1. Given functions

We have $f(x)=2x-1, \quad f: \mathbb{R}\to\mathbb{R}$ and $g(x)=\frac{x-\frac12}{x-1}, \quad g: \mathbb{R}\setminus\{1\}\to\mathbb{R}.$

We need to study the composition $f(g(x)).$

2. Find the composition

Substitute $g(x)$ into $f$: $f(g(x))=2\cdot \frac{x-\frac12}{x-1}-1.$

Now simplify: $f(g(x))=\frac{2x-1}{x-1}-1$ $=\frac{2x-1-(x-1)}{x-1}$ $=\frac{x}{x-1}.$

So, $f\circ g(x)=\frac{x}{x-1}, \quad x\neq 1.$

Thus the composed function is $h(x)=\frac{x}{x-1}, \quad h: \mathbb{R}\setminus\{1\}\to\mathbb{R}.$

3. Check whether $h(x)$ is one-one

Assume $h(x_1)=h(x_2).$ Then $\frac{x_1}{x_1-1}=\frac{x_2}{x_2-1}.$

Cross-multiplying, $x_1(x_2-1)=x_2(x_1-1).$

Expanding, $x_1x_2-x_1=x_1x_2-x_2.$

So, $-x_1=-x_2 \implies x_1=x_2.$

Hence, $h(x)$ is one-one.

4. Check whether $h(x)$ is onto

We must see whether for every $y\in\mathbb{R}$, there exists $x\in\mathbb{R}\setminus\{1\}$ such that $y=\frac{x}{x-1}.$

Solve for $x$: $y(x-1)=x$ $yx-y=x$ $x(y-1)=y$ $x=\frac{y}{y-1}.$

This is possible for all $y\neq 1$.

Now check $y=1$: $\frac{x}{x-1}=1$ $x=x-1,$ which is impossible.

So $y=1$ is not in the range.

Thus range of $h$ is $\mathbb{R}\setminus\{1\},$ not all of $\mathbb{R}$.

Hence $h$ is not onto (as codomain is $\mathbb{R}$).

5. Conclusion

The composition function $f(g(x))=\dfrac{x}{x-1}$ is:

• one-one
• not onto

Therefore, the correct option is: $\boxed{\text{A: one-one but not onto}}$

6. Comparison with stored answer

Stored correct answer: A

Our derived answer is also A, so they agree.