Question

If x is real, find values of k for which $\left|\frac{x^2+kc+1}{x^2+x+1}\right|<3$ is valid.

Answer 1

Proceed as above and note that $x^2+1={(x+\frac{1}{2})}^2+\frac{3}{4}$
which is always +ive for all real values of x .
Hence proceeding as in 51 (b), we have $3x^2+(k+2)x+3>0$ and $x^2+(2-k)x+1>0$
Now the coefficients of $x^2$ in both are +ive i.e 3 and 1 therefore the signs of the quadratics will be +ive if $B^2-4AC<0$ for both.
$\therefore {(k+2)}^2-36<0or(k+8)(k-4)<0$
and ${(2-k)}^2-4<0$ or $(k-0)(k-4)<0$
i.e $0<k<4$
The common region of the above two is $0<k<4$