Question

Find the values of $x$ which satisfy $\left|\frac{x^2}{x-1}\right|\le 1$

• A. $x\in \left[\frac{-1-\sqrt{3}}{2},\frac{-1+\sqrt{3}}{2}\right]$
• B. $x\in \left[\frac{-1-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2}\right]$
• C. $x\in \left[\frac{-1-\sqrt{7}}{3},\frac{-1+\sqrt{7}}{3}\right]$
• D. None of these

Answer 1

Answer: (B)

$x\in \left[\frac{-1-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2}\right]$

As $\left|\frac{x^2}{x-1}\right|\le 1$
$\Rightarrow \left|x^2\right|\le |x-1|,\forall x\in R-\left\{1\right\}$or $x^2\le |x-1|,\forall x\in R-\left\{1\right\}$
Thus, to find the points for which $f\left(x\right)=x^2$ is less than or equal to $g\left(x\right)=|x-1|$.
Where the two functions $f\left(x\right)$ and $g\left(x\right)$ could be plotted as shown;
Thus from the graph, $f\left(x\right)\le g\left(x\right)$ when $x\in [A,B]$ where $A$ and $B$ are points of intersection of $x^2$ and $1-x$.
$\therefore$ Solving $x^2=1-x,$ we get $x=\frac{-1-\sqrt{5}}{2}=A$
$x=\frac{-1+\sqrt{5}}{2}=B$
$\therefore$ $\left|\frac{x^2}{x-1}\right|\le 1$ is satisfied, $\forall x\in \left[\frac{-1-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2}\right]$