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:

• A. \((-\infty,0]\) and \([0,\infty)\)
• B. \([-3,-2)\) and \((5,6]\)
• C. \([1,2)\) and \([3,4)\)
• D. \([0,\infty)\) and \([0,1)\)

Hint:

For functions involving modulus and greatest integer function, always split the domain into intervals where \([x]\) remains constant.

Answer 1

The Correct Option is B

Step 1: Understand the function.
The function is $f(x)=|x|-[x]$. Here $|x|$ is the modulus and $[x]$ is the greatest integer part. A bijection must be one-one and onto, so the picture of the domain must exactly cover the codomain with no value repeated.
Step 2: Test the negative interval given in option 2.
Take $x$ in $[-3,-2)$. For negative $x$ we have $|x|=-x$. On this whole interval the greatest integer is $[x]=-3$.
Step 3: Write the simple form.
So here $f(x)=-x-(-3)=-x+3$. This is a neat straight line, which makes checking easy.
Step 4: Find the output values.
Put the ends in. At $x=-3$, $f=-(-3)+3=6$. As $x$ gets close to $-2$ from the left, $f$ gets close to $-(-2)+3=5$, but $5$ is not reached because $-2$ is not included. \[ f(x)\in(5,6] \]
Step 5: Confirm one-one.
The line $-x+3$ always falls as $x$ rises, so two different inputs never give the same output. So $f$ is one-one.
Step 6: Confirm onto and pick the answer.
The output set is exactly $(5,6]$, which matches the codomain $B$ in option 2. Every value of $B$ has a partner in $A$, so $f$ is onto too. Thus the bijection works with $A=[-3,-2)$ and $B=(5,6]$. \[ \boxed{[-3,-2)\text{ and }(5,6]} \]