Question

设α,β为二次方程x2-2kx+k+6=0的两个是跟,则对(α-1)2+(β-1)2的正确判断是

Answer 1

方程有两个实根,则△≥0

4k²-4(k+6)≥0

k²-k-6≥0

(k-3)(k+2)≥0

k≥3或k≤-2

根据韦达定理

α+β=2k

α*β=k+6

(α-1)²+(β-1)²

=(α²+β²)-2(α+β)+2

=(α+β)²-2*α*β-2*2k+2

=4k²-2(k+6)-4k+2

=4k²-6k-10

=4(k²-1.5k+0.75²)-4×0.75²-10

=4(k-0.75)²-12.25

k=3或k=-2时有最小值,为:

4×2.25²-12.25=8

没有最大值

选B