Question

The number of integral values of $k$ for which $\left|\frac{x^2+kx+1}{x^2+x+1}\right|<2$ for all real values of $x,$ is

Answer 1

$\left|\frac{x^2+kx+1}{x^2+x+1}\right|<2$
$\Rightarrow -2<\frac{x^2+kx+1}{x^2+x+1}<2$

For $x^2+x+1,$
$D=1-4=-3<0$
So, this expression is always positive.
Hence, by multiplying $(x^2+x+1)$ on both the sides of inequality,
$-2(x^2+x+1)<x^2+kx+1<2(x^2+x+1)$
This gives us two inequalities
$3x^2+(2+k)x+3>0$ and $x^2+(2-k)x+1>0$

For both these inequations to be positive for all real $x,$ discriminant must be negative.
$(2+k{)}^2-36<0,(2-k{)}^2-4<0$
$\Rightarrow (k+8)(k-4)<0,k(k-4)<0$
$\Rightarrow -8<k<4,0<k<4$
$\Rightarrow 0<k<4$
Possible integral values of $k$ are $1,2,3$