数学理试卷答案.pdf

1 福州一中 2020 届高三教学反馈理科数学检测 答案及评分参考 评分说明： 1．本解答给出了一种或几种解法供参考，如果考生的解法与本解答不同，可根据 试题的主要考查内容比照评分标准制定相应的评分细则。 2．对解答题，当考生的解答在某一步出现错误时，如果后继部分的解答未改变该 题的内容和难度，可视影响的程度决定后继部分的给分，但不得超过该部分正确解答应 给分数的一半；如果后继部分的解答有较严重的错误，就不再给分。 3．解答右端所注分数，表示考生正确做到这一步应得的累加分数。 4．只给整数分数．选择题和填空题不给中间分。 一、选择题：每小题 4 分，满分 40 分。 1. B 2. B 3. B 4. A 5. C 6. D 7. A 8. C 9. B 10. C 11. A 12. D 8. 简解： 11 1 1( 1) 1 2ac b b a= = + −  = ，    11 1 1 1 1( 1) 1 ( 1) 1 1nnn n a a n n n nc c b b b a b a a a++ + +− = − = + −  − + −  = − = ， 故数列 nc 是公差为1的等差数列，所以 的前7 项和为 7 767 2 1 352T =  +  = 。 11. 简解：设过点 ( 8,4)A − 的抛物线 2:8E y x= 的切线方程为 : 8 ( 4)l x t y+ = − ，即 84x ty t= − − () 代入 2 8yx= 得 2 8 8(4 8) 0y ty t− + + = ，由 0= 得 2 2 4 0tt− − = ， (1) 所以方程（1）有两个不相等的实数根 12,tt，且 122tt+=， 12 4tt =− ，在() 中令 0x = 得 12 88(0,4 ), (0,4 )BCtt++，设 ABC 的外接圆圆心为点 1 0 0( , )O x y ，则 0 1 ( ) 22 BCy y y= + = ，下求 0x ：线段 AB 中点横标 0 4x  =− ，纵标 0 1 44y t  =+ ， 线段 的中垂线方程为 1 1 44 ( 4)y t xt− − = − + ，令 2y = 得 2 11 0 2 1 4 2 4ttx t − + += ，由(1) 知 2 1124tt+=，故 0 3x =− ，设 的外接圆半径为 R ，则 2 29R = ，所以 的外接圆 方程为 22( 3) ( 2) 29xy+ + − = ，即 226 4 16 0x y x y+ + − − = 。 （注：当 0 2y = 时，只剩下 A,D 两个选项值得考虑，而 D 选项表示圆心为(0,2) ，舍去，故选 A） 12. 简解：由 ( ) 0fx= 得 3 x xx xea e e x=−− ，即 33(1 ) 1 11 xx xx xxa xxee ee = − = − − − + −− ，记 1 x xt e=− ，且设 ( ) 1 x xgx e=− ，一方面由 3 1att= − − + 得 2 ( 1) 3 0 ( )t a t+ − + =  ，当 0 时方程（ ）有两个不相等的实数根 12,tt，且 121t t a+ = − ， 12 3tt = ；另一方面， 由 1() x xgx e − = 知 ()gx在 ( ,1)− 上单调递减，在(1, )+ 上单调递增， 1(1) 1g e=− ， 2 (0) 1g = ，当 x → − 时， ()gx→ + ，当 x → + 时， ( ) 1gx −→ ，如图， 12 111tte   − ，且 312 312 121 , 1 1 xxx xxxtte e e− = − = − = ，因此 312 2 312(1 ) (1 )(1 )xxx xxx e e e− − − 22 1 2 2 1 2( ) 9t t t t t=   = = 。 二、填空题：每小题 4 分，满分 16 分。 13. 3 ,32   14. 9− 15. 5 2 16. 2 6 3 3( , ), ,6 4 5 5     14. 简解：当 n 为奇数时， 1nna a n+ − = − ，当n 为偶数时， 1nna a n+ −=，故 20a =  20 19 18 17 2 1( ) ( ) ( )a a a a a a= − + − + + −  19 18 17 16 3 2( ) ( ) ( )a a a a a a+ − + − + + − 1a+ (19 17 1) (18 16 2) 1 9= − + + + + + + + + = − 。 （注：逐一算出 2 3 4 19 20, , , , ,a a a a a 亦可，毕竟项数较少） 15. 简解： 如图，设点 ( , )aM h hb− ，由 1 1AMk = 得 abh ab= + ，又设点 ( , )G a t t+ ，则 22 22 () 1a t t ab + −=得 2 22 2abt ab= − ，依题设知 2th= 得 2ab= ，故 5 2e = 。 16. 简解：由3 2 2( )B C A B C+ = + + 得 2BA= ，故 0,2 0 2 ,2 0 3 ,2 A A A           −  得 ( , )64A  ； 所求为 21 6(1 tan )() tan 10tan AfA AA −=− ，记 3tan , ( ,1)3t A t=， 223() 5 tgt t += ， 2 3 6 6( ) ( )( )5 3 3g t t tt  = + − ，故 ()gt 在 36( , )33 上单调递减，在 6( ,1)3 上单调递增， 6 2 6 3 3 3( ) , (1) 1, ( )3 5 3 5g g g= = = ，故 16 tan 5tanAB− 的取值范围是 2 6 3 3,55    。 y 1yt= 2yt= 1x o 1 2x 3x 1 11 e− y x o 1A M N 2A G Q m x 3 三、解答题：共 70 分．解答应写出文字说明、证明过程或演算步骤。 17．