CAT 2022

Slot 3

QA

Question & Solution

$2x^{2}+kx+5=0$ has no real roots so D<0

$k^2-40\ <0$

$\left(k-\sqrt{40}\right)\left(k+\sqrt{40}\right)<0$

$k\in\left(-\sqrt{40},\sqrt{40}\right)$

$x^{2}+(k-5)x+1=0$ has two distinct real roots so D>0

$\left(k-5\right)^2-4>0$

$k^2-10k+21>0$

$\left(k-3\right)\left(k-7\right)>0$

$k\in\left(-\infty\ ,3\right)∪\left(7,\infty\ \right)$

Therefore possibe value of k are -6, -5, -4, -3, -2, -1, 0, 1, 2

In 9 total 9 integer values of k are possible.