Question

If $x$ is real, find the range of $y$ from the equation $x^2(y-1)-2x+(2y-1)=0$

• A. $(-\infty ,0]$
• B. $[0,\frac{7}{2}]$
• C. $[0,2]$
• D. $[0,\frac{3}{2}]$

Answer 1

The correct option is D

$[0,\frac{3}{2}]$

If $x$ is real,

$D\ge 0$

$b^2-4ac\ge 0$

$\Rightarrow 4-4(y-1)(2y-1)\ge 0$

$\Rightarrow 1-(y-1)(2y-1)\ge 0$

$\Rightarrow (y-1)(2y-1)\le 1$

$\Rightarrow 2y^2-3y\le 0$

$y(2y-3)\le 0$

$\Rightarrow (y-0)(y-(\frac{3}{2}\left)\right)\le 0$

$0\le y\le$ $\frac{3}{2}$