Key Concepts and Formulas

• Piecewise Functions: A function defined by different formulas over different intervals of its domain.
• Absolute Value Function: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$. This is fundamental for evaluating $f(|x|)$ and $|f(x)|$.
• Properties of Functions:
  • One-one (Injective): A function $h$ is one-one if distinct inputs map to distinct outputs ($h(x_1) = h(x_2) \implies x_1 = x_2$). Graphically, this means no horizontal line intersects the graph more than once.
  • Onto (Surjective): A function $h: A \to B$ is onto if every element in the codomain $B$ has at least one corresponding element in the domain $A$ (i.e., the range of $h$ is equal to the codomain $B$).

Step-by-Step Solution

We are given the function $f(x)$ and need to determine the properties of $g(x) = (f(|x|) - |f(x)|) / 2$ on the domain $[-a, a]$.

Step 1: Analyze the function $f(x)$ The function $f(x)$ is defined as: $f(x)=\left\{\begin{array}{ccc}-\mathrm{a} & \text { if } & -\mathrm{a} \leq x \leq 0 \\ x+\mathrm{a} & \text { if } & 0< x \leq \mathrm{a}\end{array}\right.$ where $a > 0$.

Let's examine the behavior of $f(x)$ in its two intervals:

• For $-a \le x \le 0$, $f(x) = -a$. This is a constant function. The range for this part is $\{-a\}$.
• For $0 < x \le a$, $f(x) = x+a$. This is a linear function with a positive slope.
  • When $x$ approaches 0 from the right ($x \to 0^+$), $f(x) \to 0+a = a$.
  • When $x = a$, $f(x) = a+a = 2a$. The range for this part is $(a, 2a]$.

Combining these, the range of $f(x)$ on $[-a, a]$ is $\{-a\} \cup (a, 2a]$.

Step 2: Determine $f(|x|)$ We need to consider the definition of $|x|$ for different intervals of $x$ in $[-a, a]$.

• Case 1: $-a \le x \le 0$ In this interval, $|x| = -x$. Since $-a \le x \le 0$, we have $0 \le -x \le a$. Therefore, $f(|x|) = f(-x) = -a$, because $-x$ falls into the interval $[0, a]$ where $f$ is defined as $x+a$ if $0<x \le a$ and as $-a$ if $-a \le x \le 0$. However, we need to be careful about the input to $f$. Since $0 \le -x \le a$, we need to check the definition of $f$ for inputs in $[0, a]$. If $-x = 0$, then $x=0$. $f(0) = -a$. If $0 < -x \le a$, then $f(-x) = (-x) + a = a-x$. So, for $-a \le x < 0$, $|x| = -x$ and $0 < |x| \le a$. Thus $f(|x|) = f(-x) = (-x) + a = a-x$. For $x=0$, $|x|=0$, so $f(|x|) = f(0) = -a$. Therefore, for $-a \le x \le 0$: $f(|x|) = \begin{cases} a-x & \text{if } -a \le x < 0 \\ -a & \text{if } x=0 \end{cases}$ This can be simplified. If $x=0$, $a-x=a$. But $f(0)=-a$. So the piecewise definition is needed. Let's re-evaluate carefully. For $-a \le x < 0$: $|x| = -x$. Since $0 < -x \le a$, $f(|x|) = f(-x) = (-x) + a = a-x$. For $x = 0$: $|x| = 0$. $f(|x|) = f(0) = -a$.

• Case 2: $0 < x \le a$ In this interval, $|x| = x$. Since $0 < x \le a$, $f(|x|) = f(x) = x+a$.

Combining these, we have: $f(|x|) = \begin{cases} a-x & \text{if } -a \le x < 0 \\ -a & \text{if } x=0 \\ x+a & \text{if } 0 < x \le a \end{cases}$

Step 3: Determine $|f(x)|$ We need to consider the sign of $f(x)$ in its domain $[-a, a]$.

• Case 1: $-a \le x \le 0$ Here, $f(x) = -a$. Since $a > 0$, $-a < 0$. So, $|f(x)| = |-a| = a$.

• Case 2: $0 < x \le a$ Here, $f(x) = x+a$. Since $0 < x \le a$ and $a > 0$, we have $a < x+a \le 2a$. Thus, $f(x) > 0$. So, $|f(x)| = |x+a| = x+a$.

Combining these, we have: $|f(x)| = \begin{cases} a & \text{if } -a \le x \le 0 \\ x+a & \text{if } 0 < x \le a \end{cases}$

Step 4: Determine $g(x) = (f(|x|) - |f(x)|) / 2$ Now we substitute the expressions for $f(|x|)$ and $|f(x)|$ into the formula for $g(x)$. We need to consider the intervals for $x$.

