Question

Let $f\left(x\right)=x^2-1$ and $g\left(x\right)=\{\begin{array}{cc}\left[|f\right(|x|\left)|\right]+|\left[f\right(x\left)\right]|,& x\in (-1,0)\cup (0,1)\\ 1& otherwise\end{array}$.
Then find the range of $\ln \left(\right[|g\left(x\right)|\left]\right),$ where $[.]$ denotes the greatest integer function

Answer 1

$f\left(x\right)=x^2-1$

When $x\in \left(-1,0\right)\cup \left(0,1\right)$ , we have:

a) $\left|x\right|\in \left(0,1\right)$

$\Rightarrow 0<\left|x\right|<1$

$\Rightarrow 0<{\left|x\right|}^2<1$

$\Rightarrow -1<{\left|x\right|}^2-1<0$

$\Rightarrow -1<f\left(\left|x\right|\right)<0$

$\Rightarrow -1<\left|f\left(\left|x\right|\right)\right|<1$

$\left[\left|f\left(\left|x\right|\right)\right|\right]=0$

b) $-1<x<0$ and $0<x<1$

$\Rightarrow 0<x^2<1$ [ squaring eliminating negative values]

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

$\Rightarrow -1<f\left(x\right)<0$

$[f\left(x\right)=x^2-1]$

$\Rightarrow \left[f\left(x\right)\right]=-1$

$\Rightarrow \left|\left[f\left(x\right)\right]\right|=1$

Thus we get $g\left(x\right)=\left[\left|f\left(\left|x\right|\right)\right|\right]+\left|\left[f\left(x\right)\right]\right|$

$=0+1$

$=1$ when $x\in \left(-1,0\right)\cup \left(0,1\right)$

Also, given $g\left(x\right)=1$ otherwise

Thus, $g\left(x\right)=1$ throughout $x\in$

Now, $\left[\left|g\left(x\right)\right|\right]=\left[\left|1\right|\right]=\left[1\right]=1\forall x\in$

Then $\ln \left(\left[\left|g\left(x\right)\right|\right]\right)=\ln \left(1\right)=0$

Range of ln $\left(\left[\left|g\left(x\right)\right|\right]\right)=\left\{0\right\}$