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

Answer 1

The Correct Option is A

Given that:

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

Here we need to get the cardinality of \(S\)

[we need to calculate the values of \( x\)  for which the given condition does not hold.]

The condition is: \(f(x) ≠ x\)

\(f(x) = \dfrac{x - 1}{x + 1}\)

Now , \(x\) for which \(f(x) ≠ x\):

\(⇒\dfrac{x - 1}{x + 1} ≠ x\)

Subtract x from both sides:

\(\dfrac{1}{x + 1} ≠ 0\)

Here, \(1\) can not be \(0\) 

So,

    \(x + 1 ≠ 0\)

\(⇒x ≠ -1\), 

[The above expression values of x for which the condition does not hold are all real numbers except for \(x = -1\).]

so \(S \) can be represented as,

\(S = {x ∈ R | x ≠ -1}\)

The cardinality of S is the number of elements in the set S. Since the set S contains all real numbers except -1, the cardinality is infinity, denoted by \(∞\). (_Ans.)