Let tan-1 x ∈(\(-\frac π2\) \(\frac π2\)) for x ∈ R Then the number of real solutions of the equation √1 + cos (2x) = √2 tan -1 (tan x) in the set (- 3π/2, - π/2) ∪ (- π/2, π/2) ∪ (π/2, 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 √1 + cos (2x) = √2 tan -1 (tan x) in the set (- 3π/2, - π/2) ∪ (- π/2, π/2) ∪ (π/2, 3π/2) is equal to
Correct Answer: 3
Approach Solution - 1
Real Solutions of the Given Equation
Given: The equation is:
\(\sqrt{1 + \cos(2x)} = \sqrt{2} \tan^{-1}(\tan x)\)
We are tasked with finding the number of real solutions of this equation in the set \( \left( -\frac{3\pi}{2}, -\frac{\pi}{2} \right) \cup \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \cup \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \).
Solution:
First, recall that the domain of \( \tan^{-1} x \) is \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), so we must consider this constraint for the equation \( \tan^{-1} (\tan x) \).
Next, consider the behavior of both sides of the equation:
- For \( \sqrt{1 + \cos(2x)} \), we know that this is a periodic function with a period of \( \pi \), as the cosine function repeats every \( \pi \). Therefore, we expect solutions within the given intervals.
- For \( \sqrt{2} \tan^{-1} (\tan x) \), we note that \( \tan^{-1} (\tan x) = x \) for \( x \in \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), but it behaves differently for values outside this range due to periodicity.
By examining the equation over the intervals \( \left( -\frac{3\pi}{2}, -\frac{\pi}{2} \right) \), \( \left( -\frac{\pi}{2}, \frac{\pi}{2} \right) \), and \( \left( \frac{\pi}{2}, \frac{3\pi}{2} \right) \), we can determine the number of solutions.
The number of real solutions of the equation in the given set is 3.
Approach Solution -2
Number of solutions = Number of intersection points = 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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