18．解:(1) (2) 当在R上递增.满足题意, 当 ∴ . ∴ ∴ 综上.a的取值范围是． 19．——青夏教育精英家教网—

18．解:(1) (2) 当在R上递增.满足题意, 当 ∴ . ∴ ∴ 综上.a的取值范围是． 19．解法一: (1) 过O作OF⊥BC于F.连接O1F. ∵OO1⊥面AC.∴BC⊥O1F. ∴∠O1FO是二面角O1-BC-D的平面角.······················· 3分 ∵OB = 2.∠OBF = 60°.∴OF =．在Rt△O1OF中.tan∠O1FO = ∴∠O1FO=60° 即二面角O1-BC-D的大小为60°·············································· 6分 (2) 在△O1AC中.OE是△O1AC的中位线.∴OE∥O1C ∴OE∥O1BC.∵BC⊥面O1OF.∴面O1BC⊥面O1OF.交线O1F．过O作OH⊥O1F于H.则OH是点O到面O1BC的距离.································· 10分 ∴OH = ∴点E到面O1BC的距离等于····················································· 12分解法二: (1) ∵OO1⊥平面AC. ∴OO1⊥OA.OO1⊥OB.又OA⊥OB.······················· 2分建立如图所示的空间直角坐标系 ∵底面ABCD是边长为4.∠DAB = 60°的菱形. ∴OA = 2.OB = 2. 则A(2.0.0).B.C(-2.0.0).O1·········· 3分设平面O1BC的法向量为=(x.y.z).则⊥.⊥. ∴.则z = 2.则x=-.y = 3. ∴=(-.3.2).而平面AC的法向量=··························· 5分 ∴ cos<.>=. 设O1-BC-D的平面角为α. ∴cosα=∴α=60°．故二面角O1-BC-D为60°．············································································· 6分 (2) 设点E到平面O1BC的距离为d. ∵E是O1A的中点.∴=(-.0.).············································· 9分则d= ∴点E到面O1BC的距离等于．··································································· 12分【查看更多】

题目列表(包括答案和解析)

解不等式:(1) (2)