Let \[A = R - \{ 3\} ,B = R - \{ 1\} ,\] and let \[f:A \to B\mid \] be defined by \[f(x) = \dfrac{{x - 2}}{{x - 3}}f\] is
A. Not one-one
B. Not onto
C. Many-one and onto
D. One-one and onto

Hint: In order to come up with a solution to this query, you'll need to solve equations. Functions with a single input and a single output are known as one-one and onto functions.
In this instance, \[f(x) = \dfrac{{x - 2}}{{x - 3}}f\] defines \[f:A \to B\].
Thus, all that is required to determine the answer for x is to substitute various values for \[x\] into the equation and see which one places an equal sign next to \[3\] on both sides of the equation.

Complete step by step solution: One-one mapping is a mapping that associates every number with its inverse. In other words,\[f:A \to B\] is a function that takes any number and returns the corresponding negative number. This can be useful in solving equations or rearranging linear sequences.
 A function \[f:A \to B\] is onto if, for every element \[b \in B\], there exists an element \[a \in A\] such that \[f(a) = b\]. A surjection is another name for onto function, and we refer to it as surjective.
Let \[{x_1},{x_2} \in A\] and let \[f\left( {{x_1}} \right) = f\left( {{x_2}} \right)\] or \[\dfrac{{{x_1} - 2}}{{{x_1} - 3}} = \dfrac{{{x_2} - 2}}{{{x_2} - 3}}\]
or \[{x_1} = {x_2}\] so, \[{\rm{f}}\] is one-one.
To find whether \[{\rm{f}}\] is onto or not, first let us find the range of f.
Let \[y = f(x) = \dfrac{{x - 2}}{{x - 3}}\] or \[x = \dfrac{{3y - 2}}{{y - 1}}\]
\[X\]- Defined if \[y \ne 1\]
i.e. The range of f is \[R - \left\{ 1 \right\}\] which the co-domain is also of \[f\].
Also, for no value of \[y,{\rm{ }}x\] can be \[3\]
i.e., if we put \[3 = x = \dfrac{{3y - 2}}{{y - 1}}\]
Multiply both sides by \[y - 1\]
\[3(y - 1) = \dfrac{{3y - 2}}{{y - 1}}(y - 1)\]
Simplify
\[3(y - 1) = 3y - 2\]
Solve\[3(y - 1) = 3y - 2\] :
This can be written as \[3y - 3 = 3y - 2\]
Subtract \[3y\] from both sides:
\[3y - 3 - 3y = 3y - 2 - 3y\]
Simplify
\[ - 3 = - 2\]
The sides are not equal:
Hence there is no solution, which implies that it is not possible.
Hence, f is onto.

So, option D is correct.

Note: Many students while solving onto functions improperly solves the operations with negative numbers and undoing multiplication of positives by division by negatives, also by distributing division to multiple factors and distributing exponents to multiple terms.