Q.23. Let . If the probability, that for all , is , where , then is equal to

Q: What is the correct answer for JEE Main Mathematics Q23?

The correct answer is D. Sol. For $ax^2 + 2\sqrt{2} bx + c > 0$ for all $x \in \mathbb{R}$, we must have $a > 0$ and discriminant $D < 0$. $D = (2\sqrt{2}b)^2 - 4ac = 8b^2 - 4ac < 0 \Rightarrow 8b^2 < 4ac \Rightarrow 2b^2 < ac$. Since $a,b,c \in \{1,2,3,4\}$, $a$ must be positive, which is always true. We need to find the number of triplets $(a,b,c)$ such that $2b^2 < ac$. Total possible outcomes = $4 \times 4 \times 4 =

Q: Q.23. Let . If the probability, that for all , is , where , then is equal to

The correct answer is D. Sol. For $ax^2 + 2\sqrt{2} bx + c > 0$ for all $x \in \mathbb{R}$, we must have $a > 0$ and discriminant $D < 0$. $D = (2\sqrt{2}b)^2 - 4ac = 8b^2 - 4ac < 0 \Rightarrow 8b^2 < 4ac \Rightarrow 2b^2 < ac$. Since $a,b,c \in \{1,2,3,4\}$, $a$ must be positive, which is always true. We need to find the number of triplets $(a,b,c)$ such that $2b^2 < ac$. Total possible outcomes = $4 \times 4 \times 4 =

Q: What are the options for this JEE Main Mathematics question?

A. 91 · B. 81 · C. 71 · D. 61

Question

Q.23. Let . If the probability, that  for all , is , where , then  is equal to

FirstInTest · Accepted Answer

Sol. For $ax^2 + 2\sqrt{2} bx + c > 0$ for all $x \in \mathbb{R}$, we must have $a > 0$ and discriminant $D < 0$. $D = (2\sqrt{2}b)^2 - 4ac = 8b^2 - 4ac < 0 \Rightarrow 8b^2 < 4ac \Rightarrow 2b^2 < ac$. Since $a,b,c \in \{1,2,3,4\}$, $a$ must be positive, which is always true. We need to find the number of triplets $(a,b,c)$ such that $2b^2 < ac$. Total possible outcomes = $4 imes 4 imes 4 = 64$. Let's list the favorable outcomes for $2b^2 < ac$: If $b=1$: $2(1)^2 < ac \Rightarrow 2 < ac$. Possible $(a,c)$ pairs: $(1,3), (1,4), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4)$. (13 pairs) If $b=2$: $2(2)^2 < ac \Rightarrow 8 < ac$. Possible $(a,c)$ pairs: $(2,4), (3,3), (3,4), (4,2), (4,3), (4,4)$. (6 pairs) If $b=3$: $2(3)^2 < ac \Rightarrow 18 < ac$. Possible $(a,c)$ pairs: $(3,4), (4,3), (4,4)$. (3 pairs) If $b=4$: $2(4)^2 < ac \Rightarrow 32 < ac$. Possible $(a,c)$ pairs: None, since max $ac = 4 imes 4 = 16$. Total favorable outcomes = $13 + 6 + 3 = 22$. Probability $P = \frac{22}{64} = \frac{11}{32}$. So $m=11, n=32$. $\mathrm{gcd}(11,32)=1$. $m+n = 11+32 = 43$. The solution in the image has a table and states 17 favorable outcomes. Let's check their table: | b | a | c | Count | |---|---|---|-------| | 1 | 1 | 3,4 | 2 | | 1 | 2 | 2,3,4 | 3 | | 1 | 3 | 1,2,3,4 | 4 | | 1 | 4 | 1,2,3,4 | 4 | | | | | Total for b=1: 2+3+4+4 = 13 | This matches my count for $b=1$. | b | a | c | Count | |---|---|---|-------| | 2 | 3 | 3,4 | 2 | | 2 | 4 | 3,4 | 2 | | | | | Total for b=2: 2+2 = 4 | My count for $b=2$ was 6 pairs: $(2,4), (3,3), (3,4), (4,2), (4,3), (4,4)$. Let's check the solution's table for $b=2$: $2b^2 < ac \Rightarrow 8 < ac$. If $a=1$, no $c$. If $a=2$, $8 < 2c \Rightarrow c > 4$. No $c$. If $a=3$, $8 < 3c \Rightarrow c > 8/3 = 2.66$. So $c=3,4$. (2 pairs: $(3,3), (3,4)$) If $a=4$, $8 < 4c \Rightarrow c > 2$. So $c=3,4$. (2 pairs: $(4,3), (4,4)$) Total for $b=2$ is $2+2=4$. This matches the solution's table. My previous list for $b=2$ was incorrect. $(2,4)$ gives $ac=8$, not $>8$. $(4,2)$ gives $ac=8$, not $>8$. So the solution's table is correct for $b=2$. | b | a | c | Count | |---|---|---|-------| | 3 | 4 | 3,4 | 2 | | | | | Total for b=3: 2 | My count for $b=3$ was 3 pairs: $(3,4), (4,3), (4,4)$. Let's check the solution's table for $b=3$: $2b^2 < ac \Rightarrow 18 < ac$. If $a=1,2,3$, no $c$. If $a=4$, $18 < 4c \Rightarrow c > 18/4 = 4.5$. No $c$. Max $c=4$. So for $b=3$, there are 0 pairs. The solution's table has 2 pairs for $b=3$ (a=4, c=3,4). This is incorrect. $4 imes 3 = 12$, $4 imes 4 = 16$. Neither is $>18$. So the solution's table is incorrect for $b=3$. Let's re-calculate carefully: $a, b, c \in \{1,2,3,4\}$. Condition: $2b^2 < ac$. Case $b=1$: $2(1)^2 < ac \Rightarrow 2 < ac$. $a=1: c=3,4$ (2 pairs) $a=2: c=2,3,4$ (3 pairs) $a=3: c=1,2,3,4$ (4 pairs) $a=4: c=1,2,3,4$ (4 pairs) Total for $b=1$: $2+3+4+4 = 13$ pairs. Case $b=2$: $2(2)^2 < ac \Rightarrow 8 < ac$. $a=1: ac < 8$, no $c$. $a=2: ac < 8$, no $c$. $a=3: c=3,4$ (2 pairs: $(3,3), (3,4)$) $a=4: c=3,4$ (2 pairs: $(4,3), (4,4)$) Total for $b=2$: $2+2 = 4$ pairs. Case $b=3$: $2(3)^2 < ac \Rightarrow 18 < ac$. $a=1,2,3,4$: Max $ac = 4 imes 4 = 16$. No pairs satisfy $ac > 18$. Total for $b=3$: 0 pairs. Case $b=4$: $2(4)^2 < ac \Rightarrow 32 < ac$. $a=1,2,3,4$: Max $ac = 4 imes 4 = 16$. No pairs satisfy $ac > 32$. Total for $b=4$: 0 pairs. Total favorable outcomes = $13 + 4 + 0 + 0 = 17$ pairs. This matches the solution's total count of 17. So the probability $P = \frac{17}{64}$. $m=17, n=64$. $\mathrm{gcd}(17,64)=1$. $m+n = 17+64 = 81$. The solution's table for $b=3$ was wrong, but the total count of 17 was correct. This is strange. It seems the table was not fully consistent with the final count. However, my re-calculation confirms 17 favorable outcomes. Final answer $m+n = 81$.

Answer

91

Answer

81

Answer

71

Answer

61

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