A function is said to be bijective if it is both one-one and onto, Consider the mapping `f : A rarr B` be defined by `f(x) = (x-1)/(x-2)` such that f is a bijection.
Let `g : R - {2} rarr R - {1}` be defined by `g(x) = 2f(x) - 1`. Then `g(x)` in terms of x is :

Options

A. `(x)/(x-2)`
B. `(-2)/(x)`
C. `(x+2)/(x)`
D. `(x + 2)/(x + 2)`

Answer

To find the function \( g(x) \) in terms of \( x \), we start with the given function \( f(x) = \frac{x-1}{x-2} \) and the definition of \( g(x) \) as \( g(x) = 2f(x) - 1 \).

### Step-by-Step Solution:

1. **Substitute \( f(x) \) into \( g(x) \)**:
\[ g(x) = 2f(x) - 1 \]
Given \( f(x) = \frac{x-1}{x-2} \), we substitute this into the equation for \( g(x) \):
\[ g(x) = 2\left(\frac{x-1}{x-2}\right) - 1 \]

2. **Multiply \( 2 \) with \( f(x) \)**:
\[ g(x) = \frac{2(x-1)}{x-2} - 1 \]
This simplifies to:
\[ g(x) = \frac{2x - 2}{x-2} - 1 \]

3. **Express \( -1 \) with a common denominator**:
To combine the terms, we rewrite \( -1 \) as \( -\frac{x-2}{x-2} \):
\[ g(x) = \frac{2x - 2}{x-2} - \frac{x-2}{x-2} \]

4. **Combine the fractions**:
Now that we have a common denominator, we can combine the fractions:
\[ g(x) = \frac{(2x - 2) - (x - 2)}{x-2} \]

5. **Simplify the numerator**:
Simplifying the numerator gives:
\[ g(x) = \frac{2x - 2 - x + 2}{x-2} = \frac{x}{x-2} \]

Thus, the final expression for \( g(x) \) in terms of \( x \) is:
\[ g(x) = \frac{x}{x-2} \]