If \[f:R \to R\] is given by \[f(x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\],then the function f is
A. many-one onto
B. many-one into
C. one-one into
D. one-one onto

Hint: To answer this question, keep in mind that a function f:A->B is a bijection if it is both one-one and onto. If there is only one image y ε B for each x ε A and each y ε B has a unique pre-image x ε A (i.e. no two elements of A have the same image in B), then f is a one-one function. Otherwise, f is a one-to-many function.

FORMULA USED:
One to one: \[f(x) = f(y) \Rightarrow x = y\forall x,y \in A\]



Complete step by step solution: Statements with two variables are known as linear equations, and they can be resolved through trial and error. The slope of the line (the y-intercept) can be used as the \[x\] -coordinate to solve a linear equation in one variable. Solving for \[x\] is a common term used for this purpose.
Certain polynomial functions have characteristics known as "many-one qualities," which state that every element in a set belongs to at least one specific member of the set.
The equation given is \[f:R \to R\], which is defined as
\[f(x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\]
As, \[f(x) = \dfrac{{{{( - x)}^2} - 4}}{{{{( - x)}^2} + 2}}\]
\[\begin{array}{l}f( - x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\\ = > f( - x) = f(x)\end{array}\]
Therefore, f is many-one
Since,
\[f(x) = \dfrac{{{x^2} - 4}}{{{x^2} + 2}}\]
On expansion, we get
\[f(x) = \dfrac{{({x^2} + 1) - 5}}{{{x^2} + 2}}\]
Now, we have to split the denominator for each term in the numerator, we get\[f(x) = \dfrac{{({x^2} + 1)}}{{{x^2} + 2}} - \dfrac{{ - 5}}{{{x^2} + 2}}\]
On canceling the similar terms in the above equation, we get
\[f(x) = 1 - \dfrac{5}{{{x^2} + 2}}\]
On simplifying the above we get
\[ \Rightarrow 1 - 5\]
\[ = - 4\]
Hence, f is many one and into.

So, Option A is correct.

Note: In order to define many-one onto functions, one must take the natural logarithm of both sides. You may obtain \[\left( x \right) = log2{\rm{ }}\left( {f\left( x \right)} \right) = log2{\rm{ }}\left( {4x + 1} \right)\] . This function can be determined by applying the inverse function theorem. \[f\] is said to be many-one onto from A into B if \[f:AB\] is given by \[f\left( x \right) = ax + by + c\] where \[A\] and \[B\] both contain at least one element.
You might conceive of this as saying that the domain of \[f\] is all real numbers between \[1\] and \[1\], inclusive. The function f is many-one onto if and only if there is a single real integer.