一道数学题

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