1. Given function

We have

$$f(x)= \begin{cases} -a, & -a\le x\le 0,\\[4pt] x+a, & 0<x\le a, \end{cases} \qquad a>0.$$

And

$$g(x)=\frac{f(|x|)-|f(x)|}{2}, \qquad x\in[-a,a].$$

We must determine whether $g:[-a,a]\to[-a,a]$ is one-one and/or onto.

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2. Find $f(|x|)$

Since $|x|\in[0,a]$, we use the second branch when $|x|>0$, and first branch at $|x|=0$.

• If $x=0$, then $|x|=0$, so $f(|x|)=f(0)=-a.$
• If $x\ne 0$, then $|x|\in(0,a]$, so $f(|x|)=|x|+a.$

Thus,

$$f(|x|)= \begin{cases} -a, & x=0,\\[4pt] |x|+a, & x\ne 0. \end{cases}$$

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3. Find $|f(x)|$

We consider the two regions of $x$.

Case 1: $-a\le x\le 0$

Then $f(x)=-a,$ so $|f(x)|=a.$

Case 2: $0<x\le a$

Then $f(x)=x+a>0,$ so $|f(x)|=x+a.$

Hence,

$$|f(x)|= \begin{cases} a, & -a\le x\le 0,\\[4pt] x+a, & 0<x\le a. \end{cases}$$

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4. Compute $g(x)$

Now evaluate $g(x)=\frac{f(|x|)-|f(x)|}{2}.$

Case 1: $-a\le x<0$

Here $|x|=-x>0$, so $f(|x|)=|x|+a=-x+a.$ Also, $|f(x)|=a.$ Therefore, $g(x)=\frac{(-x+a)-a}{2}=\frac{-x}{2}.$

Case 2: $x=0$

$f(|0|)=f(0)=-a, \qquad |f(0)|=|-a|=a.$ So $g(0)=\frac{-a-a}{2}=-a.$

Case 3: $0<x\le a$

Then $|x|=x$, so $f(|x|)=f(x)=x+a,$ and since this is positive, $|f(x)|=x+a.$ Thus, $g(x)=\frac{(x+a)-(x+a)}{2}=0.$

So the function is

$$g(x)= \begin{cases} \dfrac{-x}{2}, & -a\le x<0,\\[6pt] -a, & x=0,\\[6pt] 0, & 0<x\le a. \end{cases}$$

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5. Check whether $g$ is one-one

A function is one-one if distinct inputs give distinct outputs.

But for every $x\in(0,a]$, $g(x)=0.$ So many different values of $x$ have the same image $0$.

For example, $g\left(\frac a2\right)=0, \qquad g(a)=0.$ Since $\frac a2\ne a$, $g$ is not one-one.

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6. Check whether $g$ is onto $[-a,a]$

We find the range of $g$.

• For $-a\le x<0$, $g(x)=\frac{-x}{2}$ and since $x\in[-a,0)$, we get $g(x)\in\left(0,\frac a2\right].$ (At $x=-a$, value is $a/2$; as $x\to 0^-$, value tends to $0$ but does not equal it from this branch.)

• At $x=0$, $g(0)=-a.$

• For $0<x\le a$, $g(x)=0.$

Therefore the range is

$$\operatorname{Range}(g)=\{-a\}\cup\left[0,\frac a2\right].$$

But codomain is $[-a,a]$. Clearly values like $-\frac a2$, $a$, etc. are not attained.

Hence $g$ is not onto.

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7. Conclusion

$g$ is neither one-one nor onto.

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

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8. Comparison with stored answer

Stored correct answer: A

This matches our derived answer.