Question

函数y=1+2sinxcosx+sinx+cosx的最大值

Answer 1

y=1+2sinxcosx+sinx+cosx
=sin²x+cos²x+2sinxcosx+sinx+cosx
=(sinx+cosx)²+sinx+cosx
=(sinx+cosx+1/2)²-1/4
=[√2sin(x+45°)+1/蔽档2]²-1/4
当sin(x+45°)=1时,
y取最大宏大乱,y最仿辩大值=2+√2

Answer 2

解：y=(sinx)^2+(cosx)^2+2sinxcosx+sinx+cosx
=(sinx+cosx)^2+sinx+cosx
令t=sinx+cosx
∵sinx+cosx=√2sin(x+π/4)∈[-√2,√2]
则y=t^2+t,t∈[-√悔谈2,√2]
∵对称轴为x=-1/2
∴当t=√扰信2时，有最大碧李碰值
最大值为2＋√2