Question

The set of values of $K$ for which both the roots of the equation $4x^2-20Kx+(25K^2+15K-66)=0$ are less than $2$ is given by

• A. $(2,\infty )$
• B. $(\frac{4}{5},2)$
• C. $(-\infty ,-1)$
• D. None of these

Answer 1

Answer: (C)

$(-\infty ,-1)$

Roots of the equation $4x^2-20Kx+(25K^2+15K-66)=0$ are less than $2$
i.e $\alpha +\beta <4$ and as $a>0$ $f\left(2\right)>0$
$\Rightarrow \frac{20k}{4}<4$
$\Rightarrow k<\frac{4}{5}$ -------(1)

$f\left(2\right)=16-40k+25k^2+5k-66>0$
$\Rightarrow k^2-k-2>0$
$\Rightarrow k\in (-\infty ,-1)\cup (2,\infty )$ ----(2)

Also, $D\ge 0$

$D=400K^2-16(25K^2+15K-66)\ge 0$

$k\le \frac{22}{5}$ ... (3)

From (1), (2) and (3)
$k\in (-\infty ,-1)$
Hence, option C.