Let x=2 be a local minima of the function f(x)=2x⁴-18x²+8x+12,x∈(-4,4). If M is local maximum value of the function f in (-4,4), then M =

Solution

Maths sir · 5 steps

Factor (x-2) out of f', then use the quadratic left over to make f linear.

1. Use the given root. f'(x) = 8x³ - 36x + 8 = 4(2x³ - 9x + 2). Since x = 2 is a critical point, (x-2) is a factor, and f'(x) = 4(x-2)(2x² + 4x - 1).
2. Other critical points. 2x² + 4x - 1 = 0 gives x = (-2 ± √6)/2, about -2.22 and 0.22. Both lie in (-4, 4).
3. Find the maximum. f' is a cubic with a positive leading coefficient, so its sign runs -, +, -, + across the roots -2.22, 0.22, 2. The sign goes from + to - only at α = (√6 - 2)/2, so that is the local maximum.
4. Reduce the degree. From 2α² + 4α - 1 = 0 you get α² = (1 - 4α)/2, so 2α⁴ = (1-4α)²/2 = 1/2 - 4α + 8α². Putting this in gives f(α) = 25/2 + 4α - 10α², and using α² once more gives the line below.

   M = f(α) = 25/2 + 4α - (5 - 20α) = 15/2 + 24α

5. So. 24α = 12√6 - 24, so M = 15/2 - 24 + 12√6 = 12√6 - 33/2. This is about 12.89, and the answer is option (c).

The idea to keep. When a critical point is an ugly root of a quadratic, do not substitute it directly. Use the quadratic to replace α² again and again until f(α) is linear in α.

Answer from the NTA final answer key. Solution by PrepWiser.