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# If f: R 🡪R is defined by f(x) = |x|, then${\text{A}}{\text{. }}{{\text{f}}^{ - 1}}\left( {\text{x}} \right) = - {\text{x}} \\ {\text{B}}{\text{. }}{{\text{f}}^{ - 1}}\left( {\text{x}} \right) = \dfrac{1}{{\left| {\text{x}} \right|}} \\ {\text{C}}{\text{. }}{{\text{f}}^{ - 1}}{\text{ does not exist}} \\ {\text{D}}{\text{. }}{{\text{f}}^{ - 1}}\left( {\text{x}} \right) = \dfrac{1}{{\text{x}}} \\$  Verified
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Hint: Using the function given in the question, we check whether the function is one-one and onto in order to determine if the function is invertible, which gives us the answer.

Given Data, f(x) = |x|.

The function of mod x, shown as |x| always returns a positive value, i.e. |(+x)| or |(– x)| will give the value +x in return.

For a function f(x) to be invertible, the function must be one-one and onto.

The range of f(x) =∣x∣ is [0, ∞), (only non-negative numbers because of mod function).

While the co-domain of f(x) is given as R in the question. Hence f(x) is not onto.
(As every element in co-domain does not have a preimage in the domain, i.e. all the negative values in the co-domain are not mapped to any element in the domain because of the modulus function in the definition of f(x)).

Also, since f(x) = f (- x), f(x) is also not one-one in its domain.

Hence, f(x) is not invertible, i.e., the function ${{\text{f}}^{ - 1}}\left( {\text{x}} \right)$does not exist.
Option C is the right answer.

Note: In order to find the inverse, the function needs to be invertible. For the function to be invertible it must be both one-one and onto.
One – one function also known as an injective function is a function that maps distinct elements of its domain to distinct elements of its codomain. In other words, every element of the function's codomain is the image of at most one element of its domain.

Onto function also known as a surjective function could be explained by considering two sets, Set A and Set B which consist of elements. If for every element of B there is at least one or more than one element matching with A then it is called onto function.