向量a=(cosx+sinx,根2cosx),b=(cosx-sinx,根2sinx),f(x)=a×b 若2x2-派x

向量a=(cosx+sinx,根2cosx),b=(cosx-sinx,根2sinx),f(x)=a×b 若2x2-派x小于等于
第二问是若2x2-派x小于等于0,求函数f(x)的值域

向量a=(cosx+sinx,√（2cosx）),b=(cosx-sinx,√（2sinx）),f(x)=a×b 若2x2-πx≤0,【是不是 2x^2-πx≤0 】求函数f(x)的值域.
|向量a|=√[(cosx+sinx)^2+(2cosx)]=
=√[1+2sinxcosx+2cosx],
|向量b|=√[(cosx-sinx)^2+(2sinx)]=
=√[1-2sinxcosx+2sinx],
f(x)=a×b,
f(x)仍然是向量,方向是垂直于向量a和向量b所形成的平面、且服从右手法则.
con(向量a,向量b)=[|向量a|点乘|向量b|]/{[|向量a|][|向量b|]}=
=(cosx+sinx)(cosx-sinx)+(2cosx)(2sinx)/{√[1+2sinxcosx+2cosx]√[1-2sinxcosx+2sinx]}=
=[con^2x-sin^2x+4sinxconx]/√{[1+2sinxcosx+2cosx][1-2sinxcosx+2sinx]},
sin(向量a,向量b)=√{1-[(con^2x-sin^2x+4sinxconx)^2/[(1+2sinxcosx+2cosx)(1-2sinxcosx+2sinx)]};
|向量f(x)|=|向量a||向量b|sin(向量a,向量b)=
=√[1+2sinxcosx+2cosx]√[1-2sinxcosx+2sinx]sin(向量a,向量b)=
=√[1+2sinxcosx+2cosx-2sinxcosx-4sin^2xcos^2x-4sinxcon^2x+2sinx+4sin^2xconx+4sinxconx]sin(向量a,向量b)=
=√[1+2sinx+2cosx+4sinxcosx-4sinxcon^2x+4sin^2xconx-4sin^2xcos^2x])√{1-[(con^2x-sin^2x+4sinxconx)^2/[(1+2sinxcosx+2cosx)(1-2sinxcosx+2sinx)]},
若2x^2-πx≤0,
(√2x-π/√2)^2≤π^2/2,
-π/√2≤√2x-π/√2≤π/√2,
0≤√2x≤√2π,
0≤x≤π,
函数f(x)的值域:
|向量f(x)|=√[1+2sinx+2cosx+4sinxcosx-4sinxcon^2x+4sin^2xconx-4sin^2xcos^2x]√{1-[(con^2x-sin^2x+4sinxconx)^2/[(1+2sinxcosx+2cosx)(1-2sinxcosx+2sinx)]}=
=√{[1+2sinx+2cosx+4sinxcosx-4sinxcon^2x+4sin^2xconx-4sin^2xcos^2x]-(con^2x-sin^2x+4sinxconx)^2}=
=√{1+2sinx+2cosx+4sinxcosx-4sinxcon^2x+4sin^2xconx-4sin^2xcos^2x-(con^4x+sin^4x+16sin^2xcon^2x-2sin^2xcon^2x+8sinxcon^3x-8sin^3xconx)}=
=√{1+2sinx+2cosx+4sinxcosx-4sinxcon^2x+4sin^2xconx-4sin^2xcos^2x-con^4x-sin^4x-16sin^2xcon^2x+2sin^2xcon^2x-8sinxcon^3x+8sin^3xconx}=
=√{1+2sinx+2cosx+4sinxcosx-4sinxcon^2x+4sin^2xconx-4sin^2xcos^2x-con^4x-sin^4x-16sin^2xcon^2x+2sin^2xcon^2x-8sinxcon^3x+8sin^3xconx},
把上式整理、化简,
因为在 0≤x≤π,
0≤sinx≤1,-1≤conx≤1,
0≤sin^2x≤1,0≤sin^3x≤1,0≤sin^4x≤1,
0≤con^2x≤1,-1≤con^3x≤1,0≤con^4x≤1,
故 向量f(x)的标量值可以讨论、并且确定下来.