State which of the following functions are one-one and why?
`f:R^(+)toR^(+)` defined by f(x)`=1+x^(2)`

Answer

To determine whether the function \( f: \mathbb{R}^{+} \to \mathbb{R}^{+} \) defined by \( f(x) = 1 + x^2 \) is one-one, we will follow these steps:

### Step 1: Understand the definition of a one-one function
A function \( f \) is said to be one-one (or injective) if for any two distinct inputs \( x_1 \) and \( x_2 \) in the domain, the outputs are also distinct. Mathematically, this means:
\[ f(x_1) = f(x_2) \implies x_1 = x_2 \]

### Step 2: Set up the equation to test for injectivity
We need to check if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \). Starting with the equation:
\[ f(x_1) = f(x_2) \]
Substituting the function definition:
\[ 1 + x_1^2 = 1 + x_2^2 \]

### Step 3: Simplify the equation
By subtracting 1 from both sides, we have:
\[ x_1^2 = x_2^2 \]

### Step 4: Analyze the implications of the equation
The equation \( x_1^2 = x_2^2 \) implies:
\[ x_1 = x_2 \quad \text{or} \quad x_1 = -x_2 \]
However, since \( x_1 \) and \( x_2 \) are both in \( \mathbb{R}^{+} \) (the set of positive real numbers), the second case \( x_1 = -x_2 \) is not possible. Therefore, we conclude:
\[ x_1 = x_2 \]

### Step 5: Conclusion
Since we have shown that \( f(x_1) = f(x_2) \) leads to \( x_1 = x_2 \), we can conclude that the function \( f(x) = 1 + x^2 \) is indeed a one-one function.

### Summary
The function \( f: \mathbb{R}^{+} \to \mathbb{R}^{+} \) defined by \( f(x) = 1 + x^2 \) is one-one because for any two inputs that produce the same output, the inputs must be equal.

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