If `f (x) and g (x)` are twice differentiable functions on `(0,3) ` satisfying, `f ''(x) = g'' (x), f '(1) =4, g '(1) =6,f (2) = 3, g (2)=9,` then `f (1) -g (1)` is

Options

A. 4
B. `-4`
C. 0
D. `-2`

Answer

To solve the problem step by step, we need to analyze the given information about the functions \( f(x) \) and \( g(x) \).

### Step 1: Understanding the given information
We know that:
- \( f''(x) = g''(x) \)
- \( f'(1) = 4 \)
- \( g'(1) = 6 \)
- \( f(2) = 3 \)
- \( g(2) = 9 \)

### Step 2: Integrate the second derivatives
Since \( f''(x) = g''(x) \), we can integrate both sides with respect to \( x \):
\[ f'(x) = g'(x) + C_1 \]
where \( C_1 \) is a constant.

### Step 3: Use the initial condition for first derivatives
Now, we can use the given values of the first derivatives at \( x = 1 \):
\[ f'(1) = g'(1) + C_1 \]
Substituting the known values:
\[ 4 = 6 + C_1 \]
Solving for \( C_1 \):
\[ C_1 = 4 - 6 = -2 \]

### Step 4: Substitute \( C_1 \) back into the equation
Now we have:
\[ f'(x) = g'(x) - 2 \]

### Step 5: Integrate the first derivatives
Next, we integrate again to find \( f(x) \):
\[ f(x) = g(x) - 2x + C_2 \]
where \( C_2 \) is another constant.

### Step 6: Use the initial condition for function values
Now we will use the values of the functions at \( x = 2 \):
\[ f(2) = g(2) - 2(2) + C_2 \]
Substituting the known values:
\[ 3 = 9 - 4 + C_2 \]
This simplifies to:
\[ 3 = 5 + C_2 \]
Solving for \( C_2 \):
\[ C_2 = 3 - 5 = -2 \]

### Step 7: Substitute \( C_2 \) back into the equation
Now we have:
\[ f(x) = g(x) - 2x - 2 \]

### Step 8: Find \( f(1) - g(1) \)
To find \( f(1) - g(1) \), we substitute \( x = 1 \):
\[ f(1) = g(1) - 2(1) - 2 \]
Thus:
\[ f(1) - g(1) = -2 - 2 = -4 \]

### Final Answer
Therefore, the value of \( f(1) - g(1) \) is:
\[ \boxed{-4} \]