2000會考MC,求解答

2009-12-23 6:24 am
2000會考MC,第28題

要有解答的

回答 (2)

2009-12-24 9:15 am
✔ 最佳答案
49
y = mx+1 代入(x-2)^2 +y =1中
(x-2)^2 + (mx+1) =1
x^2 -4x+4 +m^2x^2 +2mx+1 -1 =0
(1+m^2)x^2 +(2m-4)x+4 =0
因為 L 是 C 的切線,所以只有一個實根,
△ = (2m-4)^2 - 4(1+m^2)(4) = 0
4m^2 -16m +16-16-16m^2 =0
4m(3m+4)= 0, m=0 或m =-4/3
51
cosA = 1/k, 0<A<90
∴sinA= √1-cos^2A = √(1-(1/k^2)0 = √(k^2 -1)/k
tan(A-270) = tan[360 + (A-270)] = tan(90+A)= -1/tanA= -cosA/sinA
= - [1/k]/[√(k^2 -1)/k] = -1/√(k^2 -1

53
y=1-2cos x
當y =0 在 0 =< x =< (pi)
1- 2cos x =0, cos 2x = 1/2
2x = (pi)/3, 5(pi)/3
∴ x = (pi)/6, 5(pi)/6

(pi) =180

2009-12-24 10:44:07 補充:
(pi) =180°
2010-01-07 7:40 pm


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