Let `f :R to R` be fefined as f(x) = 2x -1 and ` g : R - {1} to R` be defined as ` g(x) = (x-1/2)/(x-1)` Then the composition function ƒ(g(x)) is :

Options

A. onto but not one-one
B. both one-one and onto
C. one-one but not onto
D. neither one-one nor onto

Answer

To find the composition function \( f(g(x)) \) where \( f(x) = 2x - 1 \) and \( g(x) = \frac{x - \frac{1}{2}}{x - 1} \), we will follow these steps:

### Step 1: Write down the functions
We have:
- \( f(x) = 2x - 1 \)
- \( g(x) = \frac{x - \frac{1}{2}}{x - 1} \)

### Step 2: Substitute \( g(x) \) into \( f(x) \)
To find \( f(g(x)) \), we will replace \( x \) in \( f(x) \) with \( g(x) \):
\[ f(g(x)) = f\left(\frac{x - \frac{1}{2}}{x - 1}\right) \]

### Step 3: Apply the function \( f \)
Now we will apply the function \( f \):
\[ f(g(x)) = 2\left(\frac{x - \frac{1}{2}}{x - 1}\right) - 1 \]

### Step 4: Simplify the expression
Now we simplify the expression:
\[ = \frac{2(x - \frac{1}{2})}{x - 1} - 1 \]
\[ = \frac{2x - 1}{x - 1} - 1 \]
To combine the terms, we need a common denominator:
\[ = \frac{2x - 1 - (x - 1)}{x - 1} \]
\[ = \frac{2x - 1 - x + 1}{x - 1} \]
\[ = \frac{x}{x - 1} \]

### Step 5: Conclusion
Thus, the composition function \( f(g(x)) \) is:
\[ f(g(x)) = \frac{x}{x - 1} \]