解析:命题p对应的集合A={x|a-1 命题q对应的集合B={x|0 当A⊆B时,{■(a"-" 1≥0"," @a+1≤6"," )┤∴1≤a≤5. 当B⊆A时,{■(a"-" 1≤0"," @a+1≥6"," )┤此不等式无解. 因为p是q的既不充分又不必要条件,所以A⊈B,且B⊈A. 因此,实数a的取值范围是{a|a<1或a>5}. 答案:{a|a<1或a>5} 11.设集合A={x|-2≤x≤a},B={y|y=2x+3,x∈A},C={z|z=x2,x∈A},求B∪C=B的充要条件. 解B∪C=B⇒C⊆B. ∵A={x|-2≤x≤a}, ∴B={y|y=2x+3,x∈A}={y|-1≤y≤2a+3}. 又∵当-2≤a<0时,C={z|a2≤z≤4}; 当0≤a≤2时,C={z|0≤z≤4}; 当a>2时,C={z|0≤z≤a2}, ∴当-2≤a≤2时,C⊆B⇔4≤2a+3,即 1/2≤a≤2;