1. Rewrite the function in a simpler form

Given

$$f(x)=\frac{2^x-2^{-x}}{2^x+2^{-x}}.$$

Multiply numerator and denominator by $2^x$:

$$f(x)=\frac{2^{2x}-1}{2^{2x}+1}.$$

Let

$$t=2^{2x}>0.$$

Then

$$f(x)=\frac{t-1}{t+1}.$$

2. Check whether $f$ is one-one

We show that $f$ is strictly increasing.

Using

$$f(x)=\frac{2^{2x}-1}{2^{2x}+1},$$

let $u=2^{2x}$. Then $u'>0$ since

$$u' = 2^{2x}\cdot 2\ln 2 >0.$$

Now

$$f(x)=\frac{u-1}{u+1}.$$

Differentiate:

$$f'(x)=\frac{u'(u+1)-(u-1)u'}{(u+1)^2} =\frac{2u'}{(u+1)^2}.$$

Since $u'>0$ and $(u+1)^2>0$, we get

$$f'(x)>0 \quad \text{for all } x\in \mathbb R.$$

Hence $f$ is strictly increasing, so it is one-one.

3. Find the range of $f$

Since $t>0$,

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

Now check limiting values:

• As $x\to \infty$, $t=2^{2x}\to \infty$, so $$$$

f(x)\to 1.

- As $x\to -\infty$, $t=2^{2x}\to 0^+$, so

f(x)\to \frac{-1}{1}=-1.

$$Thus the actual range is$$

(-1,1).

$$4. **Check onto-ness with given codomain** The codomain is given as$$

(-\infty,1).

$$But the actual range is only$$

(-1,1).

So values less than or equal to $-1$ are not attained. Hence $f$ is **not onto** the codomain $(-\infty,1)$. 5. **Conclusion** - $f$ is **one-one** - $f$ is **not onto** Therefore the correct option is:

\boxed{\text{A: One-one but not onto}}

$$6. **Comparison with stored answer** Stored correct answer: **A** This matches our derived answer.$$