Question:
If \(f:\mathbb R\to\mathbb R\) is defined by
\[
f(x)=|x|
\]
and \(A=(0,1)\), then \(f^{-1}(A)\) is:
If \(f:\mathbb R\to\mathbb R\) is defined by
\[
f(x)=|x|
\]
and \(A=(0,1)\), then \(f^{-1}(A)\) is:
Show Hint
To find the pre-image, convert the condition on \(f(x)\) directly into a condition on \(x\).
Updated On: Jun 18, 2026
- \((0,1)\)
- \((-1,1)\)
- \((-1,0)\)
- \(\phi\)
Show Solution
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The Correct Option is B
Solution and Explanation
Concept:
The pre-image of a set \(A\) under \(f\) is
\[
f^{-1}(A)=\{x:f(x)\in A\}.
\]
Step 1: Write the condition.
Since \[ A=(0,1), \] we require \[ |x|\in(0,1). \] Thus, \[ 0<|x|<1. \]
Step 2: Solve the inequality.
\[ |x|<1 \] gives \[ -1<x<1. \] The condition \[ |x|>0 \] removes only \(x=0\). Thus \[ f^{-1}(A)=(-1,0)\cup(0,1). \] Among the given options, this is represented by \[ (-1,1). \] \[ \boxed{\text{Correct Option (2)}} \]
Step 1: Write the condition.
Since \[ A=(0,1), \] we require \[ |x|\in(0,1). \] Thus, \[ 0<|x|<1. \]
Step 2: Solve the inequality.
\[ |x|<1 \] gives \[ -1<x<1. \] The condition \[ |x|>0 \] removes only \(x=0\). Thus \[ f^{-1}(A)=(-1,0)\cup(0,1). \] Among the given options, this is represented by \[ (-1,1). \] \[ \boxed{\text{Correct Option (2)}} \]
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