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Question: Let \( f:A \to B \) be a function defined as \( f\left( x \right) = \dfrac{{x - 1}}{{x - 2}} \) , wh...

Let f:A→Bf:A \to B be a function defined as f(x)=x−1x−2f\left( x \right) = \dfrac{{x - 1}}{{x - 2}} , where A = R - \left\\{ 2 \right\\} and B = R - \left\\{ 1 \right\\} . Then which of the following options is correct for ff ?
(A) Invertible and f−1(y)=2y+1y−1{f^{ - 1}}\left( y \right) = \dfrac{{2y + 1}}{{y - 1}}
(B) Invertible and f−1(y)=3y−1y−1{f^{ - 1}}\left( y \right) = \dfrac{{3y - 1}}{{y - 1}}
(C) Not invertible
(D) Invertible and f−1(y)=2y−1y−1{f^{ - 1}}\left( y \right) = \dfrac{{2y - 1}}{{y - 1}}

Explanation

Solution

If f(x)=yf\left( x \right) = y then the inverse function f−1{f^{ - 1}} can be defined as f−1(y)=x{f^{ - 1}}\left( y \right) = x . So for finding the inverse you just need to put the definition of ff equal to yy . Then express the value of xx in the form of yy . Now check whether the domain of f−1{f^{ - 1}} is the set B = R - \left\\{ 1 \right\\} . If the domain satisfies then, the function is invertible.

Complete step-by-step solution:
Let’s analyse the given information first. We have a function ff which is mapped from set A to set B, where set A = R - \left\\{ 2 \right\\} which means A (values of xx ) contains all the real numbers except the integer 22 and set B = R - \left\\{ 1 \right\\} which means B (values of yy ) contains all the real numbers except integer 11 . The function ff is defined as: f(x)=x−1x−2f\left( x \right) = \dfrac{{x - 1}}{{x - 2}} .
With this information, we need to find whether or not the given function is invertible.
But first, we need to understand what does invertible mean. In mathematics, an inverse function (or anti-function) is a function that "reverses" another function: if the function ff applied to an input xx gives a result of yy , then applying its inverse function gg to yy gives the result xx , and vice versa, i.e., f(x)=yf\left( x \right) = y if and only if g(y)=xg\left( y \right) = x . The inverse function of ff is also denoted as f−1{f^{ - 1}} .
Therefore, we can say: f(x)=y⇒f−1(y)=xf\left( x \right) = y \Rightarrow {f^{ - 1}}\left( y \right) = x
Let’s put f(x)=yf\left( x \right) = y and define expression in term of yy as the variable
⇒f(x)=y=x−1x−2⇒yx−2y=x−1⇒yx−x=2y−1⇒x(y−1)=2y−1\Rightarrow f\left( x \right) = y = \dfrac{{x - 1}}{{x - 2}} \Rightarrow yx - 2y = x - 1 \Rightarrow yx - x = 2y - 1 \Rightarrow x\left( {y - 1} \right) = 2y - 1
Thus, we get: f−1(y)=x=2y−1y−1{f^{ - 1}}\left( y \right) = x = \dfrac{{2y - 1}}{{y - 1}}
And we already know that yy belongs to the set R - \left\\{ 1 \right\\} which follows domain because of the definition of f−1(y){f^{ - 1}}\left( y \right) contains (y−1)\left( {y - 1} \right) in the denominator that gives us y≠1y \ne 1 .
Hence, the function ff is invertible and f−1(y)=2y−1y−1{f^{ - 1}}\left( y \right) = \dfrac{{2y - 1}}{{y - 1}}

Thus, the option (D) is the correct answer.

Note: A function ff which is mapped from XX to YY is called an invertible function if and only if f−1{f^{ - 1}} can be mapped completely from YY to XX . Notice that for finding the inverse function of ff we just tried to express the value of the variable xx in the form of variable yy . We removed the numbers 2 and 12{\text{ and }}1 from the set of real numbers because at this point the denomination of the expression becomes zero, which is not defined and hence not allowed.