2010會考附加數都4,5,6題(5月4日)

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4)
∆=k²−4k(k−1)
=−3k²+4k
=−k(3k−4)
Set ∆≤0 and (k−1)≥0,
−k(3k−4)≤0 and k≥1
k(3k−4)≥0 and k≥1
[k≥4/3 or k≤0] and k≥1
∴k≥4/3
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6)
x−2y+3+k(2x−y−1)=0
(2k+1)x−(k+2)y−(k−3)=0
Slope=(2k+1)/(k+2)
For equal positive intercept, slope=tan135°=−1.
(2k+1)/(k+2)=−1
k=−1
Requied equation:
x−2y+3−(2x−y−1)=0
x+y−4=0
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2010-05-04 22:30:18 補充：
u right ,悪魔
k≠1(i.e,k> 0)
sorry

2010-05-04 22:35:51 補充：
↑
k>1

圖片參考：http://j.imagehost.org/0254/ScreenHunter_05_May_04_23_33.gif


http://j.imagehost.org/0254/ScreenHunter_05_May_04_23_33.gif

2010-05-05 12:06:16 補充：
更正第5題:

http://j.imagehost.org/0700/ScreenHunter_03_May_05_12_05.gif

4. 以下係我做既方法 (可信程度: <1%)
D<=0 and k-1>0 (D=Discriminant判別式)
k^2-4(k)(k-1)<=0 and k>1
-3k^2+4k<=0 and k>1
3k^2-4k>=0 and k>1
{ k<=0 or k>=4/3} and k>1
Therefore, k>1

6. 以下係我做既方法 (可信程度: <5%)
Let左L做x-2y+3+k(2x-y-1)=0之後
(1+2k)x-(2+k)+3-k=0
「通過P且有相等正截距的直線」
意思應該係「x-/y-截距一樣,兼且兩個都係正」
(1+2k)/(k-3) = (2+k)/(3-k) [左為x截距,右為y截距]
應該會計到k=3 or k=-2/3, reject k=3 [因為如果k=3, 咁條方程式既截距就變左0(非正)]

2010-05-04 21:46:07 補充：
第6題樓下岩
k應該係-1

第4題:
k應該唔可以等於1嫁bor@@
如果k=1,咁條equation咪變左直線嫁lor@@

(1+4x)^n
= 1 + n(4x) + n(n-1)/2 (4x)^2 + n(n-1)(n-2)/6 (4x)^3 + ...
= 1 + 4nx + 8n(n-1)x^2 + 32n(n-1)(n-2)x^3 / 3
4n + 8n(n-1) = 180
4n + 8n^2 - 8n = 180
2n^2 - n - 45 = 0
(n-5)(2n+9) = 0
n = 5 or n = -9/2 (rej.)
Coefficient of x^3 = 32(5)(5-1)(5-2)/3 = 640