Question

If the roots of $x^2-6kx+(2-2k+9k^2)=0$ are greater than $3$, then the range of $k$ is

• A. $(-\infty ,1)\cup (\frac{11}{9},\infty )$
• B. $(1,\frac{11}{9})$
• C. $(1,\infty )$
• D. $(\frac{11}{9},\infty )$

Answer 1

The correct option is D $(\frac{11}{9},\infty )$

Let $f\left(x\right)=x^2-6kx+(2-2k+9k^2)$, whose roots are $\alpha ,\beta$
Given: $\alpha ,\beta >3$

Conditions:
$\left(i\right)D\ge 0\Rightarrow 36k^2-4(2-2k+9k^2)\ge 0$
$\Rightarrow 36k^2-8+8k-36k^2\ge 0$
$\Rightarrow 8k\ge 8$
$\Rightarrow k\ge 1\dots \left(1\right)$

$\left(ii\right)-\frac{b}{2a}>3\text{Here,}a=1,b=-6k\Rightarrow \frac{6k}{2}>3\Rightarrow k>1\dots \left(2\right)$

$\left(iii\right)f\left(3\right)>03^2-18k+2-2k+9k^2>0\Rightarrow 9k^2-20k+11>0$
$\Rightarrow (k-1)(9k-11)>0$
$\Rightarrow k<1\text{or}k>\frac{11}{9}\dots \left(3\right)$

Taking common region from all the three regions $\left(1\right),\left(2\right)\&\left(3\right),$ we get
$k\in (\frac{11}{9},\infty )$