Consider the quadratic equation ax² – bx + c = 0, a ,b , c $\in$ N. If the given equation has two distinct real roots belonging to (1, 2) then

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1 < a < 5

a = 4

a = 3

$a \geq 5$

answer is B.

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Detailed Solution

$α - 1, β - 1, 2 - α, 2 - β \in (0, 1), use$

$AM \geq GM \frac{(α - 1) + (2 - α)}{2} \geq \sqrt{(α - 1) (2 - α)}$

$\begin{array}{l} \Rightarrow (α - 1) (2 - α) < \frac{1}{4}, (β - 1) (2 - β) < \frac{1}{4} \\ \Rightarrow 0 < (α - 1) (β - 1) (2 - α) (2 - β) < \frac{1}{16} \\ \Rightarrow 0 < (\frac{c}{a} - \frac{b}{a} + 1) (4 - \frac{2 b}{a} + \frac{c}{a}) < \frac{1}{16} \\ 0 < (a - b + c) (4 a - 2 b + c) < \frac{a^{2}}{16} \\ (a - b + c) (4 a - 2 b + c) \geq 1 \Rightarrow \frac{a^{2}}{16} > 1 \Rightarrow a \geq 5 \end{array}$