數學問題(會考題)

1.
f(x) = x^3 +2x^2 +ax +b

f(-1) = (-1)^3 + 2(-1)^2 + a(-1) + b = -1 + 2 - a + b = 0
So a - b = 1 ...... (1)

f(2) = (2)^3 + 2(2)^2 + a(2) + b = 8 + 8 + 2a + b = 0
So 2a + b = -16 ...... (2)

(1) + (2),
3a = -15
a = -5

Substitute a = -5 into (1),
(-5) - b = 1
b = -6

So f(x) = x^3 +2x^2 - 5x - 6.
A, B and E are not correct because the constants of them are +2, -2 and 0 respectively.
The constant in the expression of C = +6
The constant in the expression of D = -6

So D is the correct answer.

2.
The centre of the circle = (- (-8)/2, - (-6)/2) = (4, 3)

The slope of the line joining the centre of the circle and the mid-point of the chord
= (2 - 3)/(5 - 4) = -1

Because the line joining the centre of the circle and the mid-point of the chord is perpendicular to the chord, we have

the slope of the chord = (-1)/(-1) = 1

So the equation of the chord is
(y - 2)/(x - 5) = 1
y - 2 = x - 5
x - y - 3 = 0

So E is the correct answer.

3.
Cannot find the picture. Sorry, i can't help.

1.因x+1和x-2是f(x)的因式
則f(-1)=-1+2-a+b=0,a-b=1-------------(1)
f(2)=8+8+2a+b=0,2a+b=-16----------(2)
(1)+(2)得3a=-15,a=-5代入(1)
b=-6
即f(x)=x^3 +2x^2 -5x -6=(x+1)(x-2)(x+3)
答D

2.設此弦為A(x1,y1)B(x2,y2)
即其斜率k=(y2-y1)/(x2-x1)
因中點為(5,2)
即x1+x2=2*5=10,y1+y2=2*2=4
又因AB在圓上,所以
(x1)^2+(y1)^2-8(x1)-6(y1)+21=0--------------(1)
(x2)^2+(y2)^2-8(x2)-6(y2)+21=0--------------(2)
(2)-(1)得(x2)^2-(x1)^2+(y2)^2-(y1)^2-8(x2)+8(x1)-6(y2)+6(y1)=0
(X2-X1)(X2+X1)+(y2-y1)(y2+y1)-8(x2-x1)-6(y2-y1)=0
10(X2-X1)-8(x2-x1)+4(y2-y1)-6(y2-y1)=0<------以x1+x2=2*5=10,y1+y2=2*2=4代入後
2(X2-X1)-2(y2-y1)=0<----------設x2不等於x1,全式除以2(X2-X1)
1-(y2-y1)/(x2-x1)=0
所以k=(y2-y1)/(x2-x1)=1,且中點(5,2)在直線上,
由點斜式得y-2=1*(x-5)
即x-y-3=0
答E

3.因我不懂怎樣在這作圖,所以附助的點和線你要自己加上
設半徑為r,則圓心為C(r,r)
OP上的切點為A(r,0),OQ上的切點為B(0,r)
以下的OP,OQ,OA,OB..........為長度
OP=3,OQ=4,PQ=5
OA=OB=r,PA=PR,QB=QR(從圓外一點引圓的兩條切線長相等)
PQ=PR+RQ=QB+PA=(QO-BO)+(PO-AO)=QO+PO-2r
即5=4+3-2r
r=1,即圓心為C(1,1)
直線PQ的方程為y=-4x/3+4--------------(1)
因直線CR垂直於PQ,即斜率為3/4,且過C(1,1),由點斜式得
直線CR的方程為y-1=3/4(x-1)-----------(2)
(1)代入(2)得-4x/3+4-1=3/4(x-1)
即x=9/5
代入(1)得y=8/5
答C

第3題應該是c
睇左答案先解比你

2007-02-28 23:45:35 補充：
Let the centre of circle be (a,a)PQ=5(畢氏定理)(4-a) (3-a)=5a=1The centre of the circle=(1,1)Let the centre p.t. of the circle be SSR is vertical toPQLet the p.t R=(b,c)(b-1)/(c-1)=(-1)[(4-0)/(0-3)]b-1=[4/3(a-1)]-----(1)SR=radiusSR=11^2=(b-1)^2 (a-1)^2-----(2)Sub(1)into(2).....

2007-02-28 23:46:41 補充：
計完就找到b,代b入(2)就找到a所以答案是c