Question

Solve the equation: $[y{]}^3+2[y{]}^2=\left[y\right]+2$, where $[.]$ denotes GIF.

• A. None of the above.
• B. $x\in [-2,2)-[0,1)$
• C. $x\in (-2,2]-[0,1)$
• D. $x\in [-2,2)-(0,1]$

Answer 1

The correct option is B $x\in [-2,2)-[0,1)$

Given: $[y{]}^3+2[y{]}^2=\left[y\right]+2$
$\Rightarrow [y{]}^3+2[y{]}^2-\left[y\right]-2=0$

Let $\left[y\right]=t$
$\Rightarrow t^3+2t^2-t-2=0$
$\Rightarrow t^3-t^2+3t^2-3t+2t-2=\mathrm{0...}\left(b\right)$
$\Rightarrow t^2(t-1)+3t(t-1)+2(t-1)=\mathrm{0....}\left(a\right)$
$\Rightarrow (t-1)(t^2+3t+2)=0$
$\Rightarrow (t-1)(t^2+2t+t+2)=0$
$\Rightarrow (t-1)\left\{t\right(t+2)+1(t+2\left)\right\}=0$
$\Rightarrow (t-1)(t+2)(t+1)=0$

Either $t=1,t=-2$ or $t=-1$

For $t=1,$
$\left[y\right]=1$
Apply $\left[y\right]\le y<\left[y\right]+1$
$\therefore 1\le y<1+1$
$\Rightarrow 1\le y<2$
$\Rightarrow y\in [1,2)$

For $t=-1,$
$\left[y\right]=-1$
$\therefore -1\le y<-1+1$
$\Rightarrow -1\le y<0$
$\Rightarrow y\in [-1,0)$

For $t=-2,$
$\left[y\right]=-2$
$\therefore -2\le y<-2+1$
$\Rightarrow -2\le y<-1$
$\Rightarrow y\in [-2,-1)$

Hence, $y\in [-2,-1)\cup [-1,0)\cup [1,2)$
$\Rightarrow y\in [-2,0)\cup [1,2)$
$\Rightarrow y\in [-2,2)-[0,1)$