Problem 93 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The domain of definition of the function \(f(x)=\) \(\ln \\{x\\}+\sqrt{x-2\\{x\\}}\), where \\{\\} denotes the fractional part, is (A) \(\\{0\\} \cup[1, \infty)\) (B) \((1, \infty)\) (C) \((1, \infty)-I^{+}\) (D) None of these

Step-by-Step Solution

Answer

The domain is (C) \((1, \infty) - I^+\).

Step 1: Understand the Functional Components

The function is defined as \( f(x) = \ln \{x\} + \sqrt{x-2\{x\}} \), where \( \{x\} \) denotes the fractional part of \( x \). This means that \( \{x\} = x - \lfloor x \rfloor \), where \( \lfloor x \rfloor \) is the greatest integer less than or equal to \( x \). The range of \( \{x\} \) is \([0, 1)\).

Step 2: Evaluate the Domain of Each Component

The natural logarithm function \( \ln \{x\} \) is defined only for \( \{x\} > 0 \). Therefore, we require \( 0 < \{x\} < 1 \). Simultaneously, the square root function \( \sqrt{x - 2\{x\}} \) requires that \( x - 2\{x\} \geq 0 \). Since \( \{x\} < 1 \), this implies \( x \geq 2\{x\} \).

Step 3: Simplify the Inequality for the Square Root

Given the inequality \( x \geq 2\{x\} \), we know \( \{x\} = x - \lfloor x \rfloor \). Plug this in to get \( x \geq 2(x - \lfloor x \rfloor)\), which simplifies to \( x \geq 2x - 2\lfloor x \rfloor \) or \( x \leq 2\lfloor x \rfloor \).

Step 4: Determine the Range of \( x \)

From \( x \leq 2\lfloor x \rfloor \) and since \( \{x\} > 0 \), we deduce \( \lfloor x \rfloor = n \), where \( n \) is an integer. For \( n \leq x < n+1 \), we require \( x < 2n \), so \( n \leq x < \min(n+1, 2n) \). Thus, \( x \) can be in intervals of form \((n, 2n)\), for integer \( n \geq 1 \).

Step 5: Compile the Final Domain

The domain of \( f(x) \) is the union of all intervals where \( \{x\} > 0 \) and \( x \leq 2\lfloor x \rfloor \). These intervals are \((1, 2), (2, 4), (3, 6), \ldots\), excluding the integers. This is \((1, \infty) - I^+\), where \( I^+ \) is the set of positive integers.

Key Concepts

Domain of a function

The domain of a function is essentially the set of all possible input values (usually represented as 'x') that a function can accept without leading to undefined behavior. In mathematical terms, it's the set where the function is real and makes sense.
For our function, we need to consider both the logarithmic and square root components. Each of these has specific requirements for the domain:

• Logarithmic function \(\ln \{x\}\) implies \(\{x\} > 0\).
• Square root function \(\sqrt{x - 2\{x\}}\), necessitates \(x - 2\{x\} \geq 0\).

Combining these conditions, the domain is dictated by both conditions being fulfilled simultaneously. Observing how they interplay, it results in the domain intervals \((1, 2), (2, 4), (3, 6), \ldots\). The values specifically exclude positive integers, as they don't satisfy the condition \(\{x\} > 0\). Thus the domain becomes \((1, \infty) - I^+\), where \(I^+\) is the set of positive integers.

Logarithmic functions

Logarithmic functions are a type of mathematical function closely linked with exponential functions. They take the form \(f(x) = \ln(x)\) for natural logarithms, where the base of the logarithm is the constant \(e\) (approximately 2.71828). These functions are only defined for \(x > 0\), as logarithms of negative numbers or zero are undefined.
In the given function \(f(x) = \ln \{x\}\), the natural logarithm operates only on the fractional part of the variable \(x\). This means the function

• is only defined for \(\{x\} > 0\)
• is undefined, and thus does not exist, at integer values of \(x\) where \(\{x\} = 0\)

Understanding this helps us realize the limitations imposed on the function's domain due to the logarithmic component, contributing to excluding intervals where \(x\) is a positive integer.

Fractional part function

The fractional part function, denoted by \(\{x\}\), extracts the non-integer part of a real number \(x\). It is defined as \(\{x\} = x - \lfloor x \rfloor\), where \(\lfloor x \rfloor\) represents the greatest integer less than or equal to \(x\). This fractional component always lies within the range \([0, 1)\).
In the context of our function, the behavior of the fractional part function is crucial:

• It ensures that \(\ln \{x\}\) is only defined for \(\{x\} > 0\), meaning it eliminates integer values.
• It contributes to creating the conditions \(x - 2\{x\} \geq 0\), necessary for defining the square root function.

By understanding how the fractional part function operates, we can use it to navigate domains involving non-linear expressions, as seen with intervals like \((n, 2n)\) for integers \(n\). It's a key player in expressing the behavior of the original function and understanding how it shapes its domain.