Question

Let $ f:A \to B $ be a function defined as $ f\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 $ f $ ?
(A) Invertible and $ {f^{ - 1}}\left( y \right) = \dfrac{{2y + 1}}{{y - 1}} $
(B) Invertible and $ {f^{ - 1}}\left( y \right) = \dfrac{{3y - 1}}{{y - 1}} $
(C) Not invertible
(D) Invertible and $ {f^{ - 1}}\left( y \right) = \dfrac{{2y - 1}}{{y - 1}} $

Answer 1

Hint: If $ f\left( x \right) = y $ then the inverse function $ {f^{ - 1}} $ can be defined as $ {f^{ - 1}}\left( y \right) = x $ . So for finding the inverse you just need to put the definition of $ f $ equal to $ y $ . Then express the value of $ x $ in the form of $ y $ . Now check whether the domain of $ {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 $ f $ which is mapped from set A to set B, where set $ A = R - \left\{ 2 \right\} $ which means A (values of $ x $ ) contains all the real numbers except the integer $ 2 $ and set $ B = R - \left\{ 1 \right\} $ which means B (values of $ y $ ) contains all the real numbers except integer $ 1 $ . The function $ f $ is defined as: $ f\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 $ f $ applied to an input $ x $ gives a result of $ y $ , then applying its inverse function $ g $ to $ y $ gives the result $ x $ , and vice versa, i.e., $ f\left( x \right) = y $ if and only if $ g\left( y \right) = x $ . The inverse function of $ f $ is also denoted as $ {f^{ - 1}} $ .
Therefore, we can say: $ f\left( x \right) = y \Rightarrow {f^{ - 1}}\left( y \right) = x $
Let’s put $ f\left( x \right) = y $ and define expression in term of $ y $ as the variable
 $ \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}}\left( y \right) = x = \dfrac{{2y - 1}}{{y - 1}} $
And we already know that $ y $ belongs to the set $ R - \left\{ 1 \right\} $ which follows domain because of the definition of $ {f^{ - 1}}\left( y \right) $ contains $ \left( {y - 1} \right) $ in the denominator that gives us $ y \ne 1 $ .
Hence, the function $ f $ is invertible and $ {f^{ - 1}}\left( y \right) = \dfrac{{2y - 1}}{{y - 1}} $

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

Note: A function $ f $ which is mapped from $ X $ to $ Y $ is called an invertible function if and only if $ {f^{ - 1}} $ can be mapped completely from $ Y $ to $ X $ . Notice that for finding the inverse function of $ f $ we just tried to express the value of the variable $ x $ in the form of variable $ y $ . We removed the numbers $ 2{\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.