Question

Consider the quadratic equation $a{x}^2-bx+c=0,a,b,c\in N$, which has two distinct real roots belonging to the interval $(1,2)$.

The least value of $c$ is

• A. $4$
• B. $6$
• C. $7$
• D. $5$

Answer 1

The correct option is B $6$

Let $f\left(x\right)=a{x}^2-bx+c$

As $\alpha ,\beta$ are roots, gives

$f\left(x\right)=a(x-\alpha )(x-\beta )$

Then

$af\left(1\right)>0\Rightarrow a(1-\alpha )(1-\beta )>0$ ...(1)

$af\left(2\right)>0\Rightarrow a(2-\alpha )(2-\beta )>0$ ...(2)

From (1) and (2), we have

$af\left(1\right).af\left(2\right)>0\Rightarrow a^2(1-\alpha )(1-\beta )(2-\alpha )(2-\beta )>0$

$\Rightarrow a^2(\alpha -1)(2-\alpha )(\beta -1)(2-\beta )>0$

As $f\left(1\right)>0$ and $f\left(2\right)>0$

$\Rightarrow f\left(1\right)f\left(2\right)>0\Rightarrow f\left(1\right)f\left(2\right)\ge 1$

$\Rightarrow a^2(\alpha -1)(2-\alpha )(\beta -1)(2-\beta )\ge 1$

Now applying $A.M\ge G.M$ on $(\alpha -1)$ and $(2-\alpha )$

$\frac{(\alpha -1)+(2-\alpha )}{2}\ge \left(\right(\alpha -1\left)\right(2-\alpha ){)}^{1/2}\Rightarrow (\alpha -1\left)\right(2-\alpha )\le \frac{1}{4}$ ...(3)

Similarly $(\beta -1)(2-\beta )\le \frac{1}{4}$ ...(4)

From (3) and (4), we have
$(\alpha -1)(2-\alpha )(\beta -1)(2-\beta )<\frac{1}{16}$

As $\alpha \ne \beta$ gives $a^2>16\Rightarrow a\ge 5$

So substituting values of $b$ and $a$, we get

$\Rightarrow c>5\Rightarrow c\ge 6$