`f(x)=sqrt(x-1)+sqrt(2-x)` and `g(x)=x^(2)+bx+c` are two given functions such that `f(x)` and `g(x`) attain their maximum and minimum values respectively for same value of `x`, then

Options

A. a) `f(x)` extreme point at `x=(1)/(2)`
B. b) `f(x)` extreme point at `x=(3)/(2)`
C. c) `b=3`
D. d) `b=-3`

Answer

To solve the problem, we need to find the extreme points of the functions \( f(x) \) and \( g(x) \) and establish the relationship between them.

### Step 1: Define the functions
We have:
\[ f(x) = \sqrt{x - 1} + \sqrt{2 - x} \]
\[ g(x) = x^2 + bx + c \]

### Step 2: Find the derivative of \( f(x) \)
To find the extreme points of \( f(x) \), we first need to compute its derivative \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(\sqrt{x - 1}) + \frac{d}{dx}(\sqrt{2 - x}) \]
Using the chain rule, we get:
\[ f'(x) = \frac{1}{2\sqrt{x - 1}} - \frac{1}{2\sqrt{2 - x}} \]

### Step 3: Set the derivative equal to zero
To find the critical points, we set \( f'(x) = 0 \):
\[ \frac{1}{2\sqrt{x - 1}} - \frac{1}{2\sqrt{2 - x}} = 0 \]
This simplifies to:
\[ \frac{1}{\sqrt{x - 1}} = \frac{1}{\sqrt{2 - x}} \]

### Step 4: Square both sides
Squaring both sides gives:
\[ x - 1 = 2 - x \]

### Step 5: Solve for \( x \)
Rearranging the equation:
\[ x + x = 2 + 1 \implies 2x = 3 \implies x = \frac{3}{2} \]
Thus, \( f(x) \) attains its extreme point at \( x = \frac{3}{2} \).

### Step 6: Find the derivative of \( g(x) \)
Next, we find the derivative of \( g(x) \):
\[ g'(x) = 2x + b \]

### Step 7: Set the derivative of \( g(x) \) equal to zero
Since \( f(x) \) and \( g(x) \) attain their extreme values at the same point, we set \( g'(\frac{3}{2}) = 0 \):
\[ g'\left(\frac{3}{2}\right) = 2\left(\frac{3}{2}\right) + b = 0 \]
This simplifies to:
\[ 3 + b = 0 \]

### Step 8: Solve for \( b \)
Rearranging gives:
\[ b = -3 \]

### Conclusion
The extreme point of \( f(x) \) is at \( x = \frac{3}{2} \) and the value of \( b \) is \( -3 \). Therefore, the correct options are:
- Option B: \( f(x) \) has an extreme point at \( x = \frac{3}{2} \)
- Option D: \( b = -3 \)