Question

If $f$ and $g$ are one- one functions from $R \to R$, then,
A. \[f + g\] is one-one
B. $fg$ is one-one
C. $fog$ is one-one
D. none of these\[\]

Answer 1

Hint: Since, $f$ and $g$ are one- one functions from $R \to R$, then if ${x_1} \ne {x_2}$, then $f\left( {{x_1}} \right) \ne f\left( {{x_2}} \right)$ and $g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)$. Let $f\left( x \right) = - x$ and $g\left( x \right) = x$ and then determine whether \[f + g\] and $fg$ are one- one functions by taking examples. Similarly, prove if $fog$ is one-one using the definition of one-one function.

Complete step-by-step answer:
We are given that $f$ and $g$ are one- one functions from $R \to R$
That is, if ${x_1} \ne {x_2}$, then $f\left( {{x_1}} \right) \ne f\left( {{x_2}} \right)$ and $g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)$
Let us first see if \[f + g\] is one-one or not.
Let $f\left( x \right) = - x$ and $g\left( x \right) = x$
Then, $f\left( {{x_1}} \right) = f\left( {{x_2}} \right)$ if and only if ${x_1} = {x_2}$
Similarly, $g\left( {{x_1}} \right) = g\left( {{x_2}} \right)$ if and only if ${x_1} = {x_2}$
That is $f$ and $g$ are one-one functions.
Then, \[f + g\left( x \right) = - x + x = 0\]
For example, $f + g\left( { - 2} \right) = f + g\left( 1 \right) = 0$ but $ - 2 \ne 1$
Thus, \[f + g\] is not none-one.

Similarly, $f.g\left( x \right) = - x\left( x \right) = - {x^2}$
Let ${x_1} = 2,{x_2} = - 2$, then,
$f.g\left( 2 \right) = - {\left( 2 \right)^2} = - 4$ and $f.g\left( { - 2} \right) = - {\left( { - 2} \right)^2} = - 4$
Here, $fg\left( 2 \right) = fg\left( { - 2} \right) = - 4$ but $2 \ne - 2$
Therefore, $fg$ is not one-one.

Now, if ${x_1} \ne {x_2}$, then $g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)$ which is a given condition.
And if $g\left( {{x_1}} \right) \ne g\left( {{x_2}} \right)$ then $f\left( {g\left( {{x_1}} \right)} \right) \ne f\left( {g\left( {{x_2}} \right)} \right)$ because $f$ is a one-one function.
Hence, $fog$ is one-one.
Thus, option C is correct.

Note: If a function is one-one, then each element of the domain has a unique image. Not two different elements can map to the same element. Examples should be taken carefully such that it helps to check if the function is one-one or not.