Question

For all 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

Answer: (B)

$\frac{1}{3}$

Let $y=\frac{1-x+x^2}{1+x+x^2}=1-\frac{2x}{1+x+x^2}$
$=1-\frac{2}{\frac{1}{x}+1+x}$
Let $y=1-\frac{2}{t}$, where $t=\frac{1}{x}+1+x$
Now, $y$ is minimum, when $\frac{2}{t}$ is max $\Rightarrow$ $t$ is min.
$\therefore \frac{dt}{dx}=-\frac{1}{x^2}+1=0$
$\Rightarrow x=\pm 1$

$\Rightarrow \frac{d^{2}t}{d{x}^2}=\frac{2}{x^3}>0,\text{for}x=1$
$\therefore$ Minimum value of $y$ is $1-\frac{2}{1+1+1}=1-\frac{2}{3}$ $=\frac{1}{3}$