Question

If $x$ is real, then the minimum value of $\left[\frac{x^2-x+1}{x^2+x+1}\right]$is

• A. $\frac{1}{3}$
• B. $3$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is A

$\frac{1}{3}$

Explanation for the correct answer:

Step-1: Simplifying expression:

Let $z=$$\left[\frac{x^2-x+1}{x^2+x+1}\right]$

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

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

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

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

$z$ will be minimum when $y$ will be maximum

Step-2 : Maximum value of $y$:

Differentiating $y$ with respect to $x$ we get,

$\frac{dy}{dx}=\frac{\left(x^2+x+1\right)\left(2\right)-2x\left(2x+1\right)}{{\left(x^2+x+1\right)}^2}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \frac{d}{\mathrm{dx}}\left(\frac{u}{v}\right)=\frac{v{\displaystyle \frac{\mathrm{du}}{\mathrm{dx}}}-u{\displaystyle \frac{\mathrm{dv}}{\mathrm{dx}}}}{v^2}\right]$

$\Rightarrow$$\frac{dy}{dx}=\frac{2\left[\left(x^2+x+1\right)-\left(2x^2+x\right)\right]}{{\left(x^2+x+1\right)}^2}$

$\Rightarrow$$\frac{dy}{dx}=\frac{2\left[-x^2+1\right]}{{\left(x^2+x+1\right)}^2}$

The maximum value of $y$ occurs at $\frac{dy}{dx}=0$

$\Rightarrow$$\frac{dy}{dx}=\frac{2\left[-x^2+1\right]}{{\left(x^2+x+1\right)}^2}=0$

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

$\Rightarrow$ $x^2-1=0$

$\Rightarrow$ $x=1,-1$

$\frac{dy}{dx}$ changes sign from $+veto-ve$around $x=1$.

Therefore, at $x=1$,$y$ will attain maximum value.

When $x=1$

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

$\Rightarrow$ $z=\frac{1}{3}$

When $x=-1$

$z=\frac{{\left(-1\right)}^2-\left(-1\right)+1}{{\left(-1\right)}^2-1+1}$

$\Rightarrow$ $z=3$

As $\frac{1}{3}<3$ we can say that

$z_{min}=\frac{1}{3}$

Therefore, for real values of $x$ the minimum value of $\frac{\left[x^2-x+1\right]}{\left[x^2+x+1\right]}$is $\frac{1}{3}$.

Hence option (A) is the correct answer.