A.maths (10 points~urgent)

用戶90065

2007-10-22 12:03 am

(HKCEE 1988)
Let f(x)=x^2+2x-1 and g(x)=-x^2+2kx-k^2+6(where k is a constant).
(b) If the graphs of y=f(x) and y=g(x) intersect at only one point, find the possible values of k; and for each value of k, find the point of intersection.
(c)Find the range of values of k such that f(x)>g(x) for any real value of x.

回答 (1)

用戶195274

2007-10-22 12:20 am

✔ 最佳答案

(a) f(x) = g(x)
x^2+2x-1= -x^2+2kx-k^2+6
2x^2 + (2-2k)x + (k^2 - 7) = 0 ...(#)
Discriminant = (2-2k)^2 -4(2)(k^2 - 7) = 0
4-8k+4k^2 -8k^2 + 56 = 0
4k^2 + 8k - 60 = 0
k^2 + 2k - 15 = 0
(k+5)(k-3) = 0
k = 3 or -5
When k = 3, (#) becomes 2x^2 - 4x + 2 = 0
2(x^2 - 2x+1) = 0
2(x-1)^2 = 0
x = 1, y = f(1) = 1+2-1=2
The point of intersection when k = 3 is (1,2)
When k = -5, (#) becomes 2x^2 +12x + 18 = 0
2(x+3)^2 = 0
x = -3, y = f(-3) = 9-6-1 = 2
The point of intersection when k = -5 is (-3,2)

(c)
f(x)>g(x)
x^2+2x-1> -x^2+2kx-k^2+6
2x^2 + (2-2k)x + (k^2 - 7) > 0
Therefore, discriminant <0
4-8k+4k^2 -8k^2 + 56 < 0
4k^2 + 8k - 60 > 0
k^2 + 2k - 15 > 0
(k+5)(k-3) > 0
k<-5 or k>3

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