十條DSE程度MC,求題解

jy99920032000

2012-12-18 7:01 am

1. 如果速度減少15%,而距離不變,行走時間會
A. 增加了15%
B. 增加了17.647%
C. 減少了15%
D. 減少了17.647%

2. 陳先生可以在m日內獨立完成某些工作,而小康則需要n日獨立完成該件
工作。若他們合力工作,問完成該工作所需時間是多少?
A. 2/m+n 日
B. m+n/2 日
C. m+n/mn 日
D. mn/m+n 日

3. 半徑為r的圓與x軸和y軸相切,並經過點P(2,1)及Q(8,9),求圓的方程。
A. x^2+y^2+10x-10y-25=0
B. x^2+y^2-10x-10y+25=0
C. x^2+y^2-10x-10y=0
D. x^2+y^2-2x+2y+1=0

4. 若C為x^2+y^2+2x-8y-8=0的圓心。若圓形與x-軸相交於A及B,求三角
形CAB的面積。
A. 3
B. 12
C. 16
D. 24

5. 若直線y=mx與圓x^2+y^2+6x-2y+2=0互不相交,求m值的範圍。
A. -1<m<7
B. -7<m<1
C. m>7 or m<-1
D. m>1 or m<-7

6. 若y=f(x)的圖像變換成y=3f(2x)-2的圖像,求點(2,4)變換後的坐標。
A. (1,10)
B. (2,10)
C. (2,12)
D. (3,2)

7. 若a>b>c和a,b,c是實數,下列那一項是對的?
A. a^2>b^2>c^2
B. a/b>c/b
C. ab>ac
D. a-c>b-c

8. 一袋中有3個紅球,4個藍球,某男孩從袋中隨機地每次抽一個球,取出
後並不放回袋中,直至抽得紅球為止,求他最少要抽3次的概率。
A. 1/7
B. 2/7
C. 5/7
D. 6/7

9. 某個袋內有3個黑球,2個綠球和1個黃球,偉仔從袋中隨機重複取球。
每次只取出一個且不放回袋中,直至取得黑球為止。求他需要取球
最多三次的概率。
A. 1/20
B. 3/20
C. 19/20
D. 70/81

10. 一行5人的團隊由8男10女中挑選出,其中包括小添仔和小軒女,求他們
兩人(小添仔和小軒女)最少一人入選的概率。
A. 1/3
B. 2/3
C. 25/51
D. 26/51

更新1:

可唔可以答埋最尾哥四條?

回答 (1)

用戶10012

2012-12-19 4:50 am

✔ 最佳答案

(1) Let original speed = 100 km/h. Let distance = 100 km
so time taken = 1 hr.
New speed = 100(1 - 15%) = 85 km/h.
so new time = 100/85 = 1.17647 hr.
so % change in time = (1.17647 - 1)/1 x 100% = 17.647% increase. (B).
(2)
In 1 day, Mr Chan can finish 1/m parts of the job.
In 1 day, Hong can finish 1/n parts of the job.
So in 1 day, two together can finish (1/m + 1/n) parts of the job.
1/m + 1/n = (m + n)/mn.
So it takes mn/(m + n) days to complete the job. (D).
(3)
Since the circle touches both the x and y axis, so the x and y co - ordinates of the circle is the same and equal to the radius. Let radius be r, so center is (r,r).
Radius = sqrt [ ( r - 2)^2 + (r - 1)^2] = sqrt [ (r - 8)^2 + (r - 9)^2]
That means
(r - 2)^2 + (r - 1)^2 = (r - 8)^2 + (r - 9)^2
Solving the equation, r = 5. So center is (5,5)
So equation of circle is (x - 5)^2 + (y - 5)^2 = 5^2
x^2 + y^2 - 10x - 10y + 25 = 0 (B).
Let me take a rest and answer to others later.

2012-12-18 21:38:03 補充：
(4) Put y = 0 into the equation of the circle, x^2 + 2x - 8 = 0, (x - 2)(x + 4) = 0, x = 2 or - 4.
So the co-ordinates of A and B are (2,0) and ( - 4,0). From the equation of the circle, we know center of circle is (- 1, 4), so area of triangle ABC = [2 - (- 4)](4)/2 = 12 (B).

2012-12-18 21:43:58 補充：
(5) Sub. y = mx into equation of circle, x^2 + (mx)^2 + 6x - 2(mx) + 2 = 0
(1 + m^2)x^2 + (6 - 2m)x + 2 = 0. Since the line and the circle do not intersect, so delta < 0. That is (6 - 2m)^2 - 4(1 + m^2)(2) < 0. (3 - m)^2 - 2(1 + m^2) < 0. 9 + m^2 - 6m - 2 - 2m^2 < 0. - m^2 - 6m + 7 < 0.

2012-12-18 21:45:38 補充：
(m + 7)(m - 1) > 0. So m < - 7 or m > 1. (D)

2012-12-18 21:51:21 補充：
(6) Let y = f(x) be represented by a quadratic function with vertex ( 2, 4). So f(x) = (x - 2)^2 + 4. f(2x) = (2x - 2)^2 + 4 = 4(x - 1)^2 + 4. 3f(2x) = 3[4(x - 1)^2 + 4] = 12(x - 1)^2 + 12.
3f(2x) - 2 = 12(x - 1)^2 + 12 - 2 = 12(x - 1)^2 + 10. So the new vertex is (1, 10) (A)

2012-12-19 08:10:23 補充：
(7) If a, b and c are negative, only (D) is correct. For example a = - 2, b = - 3 and c = - 4.

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