Let `f(x)=alpha (x) beta (x) gamma (x)` for all real x, where `alpha (x), beta (X) and gamma (x)` are differentiable functions of x. If f'(2)=18 f(2), `alpha'(2)=3 alpha (2), beta' (2)=-4 beta (2) and gamma'(2)=k gamma (2)`, then the value of k is

Options

A. 14
B. 16
C. 19
D. None of these

Answer

To solve the problem, we need to find the value of \( k \) given the function \( f(x) = \alpha(x) \beta(x) \gamma(x) \) and the derivatives at \( x = 2 \).

### Step-by-Step Solution:

1. **Identify the given information:**
- \( f'(2) = 18 f(2) \)
- \( \alpha'(2) = 3 \alpha(2) \)
- \( \beta'(2) = -4 \beta(2) \)
- \( \gamma'(2) = k \gamma(2) \)

2. **Express \( f(2) \):**
\[ f(2) = \alpha(2) \beta(2) \gamma(2) \]

3. **Differentiate \( f(x) \):**
Using the product rule, we have:
\[ f'(x) = \alpha'(x) \beta(x) \gamma(x) + \alpha(x) \beta'(x) \gamma(x) + \alpha(x) \beta(x) \gamma'(x) \]

4. **Substituting \( x = 2 \):**
\[ f'(2) = \alpha'(2) \beta(2) \gamma(2) + \alpha(2) \beta'(2) \gamma(2) + \alpha(2) \beta(2) \gamma'(2) \]

5. **Plug in the values at \( x = 2 \):**
\[ f'(2) = (3 \alpha(2)) \beta(2) \gamma(2) + \alpha(2) (-4 \beta(2)) \gamma(2) + \alpha(2) \beta(2) (k \gamma(2)) \]
Simplifying this gives:
\[ f'(2) = 3 \alpha(2) \beta(2) \gamma(2) - 4 \alpha(2) \beta(2) \gamma(2) + k \alpha(2) \beta(2) \gamma(2) \]
\[ f'(2) = (3 - 4 + k) \alpha(2) \beta(2) \gamma(2) \]
\[ f'(2) = (k - 1) f(2) \]

6. **Set the equation from the given information:**
From the problem, we know that:
\[ f'(2) = 18 f(2) \]
Therefore, we equate:
\[ (k - 1) f(2) = 18 f(2) \]

7. **Divide both sides by \( f(2) \) (assuming \( f(2) \neq 0 \)):**
\[ k - 1 = 18 \]

8. **Solve for \( k \):**
\[ k = 19 \]

### Final Answer:
The value of \( k \) is \( 19 \).