Question

The minimum and maximum values of the expression $\frac{x^2}{x^2+x+1},$ $x\in R$ are

• A. $0,\frac{3}{4}$
• B. $0,\frac{4}{3}$
• C. $0,1$
• D. $0,3$

Answer 1

The correct option is B $0,\frac{4}{3}$

Let $y=\frac{x^2}{(x^2+x+1)}$
$\Rightarrow x^2(y-1)+yx+y$
Since maximum and minimum occurs at real $x$, so$\Delta \ge 0$
$y^2-4(y-1)y\ge 0$
$4y-3y^2\ge 0$
$y(4-3y)\ge 0$
$\Rightarrow y\in (0,\frac{4}{3})$
So, the minimum is $0$ and maximum is $\frac{4}{3}$.