中四符加數(聯立二元二次方程:一為線性方程和一為二次方程)

1a) f(x) = (a - x)(x - b)
= -x2 + (a + b)x - ab
f'(x) = -2x + a + b
For turning points, f'(x) = 0
x = (a + b)/2
Since f''(x) = -2 < 0
f' [(a + b)/2] is a maximum.
Maximum f(x)
= f [(a + b) / 2]
= -(a + b)2 /4 + (a + b)2 /2 - ab
= [(a + b)2 - 4ab]/4
= (a - b)2 /4
bi) (a - x)(x - b) = k
f(x) = k ... (*)
Given (*) has a double root.
By (a), the f(x) has a maximum value (a - b)2 /4.
Therefore, the graph y = f(x) and y = (a - b)2 /4 intersect at one point only.
Hence k = (a - b)2 /4
ii) Given f(x) = k has two roots with b and a.
Therefore, the graph of y = f(x) and y = k should intersect at two points.
As the graph opens downwards, k < (a - b)2 /4
Since a and b are the roots of f(x) = 0,x
k > 0 (the grpah opens downwards)
Therefore, 0 < k < (a - b)2 /4
2a) For the points of intersection, the x-coorinates satisfy both of the equations.
Substitute the eqaution of the curve into that of the straight line,
x2 + 3x + 2k = 0 ... (#)
bi) If the straight line touches the curve, discriminent of (#) = 0
32 - 4(2k) = 0
k = 9/8
ii) Substitute k = 9/8 into (#),
4x2 + 12x + 9 = 0
(2x + 3) = 0
x = -3/2
So (-3/2) + y + 9/8 = 0
y = 3/8
The coordiates of the point of intersection are (-3/2 , 3/8)