1. Key Concepts and Formulas

• One-One Function (Injective): A function $f$ is one-one if for every $y$ in the codomain, there is at most one $x$ in the domain such that $f(x) = y$. Equivalently, if $f(x_1) = f(x_2)$, then $x_1 = x_2$.
• Many-One Function: A function $f$ is many-one if there exist at least two distinct values $x_1, x_2$ in the domain such that $f(x_1) = f(x_2)$.
• Monotonicity and One-One Property: A function is one-one on an interval if it is strictly monotonic (strictly increasing or strictly decreasing) throughout that interval. If a function changes its monotonicity within an interval, it is many-one on that interval.
• First Derivative Test for Monotonicity:
  • If $f'(x) > 0$ for all $x$ in an interval, $f(x)$ is strictly increasing.
  • If $f'(x) < 0$ for all $x$ in an interval, $f(x)$ is strictly decreasing.
  • If $f'(x) = 0$ at isolated points and $f'(x)$ does not change sign, the function is still monotonic. However, if $f'(x)$ changes sign at a point, that point is a local extremum, and the function is not one-one across an interval containing such a point.
• Quotient Rule for Differentiation: If $f(x) = \frac{u(x)}{v(x)}$, then $f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$.

2. Step-by-Step Solution

Step 1: Define the function and its domain/codomain. The given function is $f: \mathbb{R} \to \mathbb{R}$ defined by $f(x) = \frac{x^2 + 2x + 1}{x^2 + 1}$.

Step 2: Calculate the first derivative, $f'(x)$, to analyze monotonicity. We use the quotient rule. Let $u(x) = x^2 + 2x + 1$ and $v(x) = x^2 + 1$. Then $u'(x) = 2x + 2$ and $v'(x) = 2x$. Applying the quotient rule: $f'(x) = \frac{(2x+2)(x^2+1) - (x^2+2x+1)(2x)}{(x^2+1)^2}$ Expand the numerator: Numerator $= (2x^3 + 2x + 2x^2 + 2) - (2x^3 + 4x^2 + 2x)$ Numerator $= 2x^3 + 2x^2 + 2x + 2 - 2x^3 - 4x^2 - 2x$ Numerator $= -2x^2 + 2$ Numerator $= 2(1 - x^2) = 2(1-x)(1+x)$ So, the derivative is: $f'(x) = \frac{2(1-x)(1+x)}{(x^2+1)^2}$

Step 3: Find the critical points by setting $f'(x) = 0$. The derivative $f'(x)$ is defined for all real $x$ since the denominator $(x^2+1)^2$ is always positive. Setting $f'(x) = 0$: $\frac{2(1-x)(1+x)}{(x^2+1)^2} = 0$ This implies $2(1-x)(1+x) = 0$. The critical points are $x=1$ and $x=-1$. These points divide the real line into three intervals: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$.

Step 4: Analyze the sign of $f'(x)$ in each interval to determine monotonicity. The sign of $f'(x)$ is determined by the sign of the numerator $2(1-x)(1+x)$, as the denominator $(x^2+1)^2$ is always positive.

• Interval $(-\infty, -1)$: Choose a test value, say $x = -2$. $f'(-2) = \frac{2(1-(-2))(1+(-2))}{((-2)^2+1)^2} = \frac{2(3)(-1)}{(5)^2} = \frac{-6}{25} < 0$. Thus, $f(x)$ is strictly decreasing on $(-\infty, -1)$.

• Interval $(-1, 1)$: Choose a test value, say $x = 0$. $f'(0) = \frac{2(1-0)(1+0)}{(0^2+1)^2} = \frac{2(1)(1)}{(1)^2} = 2 > 0$. Thus, $f(x)$ is strictly increasing on $(-1, 1)$.

• Interval $(1, \infty)$: Choose a test value, say $x = 2$. $f'(2) = \frac{2(1-2)(1+2)}{(2^2+1)^2} = \frac{2(-1)(3)}{(5)^2} = \frac{-6}{25} < 0$. Thus, $f(x)$ is strictly decreasing on $(1, \infty)$.

Step 5: Evaluate the given options based on the monotonicity analysis.

• Option (A): $f(x)$ is many-one in $(-\infty, -1)$ In $(-\infty, -1)$, $f'(x) < 0$, so $f(x)$ is strictly decreasing. A strictly decreasing function is one-one. Thus, this option is incorrect.

• Option (B): $f(x)$ is one-one in $(-\infty, \infty)$ The function changes monotonicity at $x=-1$ (decreasing to increasing) and at $x=1$ (increasing to decreasing). Since $f(x)$ is not strictly monotonic over the entire real line, it is not one-one on $(-\infty, \infty)$. For example, $f(-2) = \frac{(-2)^2+2(-2)+1}{(-2)^2+1} = \frac{4-4+1}{4+1} = \frac{1}{5}$, and $f(0) = \frac{0^2+2(0)+1}{0^2+1} = \frac{1}{1} = 1$. We can also see that $f(x)$ has a local minimum at $x=-1$ and a local maximum at $x=1$. This implies values are repeated. Thus, this option is incorrect.

• Option (C): $f(x)$ is one-one in $[1, \infty)$ but not in $(-\infty, \infty)$ In $(1, \infty)$, $f'(x) < 0$, so $f(x)$ is strictly decreasing. Since $f(x)$ is continuous, it is also strictly decreasing and thus one-one on the closed interval $[1, \infty)$. As established in Option (B), $f(x)$ is not one-one on $(-\infty, \infty)$. Therefore, this option is correct.

• Option (D): $f(x)$ is many-one in $(1, \infty)$ In $(1, \infty)$, $f'(x) < 0$, so $f(x)$ is strictly decreasing. A strictly decreasing function is one-one. Thus, this option is incorrect.

3. Common Mistakes & Tips

• Confusing one-one with onto: One-one refers to the mapping from domain to codomain (distinct inputs map to distinct outputs). Onto refers to whether every element in the codomain is mapped to by at least one element from the domain. This question is only about one-one.
• Assuming monotonicity over the entire domain: Always check the sign of the derivative in different intervals determined by critical points. A function can be one-one on sub-intervals but not on the entire domain.
• Ignoring endpoints of intervals: When checking for one-one property on closed intervals like $[a, b]$, if the function is strictly monotonic on $(a, b)$ and continuous at $a$ and $b$, it is also strictly monotonic and one-one on $[a, b]$.

4. Summary

To determine if the function $f(x) = \frac{x^2 + 2x + 1}{x^2 + 1}$ is one-one or many-one, we analyzed its monotonicity by calculating the first derivative $f'(x) = \frac{2(1-x)(1+x)}{(x^2+1)^2}$. We found critical points at $x=-1$ and $x=1$. Analyzing the sign of $f'(x)$ in the intervals $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$, we determined that $f(x)$ is strictly decreasing on $(-\infty, -1)$, strictly increasing on $(-1, 1)$, and strictly decreasing on $(1, \infty)$. Based on this, $f(x)$ is one-one on intervals where it is strictly monotonic. Specifically, it is strictly decreasing and hence one-one on $[1, \infty)$. It is not one-one on $(-\infty, \infty)$ because its monotonicity changes.

5. Final Answer The final answer is $\boxed{C}$ which corresponds to option (C).