1个回答
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1、
-1=<sina<=1,
1=<2+cosβ<=3
f(sina)>=0,即当-1=<x<=1时,f(x)>=0
f(2+cosβ)<=0,即当1=<x<=3时,f(x)<=0
所以f(1)=0
2、
f(1)=0,代入得1+b+c=0.b=-1-c
f(3)<=0,代入得9+3b+c<=0,即9+3*(-1-c)+c<=0,得c>=3
3、
f(sina)=sina*sina+b*sina+c=sina*sina+b*sina-1-b=(sina-1)(b+sina+1)
-1=<sina<=1
-2=<(sina-1)<=0
b=<(b+sina+1)<=b+2
由第二问知b<0
所以
0=<(sina-1)(b+sina+1)<=-2b
-2b=10,b=-5,c=4
f(x)=x^2-5x+4
-1=<sina<=1,
1=<2+cosβ<=3
f(sina)>=0,即当-1=<x<=1时,f(x)>=0
f(2+cosβ)<=0,即当1=<x<=3时,f(x)<=0
所以f(1)=0
2、
f(1)=0,代入得1+b+c=0.b=-1-c
f(3)<=0,代入得9+3b+c<=0,即9+3*(-1-c)+c<=0,得c>=3
3、
f(sina)=sina*sina+b*sina+c=sina*sina+b*sina-1-b=(sina-1)(b+sina+1)
-1=<sina<=1
-2=<(sina-1)<=0
b=<(b+sina+1)<=b+2
由第二问知b<0
所以
0=<(sina-1)(b+sina+1)<=-2b
-2b=10,b=-5,c=4
f(x)=x^2-5x+4
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