1. Find the domain of each basic function

Given: $f(x)=\sqrt{x}, \qquad g(x)=\sqrt{1-x}$

For square roots to be defined, the expressions inside must be non-negative.

• For $f(x)=\sqrt{x}$: $x\ge 0$ So, $D_f=[0,\infty)$

• For $g(x)=\sqrt{1-x}$: $1-x\ge 0 \implies x\le 1$ So, $D_g=(-\infty,1]$

2. Domain of $f+g$, $f-g$, and $g-f$

These require both $f(x)$ and $g(x)$ to be defined. So their common requirement is: $x\ge 0 \quad \text{and} \quad x\le 1$ Hence, $D_{f+g}=D_{f-g}=D_{g-f}=[0,1]$

3. Domain of $\dfrac{f}{g}$

We need:

• $f(x)$ defined
• $g(x)$ defined
• $g(x)\ne 0$

Now, $g(x)=\sqrt{1-x}=0 \iff 1-x=0 \iff x=1$

So for $\dfrac{f}{g}$: $0\le x\le 1, \quad x\ne 1$ Thus, $D_{f/g}=[0,1)$

4. Domain of $\dfrac{g}{f}$

We need:

• $f(x)$ defined
• $g(x)$ defined
• $f(x)\ne 0$

Now, $f(x)=\sqrt{x}=0 \iff x=0$

So for $\dfrac{g}{f}$: $0\le x\le 1, \quad x\ne 0$ Thus, $D_{g/f}=(0,1]$

5. Common domain of all the given functions

The functions listed are: $f+g,\quad f-g,\quad \frac{f}{g},\quad \frac{g}{f},\quad g-f$

Their domains are:

• $[0,1]$
• $[0,1]$
• $[0,1)$
• $(0,1]$
• $[0,1]$

The common domain is the intersection: $[0,1]\cap [0,1]\cap [0,1)\cap (0,1]\cap [0,1]=(0,1)$

So the correct option is: $\boxed{\text{C: } 0<x<1}$

6. Comparison with stored correct answer

Stored correct answer: C

Derived answer: C

They match.