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

The Correct Option is A

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