（ 12 分） 解：（1）解法一： sin sin (sin 2sin )sin 0cos cos B A B C B AB −+=， 因为sin 0B  ，故sin cos sin cos 2sin cos 0A B B A C A+ − = ···································· 2 分 故sin 2sin cos 0C C A−=，而sin 0C  ，因此 1cos 2A = ··········································· 4 分 因为0 A ，所以 3A = ······························································································· 5 分 解法二： sin ( 2 )sin 0cos cos b A b c B AB −+=，故 cos ( 2 )sin 0cos sin B b c B A b A −+=， 即 2 2 2 2 2 2 2 0a c b b c b c a b + − −+=+− ································································································· 2 分 化简得 2 2 2a b c bc= + − ，因此 ········································································· 4 分 因为 ，所以 3A = ······························································································· 5 分 （2）解法一：由 3A = 知 2 2 2a b c bc= + − ····································································· 7 分 又 3ab= ，故 2 2 23b c bc b+ − = ，即 2220b bc c+ − = ·············································· 8 分 故 2cb= ································································································································· 9 分 故 ( 3 3) 3 3b+ = + ，故 1b = ························································································ 10 分 故 2c = ， 3a = ，故 2C = ··························································································· 11 分 因此 ABC 的面积为 3 2 ··································································································· 12 分 解法二：由 3ab= ， 知 1sin 2B = ········································································ 7 分 又ba ，故 6B = ·············································································································· 8 分 故 2C = ································································································································ 9 分 故 2cb= ，故 ，即 ···································································· 10 分 故 ， ················································································································ 11 分 因此 的面积为 ··································································································· 12 分 18．（ 12 分） 解：（1）因为 AF ⊥平面 ABCD， //DE AF ，所以 DE ⊥平面 ， 故 AC DE⊥ ·························································································································· 2 分 又四边形 ABCD为菱形，故 AC BD⊥ ············································································· 3 分 又 ,DE DB 是平面 BDE 内两条相交的直线， 故 AC ⊥平面 ， 又 AC  平面 ACE ，因此平面 ACE ⊥ 平面 BDE ·························································· 4 分 4 （2）解法一：取线段 AB 中点 N ，连接 DN ，以点 D 为原点O ，分别以 ,,DN DC DE 的 方向为 ,,x y z 轴的正方向，建立空间直角坐标系O xyz− ·················································· 5 分 则点 ( 3, 1, 0)A − ， ( 3,1, 0)B ， (0, 2, 0)C ， (0, 0, 4)E ， ( 3, 0, 2)BG =− ， (0, 2, 2)BF =− ， ( 3, 1, 4)BE = − − ····················································· 6 分 设平面GBF 的法向量为 1 1 1( , , )m x y z= ，则 ( 3, 0, 2) 0, (0, 2, 2) 0, m m   − =  − = 即 11 11 3 2 0, 2 2 0, xz yz − + =− + = 可取 (2, 3, 3)m = ··············································································································· 8 分 又设平面 EBF 的法向量 2 2 2( , , )n x y z= ，则 ( 3,1, 4) 0, (0, 2, 2) 0, n n   − =  − = 即 2 2 2 22 3 4 0, 2 2 0, x y z yz  + − =− + = 可取 ( 3,1,1)n = ··················································································································· 