Question

Let \(f(x)=\dfrac{x-1}{x+1}\) ,Let \(S ={x∈R \text{ Iff } -1(x)=x \text{  does not hold} }\).The cardinality of S is 

• A. \(2\)
• B. \(3\)
• C. \(1\)
• D. a finite number, but not equal to 1,2,3
• E. \(∞\)

Answer 1

The Correct Option is A

The function is \( f(x) = \frac{x - 1}{x + 1} \), and its domain excludes \( x = -1 \).

The inverse function is \( f^{-1}(x) = \frac{x + 1}{1 - x} \), and its domain excludes \( x = 1 \).

Compute \( f(f^{-1}(x)) \): \[ f(f^{-1}(x)) = f\left(\frac{x + 1}{1 - x}\right) = \frac{\frac{x + 1}{1 - x} - 1}{\frac{x + 1}{1 - x} + 1}. \] Simplify: \[ f(f^{-1}(x)) = \frac{\frac{2x}{1 - x}}{\frac{2}{1 - x}} = x. \] Thus, \( f(f^{-1}(x)) = x \) for all \( x \in \mathbb{R} \setminus \{-1, 1\} \).

The set \( S \) contains points where \( f(f^{-1}(x)) \neq x \), which are \( x = -1 \) and \( x = 1 \). Hence: \[ S = \{-1, 1\}. \] The cardinality of \( S \) is 2.

Answer 2

Step 1: Understand the problem and given function.

We are given the function \( f(x) = \frac{x - 1}{x + 1} \), and we need to find the set \( S = \{ x \in \mathbb{R} \mid f(f^{-1}(x)) \neq x \} \). The cardinality of \( S \) is the number of elements in \( S \).

Step 2: Analyze the condition \( f(f^{-1}(x)) = x \).

By definition, for any invertible function \( f \), the composition \( f(f^{-1}(x)) \) should equal \( x \). However, this property may fail if \( f \) is not well-defined or invertible for certain values of \( x \).

Step 3: Check the domain and behavior of \( f(x) \).

The function \( f(x) = \frac{x - 1}{x + 1} \) is undefined when the denominator \( x + 1 = 0 \), i.e., at \( x = -1 \). Thus, \( f(x) \) is not defined at \( x = -1 \).

Additionally, \( f(x) \) maps \( x \to \frac{x - 1}{x + 1} \). To find \( f^{-1}(x) \), solve for \( y \) in terms of \( x \):

\[ y = \frac{x - 1}{x + 1}. \]

Rearrange:

\[ y(x + 1) = x - 1, \]

\[ yx + y = x - 1, \]

\[ yx - x = -1 - y, \]

\[ x(y - 1) = -(1 + y), \]

\[ x = \frac{-(1 + y)}{y - 1}. \]

Thus, the inverse function is:

\[ f^{-1}(x) = \frac{-(1 + x)}{x - 1}. \]

Step 4: Determine when \( f(f^{-1}(x)) \neq x \).

Substitute \( f^{-1}(x) \) into \( f(x) \):

\[ f(f^{-1}(x)) = f\left(\frac{-(1 + x)}{x - 1}\right). \]

Using the formula for \( f(x) \), substitute \( \frac{-(1 + x)}{x - 1} \) into \( f(x) = \frac{x - 1}{x + 1} \):

\[ f\left(\frac{-(1 + x)}{x - 1}\right) = \frac{\frac{-(1 + x)}{x - 1} - 1}{\frac{-(1 + x)}{x - 1} + 1}. \]

Simplify the numerator:

\[ \frac{-(1 + x)}{x - 1} - 1 = \frac{-(1 + x) - (x - 1)}{x - 1} = \frac{-1 - x - x + 1}{x - 1} = \frac{-2x}{x - 1}. \]

Simplify the denominator:

\[ \frac{-(1 + x)}{x - 1} + 1 = \frac{-(1 + x) + (x - 1)}{x - 1} = \frac{-1 - x + x - 1}{x - 1} = \frac{-2}{x - 1}. \]

Thus:

\[ f(f^{-1}(x)) = \frac{\frac{-2x}{x - 1}}{\frac{-2}{x - 1}} = \frac{-2x}{-2} = x. \]

This shows that \( f(f^{-1}(x)) = x \) holds for all \( x \neq 1 \) and \( x \neq -1 \).

Step 5: Identify the set \( S \).

The function \( f(x) \) is undefined at \( x = -1 \), and the inverse function \( f^{-1}(x) \) is undefined at \( x = 1 \). Therefore, \( f(f^{-1}(x)) \neq x \) at these two points. Thus:

\[ S = \{-1, 1\}. \]

Step 6: Determine the cardinality of \( S \).

The set \( S \) contains exactly two elements: \( -1 \) and \( 1 \). Therefore, the cardinality of \( S \) is:

\[ \text{Cardinality of } S = 2. \]

Final Answer:

\( 2 \)