高中数学题已知向量a,b满足|a
(1)设a=(x1,y1),b=(x2,y2)
ka+b=(kx1+x2,ky1+y2)
a-kb=(x1-kx2,y1-ky2)
F(k)=a*b=(x1x2,y1y2)
|a|=√(x1^2+y1^2)=1=>x1^2+y1^2=1
|b|=√(x2^2+y2^2)=1=>x2^2+y2^2=1
|ka+b|=√[(kx1+x2)^2+(ky1+y2)^2]
|a-kb|=√[(x1-kx2)^2+(y1-ky2)^2]
|ka+b|=根号3*|a-kb|
=>(kx1+x2)^2+(ky1+y2)^2=3[(x1-kx2)^2+(y1-ky2)^2]
=>(k^2-3)(x1^2+y...全部
(1)设a=(x1,y1),b=(x2,y2)
ka+b=(kx1+x2,ky1+y2)
a-kb=(x1-kx2,y1-ky2)
F(k)=a*b=(x1x2,y1y2)
|a|=√(x1^2+y1^2)=1=>x1^2+y1^2=1
|b|=√(x2^2+y2^2)=1=>x2^2+y2^2=1
|ka+b|=√[(kx1+x2)^2+(ky1+y2)^2]
|a-kb|=√[(x1-kx2)^2+(y1-ky2)^2]
|ka+b|=根号3*|a-kb|
=>(kx1+x2)^2+(ky1+y2)^2=3[(x1-kx2)^2+(y1-ky2)^2]
=>(k^2-3)(x1^2+y1^2)+(1-3k^2)(x2^2+y2^2)+8k(x1x2+y1y2)=0
=>x1x2+y1y2=(1+k^2)/(4k)
=>F(k)=(1+k^2)/(4k)
(2)F(k)=(1+k^2)/(4k)=(1-k)^2/(4k)+1/2
当k=1,F(k)有最小值1/2。
因此
x^2-2tx-1/2≤1/2
x^2-2tx-1≤0
t-√(t^2+1)≤x≤t+√(t^2+1)
∵t∈[-1,1]
∴-1-√2≤x≤1+√2。收起
