Question:
If a real valued function \(f : A \to B\) defined by \(f(x)=|x|-[x]\) is a bijection, then \(A\) and \(B\) are respectively:
If a real valued function \(f : A \to B\) defined by \(f(x)=|x|-[x]\) is a bijection, then \(A\) and \(B\) are respectively:
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For functions involving modulus and greatest integer function, always split the domain into intervals where \([x]\) remains constant.
Updated On: Jun 17, 2026
- \((-\infty,0]\) and \([0,\infty)\)
- \([-3,-2)\) and \((5,6]\)
- \([1,2)\) and \([3,4)\)
- \([0,\infty)\) and \([0,1)\)
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The Correct Option is B
Solution and Explanation
Concept:
The greatest integer function \([x]\) gives the greatest integer less than or equal to \(x\).
For the function:
\[
f(x)=|x|-[x]
\]
we analyze different intervals separately because modulus and greatest integer functions behave differently for positive and negative values.
For a function to be bijective:
• It must be one-one (injective)
• It must be onto (surjective)
Step 1: Analyze the function for negative values of \(x\).
If \(x<0\), then: \[ |x|=-x \] Thus, \[ f(x)=-x-[x] \] Take \(x\in[-3,-2)\). For this interval: \[ [x]=-3 \] Hence: \[ f(x)=-x+3 \] Now evaluate the range. At \(x=-3\), \[ f(-3)=3+3=6 \] As \(x\to -2^{-}\), \[ f(x)\to 5 \] But \(5\) is not included. Therefore: \[ f(x)\in(5,6] \]
Step 2: Check whether the function is one-one.
On the interval \([-3,-2)\), \[ f(x)=-x+3 \] which is a linear strictly decreasing function. Hence different values of \(x\) produce different function values. Therefore, the function is injective.
Step 3: Check onto property.
The obtained range is exactly: \[ (5,6] \] which matches the codomain. Thus every element of \(B\) has a pre-image in \(A\). Hence the function is surjective.
Step 4: Choose the correct option.
Thus the bijection is: \[ A=[-3,-2) \] and \[ B=(5,6] \] Hence the correct answer is: \[ \boxed{[-3,-2)\text{ and }(5,6]} \]
• It must be one-one (injective)
• It must be onto (surjective)
Step 1: Analyze the function for negative values of \(x\).
If \(x<0\), then: \[ |x|=-x \] Thus, \[ f(x)=-x-[x] \] Take \(x\in[-3,-2)\). For this interval: \[ [x]=-3 \] Hence: \[ f(x)=-x+3 \] Now evaluate the range. At \(x=-3\), \[ f(-3)=3+3=6 \] As \(x\to -2^{-}\), \[ f(x)\to 5 \] But \(5\) is not included. Therefore: \[ f(x)\in(5,6] \]
Step 2: Check whether the function is one-one.
On the interval \([-3,-2)\), \[ f(x)=-x+3 \] which is a linear strictly decreasing function. Hence different values of \(x\) produce different function values. Therefore, the function is injective.
Step 3: Check onto property.
The obtained range is exactly: \[ (5,6] \] which matches the codomain. Thus every element of \(B\) has a pre-image in \(A\). Hence the function is surjective.
Step 4: Choose the correct option.
Thus the bijection is: \[ A=[-3,-2) \] and \[ B=(5,6] \] Hence the correct answer is: \[ \boxed{[-3,-2)\text{ and }(5,6]} \]
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