If `f.R rarr R,f(x)=(x^(2)+bx+1)/(x^(2)+2x+b),(b gt 1) " and " f(x), 1/f(x)` have the same bounded set as their range, the value of b is

Options

A. `2sqrt(3)-2`
B. `2sqrt(3)+2`
C. `2sqrt(2)-2`
D. `2sqrt(2)+2`

Answer

To solve the problem, we need to find the value of \( b \) such that the functions \( f(x) \) and \( \frac{1}{f(x)} \) have the same bounded set as their range.

Given the function:
\[ f(x) = \frac{x^2 + bx + 1}{x^2 + 2x + b} \]

### Step 1: Set up the equation for \( f(x) \)
Let \( y = f(x) \). Then we can rewrite the equation as:
\[ y = \frac{x^2 + bx + 1}{x^2 + 2x + b} \]
This can be rearranged to:
\[ y(x^2 + 2x + b) = x^2 + bx + 1 \]
which simplifies to:
\[ yx^2 + 2yx + by = x^2 + bx + 1 \]
Rearranging gives:
\[ (y - 1)x^2 + (2y - b)x + (by - 1) = 0 \]

### Step 2: Ensure the discriminant is non-negative
For this quadratic in \( x \) to have real solutions, the discriminant must be non-negative:
\[ D_1 = (2y - b)^2 - 4(y - 1)(by - 1) \geq 0 \]
Expanding this:
\[ D_1 = (2y - b)^2 - 4(by^2 - y + b - 1) \geq 0 \]

### Step 3: Set up the equation for \( \frac{1}{f(x)} \)
Now consider \( \frac{1}{f(x)} \):
\[ \frac{1}{f(x)} = \frac{x^2 + 2x + b}{x^2 + bx + 1} \]
Let \( z = \frac{1}{f(x)} \). Then:
\[ z(x^2 + bx + 1) = x^2 + 2x + b \]
Rearranging gives:
\[ (z - 1)x^2 + (bz - 2)x + (z - b) = 0 \]

### Step 4: Ensure the discriminant is non-negative for \( \frac{1}{f(x)} \)
The discriminant for this quadratic must also be non-negative:
\[ D_2 = (bz - 2)^2 - 4(z - b)(z - 1) \geq 0 \]

### Step 5: Compare the discriminants
Since \( f(x) \) and \( \frac{1}{f(x)} \) have the same range, the discriminants \( D_1 \) and \( D_2 \) must be identical. Thus, we can equate the coefficients of the corresponding terms.

### Step 6: Solve for \( b \)
By comparing the coefficients of \( D_1 \) and \( D_2 \), we get:
\[ b^2 - 4 = 4 - 4b \]
Rearranging gives:
\[ b^2 + 4b - 8 = 0 \]
Using the quadratic formula:
\[ b = \frac{-4 \pm \sqrt{(4)^2 - 4 \cdot 1 \cdot (-8)}}{2 \cdot 1} \]
\[ b = \frac{-4 \pm \sqrt{16 + 32}}{2} \]
\[ b = \frac{-4 \pm \sqrt{48}}{2} \]
\[ b = \frac{-4 \pm 4\sqrt{3}}{2} \]
\[ b = -2 \pm 2\sqrt{3} \]
Since \( b > 1 \), we take:
\[ b = -2 + 2\sqrt{3} \]

### Final Answer
Thus, the value of \( b \) is:
\[ \boxed{2\sqrt{3} - 2} \]