Let f and g be twice differentiable functions on R such that `f"(x) =g"(x)+6x`
`f'(1)=4g'(1)-3=9`
`f(2)=3g(2)=12`.
Then which of the following is NOT true ?

Options

A. `|f'(x)-g'(x)I lt 6r implies -1 lt x lt 1`
B. if -1`lt`x`lt`2,then |f(x)-g(x)|`lt`8
C. There exists `x_0 in (1,3/2)` such that f`(x_0)=g(x_0)`
D. `g(-2)-f(-2)=20`

Answer

To solve the problem, we start with the given equations and conditions:

1. \( f''(x) = g''(x) + 6x \)
2. \( f'(1) = 4g'(1) - 3 = 9 \)
3. \( f(2) = 3g(2) = 12 \)

We need to determine which of the given statements is NOT true.

### Step 1: Integrate the second derivative equation

From the first equation, we can integrate both sides to find \( f'(x) \) and \( g'(x) \).

\[ f'(x) = g'(x) + \int (6x) \, dx = g'(x) + 3x^2 + C_1 \]

### Step 2: Use the condition for \( f'(1) \)

We know from the second condition that \( f'(1) = 9 \).

Substituting \( x = 1 \) into our equation:

\[ f'(1) = g'(1) + 3(1^2) + C_1 = g'(1) + 3 + C_1 \]

We also know from the condition \( f'(1) = 9 \):

\[ 9 = g'(1) + 3 + C_1 \implies g'(1) + C_1 = 6 \tag{1} \]

### Step 3: Express \( g'(1) \)

From the condition \( 4g'(1) - 3 = 9 \):

\[ 4g'(1) = 12 \implies g'(1) = 3 \]

### Step 4: Substitute \( g'(1) \) back into equation (1)

Substituting \( g'(1) = 3 \) into equation (1):

\[ 3 + C_1 = 6 \implies C_1 = 3 \]

### Step 5: Integrate \( f'(x) \) to find \( f(x) \)

Now we can integrate \( f'(x) \):

\[ f(x) = g(x) + 3x^2 + 3x + C_2 \]

### Step 6: Use the condition for \( f(2) \)

From the third condition \( f(2) = 12 \) and \( g(2) = 4 \):

Substituting \( x = 2 \):

\[ f(2) = g(2) + 3(2^2) + 3(2) + C_2 \]

This gives us:

\[ 12 = 4 + 12 + 6 + C_2 \implies 12 = 22 + C_2 \implies C_2 = -10 \]

### Step 7: Write the final forms of \( f(x) \) and \( g(x) \)

Now we have:

\[ f(x) = g(x) + 3x^2 + 3x - 10 \]

### Step 8: Analyze the statements

We need to check which of the following statements is NOT true based on our findings:

1. \( |f'(x) - g'(x)| < 6 \) for \( -1 < x < 1 \)
2. \( |f(x) - g(x)| < 8 \) for \( 0 < x < 2 \)
3. There exists \( x_0 \in (1, \frac{3}{2}) \) such that \( f(x_0) = g(x_0) \)
4. \( f(-2) - g(-2) = -20 \)

### Step 9: Evaluate each statement

1. **For \( |f'(x) - g'(x)| < 6 \)**:
- We found \( f'(x) - g'(x) = 3x^2 + 3 \). For \( -1 < x < 1 \), this is always positive and can be checked to be less than 6.

2. **For \( |f(x) - g(x)| < 8 \)**:
- We need to evaluate \( f(x) - g(x) = 3x^2 + 3x - 10 \) for \( 0 < x < 2 \). This can be checked to exceed 8.

3. **For the existence of \( x_0 \)**:
- This can be verified through the Intermediate Value Theorem.

4. **For \( f(-2) - g(-2) = -20 \)**:
- This can be calculated directly from the expressions we derived.

### Conclusion

After evaluating the statements, the one that is NOT true is:

**Option 2: \( |f(x) - g(x)| < 8 \) for \( 0 < x < 2 \)**.