Problem 5 - A Textbook of Differential Calculus

Question

Let \(f(x)=\sqrt{|x|-\\{x\\}}\) (where \\{\\} denotes the fractional part of \(x\) and \(X, Y\) and its domain and range respectively, then (a) \(f: X \rightarrow Y: y=f(x)\) is one-one function (b) \(X \in\left(-\infty,-\frac{1}{2}\right] \cup[0, \infty)\) and \(Y \in\left[\frac{1}{2}, \infty\right)\) (c) \(X \in\left(-\infty,-\frac{1}{2}\right] \cup[0, \infty)\) and \(Y \in[0, \infty)\) (d) None of the above

Step-by-Step Solution

Answer

Option (c): \(X = (-\infty, -1] \cup [0, \infty)\) and \(Y = [0, \infty)\) is correct.

Step 1: Understanding the Function Components

The function given is \(f(x) = \sqrt{|x| - \{x\}}\). Here, \(|x|\) represents the absolute value of \(x\), and \(\{x\}\) denotes the fractional part of \(x\). The fractional part is defined as \(\{x\} = x - \lfloor x \rfloor\), where \(\lfloor x \rfloor\) is the greatest integer less than or equal to \(x\).

Step 2: Evaluating the Domain of \(f(x)\)

We need \(|x| - \{x\} \geq 0\) for \(f(x)\) to be real, because the square root function is only defined for non-negative numbers. Thus, \(|x| - (x - \lfloor x \rfloor) \geq 0\), simplifying to \(|x| - x + \lfloor x \rfloor \geq 0\). Two cases arise: for \(x \geq 0\) and \(x < 0\).

Step 3: Solving for \(x \geq 0\)

If \(x \geq 0\), then \(|x| = x\). Hence, the condition simplifies to \(\lfloor x \rfloor \geq 0\), which is always true for \(x \geq 0\). Therefore, \(x \in [0, \infty)\).

Step 4: Solving for \(x < 0\)

If \(x < 0\), then \(|x| = -x\). The condition now is \(-x + x - \lfloor x \rfloor \geq 0\). Simplifying, we have \(\lfloor x \rfloor \leq 0\). Since \(x\) is negative, it implies \(\lfloor x \rfloor \leq -1\). Thus, valid \(x\) are \((-\infty, -1]\).

Step 5: Defining the Complete Domain \(X\)

Combining both intervals, we get the domain of \(f(x)\) as \(X = (-\infty, -1] \cup [0, \infty)\).

Step 6: Finding the Range \(Y\)

For any \(x \in X\), we have \(f(x) = \sqrt{\lfloor x \rfloor}\) because \(|x| - \{x\} = \lfloor x \rfloor\). Since \(\lfloor x \rfloor\) is an integer, it is clear that the minimum value \(\lfloor x \rfloor\) can be is \(0\), thus the range \(Y = [0, \infty)\).

Step 7: Determining Which Option is Correct

Comparing with the given options: - (a) says the function is one-one, which it is not since many \(x\) values can give the same integer value, thus the same output.- (b) suggests domains and range that do not match our analysis.- (c) matches our determined domain and range.Therefore, option (c) is consistent with our findings.

Key Concepts

Domain and Range

In differential calculus, the domain and range of a function are fundamental concepts that help to understand the behavior of that function. The domain of a function refers to all the possible input values (usually denoted as 'x') that the function can accept without causing any mathematical issues, such as division by zero or taking the square root of a negative number. It defines the 'starting' side of a function. In our exercise with the function \(f(x) = \sqrt{|x| - \{x\}}\), to determine the domain, we need to ensure the expression inside the square root is non-negative.

Solving \(|x| - \{x\} \geq 0\), we consider two cases.

• For \(x \geq 0\), \(|x| = x\) and the condition is always true.
• For \(x < 0\), \(|x| = -x\), leading to the constraint \(\lfloor x \rfloor \leq -1\).

This gives a domain of \(X = (-\infty, -1] \cup [0, \infty)\).

The range of a function is the set of all possible outputs or results once inputs are fed into the function. For the given \(f(x)\), once the domain is established, the range derived from the function \(f(x) = \sqrt{\lfloor x \rfloor}\) reveals that all non-negative integers, starting from 0, correspond to range values. Thus, the range is \([0, \infty)\).

Fractional Part Function

The fractional part function, denoted as \(\{x\}\), extracts the non-integer part of a number. It's defined mathematically as \(\{x\} = x - \lfloor x \rfloor\). This function is significant because it helps in distinguishing between the integer and decimal components of any real number.

For any number \(x\), \(\lfloor x \rfloor\) represents the greatest integer less than or equal to \(x\). The fractional part, therefore, lies in the interval \([0, 1)\), making it a vital segment while solving complex mathematical problems involving real numbers.

In our exercise, understanding \(\{x\}\) is crucial. It aids in the simplification of the expression \(|x| - \{x\}\), ensuring that the result remains within bounds that the square root function can evaluate without issues, allowing us to correctly identify the domain and range of a given function.

One-One Function

A function is considered 'one-one' or injective if it assigns a unique output for every unique input. This means that for every \(x_1\) and \(x_2\) in the domain, if \(f(x_1) = f(x_2)\), then \(x_1 = x_2\).

However, in the context of our function \(f(x) = \sqrt{|x| - \{x\}}\), the function does not satisfy the one-one condition. This is due to the fact that multiple values of \(x\) can result in the same integer when applying the ceiling function. Thus, different \(x\) values can yield the same output after evaluating the function. For instance, any fractional number within certain intervals will round down to the same integer value as part of finding \(\lfloor x \rfloor\).

• Example: For \(-1.5\) and \(-1.1\), both have the floor value of \(-2\).

As a result, the function cannot be considered one-one, because it does not maintain a unique correspondence between input values and output values, which is evident from the exercise solution provided.