Question

A real valued function \(f\) defined by \[ f(x)=|x|-x \] is

• A. an injection but not surjection, if \([0,\infty)\) is its domain and \((-\infty,0]\) is its codomain
• B. a bijection, if \((-\infty,0]\) is its domain and also codomain
• C. a bijection, if \([0,\infty)\) is its domain and also codomain
• D. a surjection but not injection, if \(R\) is its domain and \([0,\infty)\) is its codomain

Hint:

For modulus functions, always split into cases \(x\geq0\) and \(x& lt;0\) before checking injective or surjective nature.

Answer 1

The Correct Option is D

Concept: To determine whether a function is injective, surjective or bijective, we first simplify the function by considering modulus definition. \[ |x|=\begin{cases} x,& x\geq0\\ -x,& x& lt;0 \end{cases} \] Thus function behavior changes over intervals.

Step 1: Simplify the function.
Substituting modulus definition, For \(x\geq0\) \[ f(x)=x-x=0 \] For \(x& lt;0\) \[ f(x)=-x-x=-2x \] Hence \[ f(x)= \begin{cases} 0,& x\geq0\\ -2x,& x& lt;0 \end{cases} \]

Step 2: Check each option.
Option A: Domain \([0,\infty)\) Then \[ f(x)=0 \] constant function cannot be injective. False. Option B: Domain \((-\infty,0]\) Different x values give different outputs but codomain mismatch. False. Option C: Domain \([0,\infty)\) Again constant function. Not bijection. False. Option D: Domain \(R\) Codomain \([0,\infty)\) All positive values attained. For example \[ f(-1)=2,\qquad f(-2)=4 \] Range becomes \[ [0,\infty) \] So surjective. But \[ f(1)=0,\qquad f(2)=0 \] Not injective. Hence correct. \[ \boxed{\text{Option (4)}} \]