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莆田一中2021届高三模拟考数学参考答案1-4：BBAC5-8：CCBD9．BD10．CD11．ABC12．ACD13．118．514．解析：法一：由题意21(1)C,2nnann++==34542222221226357111111111111CCC

CCC222222121223344556677naaaaaa+++++=+++++=+++++=．法二：直接算11111101213611572+++++=15．,,,,,3333−等16．8423+17．解：（Ⅰ）中，

，由正弦定理得；.............................................................................1分所以，...........................................................

.....2分化简得；..............................................................................................

.....3分又，所以，所以；....................................................................................................

..................4分又，所以；..............................................................................................

..............................5分（Ⅱ）中，，，由余弦定理得，所以；...........................................................

.................................................................7分所以，...........................................

.............................8分求得；...........................................................................................

..9分所以.....................................................................10分18．解：（1）选①Sn=n（n+），可得a1=S1=

1+，............................................................1分解得a1=2，..................................................

.................................................................................2分即Sn=n2+n，则a1+a2=6，............................

.................................................................................3分即a2=4，d=a2-a1=2，...........................

.........................................................................................4分所以an=2+2（n-1）=2n；...............................

.............................................................................5分选②S2=a3，a4=a1a2，可得2a1+d=a1+2d，a1+3d=a1（a1+d），..................

............................2分解得a1=d=2，..............................................................................................................

................4分所以an=2+2（n-1）=2n；...................................................................................

.......................5分选③a1=2，a4是a2，a8的等比中项，可得a42=a2a8，..............................................................1分即

（2+3d）2=（2+d）（2+7d），...........................................................................................2分解得d=2（d=0舍去），........

....................................................................................................4分所以an=2+2（n-1）=2n；.....................

....................................................................................5分（2）法一：由（1）知nnnb422==，....................

............................................................6分1cos2abcB+=1sinsinsincos2ABCB+=1sincoscossinsinsincos2BCBCBCB++=1sincossin2BCB=−

(0,)Bsin0B1cos2C=−(0,)C23C=ABC2a=3b=22212cos49223()192cababC=+−=+−−=19c=22291944cos2231919bcaAbc+−+−===243sin1()1919A=−=3483sin22

sincos2191919AAA===则4246444221+++=−−nTnnnn，两边同时乘于41得1246444241321+++=−−−nTnnnn............................

...............................................8分两式相减得nnnTnnnnnn2)14(38241)44(122424242424321−−=−−−=−+++=−−...

................................................................11分即389)14(32nTnn−−=或（932)34(81−−=+nTnn）............

..........................................12分法二：由（1）知nnnb422==，......................................

...................................................6分则4246444221+++=−−nTnnnn，则12424644124−+++=nnnnT................

....................................................................................7分令1242464412−+++=nnnP，则n

nnP424644424132+++=..................................................................................................8分两式相减得nnnnnnnnnnP424113842

4114112424141411124312−−=−−−=−+++=−............................................................................10分得nnnnP438411932

−−=...................................................................................................11分即389)14(32nT

nn−−=或（932)34(81−−=+nTnn）......................................................12分21解析：（1）由22222121914ceaababc==+=

=+，..................................................2分解得22a=，23b=，.........................................

...................3分故椭圆C的方程为：22143xy+=．...................................................4分（2）设点00(,)Pxy，直线PA，PB的斜率分别为1k，2k，则212314kke=−=−．.......

..............................................................................................................6

分又1:(2)PAykx=+，令4x=，得1(4,6)Mk，..........................................................................7分同理：1:(2)PBykx=−，令4x=，得2(4,2)Nk．

..................................................................8分由椭圆的对称性，知定点在x轴上，设圆过定点(,0)Rm，则1266144RMRNkkkkmm==−−−，...............

....................................................................................10分解得1m=或7m=．...................................

......................................................................................11分故以MN为直径的圆，过x轴上两个不同的定点(1，0)和（7，0）．..

.....................12分法二:由（1）知A（-2，0），B（2，0），设点P（x0,y0）设直线AP的方程为2−=myx，令4=x得，my6=，即)6,4(mM...............................................

.....................................................5分联立−==+213422myxyx消去x，得：()0124322=−+myym得：4312020+=+mmy，即431220+=mmy

................................................................................................6分代入4−=myx，得4386

220+−=mmx即P++−4312,4386222mmmm.....................................................................................

.................................7分所以直线BP：)2(43−−=xmy，................................................................

.......................................8分令4=x得，my23−=，即)23,4(mN−.....................................

......................................................9分设Q（x,y）是以MN为直径的椭圆上的任一点，则0=NQMQ，即07623823,46,422

=+−+−+=+−−−ymmxyxmyxmyx.........................................10分令y=0，得x=1或7...

............................................................................................................................

.......11分故以MN为直径的圆，过x轴上两个不同的定点(1，0)和（7，0）．.......................12分