Question

The value of $k$ for which both the roots of the equation $4x^2-20kx+\left(25k^2+15k-66\right)=0$ are less than $2$ , lies in

• A. $\left(\frac{4}{5},2\right)$
• B. $\left(0,2\right)$
• C. $\left(-1,-\frac{4}{5}\right)$
• D. $\left(-\in fty,-1\right)$

Answer 1

Answer: (D)

Let, $f\left(x\right)=4x^2-20kx+\left(25k^2+15k-66\right)=0\dots \dots \dots \dots \dots ..(i)$
Let the roots of $f\left(x\right)=0be\alpha ,\beta$
Since $\alpha ,\beta$ are real.
$\therefore \Delta \ge 0$
$\Rightarrow 400k^2-4.4\left(25k^2+15k-66\right)\ge 0$
$\Rightarrow -15k+66\ge 0\Rightarrow k\le \frac{22}{5}\dots \dots \dots \dots ..\left(ii\right)$
We have $\alpha ,\beta <2$
$\therefore \alpha +\beta <4$
$\Rightarrow -\frac{\left(-20k\right)}{4}<4\Rightarrow k<\frac{4}{5}\dots \dots \dots \dots \dots \left(iii\right)$
$f\left(x\right)=4\left(x-\alpha \right)\left(x-\beta \right)$
$\therefore f\left(2\right)=4\left(2-\alpha \right)\left(2-\beta \right)=4\left(+\right)\left(+\right)=+ve$
$\therefore f\left(2\right)=16-40k+\left(25k^2+15k-66\right)>0$
$\Rightarrow 25k^2-25k-50>0\Rightarrow k^2-k-2>0$
$\Rightarrow \left(k+1\right)\left(k-2\right)>0\Rightarrow k<-1ork>2\dots \dots \dots \dots \left(iv\right)$
Combining $\left(ii\right),\left(iii\right)\&\left(iv\right),wegetk\in \left(-\in fty,-1\right)$