• Interval 1: $-a \le x < 0$ From Step 2, $f(|x|) = a-x$. From Step 3, $|f(x)| = a$. So, $g(x) = ( (a-x) - a ) / 2 = (-x) / 2 = -x/2$.

• Interval 2: $x = 0$ From Step 2, $f(|x|) = f(0) = -a$. From Step 3, $|f(x)| = |f(0)| = |-a| = a$. So, $g(x) = ( (-a) - a ) / 2 = (-2a) / 2 = -a$.

• Interval 3: $0 < x \le a$ From Step 2, $f(|x|) = x+a$. From Step 3, $|f(x)| = x+a$. So, $g(x) = ( (x+a) - (x+a) ) / 2 = 0 / 2 = 0$.

Combining these results, we get the piecewise definition of $g(x)$ on $[-a, a]$: $g(x)=\left\{\begin{array}{ccc}-x/2 & \text { if } & -a \leq x < 0 \\ -a & \text { if } & x=0 \\ 0 & \text { if } & 0< x \leq a\end{array}\right.$

Step 5: Analyze the properties of $g(x)$ (One-one and Onto)

• One-one (Injective) Property: Let's examine the function $g(x)$ in each interval:

  • For $-a \le x < 0$, $g(x) = -x/2$. This is a linear function with a negative slope. For any two distinct values $x_1, x_2$ in $[-a, 0)$ with $x_1 \ne x_2$, we have $-x_1/2 \ne -x_2/2$. So, $g(x)$ is one-one on $[-a, 0)$.
  • At $x=0$, $g(0) = -a$.
  • For $0 < x \le a$, $g(x) = 0$. This is a constant function. All values in $(0, a]$ map to $0$. For example, $g(0.1) = 0$ and $g(0.2) = 0$, but $0.1 \ne 0.2$. Therefore, $g(x)$ is not one-one on $(0, a]$.

  Since $g(x)$ is not one-one on its entire domain $[-a, a]$ (specifically, on the interval $(0, a]$), the function $g$ is not one-one.

• Onto (Surjective) Property: The codomain of $g(x)$ is given as $[-a, a]$. We need to determine the range of $g(x)$ and check if it equals $[-a, a]$.

  Let's find the range of $g(x)$ for each part of its definition:

  • For $-a \le x < 0$, $g(x) = -x/2$.

    • When $x = -a$, $g(-a) = -(-a)/2 = a/2$.
    • As $x$ approaches $0$ from the left ($x \to 0^-$), $g(x) = -x/2 \to 0$. The range for this part is $(0, a/2]$.

  • At $x=0$, $g(0) = -a$.

  • For $0 < x \le a$, $g(x) = 0$. The range for this part is $\{0\}$.

  Combining the ranges from these parts, the overall range of $g(x)$ on $[-a, a]$ is: Range of $g = (0, a/2] \cup \{-a\} \cup \{0\} = \{-a\} \cup [0, a/2]$.

  The codomain is $[-a, a]$. The range of $g$ is $\{-a\} \cup [0, a/2]$. Since the range $\{-a\} \cup [0, a/2]$ is not equal to the codomain $[-a, a]$ (for example, the value $a/2 + \epsilon$ for small $\epsilon > 0$ is in the codomain but not in the range), the function $g$ is not onto.

Common Mistakes & Tips

• Careful handling of absolute values: Ensure that the definition of $|x|$ is applied correctly based on the sign of $x$, and similarly for $|f(x)|$ based on the sign of $f(x)$. Pay close attention to the endpoints of intervals.
• Piecewise definitions for $f(|x|)$ and $|f(x)|$: When constructing the piecewise definitions for $f(|x|)$ and $|f(x)|$, ensure that the correct interval for the input to $f$ (in the case of $f(|x|)$) or the correct condition for the sign of $f(x)$ (in the case of $|f(x)|$) is used.
• Domain of $g(x)$: When analyzing $g(x)$, break it down into the same intervals as derived from the definitions of $f(|x|)$ and $|f(x)|$. Evaluate $g(x)$ for each part and then check the one-one and onto properties for the entire domain $[-a, a]$.
• Range vs. Codomain: For a function to be onto, its range must be exactly equal to its codomain.

Summary

The function $g(x)$ was constructed by carefully evaluating $f(|x|)$ and $|f(x)|$ over the domain $[-a, a]$, considering the piecewise definition of $f(x)$ and the properties of the absolute value function. This led to a piecewise definition for $g(x)$. Upon analyzing this definition, it was found that $g(x)$ is not one-one because the interval $(0, a]$ maps to a single value (0), and $g(x)$ is not onto because its range, $\{-a\} \cup [0, a/2]$, is a proper subset of the codomain $[-a, a]$.

The final answer is \boxed{A}.