Question

The number of integer values of the parameter $k$ in the inequality $\left|\frac{x^2+kx+1}{x^2+x+1}\right|<3$ satisfied for all real values of $x$ are

Answer 1

$\left|\frac{x^2+kx+1}{x^2+x+1}\right|<3$
$-3<\frac{x^2+kx+1}{x^2+x+1}<3$
Since, $x^2+x+1={(x+\frac{1}{2})}^2+\frac{3}{4}>0,$
$\therefore -3(x^2+x+1)<x^2+kx+1<3(x^2+x+1)$ .................(1)
${4x}^2+(k+3)x+4>0$ and ${2x}^2-(k-3)x+2>0$ ...................(2)
$\because$ $4>0$ and $2>0$
Thus, inequality $\left(1\right)$ will be valid.
If ${(k+3)}^2-\mathrm{4.4.4}<0$ or $-11<k<5$.............. (3)
and the inequality $\left(2\right)$ will be valid.
If ${(k-3)}^2-\mathrm{4.2.2}<0$ or $-1$ .............(4)
The condition (3) and (4) will hold simultaneously if $-1<k<5$.
$k=0,1,2,3,4$.
So, $5$ values.