Differential EquationHard

Question

Let f (x) be a polynomial of degree four having extreme values at x = 1 and x = 2. If = 3, then f (2) is equal to :

Options

A.0
B.4
C.- 8
D.- 4

Solution

Clearly f(x) = ax4 + bx3 + cx2 + dx + e
Now,
∴  Clearly d = e = 0
Now, (1 + ax2 + bx + c) = 3
∴  c = 2
hence,
f(x) = ax4 + bx3 + 2x2
∴  f′(x) = 4ax3 + 3bx2 + 4x
= x(4ax2 + 3bx + 4)
Now, x = 1 and x = 2 are also solutions
∴    3 = and 2 = ⇒ a = and b = - 2
∴  f(x) = - 2x3 + 2x2
f(2) = 8 - 16 + 8 = 0

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