第3个回答 2012-07-07
y=sinx+√(2+(cosx)^2)
=sinx+√(3-(sinx)^2)
=t+√(3-t^2)……t=sinx∈[-1,1]
显然y>0
3-t^2=(y-t)^2=t^2-2yt+y^2
f(t)=2t^2-2yt+y^2-3=0
在[-1,1]中至少有一个实数根
y1=f(t)顶点为(y/2,y^2/2-3)
△=4y^2-8(y^2-3)=-4(y^2-6)>=0,-√6<=y<=√6^
1。若y/2<=1即y<=2
则f(1)=2-2y+y^2-3=y^2-2y-1>=0或f(-1)=2+2y+y^2-3=y^2+2y-1>=0
y<=1-√2(舍去)或y>=1+√2或y<=-1-√2(舍去)或y>=-1+√2
∴-1+√2<=y<=2
2。若y/2>=1即y>=2
则f(1)=2-2y+y^2-3=y^2-2y-1<=0且f(-1)=2+2y+y^2-3=y^2+2y-1>=0
1-√2<=y<=1+√2且[y<=-1-√2(舍去)或y>=-1+√2]
∴-1+√2<=y<=1+√2
∴2<=y<=1+√2
故 -1+√2<=y<=1+√2(注意1+√2<√6)
y[min]+y[max]=-1+√2+1+√2=2√2