10 分 故 26cos , 5 mn mn mn    = = ····························································································· 11 分 因此二面角G BF E−−的正弦值为 1 5 ·············································································· 12 分 解法二：前同法一，得平面 的法向量为 (2, 3, 3)m = ········································· 8 分 点 E 到平面 的距离 30cos , 10 EB m h EB EB m m  =    = = ······························ 10 分 作 EQ FB⊥ 于点Q ，由 22FB FE==， 25EB = 得 30 2EQ = ················ 11 分 因此二面角 的正弦值为 h EQ ，即 1 5 ···························································· 12 分 19．（ 12 分） 解：（1）依题意知 23 9 9a −=，所以 32a = ···························································· 2 分 故椭圆 E 的方程为 22 118 9 xy+= ························································································ 3 分 （2）设直线 : ( 3)AB y x m m= − +  ·············································································· 4 分 把 y x m= − + 代入 22 :118 9 xyE +=得 223 4 2 18 0 ( )x mx m− + − =  ······················ 6 分 由判别式 0 得3 3 3m ，此时方程() 有两个不相等的实数根 12,xx， A B C D（o） E F G x N z y 5 且 2 1 2 1 2 4 2 18,33 mmx x x x −+ = = ····················································································· 7 分 可得 24 273AB m=− ····································································································· 8 分 2 (6 )2BC m=− ·············································································································· 9 分 由 4 3AB BC=可得 227 (6 ) 2mm− − = ······························································· 10 分 记 2( ) 27 (6 )f m m m= − − ，其中3 3 3m ， 则函数 ()fm在 (3,3 3) 上单调递减 ················································································· 11 分 且 (5) 2f = ，故 5m = ，因此直线 AB 的方程为 5yx= − + ····································· 12 分 20．（ 12 分） 解：（1）方差 2 1 (25 4 0 4 9 9 9 0 1 1) 6.210s = + + + + + + + + + = ································ 2 分 （2）在第二种试验中服药有效的白鼠有 4 只，服药无效的白鼠有 6 只， 故 的可能取值为1,2,3,4 ··································································································· 3 分 16 46 7 10 4( 1) 120 CCP C = = = ， 25 46 7 10 36( 2) 120 CCP C = = = ， 34 46 7 10 60( 3) 120 CCP C = = = ， 43 46 7 10 20( 4) 120 CCP C = = = ， 因此 的分布列为  1 2 3 4 P 1 30 3 10 1 2 1 6 ················································································································································· 6 分 14() 5E  = ······························································································································ 7 分 （3）（ ⅰ） 1 990 9at=− ······································································································ 8 分 依题设知 1 9(1 ) (1000 )100 10n n n ta a a+ = − + − ， 即 1 10 900 (0 10)100nn ta a t+ −= +   ·················································································· 9 分 （ⅱ） 21 10 10900 (990 9 ) 900100 100 tta a t−−= + = − + ， 由 2 950a  可得(990 9 )(10 ) 5000tt− −  ······································································ 10 分 记函数 ( ) (990 9 )(10 )f t t t= − − ，其中0 10t ， 则函数 在 (0,10) 上单调递减， 且 (4) 5724, (5) 4725ff== ···························································································· 11 分 故最大的正整数 4t = ·········································································································· 12 分 21．