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

Answer: (A)

a finite number, but not equal to 1,2,3

Answer 2

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.)