1. Given functions

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

We need to study the composite function $f\circ g.$

2. Find the composite function

By definition, $(f\circ g)(x)=f(g(x))=g(x)-1.$

Substitute $g(x)$: $(f\circ g)(x)=\frac{x^2}{x^2-1}-1.$

Simplify:

So, $(f\circ g)(x)=\frac{1}{x^2-1}, \qquad x\in \mathbb{R}\setminus\{-1,1\}.$

3. Check whether $f\circ g$ is one-one

A function is one-one if different inputs always give different outputs.

Here, $(f\circ g)(x)=\frac{1}{x^2-1}.$

Since the expression depends on $x^2$, we have $(f\circ g)(x)=(f\circ g)(-x)$ for every $x$ in the domain where both are defined.

For example, $(f\circ g)(2)=\frac{1}{4-1}=\frac13,$ $(f\circ g)(-2)=\frac{1}{4-1}=\frac13.$

But $2\neq -2$, so the function is not one-one.

4. Check whether $f\circ g$ is onto

We must find the range of $y=\frac{1}{x^2-1}, \qquad x\in \mathbb{R}\setminus\{-1,1\}.$

Let $t=x^2.$ Since $x\in \mathbb{R}\setminus\{-1,1\}$, we have $t\ge 0, \quad t\neq 1.$

Then $y=\frac{1}{t-1}.$

Now analyze possible values:

• If $0\le t<1$, then $t-1\in[-1,0)$, so $y=\frac{1}{t-1}\in(-\infty,-1].$ In particular, at $t=0$, $y=-1.$

• If $t>1$, then $t-1>0$, so $y\in(0,\infty).$

Thus the range is $(-\infty,-1]\cup(0,\infty).$

So values in $(-1,0]$ are not attained. For example, there is no $x$ such that $\frac{1}{x^2-1}=-\frac12,$ because that would give $x^2-1=-2 \implies x^2=-1,$ which is impossible over $\mathbb{R}$.

Hence $f\circ g$ is not onto as a function into $\mathbb{R}$.

5. Conclusion

The function $f\circ g$ is:

• not one-one
• not onto

Therefore, the correct option is $\boxed{\text{D: neither one-one nor onto}}.$

6. Comparison with stored correct answer

Stored correct answer: D

Our derived answer is also D, so they agree.