Question

Consider a quadratic polynomial $f\left(x\right)=x^2-4ax+5a^2-6a.$
$p:$ denotes smallest positive integral value of $a$, for which $f\left(x\right)$ is positive for every real $x$, and
$q:$ denotes largest distance between the roots of the equation $f\left(x\right)=0$
The value of $p+q$ is

• A. $9$
• B. $11$
• C. $13$
• D. $15$

Answer 1

The correct option is C $13$

We have, $f\left(x\right)=x^2-4ax+5a^2-6a$
For $f\left(x\right)>0\forall x\in R$
$D<0$
$\Rightarrow 16a^2-4(5a^2-6a)<0$
$\Rightarrow a^2-6a>0$
$\Rightarrow a\in (-\infty ,0)\cup (6,\infty )$
$\therefore p=7$
Now, the distance between the roots is given as
$|x_1-x_2|=\sqrt{(x_1+x_2{)}^2-4x_1{x}_2}$
where, $x_1$ and $x_2$ are the roots of the given equation.
Now, $|x_1-x_2|=\sqrt{16a^2-4(5a^2-6a)}$
$[\because x_1+x_2=4a,x_1{x}_2=5a^2-6a]$
$\Rightarrow |x_1-x_2|=2\sqrt{-a^2+6a}=2\sqrt{9-(a-3{)}^2}$
For $|x_1-x_2{|}_{max}\Rightarrow (a-3{)}^2=0$
$\therefore |x_1-x_2{|}_{max}=6$.
$\therefore q=6,\therefore p+q=13$.