Question

It is given that at x = 1, the function \(f(x) = x^4 - 62x^2 + ax + 9\), attains its maximum value in the interval \([0, 2]\). Then, the value of 'a' is

• A. 12
• B. 120
• C. 100
• D. 20

Hint:

When a problem states that an extremum (max or min) of a differentiable function occurs at an interior point of an interval, immediately set the first derivative to zero at that point. This is often the key to solving for unknown parameters in the function.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a differentiable function to have a local maximum or minimum at an interior point of an interval, its first derivative at that point must be zero. This is a necessary condition for an extremum at an interior point.

Step 2: Key Formula or Approach:
1. Find the first derivative of the function, \(f'(x)\).
2. Since the maximum value on the interval \([0, 2]\) occurs at \(x=1\) (which is an interior point), we must have \(f'(1) = 0\).
3. Solve the equation \(f'(1) = 0\) for the unknown parameter 'a'.

Step 3: Detailed Explanation:
The given function is \(f(x) = x^4 - 62x^2 + ax + 9\).
First, we compute the derivative of \(f(x)\) with respect to \(x\): \[ f'(x) = \frac{d}{dx}(x^4 - 62x^2 + ax + 9) \] \[ f'(x) = 4x^3 - 124x + a \] We are given that the function attains its maximum value in the interval \([0, 2]\) at the point \(x=1\). Since \(x=1\) is an interior point of this interval (i.e., not an endpoint), it must be a critical point of the function. For a differentiable function like this polynomial, this means the first derivative must be zero at that point. \[ f'(1) = 0 \] Now we substitute \(x=1\) into the derivative and set it to zero: \[ 4(1)^3 - 124(1) + a = 0 \] \[ 4 - 124 + a = 0 \] \[ -120 + a = 0 \] \[ a = 120 \] To be certain it's a maximum, we can use the second derivative test. \(f''(x) = 12x^2 - 124\). At \(x=1\), \(f''(1) = 12 - 124 = -112\), which is negative, confirming a local maximum.
Step 4: Final Answer:
The value of 'a' is 120.