a-maths!!!!

設f(x)=x²-4x+2及g(x)=-x²-kx+2k,其中k為實數。

(a)(i)若y=f(x)的圖像與y=g(x)的圖像相交於一點,試求k可取的範圍。

Ans: that mean f(x)-g(x)=0=>2x²+(k-4)x+(2+2k)=0 , and Δ=0

Δ=0
b²-4ac=0
(k-4)²-4(2)(2+2k)=0
k²-8k+16-16-16k=0
k²-24k=0
k(k-24)=0
k=24 and k=0//

(ii)對於每一個k值,求出交點

if k=24, x= -b/2a= -(k-4)/2(2)= -20/4 =-5//

if k=0 , x= -b/2a= -(k-4)/2(2)=4/4=1//

(b)若對於所有實數x,均有f(x)>g(x),試求k可取值範圍。

ans :

f(x)>g(x) => f(x)-g(x)>0 => Δ<0

Δ<0
b²-4ac<0
(k-4)²-4(2)(2+2k)<0
k²-8k+16-16-16k<0
k²-24k<0
k(k-24)<0
0<24//

2007-08-20 18:35:13 補充：
for the last sentance ,,it is "0<24"not "0<24"

(a)(i)y=x²-4x+2.....(1)
.......y=-x²-kx+2k...(2)
(1)-(2),2x²+(k-4)x+2-2k=0........(3)
since y=f(x)的圖像與y=g(x)的圖像相交於一點
so D=0,(k-4)²-8(2-2k)=0
.........................k=0 or k=-8
(ii)when k=0,consider(3),2x²-4x+2=0
.......................................x=1
when x=1,consider(1)or(2),y=-1
when k=-8,consider(3),2x²-12x+18=0
................................................x=3
when x=3,consider(1)or(2),y=-1
即k=0,(1,-1),..k=-8,(3,-1)
(b)因為f(x)>g(x),所以佢地相交唔到
consider(3)D>0,.......(k-4)²-8(2-2k)>0
........................k<0 and k>-8 ///or/// k>0 and k<-8(rejected)
so f(x)>g(x),k可取值範圍係0>k>-8
唔夠清楚可以再問

2007-08-25 12:01:37 補充：
f(x)-g(x)那裡他唔係計錯左咩?(2-2k)
我想問我個答案錯???~;~'