Question

If the domain and the range of the real-valued function \[ f(x)=\frac{1}{\sqrt{|x|-[x]}} \] are \(A\) and \(B\), respectively, then \[ A\cap B = \ ? \] (Here, \(R^{+}\) denotes the set of positive real numbers and \(Z^{+}\) denotes the set of positive integers.) 

• A. \(R^{+}-Z^{+}\)
• B. \(R^{+}\)
• C. \(R-Z\)
• D. \(R-(Z^{+}\cup\{0\})\)

Hint:

Whenever a function contains both modulus and greatest integer function, always split the problem into cases: \[ x\geq0 \qquad \text{and}\qquad x& lt;0 \] For expressions inside square roots in denominator, remember the quantity must be strictly positive, not merely non-negative.

Answer 1

The Correct Option is A

Concept: To solve this problem, we must carefully determine both the domain and the range of the given function. The function is \[ f(x)=\frac{1}{\sqrt{|x|-[x]}} \] where \([x]\) denotes the greatest integer function (also called floor function), defined as the greatest integer less than or equal to x. Since the expression is inside a square root in the denominator, two conditions are necessary:
• Quantity inside square root must be strictly positive.
• Denominator cannot become zero. Thus we require \[ |x|-[x]& gt;0 \] After finding the domain A, we determine all possible output values to obtain range B, and finally compute the intersection \(A\cap B\).

Step 1: Find the domain by applying the condition for existence of the function.
Since the denominator contains a square root, we must satisfy \[ |x|-[x]& gt;0 \] To analyze this properly, we divide into two cases. Case 1: When \(x\geq0\)
For non-negative values of x, absolute value behaves as \[ |x|=x \] So the expression becomes \[ x-[x] \] But we know that \[ x-[x]=\{x\} \] where \(\{x\}\) denotes the fractional part of x. The fractional part always satisfies \[ 0\leq \{x\}& lt;1 \] Now our condition requires \[ \{x\}& gt;0 \] This means x cannot be an integer. Hence for positive side: \[ x& gt;0,\qquad x\notin Z^{+} \] Also at \(x=0\) \[ |0|-[0]=0 \] which makes denominator zero. So \(x=0\) is also excluded. Thus allowed values here are \[ x\in R^{+}-Z^{+} \] Case 2: When \(x& lt;0\)
For negative values, \[ |x|=-x \] Hence expression becomes \[ -x-[x] \] Now let \[ x=-2.7 \] Then \[ [x]=-3 \] Thus \[ |x|-[x]=2.7-(-3)=5.7 \] which is positive. Similarly, for a negative integer, \[ x=-2 \] then \[ [x]=-2 \] Hence \[ |x|-[x]=2-(-2)=4 \] Again positive. Therefore every negative real number satisfies the condition. So all negative real numbers belong to the domain. That gives \[ x& lt;0 \] Combining both cases: \[ A=R-(Z^{+}\cup\{0\}) \] Thus domain is \[ A=R-(Z^{+}\cup\{0\}) \]

Step 2: Now determine the range of the function.
We study all possible values of \[ |x|-[x] \] because output depends directly on this expression. Recall \[ f(x)=\frac{1}{\sqrt{|x|-[x]}} \] Again divide into cases. For \(x\geq0\)
We obtained \[ |x|-[x]=x-[x]=\{x\} \] Since positive integers are excluded, \[ 0& lt;\{x\}& lt;1 \] Thus denominator takes values in interval \[ (0,1) \] Hence function values become \[ f(x)=\frac{1}{\sqrt{t}},\qquad0& lt;t& lt;1 \] This gives \[ f(x)& gt;1 \] So one part of range is \[ (1,\infty) \] For \(x& lt;0\)
From earlier discussion, \[ |x|-[x] \] takes values greater than or equal to 2. Hence \[ |x|-[x]\in[2,\infty) \] So function becomes \[ f(x)=\frac{1}{\sqrt{t}},\qquad t\geq2 \] Thus \[ 0& lt;f(x)\leq\frac{1}{\sqrt2} \] So second part of range is \[ \left(0,\frac{1}{\sqrt2}\right] \] Therefore total range is \[ B=\left(0,\frac{1}{\sqrt2}\right]\cup(1,\infty) \]

Step 3: Find intersection \(A\cap B\).
We know \[ A=R-(Z^{+}\cup\{0\}) \] and \[ B=\left(0,\frac{1}{\sqrt2}\right]\cup(1,\infty) \] Observe carefully:
• Range contains only positive real numbers.
• Positive integers like 2, 3, 4, ... belong to range.
• But positive integers are excluded from domain. Hence intersection contains all positive real numbers except positive integers. So \[ A\cap B=R^{+}-Z^{+} \] Thus final answer is \[ \boxed{A\cap B=R^{+}-Z^{+}} \]