Question

若關于x的不等式（1+k²）x≤k⁴+4的解集是M，則對任意常數k，總有（    ）

A2∈M,0∈M                    B2M,0M

C2∈M,0M                    D2M,0∈M

Answer 1

思路解析：由(1+k²)x≤k⁴+4,得x≤,

令f(k)=,再令k²+1=t(t≥1),則k²=t-1,

f(k)==t+-2≥-2＞4-2=2.(當且僅當t=5t,即t=時“=”成立).

所以2∈M,0∈M.

答案：A