,2,3,4}a=4\Rightarrow c\in\{1,2,3,4\} (4). Subtotal =13=13. - b=2b=2 (2b2=82b^2=8, need ac>8ac>8): a=3⇒c∈{3,4}a=3\Rightarrow c\in\{3,4\} (2); a=4⇒c∈{3,4}a=4\Rightarrow c\in\{3,4\} (2). Subtotal =4=4. - b=3,4b=3,4 (2b2=18,322b^2=18,32): no acac up to 1616 qualifies. Subtotal =0=0. Step 3: Favourable =13+4=17=13+4=17; total =43=64=4^3=64. So
P=1764=mn.P=\frac{17}{64}=\frac{m}{n}.
 Step 4:
m+n=17+64=81.m+n=17+64=81.
 Correct answer: 81
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