Question:
If \(f:(-\infty,0)\to \mathbb{R}\) is defined by
\[
f(x)=\frac{[x]}{|x|}
\]
then \(f(x):x\in(-\infty,0)\) is:
If \(f:(-\infty,0)\to \mathbb{R}\) is defined by
\[
f(x)=\frac{[x]}{|x|}
\]
then \(f(x):x\in(-\infty,0)\) is:
Show Hint
For negative inputs:
\[
|x|=-x
\]
This simplification is very useful in range problems involving modulus.
Updated On: Jun 17, 2026
- \((-\infty,0)\)
- \([-1,0)\)
- \((-2,-1]\)
- \((-\infty,-1]\)
Show Solution
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The Correct Option is D
Solution and Explanation
Concept:
For negative numbers:
\[
|x|=-x
\]
Hence the given function becomes:
\[
f(x)=\frac{[x]}{-x}
\]
To determine the range, we analyze the function interval by interval.
Step 1: Take \(x\in[-n,-n+1)\).
For this interval: \[ [x]=-n \] where \(n\in\mathbb{N}\). Therefore: \[ f(x)=\frac{-n}{-x} =\frac{n}{x} \] Since \(x<0\), the values remain negative.
Step 2: Determine extreme values.
For \(x\in[-n,-n+1)\), At \(x=-n\), \[ f(x)=\frac{-n}{n}=-1 \] As \(x\to(-n+1)^-\), \[ f(x)\to \frac{-n}{n-1} \] which is less than \(-1\). Thus values extend continuously below \(-1\).
Step 3: Combine all intervals.
For different values of \(n\), the function covers: \[ (-\infty,-1] \] Thus the complete range is: \[ \boxed{(-\infty,-1]} \]
Step 1: Take \(x\in[-n,-n+1)\).
For this interval: \[ [x]=-n \] where \(n\in\mathbb{N}\). Therefore: \[ f(x)=\frac{-n}{-x} =\frac{n}{x} \] Since \(x<0\), the values remain negative.
Step 2: Determine extreme values.
For \(x\in[-n,-n+1)\), At \(x=-n\), \[ f(x)=\frac{-n}{n}=-1 \] As \(x\to(-n+1)^-\), \[ f(x)\to \frac{-n}{n-1} \] which is less than \(-1\). Thus values extend continuously below \(-1\).
Step 3: Combine all intervals.
For different values of \(n\), the function covers: \[ (-\infty,-1] \] Thus the complete range is: \[ \boxed{(-\infty,-1]} \]
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