Let `f:R rarr R` be a function such that `f(x)=(x^(2)+2x+1)/(x^(2)+1)`.Then

Options

A. `f(x)` is many-one in `(-oo,-1)`
B. `f(x)` is one - one in `(-oo,oo)`
C. `f(x)` is many-one in `(1,oo)`
D. `f(x)` is one - one in `[1,oo)` but not in `(-oo,oo)]`

Answer

To determine whether the function \( f(x) = \frac{x^2 + 2x + 1}{x^2 + 1} \) is one-to-one (1-1), we will analyze the function step by step.

### Step 1: Simplify the Function
First, we can simplify the function \( f(x) \):

\[ f(x) = \frac{x^2 + 2x + 1}{x^2 + 1} = \frac{(x + 1)^2}{x^2 + 1} \]

### Step 2: Analyze the Function
Next, we will analyze the behavior of the function. We can rewrite \( f(x) \) as:

\[ f(x) = \frac{(x + 1)^2}{x^2 + 1} \]

### Step 3: Find the Derivative
To check if the function is one-to-one, we can find the derivative \( f'(x) \) and analyze its sign.

Using the quotient rule, we have:

\[ f'(x) = \frac{(x^2 + 1) \cdot 2(x + 1) - (x + 1)^2 \cdot 2x}{(x^2 + 1)^2} \]

Simplifying the numerator:

\[ = 2(x + 1)(x^2 + 1) - 2x(x + 1)^2 \]
\[ = 2(x + 1)(x^2 + 1 - x(x + 1)) \]
\[ = 2(x + 1)(x^2 + 1 - x^2 - x) = 2(x + 1)(1 - x) \]

Thus, we have:

\[ f'(x) = \frac{2(x + 1)(1 - x)}{(x^2 + 1)^2} \]

### Step 4: Analyze the Sign of the Derivative
Now, we analyze the sign of \( f'(x) \):

- \( f'(x) > 0 \) when \( x < 1 \) (since \( x + 1 > 0 \) for all \( x > -1 \))
- \( f'(x) < 0 \) when \( x > 1 \)

This means that the function is increasing on the interval \( (-\infty, 1) \) and decreasing on the interval \( (1, \infty) \).

### Step 5: Conclusion
Since the function increases up to \( x = 1 \) and then decreases, it is not one-to-one over the entire real line \( (-\infty, \infty) \). However, it is one-to-one on the interval \( [1, \infty) \).

Thus, the function \( f(x) \) is one-to-one on the interval \( [1, \infty) \) but not on the entire domain of real numbers.

### Final Answer
The function \( f(x) \) is one-to-one on the interval \( [1, \infty) \).

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