Question

Let \[ f(x)=\frac{|x|-1}{|x-1|} \] be a real valued function from \( \mathbb{R}-\{1\} \) to \(B\). If \(f\) is a surjection, then \(B=\)

• A. \(\left[-\dfrac{1}{2},\dfrac{1}{2}\right]\)
• B. \([-2,2]\)
• C. \([0,1]\)
• D. \([-1,1]\)

Hint:

To find the codomain of a surjective function, first determine its range. For modulus functions, split the domain at points where the expressions inside the modulus become zero.

Answer 1

The Correct Option is D

Concept: If \(f:A\to B\) is surjective, then every element of \(B\) must be an image of some element of \(A\). Therefore, \(B\) is the range of the function \(f\).

Step 1: Find \(f(x)\) for \(x\lt 0\). For \(x\lt 0\), \[ |x|=-x,\qquad |x-1|=1-x. \] Hence, \[ f(x)=\frac{-x-1}{1-x} =\frac{x+1}{x-1}. \] Let \[ y=\frac{x+1}{x-1}. \] Then, \[ x=\frac{y+1}{y-1}. \] Since \(x\lt 0\), we get \[ -1\lt y\lt 1. \] Thus, the range for \(x\lt 0\) is \((-1,1)\).

Step 2: Find \(f(x)\) for \(0\le x\lt 1\). For \(0\le x\lt 1\), \[ |x|=x,\qquad |x-1|=1-x. \] Therefore, \[ f(x)=\frac{x-1}{1-x}=-1. \] Hence, \(-1\) belongs to the range.

Step 3: Find \(f(x)\) for \(x\gt 1\). For \(x\gt 1\), \[ |x|=x,\qquad |x-1|=x-1. \] Therefore, \[ f(x)=\frac{x-1}{x-1}=1. \] Hence, \(1\) belongs to the range.

Step 4: Determine the range of \(f\). Combining all cases, \[ \text{Range}(f)=(-1,1)\cup\{-1\}\cup\{1\}=[-1,1]. \] Since \(f\) is surjective onto \(B\), \[ B=[-1,1]. \] \[ \boxed{B=[-1,1]} \]