The domain of the function $f(x) = \sqrt {\cos x}$ is
- $\left[0, \frac {\pi}{2}\right]$
- $\left[0, \frac {\pi}{2}\right] \cup \left[\frac {3 \pi}{2},2\pi \right]$
- $\left[\frac {3 \pi}{2},2\pi \right]$
- $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
The Correct Option is B
Solution and Explanation
Given that;
\(f(x)=\sqrt{cosx}\)
\(\therefore \) \(cosx≥0\)
i.e., \(x\in[2n\pi+0,2n\pi+\frac{\pi}{2}]\cup[2n\pi+\frac{3\pi}{2},2n\pi+2\pi]\)
where,\(n\in{z}\)
For n=0
\(x\in[0,\frac{\pi}{2}\cup[\frac{3\pi}{2},2\pi]\)
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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