题目内容
值域为集合{5,10},其对应关系为y=x2+1的函数个数为( )
| A.1 | B.4 | C.7 | D.9 |
令x2+1=5可得,x=2或x=-2,令x2+1=10可得x=3或x=-3
故满足条件的函数有y=x2+1得定义域得可能情况有:x∈{2,3},x∈{2,-3},x∈{-2,3},x∈{-2,-3},x∈{2,-2,3},x∈{2,-2,-3},x∈{2,3,-3},x∈{-2,3,-3},x∈{2,-2,-3,3}共9个
故选:D
故满足条件的函数有y=x2+1得定义域得可能情况有:x∈{2,3},x∈{2,-3},x∈{-2,3},x∈{-2,-3},x∈{2,-2,3},x∈{2,-2,-3},x∈{2,3,-3},x∈{-2,3,-3},x∈{2,-2,-3,3}共9个
故选:D
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