（ 12 分） 解：（1） ( ) 1 2xf x e ax = − − ，当 1 2a = 时， ( ) 1xf x e x = − − ···································· 1 分 ( ) 1 0xf x e = −  ，所以函数 ()fx 在(0, )+ 上单调递增， 6 当 0x  时， ( ) (0) 0f x f=，所以函数 ()fx在 (0, )+ 上单调递增， 当 时， ( ) (0) 0f x f=，即 ( ) 0fx ···································································· 2 分 （2）等价于 21 2 ln 0xe ax x x x− − − −  ，记 2( ) 1 2 ln ( 0)xg x e ax x x x x= − − − −  ， 首先 (1) 0g  ，即 2 2 ea − ，而 2012 e −，故整数 0a  ······································· 5 分 另法：对于 ，记 ， 当 ( 0,1x 时， 2 ln 0xx−，此时 ( ) 1 2xg x e ax x − − − 设 ( ) 1 (2 1) ( 0)xM x e a x x= − − +  ， ( ) (2 1) ( 0)xM x e a x = − +  ， 若 0a  ，记  min 1, ln(2 1)ba=+，则 ( )0,xb 时， ( ) (2 1) 0xM x e a = − +  ， ( ) ( ) (0) 0g x M x M  = ，与 ( ) 0gx 矛盾，因此 0a  ············································· 5 分 （注：事实上，本题的第（2）小题中“a 是整数”的条件可为“a 是实数”，结论不变） 下证 0a = 时 ( ) 0gx 恒成立，即 21 ln 0xe x x x− − −  ··············································· 6 分 证法一： 等价于 22 11ln 0 xe xxxx− − −  ····························································· 7 分 设 22 11( ) ln ( 0) xeh x x xxxx= − − −  ， 则 3 3 3 ( 2) ( 2)( 1) ( 2)( 1 )() xxe x x x x e xhx x x x − − + − − − = − = ··············································· 9 分 由（1）知 10xex− −  ，故 ()hx在( )0,2 上单调递减，在( )2,+ 上单调递增， 所以 2 3 4ln 2( ) (2) 4 eh x h −−= ······················································································ 10 分 因 ln 2 1 ，故 223 4ln 2 7 044 ee− − −，即 ( ) 0hx  ··············································· 11 分 所以存在最大整数 ，使得 2( ) lnf x x x x + 对一切 0x  恒成立 ····················· 12 分 证法二：要证 21 lnxe x x x + + ，先证 2 3 41 1 11 ( 0)2 6 24 xe x x x x x + + + +  ········ 7 分 设 2 3 41 1 1( ) 1 ( 0)2 6 24 xM x e x x x x x= − − − − −  ， 2311( ) 1 26 xM x e x x x = − − − − ， 21( ) 1 2 xM x e x x = − − − ， ( ) 1xM x e x = − − ，由（ 1）知 ，即 ( ) 0Mx  ， 故 ()Mx 在( )0,+ 上单调递增，当 0x  时， ( ) (0) 0M x M =， 故 ()Mx 在 上单调递增，当 时， ( ) (0) 0M x M=， 故 ()Mx在 上单调递增，当 时， ( ) (0) 0M x M=， 即 2 3 41 1 11 ( 0)2 6 24 xe x x x x x + + + +  ········································································· 8 分 注意到 2 3 4 2 21 1 1 1 1 11 1 ( )2 6 24 2 6 24x x x x x x x x+ + + + = + + + + ， 设 21 1 1( ) ln ( 0)2 6 24G x x x x x= + + −  ， 7 222 12 ( 1) 13 ( 1 13)( 1 13)() 12 12 12 x x x x xGx + − + − + + + − = = = ， 故 ()Gx在( )0, 13 1− 上单调递减，在( )13 1,− + 上单调递增 ································· 9 分 设 0 13 1x =−，易知 02 xe，故 2 00 1 1 1 12 6 24xx+ +  ，且 0ln ln 1xe=， 故 0( ) ( ) 0G x G x ··········································································································· 10 分 因此 21 1 1 ln2 6 24x x x+ +  ，即 2 3 4 21 1 11 1 ln2 6 24x x x x x x x+ + + +  + + ············ 11 分 所以存在最大整数 0a = ，使得 2( ) lnf x x x x + 对一切 0x  恒成立 ····················· 12 分 22.