If $\left|\frac{x^2+k x+1}{x^2+x+1}\right| < 3$ for all real $x$, then $k$ is in the interval
If $\left|\frac{x^2+k x+1}{x^2+x+1}\right| < 3$ for all real $x$, then $k$ is in the interval
$(-\infty,-1)$
$(-1,6)$
$(-1,5)$
$(6, \infty)$
Solution
The given inequality is $\left|\frac{x^2+k x+1}{x^2+x+1}\right| < 3$.
Let's first write this inequality as two separate inequalities:
1) $\frac{x^2+k x+1}{x^2+x+1} < 3$
2) $\frac{x^2+k x+1}{x^2+x+1} > -3$
Solving the first inequality, we get:
$x^2+k x+1 < 3x^2+3x+3$
This simplifies to:
$2x^2+(k-3)x+2 > 0$
This is a quadratic inequality. For this inequality to hold for all real $x$, the discriminant of the quadratic equation must be negative. The discriminant of a quadratic equation $ax^2+bx+c$ is given by $b^2-4ac$. So, we have:
$(k-3)^2-4(2)(2) < 0$
$k^2-6k+9-16 < 0$
$k^2-6k-7 < 0$
Factoring the quadratic, we get:
$(k-7)(k+1) < 0$
Solving this inequality, we find that $k$ lies in the interval $(-1,7)$.
Now, solving the second inequality, we get:
$x^2+k x+1 > -3x^2-3x-3$
This simplifies to:
$4x^2+(k+3)x+4 > 0$
Again, for this inequality to hold for all real $x$, the discriminant of the quadratic equation must be negative. So, we have:
$(k+3)^2-4(4)(1) < 0$
$k^2+6k+9-16 < 0$
$k^2+6k-7 < 0$
Factoring the quadratic, we get:
$(k-1)(k+7) < 0$
Solving this inequality, we find that $k$ lies in the interval $(-7,1)$.
Since $k$ must satisfy both inequalities, the intersection of the two intervals $(-1,7)$ and $(-7,1)$ gives the final solution. The intersection is $(-1,1)$.
Therefore, the value of $k$ lies in the interval $(-1,1)$. So, the correct option is C) $(-1,1)$.