Question

Which of the following function is inverse function
A) \[f\left( x \right) = \dfrac{1}{{x - 1}}\]
B) \[f\left( x \right) = {x^2}\] , for all x
C) \[f\left( x \right) = {x^2},x \geqslant 0\]
D) \[f\left( x \right) = {x^2},x \leqslant 0\]

Answer 1

Hint: The rule of inverse function is, for a function \[f:X \to Y\] to have an inverse, it must have the property that for every y in Y, there is exactly one x in X such that \[f\left( x \right) = y\] . This property ensures that a function \[g:Y \to X\] exists with the necessary relationship with f. The domain of the function becomes the range of the inverse of the function. The range of the function becomes the domain of the inverse of the function, hence consider all the options and solve for one-to-one function, such that if \[f\left( x \right)\] is one-one, then the inverse exists.

Complete step by step solution:
To find the inverse function from the following options, let us solve all the options and find the solution, which satisfy the inverse of a function.
Option A: \[f\left( x \right) = \dfrac{1}{{x - 1}}\] , \[x \ne 1\]
Let, \[f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\]
From the given function, we get:
 \[ \Rightarrow \dfrac{1}{{{x_1} - 1}} = \dfrac{1}{{{x_2} - 1}}\]
 \[ \Rightarrow {x_1} - 1 = {x_2} - 1\]
Hence, we get:
 \[ \Rightarrow {x_1} = {x_2} - 1 + 1\]
 \[ \Rightarrow {x_1} = {x_2}\]
Hence, \[f\left( x \right)\] is one-one.
Let, \[y = \dfrac{1}{{x - 1}}\]
 \[ \Rightarrow x - 1 = \dfrac{1}{y}\]
As, we need to solve for x we have:
 \[ \Rightarrow x = \dfrac{1}{y} + 1\] ,
Therefore, \[Range \in R - \left\{ 0 \right\}\] i.e., f is onto, so \[{f^{ - 1}}\] (f inverse) exists.
Option B: \[f\left( x \right) = {x^2}\] , for all x
Let, \[f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\]
From the given function, we get:
 \[ \Rightarrow {x_1}^2 - {x_2}^2\]
 \[ \Rightarrow {x_1} = \pm {x_2}\]
Therefore, f is not one-one, so \[{f^{ - 1}}\] (f inverse) does not exist.
Option C: \[f\left( x \right) = {x^2},x \geqslant 0\]
Let, \[f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\]
From the given function, we get:
 \[ \Rightarrow {x_1}^2 = {x_2}^2\]
 \[ \Rightarrow {x_1} = \pm {x_2}\]
 \[ \Rightarrow {x_1} = {x_2}\left( {x > 0} \right)\]
Hence, f is one-one.
Now, let \[y = {x^2}\] ,
 \[x = \sqrt y \]
 \[ \Rightarrow y > 0\]
Therefore, f is not onto, so \[{f^{ - 1}}\] (f inverse) does not exist.
Option D: \[f\left( x \right) = {x^2},x \leqslant 0\]
Let, \[f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\]
From the given function, we get:
 \[ \Rightarrow {x_1}^2 = {x_2}^2\]
 \[ \Rightarrow {x_1} = \pm {x_2}\]
 \[ \Rightarrow {x_1} = {x_2}\left( {as,x \leqslant 0} \right)\]
Hence, f is one-one.
Range \[ \Rightarrow y \geqslant 0\]
Therefore, f is not onto, so \[{f^{ - 1}}\] (f inverse) does not exist.
Hence, option A is the right answer.
So, the correct answer is “Option A”.

Note: One to One Function Inverse: If f is a function defined as \[y = f\left( x \right)\] , then the inverse function of f is \[x = {f^{ - 1}}\left( y \right)\] i.e., \[{f^{ - 1}}\] defined from y to x. The inverse function co-domain of f is the domain of \[{f^{ - 1}}\] and the domain of f is the co-domain of \[{f^{ - 1}}\] . Only one-to-one functions have its inverse since each element from the range corresponds to one and only one domain element.
One to one function basically denotes the mapping of two sets. A function g is one-to-one if every element of the range of g corresponds to exactly one element of the domain of g. One-to-one is also written as 1-1. If for every element of B, there is at least one or more than one element matching with A, then the function is said to be onto.