（10 分） 解：（1） 22 1 : ( 3) 3C x y+ − = ························································································· 2 分 22 2 :( 1) 1C x y−+= ············································································································· 4 分 （2）解法一：曲线 化为极坐标方程为 2 3sin= ·············· 5 分 设点 12( , ), ( , )AB    ，依题设知 122 3sin , 2cos   == ································ 6 分 所以 2 3sin 2cos 4 sin( )6AB   = − = − ····························································· 7 分 由 22AB = 知 2sin( )62  − =  ·················································································· 8 分 因为 5 6 6 6   −  −  ，故 64  −=或 3 64  −= ·················································· 9 分 因此 5 12  = 或 11 12  = ····································································································· 10 分 解法二：圆心 1(0, 3)C 到直线 : tanl y x=的距离 1 3 cosh = ， 圆心 2 (1,0)C 到直线 的距离 2 sinh = ····················································· 5 分 所以 22 3 3cos 2 3sinOA = − = ， 22 1 sin 2 cosOB = − = ··················· 6 分 故 2 3sin 2cos 4 sin( )6AB   = − = − ································································· 7 分 由 0 知 5 6 6 6   −  −  ， 结合 4 sin( ) 2 26  −= 知 2sin( )62  −= ································································ 8 分 因此 64  −=或 3 64  −= ，所以 5 12  = 或 11 12  = ·········································· 10 分 23.（10 分） 解：（1）等价于 22( 2) (3 1)xx+  − ，即 28 10 3 0xx− −  ··········································· 2 分 故不等式 ( ) ( )f x g x 的解集为 13 42xx−   ···························································· 3 分 （2）解法一：3 ( ) ( ) 4f x g x ax+  + 即3 2 3 1 4x x ax+ + −  + ， （ⅰ）当 2x − 时， 69ax x − − ，故 9 6a x − − ························································· 4 分 8 而 9 6x−−在( ,2− − 上单调递增，故 max 93( 6) 2x− − = − ，故 3 2a − ························ 5 分 （ⅱ）当 12 3x−   时， 30ax − ··················································································· 6 分 记 ( ) 3h x ax=−，等价于 ( 2) 0 1( ) 03 h h −  ，即 3 92 a−   ···················································· 7 分 （ⅲ）当 1 3x  时， 61ax x+，即 1 6a x+ ································································· 8 分 而 1 6x + 在 1,3 + 上单调递减，当 x → + 时， 1 66x ++→ ，故 6a  ··················· 9 分 综上，实数 a 的取值范围是 3 ,62 −  ··············································································· 10 分 解法二：3 ( ) ( ) 4f x g x ax+  + 即3 2 3 1 4x x ax+ + −  + ， 在同一直角坐标系中作 ( ) 3 2 3 1h x x x= + + − 和 ( ) 4M x ax=+的图象 ····················· 4 分 16 5, 3 1( ) 7, 2 3 6 5, 2 xx h x x xx  + = −    − −  −  ·························································································· 5 分 其图象为： 图中点 ( 2,7)A − ，点 1( ,7)3B ，点 (0,4)C ， 3 ,92CA CBkk= − = ································· 7 分 要使 恒成立， 必须06a或 3 02 a−   ······························································································· 9 分 因此实数 的取值范围是 ··················································································· 10 分 A B C y O x