Let `f(x)a n dg(x)` be differentiable for `0lt=xlt=2` such that `f(0)=2,g(0)=1,a n df(2)=8.` Let there exist a real number `c` in `[0,2]` such that `f^(prime)(c)=3g^(prime)(c)dot` Then find the value of `g(2)dot`

Answer

To solve the problem step by step, we will utilize the Mean Value Theorem (MVT) and the given conditions.

### Step 1: Apply the Mean Value Theorem to \( f(x) \)

According to the Mean Value Theorem, if \( f(x) \) is continuous on the closed interval \([0, 2]\) and differentiable on the open interval \((0, 2)\), then there exists at least one number \( c \) in \((0, 2)\) such that:

\[ f'(c) = \frac{f(2) - f(0)}{2 - 0} \]

### Step 2: Substitute the given values for \( f(2) \) and \( f(0) \)

From the problem, we know:
- \( f(0) = 2 \)
- \( f(2) = 8 \)

Now substituting these values into the equation:

\[ f'(c) = \frac{8 - 2}{2} = \frac{6}{2} = 3 \]

### Step 3: Relate \( f'(c) \) and \( g'(c) \)

The problem states that there exists a real number \( c \) in \([0, 2]\) such that:

\[ f'(c) = 3g'(c) \]

From Step 2, we found that \( f'(c) = 3 \). Therefore, we can set up the equation:

\[ 3 = 3g'(c) \]

### Step 4: Solve for \( g'(c) \)

Dividing both sides by 3:

\[ g'(c) = 1 \]

### Step 5: Apply the Mean Value Theorem to \( g(x) \)

Now, we apply the Mean Value Theorem to \( g(x) \) over the interval \([0, 2]\):

\[ g'(c) = \frac{g(2) - g(0)}{2 - 0} \]

### Step 6: Substitute the known value for \( g(0) \)

From the problem, we know:

- \( g(0) = 1 \)

Substituting this into the equation gives:

\[ g'(c) = \frac{g(2) - 1}{2} \]

### Step 7: Substitute the value of \( g'(c) \)

We found that \( g'(c) = 1 \). Now substituting this value into the equation:

\[ 1 = \frac{g(2) - 1}{2} \]

### Step 8: Solve for \( g(2) \)

Multiplying both sides by 2:

\[ 2 = g(2) - 1 \]

Now, adding 1 to both sides:

\[ g(2) = 3 \]

### Final Answer

Thus, the value of \( g(2) \) is:

\[ \boxed{3} \]