2015眉山一诊 四川省眉山市2015届高三第一次诊断性考试 数学理 扫描版含答案

眉山市高中2015届第一次诊断性考试数学（理工类）参考答案一、选择题：16、解（1）∵(2)cos cos a c B b C -=，由正弦定理得：(2sin sin )cos sin cos A C B B C -=， ∴2sin cos sin cos cos sin sin()sin A B C B C B B C A =+=+=∵0A π<<，∴sin 0A > ∴2cos 1B =，1cos 2B = 又0B π<<∴3B π=； ………………………………………………………………………………… 6分（2）方法一：∵3a =，ABC △13sin 23c π⨯=∴2c =， … 8分22223223cos73b π=+-⨯⨯=，即b =， ………………………………………… 9分cos A ==， …………………………………………………………10分∴cos()BA AC bc A π=-2(1==-. …………………………………………12分方法二：2()BA AC BA BC BA BA BC BA ⋅=-=⋅-221cos ,23212BA BC BA BC BA =⋅⋅〈〉-=⨯⨯-=- ………………………………12分17、解（1）设“小王通过招聘考核”为事件A ，则P(A)＝133334520⨯⨯=所以小王通过招聘考核的概率为320……………………………………………………4分 （2）X 的可能取值为0元，1200元，2200元，3600元 ……………………………5分12(0)133P X ==-=，131(1200)(1)3412P X ==⨯-=，1331(2200)(1)34510P X ==⨯⨯-=1333(3600)34520P X ==⨯⨯= (9)分数学期望为()01200220036008603121020E X =⨯+⨯+⨯+⨯=（元） ……12分 18、解（1）设等比数列{}n a 的首项为1a ，公比为q ，以题意有：3242(2)a a a +=+代入23428a a a ++=，得38a =∴311231208a q a q a a q ⎧+=⎪⎨==⎪⎩ ……………………………………………………………………… 3分解之得：11322122a a q q =⎧=⎧⎪⎨⎨==⎩⎪⎩或 (5)分又∵{}n a 单调递增，∴12,2,a q ==∴2n n a = ………………………………………………………………………………… 6分(2)22log 22n n n n b n ==⋅ …………………………………………………………… 7分 ∴231222322nn s n =⨯+⨯+⨯++⨯①∴23412122232(1)22n n n s n n +=⨯+⨯+⨯++-⨯+⨯②∴②-①得：12322222n nn s n +=⨯-----=12(21)221n n n +-⨯--=11222n n n ++-+⋅+ (9)分由12500n n s n +-⋅+<得12520n +-+<，∴12n +>52.又当4n ≤时，152232n +≤=＜52 当5n ≥时，162264n +≥=﹥52故使12500n n s n +-⋅+<成立的正整数n 的最小值为5 ………………………………12分19、D O F EC1B 1A 1CBAG HEA（Ⅰ）证明：如图1，连接OA 1，O 为AB 的中点，且14AF AB =所以，AF=FO ，又E 为A A 1的中点 所以，EF ∥OA 1 ····························································································· ·············· 2分 在三棱柱ABC －A 1B 1C 1中，A 1B 1∥AB 且A 1B 1=AB 因为，O 、D 分别为AB 、 A 1B 1中点 所以，OB ∥A 1D 且OB =A 1D 所以，OBDA 1为平行四边形 所以，OA 1∥BD ···························································································· ·············· 3分 所以，EF ∥BD ，又EF ⊄平面BDC ，BD ⊂平面BDC 所以，EF ∥平面BDC 1． ·············································································· ·············· 4分 （Ⅱ）证明：如图1，因为，AA 1⊥平面ABC ，OC ⊂平面ABC 所以，AA 1⊥OC ···························································································· ·············· 5分 因为，AB=BC ，O 为AB 中点所以，OC ⊥AB ，又AB 、AA 1⊂平面ABB 1 A 1，AB AA 1=A ················· ·············· 6分 所以，OC ⊥平面ABB 1 A 1，又OC ⊂平面OCC 1D 所以，平面OCC 1D ⊥平面ABB 1 A 1． ························································· ·············· 8分 （Ⅲ）解法一，如图2建立空间直角坐标系O —xyz ，设AB=2 则1(0,1,0),(0,1,2),(0,1,1)A A E ---1(0,1,0),(0,0,2)C B D ································································ ·············· 9分 所以，1(3,1,2),(0,2,1),(0,1,2)BC BE BD =-=-=- 设平面EBC 1的法向量为1111(,,)n x y z =则1111111132020n BC x y z n BE y z ⎧⋅=-+=⎪⎨⋅=-+=⎪⎩取1(3,1,2)n =- ························································································· ············ 10分设平面DBC 1的法向量为2222(,,)n x y z =则2122221132020n BC x y z n BD y z ⎧⋅=-+=⎪⎨⋅=-+=⎪⎩取1(0,2,1)n =······························································································· ············ 11分所以，12cos ,5n n <>== 故，所求二面角E －BC 1－D ········································ ············ 12分 （Ⅲ）解法二，如图1，在三棱柱ABC －A 1B 1C 1中 因为，O 、D 分别为AB 、 A 1B 1的中点所以，OD 平行且等于AA 1，AA 1平行且等于CC 1， 所以，CODC 1为平行四边形所以，C 1D ∥CO ，由（Ⅱ）知，OC ⊥平面ABB 1 A 1 所以，C 1D ⊥平面ABB 1 A 1所以，面C 1DB ⊥平面ABB 1A 1 ···································································· ·············· 9分 过E 作EG ⊥BD 于G ，过G 作GH ⊥B C 1于H ，连接EH 所以，EG ⊥平面BDC 1 所以，EG ⊥GH ，EG ⊥BC 1 所以，BC 1⊥平面EGH所以，BC 1⊥EH所以，GHE ∠为所求二面角E －BC 1－D 的平面角 ··································· ············ 10分 设AB=2，连接DE所以，所以，1141122BDE S EG ∆=---=，所以，5EG =，所以，5BG =因为，11GH BH C D C B=，又11C D C B =,所以GH =所以，EH = ······················································································ ············ 11分∴cos GH GHE EH ∠==E －BC 1－D··· ············ 12分 20、解：（Ⅰ）由已知知道函数()f x 的定义域为{|0}x x > ··············· ············· 1分当1a =-时，()ln f x x x =-+，所以/11()1xf x x x-=-+= ·········· ············· 2分当01x <<时，/()0f x >；当1x >时，/()0f x <所以，()f x 的单调增区间为(0,1)，减区间为(1,)+∞． ··············· ············· 4分（Ⅱ）因为，/1()f x a x =+，令/()0f x =解得1x a =-················· ············· 5分由/()0f x >解得10x a <<-，由/()0f x <解得1x e a-<<从而()f x 的单调增区间为1(0,)a -，减区间为1(,)e a- ················· ············· 6分所以，max 11()()1ln()3f x f a a=-=-+-=-解得，2a e =-． ··············································································· ············· 8分 （Ⅲ）由（Ⅰ）知当1a =-时，max ()(1)1f x f ==-，所以，|()|f x ≥1 ·············································································· ············· 9分令ln 1()2x g x x =+，则/21ln ()xg x x -= 当0x e <<时，/()0g x >；当x e >时，/()0g x < 从而()g x 在(0,)e 上单调递增，在(,)e +∞上单调递减所以，max 11()()12g x g e e ==+< ······················································ ············ 11分所以，|()|f x >()g x ，即|()|f x >ln 12x x + 所以，方程|()|f x =ln 12x x +没有实数根． ····································· ············ 13分 21、解:(Ⅰ)依题意,22222(4)(2)(2)()4(4)4(4)m x m x x f x x x --+'==++ ·················· ················· 1分① 当0m >时,()022,()02f x x f x x ''≥⇒-≤≤<⇒<-或2x >所以()f x 在[2,2]-上单调递增;在(,2),(2,)-∞-+∞上单调递减 ··· ················· 2分 ② 当0m <时,()022,()02f x x f x x ''≤⇒-≤≤>⇒<-或2x >所以()f x 在[2,2]-上单调递减;在(,2),(2,)-∞-+∞上单调递增． ················· 3分 (Ⅱ)当2,22m x <--≤≤时,||111()2222x m x m xm g x --⎛⎫⎛⎫⎛⎫===⋅ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭在[2,2]-上单调递减 ······················ ················· 4分 由(Ⅰ)知,()f x 在[2,2]-上单调递减 ·················································· ················· 5分 所以21()()()24162xm mx F x f x g x x ⎛⎫=+=+ ⎪+⎝⎭在[2,2]-上单调递减 ······ ················· 6分2max ()(2)4221616m m m mF x F +=-=⨯-=-∴ ·············································· ················· 7分 2min ()(2)216m mF x F -==+． ·································································· ················· 8分 (Ⅲ)当2m ≥,1[2,)x ∈+∞时,11121()()416mxh x f x x ==+,由(Ⅰ)知1()h x 在[2,)+∞上单调递减,从而1()(0,(2)]h x f ∈,即1()0,16m h x ⎛⎤∈ ⎥⎝⎦················································ ················· 9分 当2m ≥,22x <时,222||22111()()2222x m m x mx h x g x --⎛⎫⎛⎫⎛⎫====⋅ ⎪⎪ ⎪⎝⎭⎝⎭⎝⎭,在(,2)-∞上单调递增, 从而2()(0,(2))h x g ∈,即221()0,2m h x -⎛⎫⎛⎫∈ ⎪ ⎪ ⎪⎝⎭⎝⎭········································· ···················· 10分对于任意的1[2,)x ∈+∞,总存在唯一的2(,2)x ∈-∞,使得12()()h x h x =成立, 只需21162m m -⎛⎫< ⎪⎝⎭,即210162m m -⎛⎫-< ⎪⎝⎭成立即可. ········································· ············ 11分记函数21()162m m H m -⎛⎫=- ⎪⎝⎭,易知21()162m m H m -⎛⎫=- ⎪⎝⎭在[2,)+∞上单调递增,且(4)0H =··········································································································· ·········· 13分 所以m 的取值范围为[2,4)． ·························································· ·········· 14分。