Question:medium

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

Show 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.
Updated On: Feb 18, 2026
  • 12
  • 120
  • 100
  • 20
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Concept Overview:
For a differentiable function, a necessary condition for a local maximum or minimum at an interior point is that its first derivative equals zero at that point.

Step 2: Methodology:
1. Determine the function's first derivative, \(f'(x)\).
2. Given that the maximum on \([0, 2]\) occurs at \(x=1\), then \(f'(1) = 0\).
3. Solve \(f'(1) = 0\) to find 'a'.

Step 3: Detailed Solution:
Given \(f(x) = x^4 - 62x^2 + ax + 9\), we find the derivative: \[ f'(x) = \frac{d}{dx}(x^4 - 62x^2 + ax + 9) \]\ \[ f'(x) = 4x^3 - 124x + a \]\ Since the maximum occurs at \(x=1\) within the interval \([0, 2]\), and \(x=1\) is an interior point, it's a critical point where the derivative is zero. Thus: \[ f'(1) = 0 \]\ Substituting \(x=1\) into the derivative: \[ 4(1)^3 - 124(1) + a = 0 \]\ \[ 4 - 124 + a = 0 \]\ \[ -120 + a = 0 \]\ \[ a = 120 \]\ Confirming it's a maximum using the second derivative test: \(f''(x) = 12x^2 - 124\). At \(x=1\), \(f''(1) = 12 - 124 = -112\), which is negative, indicating a local maximum.
Step 4: Solution:
The value of 'a' is 120.
Was this answer helpful?
0