Suppose fis defined from `R -> [-1,1]` as `f(x)=(x^2-1)/(x^2+1)` where R is the set of real number .then the statement which does not hold is

Options

A. f is many-one onto
B. f increases for `xgt0` and decreases for `xlt0`
C. minimum value is not attained even though f is bounded
D. the area included by the curve `y-f(x)` and the line `y=1` is `pi` sq units

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x^2 - 1}{x^2 + 1} \) and determine which statement about this function does not hold true.

### Step 1: Check if the function is even
We start by checking if the function is even by evaluating \( f(-x) \).

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

**Hint:** A function is even if \( f(-x) = f(x) \) for all \( x \).

### Step 2: Determine if the function is one-to-one
Since \( f(x) = f(-x) \), this indicates that the function is not one-to-one (injective). For a function to be one-to-one, each output must correspond to exactly one input.

**Hint:** A function is one-to-one if \( f(a) = f(b) \) implies \( a = b \).

### Step 3: Analyze the range of the function
Next, we will analyze the range of \( f(x) \). The function can be rewritten as:

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

Since \( x^2 + 1 \) is always positive and increases as \( |x| \) increases, the term \( \frac{2}{x^2 + 1} \) decreases. Therefore, the function \( f(x) \) is bounded between -1 and 1.

**Hint:** To find the range, analyze the behavior of the function as \( x \) approaches positive and negative infinity.

### Step 4: Find the minimum value of the function
To find the minimum value of \( f(x) \), we can evaluate it at \( x = 0 \):

\[ f(0) = \frac{0^2 - 1}{0^2 + 1} = \frac{-1}{1} = -1 \]

As \( x \) approaches ±∞, \( f(x) \) approaches 1. Thus, the minimum value of \( f(x) \) is -1, which is attained at \( x = 0 \).

**Hint:** To find minimum or maximum values, consider critical points and endpoints of the function.

### Conclusion
Based on our analysis:
- The function is even.
- The function is not one-to-one.
- The function is bounded between -1 and 1.
- The minimum value of the function is -1, and it is attained at \( x = 0 \).

Thus, the statement that does not hold is that the minimum value does not attain, even though the function is bounded.

### Final Answer
The incorrect statement is: "The minimum value does not attain even though \( f \) is bounded."