Question

If the domain and the range of the real valued function \[ f(x)=\frac{1}{\sqrt{|x|-[x]}} \] are A and B, then \(A\cap B=\) (\(R^{+}\) is set of positive real numbers and \(Z^{+}\) is 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: \[ x\geq0 \qquad \text{and}\qquad x<0 \] 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 \[ |x|-[x]>0 \] 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 \[ |x|-[x]>0 \] 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 \[ 0\leq \{x\}<1 \] Now our condition requires \[ \{x\}>0 \] This means x cannot be an integer. Hence for positive side: \[ x>0,\qquad x\notin Z^{+} \] 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 \(x<0\)
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 \[ x<0 \] 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, \[ 0<\{x\}<1 \] Thus denominator takes values in interval \[ (0,1) \] Hence function values become \[ f(x)=\frac{1}{\sqrt{t}},\qquad0<t<1 \] This gives \[ f(x)>1 \] So one part of range is \[ (1,\infty) \]

For \(x<0\)
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 \[ 0<f(x)\leq\frac{1}{\sqrt2} \] 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 \[ \boxed{A\cap B=R^{+}-Z^{+}} \]