Question

If ${ax}^2-bx+c=0$ have two distinct roots lying in the interval $(0,1);a,b,cϵN$, then the least value of ${\log }_{5}abc$, is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

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

It has two distinct real roots lying between 0 and 1.

$f\left(x\right)$ can be written as,$f\left(x\right)=a(x-\alpha )(x-\beta )$, where $\alpha$ and $\beta$ are the roots of the equation.

From the graph it can be observed that $f\left(0\right)f\left(1\right)>0$

As a,b and c are integers $f\left(0\right)f\left(1\right)>1$

$f\left(0\right)=a\alpha \beta$

$f\left(1\right)=a(1-\alpha )(1-\beta )$

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

Using $AM\ge GM$,we get

$\frac{\alpha +1-\alpha +\beta +1-\beta }{4}>\left(\alpha \beta \right(1-\alpha \left)\right(1-\beta ){)}^{\frac{1}{4}}$

AM and GM cannot be equal because the equation has distinct roots

$\frac{a^2}{16}>a^2\left(\alpha \beta \right(1-\alpha \left)\right(1-\beta \left)\right)$.

$\Rightarrow \frac{a^2}{16}>1$

$\Rightarrow a>4$

Therefore the minimum value of a is 5.

The discriminant of the equation, $\Delta =b^2-4ac\ge 0$

$\Rightarrow b^2>20c$

Taking the minimum value of c=1,

we get $b^2>20$

$\Rightarrow$ Minimum Value of b = 5.

$\Rightarrow abc=25$

$\Rightarrow lo{g}_{5}abc=2$

Therefore,Option B is correct.