Problem 18 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The functions \(f(x)=\log (x-1)-\log (x-2)\) and \(g(x)=\) \(\log \left(\frac{x-1}{x-2}\right)\) are identical when \(x\) lies in the interval (A) \([1,2]\) (B) \([2, \infty]\) (C) \((2, \infty)\) (D) \((-\infty, \infty)\)

Step-by-Step Solution

Answer

(C) (2, \infty)

Step 1: Simplify the Function f(x)

The function is given as \( f(x) = \log (x-1) - \log (x-2) \). Recall the logarithmic identity: \( \log a - \log b = \log \left( \frac{a}{b} \right) \). Applying this identity, we rewrite \( f(x) \) as \( \log \left( \frac{x-1}{x-2} \right) \).

Step 2: Compare with Function g(x)

The function \( g(x) \) is given as \( \log \left( \frac{x-1}{x-2} \right) \). Since \( f(x) \) is simplified to \( \log \left( \frac{x-1}{x-2} \right) \), both \( f(x) \) and \( g(x) \) are, in fact, identical.

Step 3: Determine the Domain of the Functions

The expression \( \frac{x-1}{x-2} \) is defined when both \( x-1 \) and \( x-2 \) are positive, and \( x-2 eq 0 \). Therefore, \( x > 2 \). For logarithm to be defined, \( \frac{x-1}{x-2} > 0 \), so the domain is \((2, \infty)\).

Step 4: Match the Domain with the Given Options

Given the options, the correct domain where \( f(x) \) and \( g(x) \) are identical is \((2, \infty)\), which corresponds to option (C).

Key Concepts

Logarithmic Identity

When working with logarithmic functions, one powerful tool is the logarithmic identity. This identity states that \( \log a - \log b = \log \left( \frac{a}{b} \right) \). Essentially, if you are given the difference of two logarithms with the same base, you can express it as a single logarithm.
This property simplifies complex expressions, making them easier to work with. For example, in the function \( f(x) = \log (x-1) - \log (x-2) \), the identity allows you to rewrite it as a single log function: \( \log \left( \frac{x-1}{x-2} \right) \).
This is especially helpful in comparing functions or solving equations involving logarithms, as it reduces the number of log terms you have to work with.

Domain

The domain of a function refers to the set of input values (\( x-values \)) for which the function is defined. For logarithmic functions, it is crucial to remember that the argument inside the log must be positive. Specifically, \( \log(x) \) is only defined for \( x > 0 \).
In our original problem, we have the expression \( \log \left( \frac{x-1}{x-2} \right) \). To find its domain, we need to ensure that \( \frac{x-1}{x-2} > 0 \). Let's break it down further:

• For \( x-2 eq 0 \), which means \( x eq 2 \).
• Both \( x-1 \) and \( x-2 \) must be positive, leading to \( x > 2 \).

Putting these conditions together, the domain of the function is strictly greater than 2, hence \( x \in (2, \infty) \). This ensures the expression inside the logarithm is valid.

Interval

An interval in mathematics is a range of numbers between two set values. We use it to describe where a function behaves in particular ways or is defined. There are different types of intervals, such as:

• Open intervals \((a, b)\), indicating all numbers between \( a \) and \( b \) but not including \( a \) or \( b \).
• Closed intervals \([a, b]\), including the endpoints \( a \) and \( b \).
• Semi-open or half-closed intervals \((a, b]\) or \([a, b)\), including one endpoint but not the other.

In the context of the functions given, the interval \((2, \infty)\) implies that the function values are defined and identical for all real numbers greater than 2, but not including 2 itself. This helps us determine a valid solution set for the functions to ensure that they are identical throughout their domain.