Question

Let $f:R\to R$be defined as $f\left(x\right)=2x-1$and $g:R-\left\{1\right\}\to R$ be defined as $g\left(x\right)=\frac{\left(x-\frac{1}{2}\right)}{(x-1)}$. Then the composition function $f\left(g\right(x)$is:

• A. Both one-one and onto
• B. onto but not one-one
• C. Neither one-one nor onto
• D. one-one but not onto

Answer 1

The correct option is D

one-one but not onto

Determine the composition function$f\left(g\right(x)$

Step 1: Determining the relation of $f\left(g\right(x)$

We have, $f\left(x\right)=2x-1$, $g\left(x\right)=\frac{\left(x-\frac{1}{2}\right)}{(x-1)}$

Now,

$$\begin{gathered} f\left(g\right(x\left)\right)=2g\left(x\right)-1 \\ =2\frac{\left(x-{\displaystyle \frac{1}{2}}\right)}{(x-1)}-1 \\ =\frac{2(2x-1)-2(x-1)}{2(x-1)} \\ =\frac{4x-2-2x+2}{2x-2} \\ =\frac{x}{x-1} \end{gathered}$$

So, the range of $f\left(g\right(x)$ is $R-\left\{1\right\}$

Codomain is $R.$

Hence, $f\left(g\right(x)$ is not onto as the range and codomain are not the same.

Step 2: Determining the relation

If the function is one-one, then the function is always increasing or decreasing in its domain.

$$\begin{gathered} f\left(g\right(x\left)\right)=\frac{x}{x-1} \\ \Rightarrow f\text{'}g\left(x\right)=\frac{(x-1)-x\left(1\right)}{{(x-1)}^2} \\ =\frac{-1}{{(x-1)}^2} \end{gathered}$$

We can conclude that $f\text{'}\left(g\right(x)$ is always decreasing as there is a negative sign.

So, the function is one-one.

Therefore, the correct answer is option (D).