數學問題! 星期一考A-Level! 急幫忙!

(A) (1 - 2x^2) / x^2 = 1/x^2 - 2 = x^(-2) - 2
∫(1 - 2x^2) / x^2 dx
= ∫ x^(-2) - 2 dx
= -1/x - 2x + C where C is a constant
(B)(a) y = x + 2 ... (1)
4y + 3x^2 = 7 ... (2)
Sub (1) into (2), 4(x + 2) + 3x^2 = 7
3x^2 + 4x + 1 = 0
(3x + 1)(x + 1) = 0
3x + 1 = 0 or x + 1 = 0
x = -1/3 or x = -1
When x = -1/3, y = -1/3 + 2 = 5/3
When x = -1, y = -1 + 2 = 1
Therefore the points of intersection are (-1/3, 5/3) and (-1, 1)
(b) Let the line be y = f(x) = 3x + c
Sub into 4y + 3x^2 = 7 gives 4(3x + c) + 3x^2 = 7
3x^2 + 12x + (4c - 7) = 0 ... (3)
Since there is only one point of intersection, the discriminant of the quadratic equation = 0
12^2 - 4(3)(4c - 7) = 0
144 = 12(4c - 7)
4c - 7 = 12
c = 19/4
Therefore f(x) = 3x + 19/4
(c) The quadratic equation in (3) becomes 3x^2 + 12x + 12 = 0
x^2 + 4x + 4 = 0
(x + 2)^2 = 0
x = -2
y = 3*(-2) + 19/4 = -5/4
The point of intersection is (-2, -5/4)