3sin^2a+2cos^2b=2sina 则sin^2a+cos^b的取值范围是我做了[0,1/2] 答案是[0,4/9]
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3sin^2a+2cos^2b=2sina 则sin^2a+cos^b的取值范围是
我做了[0,1/2] 答案是[0,4/9]
因为sina有范围.由题设3sin^2a
因为sin^2b>=0,sin^2a>=0;
所以:2sin^2b+3sin^2a>=0.
即:2sina>=0,得到:sina>=0.
∵2sin^2b+3sin^2a=2sina,
∴sin^2b=sina-(3/2)sin^2a>=0
进一步:
Sina(1-3/2sina)>=0
1-3/2sina>=0,得到:...
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因为sin^2b>=0,sin^2a>=0;
所以:2sin^2b+3sin^2a>=0.
即:2sina>=0,得到:sina>=0.
∵2sin^2b+3sin^2a=2sina,
∴sin^2b=sina-(3/2)sin^2a>=0
进一步:
Sina(1-3/2sina)>=0
1-3/2sina>=0,得到:sina<=2/3.
即sina的取值范围为:[0,2/3].
则:
m=sin^2a+sin^2b
=sin^2a+sina-(3/2)sin^2a
=-(1/2)sin^2a+sina
=-(1/2)(sin^2a-2sina+1)+1/2
=-(1/2)(sina-1)^2+1/2.
因为0<=sina<=2/3.所以:
当sina=2/3,m有最大值m=4/9
当sina=0,m有最小值m=0。
所以m的取值范围为:[0,4/9].
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