Question

For al real $x$, the minimum value of $\frac{1-x+x^2}{1+x+x^2}$ is

• A. $0$
• B. $\frac{1}{3}$
• C. $1$
• D. $3$

Answer 1

The correct option is B $\frac{1}{3}$

Let $z=\frac{1-x+x^2}{1+x+x^2}$

$\Rightarrow z+zx+z{x}^2=1-x+x^2$

$\Rightarrow z{x}^2-x^2+zx+z+z-1=0$

$\Rightarrow x^2(z-1)+x(z+1)+(z-1)=0$

For real $x,B^2-4AC\ge 0$

$\Rightarrow {(z+1)}^2-4(z-1)(z-1)\ge 0$

$\Rightarrow z^2+2z+1-4z^2+8z-4\ge 0$

$\Rightarrow -3z^2+10z-3\ge 0\Rightarrow -3z^2+9z+z-3\ge 0$

$\Rightarrow -3z(z-3)+1(z-3)\ge 0$

$\Rightarrow (z-3)(-3z+1)\ge 0\Rightarrow \frac{1}{3}\le z\le 3.$

$\therefore$minimum valve of $z=\frac{1}{3}.$