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Quadratic ax^2-bx+c, Roots in (1,2): Find a |

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

Question
Consider ax2−bx+c=0ax^2-bx+c=0 with a,b,c∈Na, b, c\in\mathbb{N}. If it has two distinct real roots in (1,2)(1,2), then
A1<a<51<a<5
Ba≥5a\ge 5correct
Ca=4a=4
Da=3a=3
Solution
Step 1: Let the roots be α,β∈(1,2)\alpha,\beta\in(1,2), with α+β=ba\alpha+\beta=\frac{b}{a} and αβ=ca\alpha\beta=\frac{c}{a}. Then α−1, 2−α, β−1, 2−β\alpha-1,\ 2-\alpha,\ \beta-1,\ 2-\beta all lie in (0,1)(0,1). Step 2: Since (α−1)+(2−α)=1(\alpha-1)+(2-\alpha)=1, AM–GM gives (α−1)(2−α)≤14(\alpha-1)(2-\alpha)\le\frac14, and likewise for β\beta. Multiplying (strict, as the roots differ):
(α−1)(2−α)(β−1)(2−β)<116.(\alpha-1)(2-\alpha)(\beta-1)(2-\beta)<\frac{1}{16}.
 Step 3: Regroup and write through the coefficients:
(α−1)(β−1)=αβ−(α+β)+1=ca−ba+1=a−b+ca,(\alpha-1)(\beta-1)=\alpha\beta-(\alpha+\beta)+1=\frac{c}{a}-\frac{b}{a}+1=\frac{a-b+c}{a},
(2−α)(2−β)=4−2(α+β)+αβ=4−2ba+ca=4a−2b+ca.(2-\alpha)(2-\beta)=4-2(\alpha+\beta)+\alpha\beta=4-\frac{2b}{a}+\frac{c}{a}=\frac{4a-2b+c}{a}.
 Step 4: So
(a−b+c)(4a−2b+c)a2<116.\frac{(a-b+c)(4a-2b+c)}{a^2}<\frac{1}{16}.
 As a,b,c∈Na,b,c\in\mathbb N, both a−b+ca-b+c and 4a−2b+c4a-2b+c are positive integers ⇒\Rightarrow their product is ≥1.\ge 1. Step 5:
1a2≤(a−b+c)(4a−2b+c)a2<116 ⇒ a2>16 ⇒ a≥5.\frac{1}{a^2}\le\frac{(a-b+c)(4a-2b+c)}{a^2}<\frac{1}{16}\ \Rightarrow\ a^2>16\ \Rightarrow\ a\ge 5.
 Correct answer: (2)
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