State which of the following functions are one-one and why?
`f:R.{1}toR` defined by `f(x)=(x+1)/(x-1)`

Answer

To determine whether the function \( f: \mathbb{R} \setminus \{1\} \to \mathbb{R} \) defined by \( f(x) = \frac{x+1}{x-1} \) is one-one, we will follow these steps:

### Step 1: Assume \( f(x_1) = f(x_2) \)
We start by assuming that the function values at two different inputs are equal:
\[ f(x_1) = f(x_2) \]
This means:
\[ \frac{x_1 + 1}{x_1 - 1} = \frac{x_2 + 1}{x_2 - 1} \]

### Step 2: Cross-multiply
To eliminate the fractions, we cross-multiply:
\[ (x_1 + 1)(x_2 - 1) = (x_2 + 1)(x_1 - 1) \]

### Step 3: Expand both sides
Expanding both sides gives:
\[ x_1 x_2 - x_1 + x_2 - 1 = x_1 x_2 - x_2 + x_1 - 1 \]

### Step 4: Simplify the equation
Now, we simplify the equation by subtracting \( x_1 x_2 - 1 \) from both sides:
\[ -x_1 + x_2 = -x_2 + x_1 \]
Rearranging this leads to:
\[ -x_1 + x_2 + x_2 - x_1 = 0 \]
This simplifies to:
\[ 2x_2 = 2x_1 \]

### Step 5: Conclude that \( x_1 = x_2 \)
Dividing both sides by 2 gives:
\[ x_1 = x_2 \]
Since we started with the assumption that \( f(x_1) = f(x_2) \) and derived that \( x_1 = x_2 \), we conclude that the function is one-one.

### Final Conclusion
Thus, the function \( f(x) = \frac{x+1}{x-1} \) is a one-one function.

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