Question

Let $f(x) = \begin{cases} -a & \text{if } -a \leq x \leq 0, \\\ x + a & \text{if } 0<x \leq a \end{cases}$ where $a>0$ and $g(x) = (f(|x|) - |f(x)|)/2$. Then the function $g : [-a, a] \to [-a, a]$ is:

• A. neither one-one nor onto.
• B. both one-one and onto.
• C. one-one.
• D. onto.

Answer 1

Answer: (A)

neither one-one nor onto.

Answer 2

Given the piecewise function: $f(x) = \begin{cases} -a & \text{if } -a \le x \le 0 \\\ x + a & \text{if } 0 < x \le a \end{cases}$

and the function: $g(x) = \frac{f(|x|) - |f(x)|}{2}.$

We will analyze the behavior of $g(x)$ over the domain $[-a, a]$.

Case 1: $x \in [-a, 0]$ In this interval, $|x| = -x$ and $f(x) = -a$.

Thus: $f(|x|) = -a \quad \text{and} \quad |f(x)| = | - a | = a.$

Substituting into the expression for $g(x)$: $g(x) = \frac{-a - a}{2} = -a.$

Case 2: $x \in (0, a]$ In this interval, $|x| = x$ and $f(x) = x + a$.

Thus: $f(|x|) = x + a \quad \text{and} \quad |f(x)| = |x + a| = x + a.$

Substituting into the expression for $g(x)$: $g(x) = \frac{(x + a) - (x + a)}{2} = 0.$ Behavior of $g(x)$: - For $x \in [-a, 0]$, $g(x) = -a$. - For $x \in (0, a]$, $g(x) = 0$.

Since $g(x)$ takes only two distinct values ($-a$ and $0$) over the entire interval $[-a, a]$, it is clear that: - $g(x)$ is not one-one (injective) because different inputs give the same output. - $g(x)$ is not onto (surjective) because it does not cover the entire range $[-a, a]$.

Therefore: $g(x) \text{ is neither one-one nor onto.}$