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Step 1: Write down the given functions
We have two functions:
1) $f(x) = 2x - 1$, with domain and codomain as $R$.
2) $g(x) = \frac{x - \frac{1}{2}}{x - 1}$, with domain $R \setminus \{1\}$ and codomain as $R$.
Step 2: Form the composition $f(g(x))$
The composition $f(g(x))$ is calculated by substituting $g(x)$ into $f$. Thus,
$f(g(x)) = 2 \bigl(g(x)\bigr) - 1.$
Since $g(x) = \frac{x - \frac{1}{2}}{x - 1}$, we multiply by 2 and then subtract 1:
$2 \cdot \frac{x - \frac{1}{2}}{x - 1} \;=\; \frac{2x - 1}{x - 1}.$
Therefore,
$f(g(x)) \;=\; \frac{2x - 1}{x - 1} \;-\; 1.
Next, find a common denominator:
$\frac{2x - 1}{x - 1} \;-\; 1 \;=\; \frac{2x - 1}{x - 1} \;-\; \frac{x - 1}{x - 1} \;=\; \frac{2x - 1 - (x - 1)}{x - 1}.$
Simplify the numerator:
$2x - 1 - (x - 1) \;=\; 2x - 1 - x + 1 \;=\; x.
Hence,
$f(g(x)) \;=\; \frac{x}{\,x - 1\,} \;=\; 1 \;+\; \frac{1}{\,x - 1\,}.$
Step 3: Graphical representation (as given)
Below is a reference graph of $1 + \frac{1}{\,x-1\,}$:
Step 4: Check injectivity (one-one)
A function is one-one (injective) if different inputs always map to different outputs. Let $x_1$ and $x_2$ be two different values in the domain $R \setminus \{1\}$. Suppose
$\frac{x_1}{x_1 - 1} = \frac{x_2}{x_2 - 1}.$
Cross-multiplying,
$x_1(x_2 - 1) = x_2(x_1 - 1).$
This expands and simplifies to $x_1 x_2 - x_1 = x_1 x_2 - x_2$. Rearranging yields $-\,x_1 = -\,x_2 \implies x_1 = x_2.$
Thus, $f(g(x))$ does not take the same value for two different inputs, proving it is injective.
Step 5: Check surjectivity (onto)
A function is onto (surjective) if for every possible real value $y$ in the codomain, there exists $x$ in the domain $R \setminus \{1\}$ such that $f(g(x)) = y$. Here, $f(g(x)) = \frac{x}{x-1}$.
Consider $y = 1$. We want $\frac{x}{x-1} = 1.$ This implies:
$x = x - 1
\;\;\Longrightarrow\;\;
0 = -1,
which is a contradiction. Hence, $y = 1$ is not attained by the function. Therefore, it does not cover all real numbers, so it is not onto $\bigl(R\bigr)$.
Conclusion
Since $f(g(x))$ is injective but not surjective onto all of $R$, the correct option is "one-one but not onto."
