Question

Let f(x) be a polynomial of degree 3 if the curve y = f(x) has relative extremities at x = ± 2/√3 and passes through (0, 0) and (1, -2) dividing the circle x² + y² = 4 in two parts. Then the integral part of areas of these two parts is ______________.

Answer 1

Since y = f(x) has relative extremities at x = ± 2/√3 these points are critical points and hence they must be roots of f ′(x) = 0 (Clearly f is differentiable everywhere).

Therefore,

f(x) = a(x³/3 - 4x/3) + b

This passes through (0, 0) and (1, -2). So, b = 0 and

a(1/3 - 4/3) = - 2 ⇒ a = 2

Therefore, f(x) = 2x/3(x² - 4).

Hence, f(x) meets the x-axis at (0, 0), (-2, 0) and (2, 0). Since f(-x) = -f(x), the curve y = f(x) is symmetrical about the origin.

Required area = 2π. So, integral part = 6