Question

The value of $k$ for which both the roots of the equation $4x^{2}-20kx+\left(25 k^{2} + 15 k - 66\right)=0$ 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, $f\left(x\right)=4x^{2}-20kx+\left(25 k^{2} + 15 k - 66\right)=0\ldots \ldots \ldots \ldots \ldots ..\left(\right.i\left.\right)$
Let the roots of $f\left(x\right)=0 \, be \, \alpha , \, \beta $
Since $\alpha , \, \beta $ are real.
$\therefore \, \Delta \geq 0$
$\Rightarrow 400k^{2}-4.4\left(25 k^{2} + 15 k - 66\right)\geq 0$
$\Rightarrow \, -15k+66\geq 0 \, \Rightarrow \, k\leq \frac{22}{5}\ldots \ldots \ldots \ldots ..\left(i i\right)$
We have $\alpha , \, \beta < 2$
$\therefore \, \, \alpha +\beta < 4$
$\Rightarrow \, \, -\frac{\left(- 20 k\right)}{4} < 4 \, \, \Rightarrow k < \frac{4}{5}\ldots \ldots \ldots \ldots \ldots \left(i i i\right)$
$f\left(x\right)=4\left(x - \alpha \right)\left(x - \beta \right)$
$\therefore \, \, \, f\left(2\right)=4\left(2 - \alpha \right)\left(2 - \beta \right)=4\left(+\right)\left(+\right)=+ve$
$\therefore \, \, f\left(2\right)=16-40k+\left(25 k^{2} + 15 k - 66\right)>0$
$\Rightarrow 25k^{2}-25k-50>0 \, \, \, \Rightarrow \, \, k^{2}-k-2>0$
$\Rightarrow \, \, \left(k + 1\right)\left(k - 2\right)>0 \, \, \, \Rightarrow \, k < -1 \, \, or \, k>2\ldots \ldots \ldots \ldots \left(i v\right)$
Combining $\left(i i\right), \, \left(i i i\right) \, \& \, \left(i v\right), \, we \, get \, k\in \left(- \in fty , - 1\right)$