Let the sets A and B denote the domain and range respectively of the function f(x) = 1/√(⌈x⌉ - x), w

Q: Wait, why do we exclude all integers from the domain? Can't the ceiling function handle integers?

Great question! The ceiling function is perfectly happy with integers, but our function $f(x)$ is not. If $x$ is an integer, $\lceil x \rceil = x$, making the denominator $\sqrt{0} = 0$. Division by zero is undefined in the real number system, so we must strictly exclude all integers from the domain.

Q: I'm confused about the substitution. Is $\lceil x \rceil - x$ always equal to $1 - \{x\}$?

Only for non-integers! Remember, the fractional part $\{x\}$ is defined as $x - \lfloor x \rfloor$. For non-integers, $\lceil x \rceil = \lfloor x \rfloor + 1$. Substituting this, we get $(\lfloor x \rfloor + 1) - x = 1 - (x - \lfloor x \rfloor) = 1 - \{x\}$. It is a beautiful simplification that turns a complex ceiling expression into a simple fractional part.

Q: Why is the range $(1, \infty)$? If the denominator is between $0$ and $1$, shouldn't the reciprocal be between $1$ and $\infty$?

Exactly! You have got the intuition right. Since $0 < \{x\} < 1$, it follows that $0 < 1 - \{x\} < 1$. Taking the square root preserves this inequality: $0 < \sqrt{1 - \{x\}} < 1$. When you take the reciprocal of a number strictly between $0$ and $1$, the result is always greater than $1$. Thus, the range is $(1, \infty)$.

Question

Let the sets $A$ and $B$ denote the domain and range respectively of the function $f(x) = \frac{1}{\sqrt{\lceil x \rceil - x}}$, where $\lceil x \rceil$ denotes the smallest integer greater than or equal to $x$. Then among the statements
(S1): $A \cap B = (1, \infty) - \mathbb{N}$ and
(S2): $A \cup B = (1, \infty)$

Options:

(A) Only (S2) is true

(B) Only (S1) is true

(C) Neither (S1) nor (S2) is true

(D) Both (S1) and (S2) are true

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The function is defined as $f(x) = \frac{1}{\sqrt{\lceil x \rceil - x}}$. At first glance, it appears intimidating due to the jagged nature of the ceiling function, $\lceil x \rceil$.

However, complexity is often just a mask for elegance. Let us peel back that mask systematically.

### Phase 1: The Domain - Finding the Safe Zones

Every function has its boundaries, or 'no-go' zones. For our function, we face two primary constraints: we cannot take the square root of a negative number, and we cannot divide by zero.

This implies that the expression inside the square root must be strictly positive:
$$\lceil x \rceil - x > 0$$

Visualize the graph of $\lceil x \rceil - x$. It is a sawtooth pattern that hits zero exactly at every integer. By definition, $\lceil x \rceil$ is the smallest integer greater than or equal to $x$.

When $x$ is an integer, $\lceil x \rceil = x$, resulting in a denominator of zero. Since division by zero is undefined, we must exclude all integers from our domain.

Thus, our domain $A$ is the set of all real numbers excluding integers:
$$A = \mathbb{R} - \mathbb{Z}$$

### Phase 2: The Algebraic Transformation

To determine the range, we simplify the expression $\lceil x \rceil - x$. We express $x$ as the sum of its integer part and its fractional part:
$$x = \lfloor x \rfloor + \{x\}$$

For any non-integer $x$, the ceiling function is defined as:
$$\lceil x \rceil = \lfloor x \rfloor + 1$$

Substituting these into our expression, we obtain:
$$\lceil x \rceil - x = (\lfloor x \rfloor + 1) - (\lfloor x \rfloor + \{x\})$$

The integer parts $\lfloor x \rfloor$ cancel out perfectly, leaving us with the simplified expression:
$$\lceil x \rceil - x = 1 - \{x\}$$

### Phase 3: The Range - The Blow-up Effect

We have reduced our function to:
$$f(x) = \frac{1}{\sqrt{1 - \{x\}}}$$

By definition, for any non-integer $x$, the fractional part $\{x\}$ is strictly between $0$ and $1$. Consequently, the term $1 - \{x\}$ is also strictly between $0$ and $1$.

