Browse Questions

# For all real values of $x$, the minimum value of $\large\frac{1-x+x^2}{1+x+x^2}$ is

$(A)\; 0 \quad (B)\; 1 \quad (C)\; 3 \quad (D)\; \frac{1}{3}$

Toolbox:
• $\large\frac{d}{dx}$$(x^n)=nx^{n-1} Step 1: Let y=\large\frac{1-x+x^2}{1+x+x^2} \large\frac{dy}{dx}=\frac{(-1+2x)(1+x+x^2)-(1-x+2x)(1+2x)}{(1+x+x^2)^2} Numerator of \large\frac{dy}{dx}=$$(-1+2x)(1+x+x^2)-(1+2x)(1-x+x^2)$
$\Rightarrow (-1-x-x^2)+(2x+2x^2+2x^3)-(1-x+x^2)-(2x-2x^2+2x^3)$
$\Rightarrow -2+2x^2=2(x^2-1)$
$\Rightarrow 2(x-1)(x+1)$
Step 2:
$\large\frac{dy}{dx}=\frac{2(x-1)(x+1)}{(x^2+x+1)^2}$