$f\left(x\right) = \frac{x}{e^{x}-1} + \frac{x}{2} + 2 \cos^{3} \frac{x}{2} $ on $R-\left\{0\right\} $ is
- One one function
- Bisection
- Algebraic function
- Even function
The Correct Option is D
Solution and Explanation
$f(x)=\frac{x}{e^{x}-1}+\frac{x}{2}+2 \cos ^{3} \frac{x}{2}$ on
$R-\{0\}$.
$\because f(-x) =\frac{-x}{e^{-x}-1}-\frac{x}{2}+2 \cos ^{3}\left(-\frac{x}{2}\right)$
$=\frac{-x e^{x}}{1-e^{x}}-\frac{x}{2}+2 \cos ^{3} \frac{x}{2}$
$=\frac{x e^{x}}{e^{x}-1}-\frac{x}{2}+2 \cos ^{3} \frac{x}{2}$
$=\frac{x\left(e^{x}-1+1\right)}{e^{x}-1}-\frac{x}{2}+2 \cos ^{3} \frac{x}{2}$
$=x+\frac{x}{e^{x}-1}-\frac{x}{2}+2 \cos ^{3} \frac{x}{2}$
$=\frac{x}{e^{x}-1}+\frac{x}{2}+2 \cos ^{3} \frac{x}{2}=f(x)$
$\because f(-x)=f(x), \forall x \in R-\{0\}$
$\therefore f(x)$ is an even function.
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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
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