Question

Which of the following function is (are) injective map (s)?
A. \[f(x) = {x^2} + 2,x \in ( - \infty ,\infty )\]
B. \[f(x) = |x + 2|,x \in [ - 2,\infty )\]
C. \[f(x) = (x - 4)(x - 5),x \in ( - \infty ,\infty )\]
D. \[f(x) = \dfrac{{4{x^2} + 3x - 5}}{{4 + 3x - 5{x^2}}},x \in ( - \infty ,\infty )\]

Answer 1

Hint: The injective map is an injective function, which means that for each set of elements \[x\] and \[y\] in the function's domain, there exists a single element \[z\] such that \[f\left( x \right){\rm{ }} = {\rm{ }}g\left( y \right)\].
Whenever,\[f\left( x \right) = f\left( y \right)\], then\[x = y\]. And equivalently, if \[x{\rm{ }} \ne {\rm{ }}y\], then \[f\left( x \right){\rm{ }} \ne {\rm{ }}f\left( y \right)\] formally, it is stated as, if \[f\left( x \right){\rm{ }} = {\rm{ }}f\left( y \right)\] implies \[x = y\], then \[f\] is one-to-one mapped, or \[f\] is \[1 - 1\].

Complete step by step solution: A function is one-to-one or injective if it does not map two different elements in the domain to the same element in the range. An injective function is called an injection. An injection may also be called a one-to-one (or\[1-1\]) function; some people consider this less formal than "injection''.
The function \[f(x) = {x^2} + 2,x \in ( - \infty ,\infty )\] is not injective as \[f(1) = f( - 1)\]
Apply rule \[{1^a} = 1\]:
\[{1^2} = 1\]
\[f = 1 + 2\]
Add the numbers:
\[1 + 2 = 3\]
\[f = 3\]
Follow the PEMDAS order of operations
Calculate exponents\[{( - 1)^2}\]:
\[ = 1 + 2\]
Add and subtract (left to right)\ [1 + 2\]:
\[ = 3\]
But, \[1 \ne - 1\].
The function \[f(x) = (x - 4)(x - 5),x \in ( - \infty ,\infty )\] is not one-one as \[f(4) = f(5)\], but \[4 \ne 5\].
 The function,
\[f(x) = \dfrac{{4{x^2} + 3x - 5}}{{4 + 3x - 5{x^2}}}x \in ( - \infty ,\infty )\] is also not injective as\[f(1) = f( - 1)\], but \[1 \ne - 1\].
For the function,
\[f(x) = |x + 2|,x \in [ - 2,\infty )\].
Let\[f(x) = f(y),x,y \in [ - 2,\infty )\]
\[ \Rightarrow x + 2 = y + 2\]
\[ \Rightarrow x = y\]
 \[{\rm{f}}\] is an injection.

So, option B is correct.

Note: Most students make a mistake when they try to find an injective map. This occurs when they use the order of operations incorrectly. For example, if \[f\left( x \right)\]is defined as \[{x^2} + 2\], then students may try to apply the inverse function—in other words, working backwards from \[x\] to get \[f\left( { - 2} \right)\]. However, this does not work because \[ - 1\] and \[1\] are not invertible functions. The correct way to find an injective map is by using PEMDAS order of operations: first parentheses, exponents (left-to-right), median theta operator, multiplicative operators (left-to-right), additive operators (left-to-right).