If $x$ is real, then the minimum value of $y=\frac{x^2-x+1}{x^2+x+1}$ is
If $x$ is real, then the minimum value of $y=\frac{x^2-x+1}{x^2+x+1}$ is
- $3$
- $\frac{1}{3}$
- $\frac{1}{3}$
- $2$
Solution
Let $y=\frac{1-x+x^2}{1+x+x^2}$
On differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{d y}{d x} & =\frac{\left(1+x+x^2\right)(-1+2 x)-\left(1-x+x^2\right)(1+2 x)}{\left(1+x+x^2\right)^2} \\
& =\frac{-1+2 x-x+2 x^2-x^2+2 x^3-1-2 x+x}{\left(1+x+x^2\right)^2} \\
& =\frac{-2+2 x^2-x^2-2 x^3}{\left(1+x+x^2\right)^2} \\
& =\frac{2 x^2-2}{\left(1+x+x^2\right)^2}
\end{aligned}
$
$
=\frac{2\left(x^2-1\right)}{\left(1+x+x^2\right)^2}
$
Put $\frac{d y}{d x}=0 \Rightarrow x^2=1 \Rightarrow x= \pm 1$
Now, $\frac{d^2 y}{d x^2}=\frac{\left.\left(1+x+x^2\right)(1+2 x)\right]}{\left(1+x+x^2\right)^4}$
$
\begin{aligned}
& =\frac{4\left(1+x+x^2\right)\left[\left(1+x+x^2\right) x-\left(x^2-1\right)(1+2 x)\right]}{\left(1+x+x^2\right)^4} \\
& =\frac{4\left[x+x^2+x^3-x^2-2 x^3+1+2 x\right]}{\left(1+x+x^2\right)^3} \\
& =\frac{4\left(1+3 x-x^3\right)}{\left(1+x+x^2\right)^3}
\end{aligned}
$
At $x=1$,
$
\begin{aligned}
\left(\frac{d^2 y}{d x^2}\right)_{x=1} & =\frac{4\left[1+3(1)-1^3\right]}{\left(1+1+1^2\right)^3} \\
& =\frac{4(3)}{3^3}=\frac{4}{9}>0
\end{aligned}
$
At $x=-1$,
$
\begin{aligned}
& \left(\frac{d^2 y}{d x^2}\right)_{x=-1}=\frac{4\left[1+3(-1)-(-1)^3\right]}{\left[1+(-1)+(-1)^2\right]^3} \\
& =\frac{4(1-3+1)}{(1-1+1)^3}=4(-1)=-4 < 0
\end{aligned}
$
$\therefore$ By second derivative test, $f$ is minimum at $x=1$ and the minimum value is given by
$
y=\frac{1-1+1}{1+1+1}=\frac{1}{3}
$
Asked in: AP EAMCET 2014
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