Question:
Let tan-1 x ∈(\(-\frac π2\) \(\frac π2\)) for x ∈ R. Then the number of real solutions of the equation \(\sqrt{1 + \cos (2x)} = \sqrt2 \tan^{-1}(\tan x)\) in the set \((-\frac{3π}{2}, -\frac{π}{2}) ∪ (-\frac{π}{2},\frac{π}{2}) ∪ (\frac{π}{2},\frac{3π}{2})\) is equal to
Let tan-1 x ∈(\(-\frac π2\) \(\frac π2\)) for x ∈ R. Then the number of real solutions of the equation \(\sqrt{1 + \cos (2x)} = \sqrt2 \tan^{-1}(\tan x)\) in the set \((-\frac{3π}{2}, -\frac{π}{2}) ∪ (-\frac{π}{2},\frac{π}{2}) ∪ (\frac{π}{2},\frac{3π}{2})\) is equal to
Updated On: Sep 2, 2024
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Solution and Explanation
Given :
\(\sqrt{1 + \cos (2x)} = \sqrt2 \tan^{-1}(\tan x)\)
\(⇒|\cos x|=\tan^{-1}(\tan x)\)
Number of solutions = Number of intersection points = 3.
So, the correct answer is 3.
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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.
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