The function `f(x) = x^4 -62 x^2 + ax + 9` attains its maximum value on the interval [0,2] at x = 1. Then the value of a is

Options

A. 120
B. `-120`
C. 52
D. none of these

Answer

To solve the problem, we need to find the value of \( a \) such that the function \( f(x) = x^4 - 62x^2 + ax + 9 \) attains its maximum value at \( x = 1 \) in the interval [0, 2].

### Step-by-Step Solution:

1. **Find the derivative of the function**:
To find the maximum or minimum values, we first need to find the critical points by taking the derivative of the function and setting it to zero.
\[ f'(x) = \frac{d}{dx}(x^4 - 62x^2 + ax + 9) = 4x^3 - 124x + a \]

2. **Set the derivative to zero at \( x = 1 \)**:
Since the maximum occurs at \( x = 1 \), we set the derivative equal to zero at this point:
\[ f'(1) = 4(1)^3 - 124(1) + a = 0 \]
Simplifying this, we get:
\[ 4 - 124 + a = 0 \]

3. **Solve for \( a \)**:
Rearranging the equation gives:
\[ a - 120 = 0 \implies a = 120 \]

4. **Conclusion**:
Therefore, the value of \( a \) is \( 120 \).

### Final Answer:
\[ \boxed{120} \]