Let f : R R be a polynomial function of degree four having extreme values at x = 4 and x = 5. If x 0 f(x)x2 = 5, then f(2) is equal to :
- 8
- 10
- 12
- 14
Answer
(2) 10
Solution
x 0 f(x)x2 = 5 is finite, f(x) has no constant or x term and the x2 coefficient is 5:
f(x) = 5x2 + ax3 + bx4
f'(x) = 10x + 3ax2 + 4bx3 = x(4bx2 + 3ax + 10)
Extreme values at x = 4, 5 4 and 5 are roots of 4bx2 + 3ax + 10 = 0
Product of roots: 104b = 20 b = 18
Sum of roots: - 3a4b = 9 3a = - 92 a = - 32
f(x) = 5x2 - 32x3 + 18x4
f(2) = 20 - 12 + 2 = 10
Answer: (2)