If f(x) = loge (1 − x) and g(x) = [x],

Given:

f(x) = loge (1 − x) and g(x) = [x]

Clearly, f(x) = loge (1 − x)  is defined for all ( 1 -">−- x)  > 0.

$\Rightarrow 1>x$

$\Rightarrow x<1$

$\Rightarrow x \in(-\infty, 1)$

Thus, domain (f ) = ( -">−- ∞, 1)

Again,

g(x) = [x] is defined for all x ∈ R.

Thus, domain (g) = R

$\therefore$ Domain $(f) \cap$ Domain $(g)=(-\infty, 1) \cap \mathrm{R}$

$=(-\infty, 1)$

Hence,

(i ) $(f+g):(-\infty, 1) \rightarrow R$ is given by $(f+g)(x)=f(x)+g(x)=\log _{e}(1-x)+[x]$.

(ii) $(f g):(-\infty, 1) \rightarrow R$ is given by $(f g)(x)=f(x) \cdot g(x)=\log _{e}(1-x)[x]=[x] \log _{e}(1-x)$.

(iii) Given:

$g(x)=[x]$

If $[x]=0$,

$x \in(0,1)$

Thus,

domain $\left(\frac{f}{g}\right)=$ domain $(f) \cap$ domain $(g)-\{x: g(x)=0\}$

$\frac{f}{g}:(-\infty, 0) \rightarrow R$ is defined by $\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\log _{e}(1-x)}{[x]} .$

(iv) Given:

$f(x)=\log _{e}(1-x)$

$\Rightarrow \frac{1}{f(x)}=\frac{1}{\log _{e}(1-x)}$

$\frac{1}{f(x)}$ is defined if $\log _{e}(1-x)$ is defined and $\log _{e}(1-x) \neq 0$

$\Rightarrow(1-x)>0$ and $(1-x) \neq 0$

$\Rightarrow x<1$ and $x \neq 0$

$\Rightarrow x \in(-\infty, 0) \cup(0,1)$

Thus, d omain $\left(\frac{g}{f}\right)=(-\infty, 0) \cup(0,1)=(-\infty, 1)$.

$\frac{g}{f}:(-\infty, 0) \cup(0,1) \rightarrow R$ defined by $\left(\frac{g}{f}\right)(x)=\frac{g(x)}{f(x)}=\frac{[x]}{\log _{e}(1-x)}$

$(f+g)(-1)=f(-1)+g(-1)$

$=\log _{e}\{1-(-1)\}+[-1]$

$=\log _{e} 2-1$

Hence, $(f+g)(-1)=\log _{e} 2-1$

$(f g)(0)=\log _{e}(1-0) \times[0]=0$

$\left(\frac{f}{g}\right)\left(\frac{1}{2}\right)=$ does not exist.

$\left(\frac{g}{f}\right)\left(\frac{1}{2}\right)=\frac{\left[\frac{1}{2}\right]}{\log _{e}\left(1-\frac{1}{2}\right)}=0$