Which of the following functions is inverse of itself?
- $ f\left(t\right) = \frac{\left(1-t\right)}{\left(1+t\right)} $
- $ f\left(t\right) = \frac{\left(1-t^{2}\right)}{\left(1+t^{2}\right)} $
- $ f\left(t\right) = 4^{log\, t} $
- $ f(t) = 2^t $
The Correct Option is A
Solution and Explanation
$\therefore t=f^{-1}(y)$
Now, $y=f (t) =\frac{1-t}{1+t}$
$\Rightarrow y+ty=1-t$
$\Rightarrow t+ty=1-y$
$\Rightarrow t=\frac{1-y}{1+y}$
i.e., $f^{-1}(y) =\frac{1-y}{1+y}$
or $f^{-1}(t)=\frac{1-t}{1+t}$
Thus, this function is inverse of itself
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Concepts Used:
Functions
A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. Let A & B be any two non-empty sets, mapping from A to B will be a function only when every element in set A has one end only one image in set B.
Kinds of Functions
The different types of functions are -
One to One Function: When elements of set A have a separate component of set B, we can determine that it is a one-to-one function. Besides, you can also call it injective.
Many to One Function: As the name suggests, here more than two elements in set A are mapped with one element in set B.
Moreover, if it happens that all the elements in set B have pre-images in set A, it is called an onto function or surjective function.
Also, if a function is both one-to-one and onto function, it is known as a bijective. This means, that all the elements of A are mapped with separate elements in B, and A holds a pre-image of elements of B.
Read More: Relations and Functions