Let \(R\) be the set of all real numbers and \(f : R \rightarrow R\) be given by \(f(x) = 3x^2 + 1\).Then the set \(f ^{-1}\left(\left[1, 6\right]\right)\) is
- $\left\{-\sqrt{\frac{5}{3}},0, \sqrt{\frac{5}{3}}\right\}$
- $\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]$
- $\left[-\sqrt{\frac{1}{3}}, \sqrt{\frac{1}{3}}\right]$
- $\left(-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right)$
The Correct Option is B
Solution and Explanation
Given, \(y=3 x^{2}+1\)
\(\Rightarrow 3 x^{2} =y-1\)
\(\Rightarrow x^{2}=\frac{y-1}{3}\)
\(\Rightarrow x=\pm \sqrt{\frac{y-1}{3}}\)
\(\therefore f^{-1}(x)=\pm \sqrt{\frac{x-1}{3}}\)
When \(x \in[1,6]\)
Then, \(f^{-1}(x) \in\left[-\sqrt{\frac{5}{3}}, \sqrt{\frac{5}{3}}\right]\)
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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