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Question: Let A = {x ∈ R : x is not a positive integer} Define a function \(f:A\to R\) as \(f(x)=\dfrac{2x}{x-...

Let A = {x ∈ R : x is not a positive integer} Define a function f:A→Rf:A\to R as f(x)=2xx−1f(x)=\dfrac{2x}{x-1} then f is
a) injective but nor surjective b) not injective a) surjective but not injective a) neither injective nor surjective \begin{aligned} & \text{a) }\text{injective but nor surjective} \\\ & \text{b) not }\text{injective} \\\ & \text{a) }\text{surjective but not injective} \\\ & \text{a) }\text{neither injective nor surjective} \\\ \end{aligned}

Explanation

Solution

First we will try to find if we can get a value of f(x) in R for which x is not in A. and hence the function f(x) does not cover the whole range R. So it is not surjective. For injective we will check that if f(x) = f(y) then we get x = y.

Complete step-by-step answer:
Now consider f(x)=2xx−1f(x)=\dfrac{2x}{x-1}
Let us consider point y = 4. Now 4 is a real number hence belongs to the range of R.
Now let us check if there is any point x in set A such that f (x) = 4.
Hence if there is such x we will have
4=2xx−1 ⇒4(x−1)=2x ⇒4x−4=2x ⇒4x−2x=4 ⇒2x=4 ⇒x=2 \begin{aligned} & 4=\dfrac{2x}{x-1} \\\ & \Rightarrow 4\left( x-1 \right)=2x \\\ & \Rightarrow 4x-4=2x \\\ & \Rightarrow 4x-2x=4 \\\ & \Rightarrow 2x=4 \\\ & \Rightarrow x=2 \\\ \end{aligned}
But now x = 2 is a positive integer and by definition of A we have 2 does not belongs to A
Hence there is no x in A such that f(x) = 4.
Hence the function is not surjective.
Now let us check if the function is injective
Let us say the function has two values the same that is f(x) = f(y). Then if the function is injective this will be true only if x = y. if x and y are distinct and still f(x) = f(y) then simply the function is not one-one.
Now consider f(x) = f(y), hence we get.
2xx−1=2yy−1 ⇒xx−1=yy−1 \begin{aligned} & \dfrac{2x}{x-1}=\dfrac{2y}{y-1} \\\ & \Rightarrow \dfrac{x}{x-1}=\dfrac{y}{y-1} \\\ \end{aligned}
Now using the property of dividendo we get
⇒x−(x−1)x−1=y−(y−1)y−1 ⇒1x−1=1y−1 ⇒y−1=x−1 ⇒y=x \begin{aligned} & \Rightarrow \dfrac{x-(x-1)}{x-1}=\dfrac{y-(y-1)}{y-1} \\\ & \Rightarrow \dfrac{1}{x-1}=\dfrac{1}{y-1} \\\ & \Rightarrow y-1=x-1 \\\ & \Rightarrow y=x \\\ \end{aligned}
Hence we have if f(x) = f(y) then x = y.
Hence the function is one-one or injective.

So, the correct answer is “Option A”.

Note: Here injective means one-one and surjective means onto, not to be confused in these two words. Also while trying to prove that the function is not surjective try and check if there is any value of x which is not in domain and for which the value of function is still in R. here we could see that for x = 2 is not in A and we get f(2) is in R. hence we could easily contradict the statement.