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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

ax^2-bx+c=0

with

a, b, c\in\mathbb{N}

. If it has two distinct real roots in

(1,2)

, then

1<a<5

a\ge 5

correct

a=4

a=3

Solution

Step 1: Let the roots be

\alpha,\beta\in(1,2)

, with

\alpha+\beta=\frac{b}{a}

and

\alpha\beta=\frac{c}{a}

. Then

\alpha-1,\ 2-\alpha,\ \beta-1,\ 2-\beta

all lie in

(0,1)

. Step 2: Since

(\alpha-1)+(2-\alpha)=1

, AM–GM gives

(\alpha-1)(2-\alpha)\le\frac14

, and likewise for

\beta

. Multiplying (strict, as the roots differ):

(\alpha-1)(2-\alpha)(\beta-1)(2-\beta)<\frac{1}{16}.

Step 3: Regroup and write through the coefficients:

(\alpha-1)(\beta-1)=\alpha\beta-(\alpha+\beta)+1=\frac{c}{a}-\frac{b}{a}+1=\frac{a-b+c}{a},

(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

\frac{(a-b+c)(4a-2b+c)}{a^2}<\frac{1}{16}.

a,b,c\in\mathbb N

, both

a-b+c

and

4a-2b+c

are positive integers

\Rightarrow

their product is

\ge 1.

Step 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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