As $\{x\}$ approaches $0$, the denominator approaches $0$, causing the function value to shoot toward infinity. As $\{x\}$ approaches $1$, the denominator approaches $1$, and the function value approaches $1$.

Therefore, the range $B$ is the interval:
$$B = (1, \infty)$$

### Phase 4: The Final Verdict

We have identified our sets: $A = \mathbb{R} - \mathbb{Z}$ and $B = (1, \infty)$.

Statement (S1) concerns the intersection $A \cap B$. We take the interval $(1, \infty)$ and remove all integers. The integers greater than $1$ are $\{2, 3, 4, \dots\}$, which are the natural numbers $\mathbb{N}$.

Thus, $A \cap B = (1, \infty) - \mathbb{N}$. This confirms that **(S1) is true**.

Statement (S2) concerns the union $A \cup B$. The set $A$ includes negative numbers (e.g., $-0.5$), which are not contained in $(1, \infty)$.

Because the union includes these negative values, it is much larger than $(1, \infty)$. Therefore, **(S2) is false**.

Select Answer:

Visualized Solution

Defining the Function $f (x)$

Condition for Domain $A$

Analyzing $⌈ x ⌉ - x$

Concluding Domain $A$

Relating to Fractional Part

Simplifying the Expression

Bounding the Denominator

Concluding Range $B$

Evaluating Statement (S1)

Proving Statement (S1) is True

Evaluating Statement (S2)

Final Conclusion

The Sigma Insight: Domain and Range of a Function

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Phase 4

Similar Questions

Let $f : (1, 3) \to R$ be a function defined by $f (x) = \frac{x [ x ]}{x ^{2} + 1}$ , where $[x]$ denotes the greatest integer $\leq x$ . Then the range of $f$ is:

The domain of the function $f (x) = \frac{1}{∣ x ∣ - x}$ is

If the functions are defined as $f (x) = x$ and $g (x) = 1 - x$ , then what is the common domain of the following functions: $f + g, f - g, f / g, g / f, g - f$ where $(f \pm g) (x) = f (x) \pm g (x), (f / g) (x) = \frac{f ( x )}{g ( x )}$

If the function $f : R - {1, - 1} \to A$ defined by $f (x) = \frac{x ^{2}}{1 - x ^{2}}$ , is surjective, then $A$ is equal to

If the domain of the function $f (x) = \frac{[ x ]}{1 + x ^{2}}$ , where $[x]$ is greatest integer $\leq x$ , is $(2, 6)$ , then its range is

If $f (x) = sin ln (\frac{4 - x ^{2}}{1 - x})$ , then domain of $f (x)$ is .... and its range is .........

Let $f : R \to R$ be defined by $f (x) = \frac{x}{1 + x ^{2}}, x \in R$ . Then the range of $f$ is :

The domain of the function $f (x) = \frac{1}{[ x ] ^{2} - 3 [ x ] - 10}$ is (where $[x]$ denotes the greatest integer less than or equal to $x$ )

The domain of definition of the function $y = \frac{1}{l o g _{10} ( 1 - x )} + x + 2$ is

Let $[x]$ denote the greatest integer $\leq x$ , where $x \in R$ . If the domain of the real valued function $f (x) = \frac{∣ [ x ] ∣ - 2}{∣ [ x ] ∣ - 3}$ is $(- \infty, a) \cup [b, c) \cup [4, \infty)$ , $a < b < c$ , then the value of $a + b + c$ is:

Select Answer:

Visualized Solution

Defining the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x)

Condition for Domain .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ⌈x⌉−x

Concluding Domain .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A