Question

If $f\left(x\right)=ln(x^2-x+2);R^{+}\to R$ and $g\left(x\right)=\left\{x\right\}+1;[1,2]\to [1,2]$, where $\left\{x\right\}$ denotes fractional part of $x$. Find the domain and range of $f\left(g\right(x\left)\right)$ when defined.

Answer 1

Given,

$f\left(x\right)=\ell n\left(x^2-x+2\right);R^{+}\to R$

$g\left(x\right)=\left\{x\right\}+1;\left[1,2\right]\to \left[1,2\right]$

Dom $f\left(g\left(x\right)\right)\&$ Range $f\left(g\left(x\right)\right)=?$

$f\left(g\left(x\right)\right)={\left(g\left(\right)\right)}^2-g\left(x\right)+2$

Domain of $+\left(g\right(x\left)\right)=$ Domain of $\left(g\right(x\left)\right)=[1,2]$

and $g\left(x\right)=\left[x\right]+1$

$$\begin{array}{c}=\alpha +1\\ \because \alpha =\left[x\right]\end{array}$$

$$\begin{array}{c}f\left(g\left(x\right)\right)=\ell n{\left(\alpha +1\right)}^2-\left(\alpha +1\right)+2\\ =\ell n\left({\alpha }^2+2\alpha +1-\alpha -1+2\right)\\ =\ell n\left({\alpha }^2+\alpha +2\right)\end{array}$$

Now,

$$\begin{array}{c}\alpha =\left[x\right]\\ 0⩽{\alpha }^2<1----\left(i\right)\\ 0⩽\alpha <1----\left(ii\right)\end{array}$$

Adding (i) and (ii)

$$\begin{array}{c}0⩽{\alpha }^2+\alpha <2\\ 2⩽{\alpha }^2+\alpha 2<4\\ Range\left[\log 2,\log 4\right]\\ \log 2⩽\log {\alpha }^2+\alpha +2<\log 4\end{array}$$