Let `f(x)a n dg(x)` be differentiable for `0lt=xlt=1,` such that `f(0)=0,g(0)=0,f(1)=6.` Let there exists real number `c` in (0,1) such that `f^(prime)(c)=2g^(prime)(c)dot` Then the value of `g(1)` must be (a) 1 (b) 3 (c) `-2` (d) `-1`

Options

A. 1
B. 3
C. `-2`
D. `1-`

Answer

To solve the problem step by step, we will analyze the information given and apply the necessary mathematical concepts.

### Step 1: Understand the Given Information
We know that:
- \( f(x) \) and \( g(x) \) are differentiable for \( 0 < x < 1 \).
- \( f(0) = 0 \), \( g(0) = 0 \), and \( f(1) = 6 \).
- There exists a real number \( c \) in \( (0, 1) \) such that \( f'(c) = 2g'(c) \).

### Step 2: Apply Rolle's Theorem
Since \( f(x) \) and \( g(x) \) are continuous on the closed interval \([0, 1]\) and differentiable on the open interval \( (0, 1) \), we can apply Rolle's Theorem. According to this theorem, there exists at least one point \( c \) in \( (0, 1) \) such that:
\[ f'(c) = 2g'(c) \]

### Step 3: Rearranging the Equation
From the condition \( f'(c) = 2g'(c) \), we can rearrange this equation:
\[ f'(c) - 2g'(c) = 0 \]
This implies that the function \( F(x) = f'(x) - 2g'(x) \) has at least one root in the interval \( (0, 1) \).

### Step 4: Integrate the Function
Now, we integrate \( F(x) \):
\[ F(x) = f'(x) - 2g'(x) \]
Integrating both sides from \( 0 \) to \( 1 \):
\[ \int_0^1 F(x) \, dx = \int_0^1 f'(x) \, dx - 2 \int_0^1 g'(x) \, dx \]
Using the Fundamental Theorem of Calculus:
\[ F(1) - F(0) = f(1) - f(0) - 2(g(1) - g(0)) \]
Substituting the known values:
\[ f(1) = 6, \quad f(0) = 0, \quad g(0) = 0 \]
This simplifies to:
\[ 6 - 0 - 2(g(1) - 0) = 0 \]
Thus:
\[ 6 - 2g(1) = 0 \]

### Step 5: Solve for \( g(1) \)
From the equation \( 6 - 2g(1) = 0 \):
\[ 2g(1) = 6 \]
Dividing both sides by 2:
\[ g(1) = 3 \]

### Conclusion
The value of \( g(1) \) must be \( 3 \).

### Final Answer
The answer is \( (b) \, 3 \).