2023届绵阳二诊理科数学答案_2.2025数学总复习_数学高考模拟题_2023年模拟题_老高考_四川省绵阳市高2023届高三第二次诊断性考试绵阳二诊数学

绵阳市高中 2020 级第二次诊断性考试 理科数学参考答案及评分意见 一、选择题：本大题共12小题，每小题5分，共60分． DACBD ABCAD CA 二、填空题：本大题共4小题，每小题5分，共20分． 1 13．9 14． 15． 5 16．[1，3) 3 三、解答题：本大题共6小题，共70分． 17．解：（1）由3acosCa2sinC 3b， 可得3sinAcosCasinAsinC 3sinB ，··················································2分 又3sinB3sin(AC)3(sinAcosCcosAsinC) ，····································4分 ∴asinA3cosA，5分  且A ，可得a 3；·······································································6分 3   1 1 （2）由BAAC cbcos(A) cb ，可得cb1，······················8分 2 2 由余弦定理c2 b2 a2 2bccosA4.····················································9分  1   ∵AT  (ABAC)， 2  1     平方可得(AT)2  [(AB)2 (AC)2 2ABAC] ，··································10分 4  1 5 5 即(AT)2  (c2 b2 2bccosA) ，所以AT  ．·······························12分 4 4 2 5 18．解：（1）因为0.92<0.99，根据统计学相关知识，R2越大，意味着残差平方和(y yˆ)2 i i1 越小，那么拟合效果越好，因此选择非线性回归方程②yˆ mˆx2 nˆ进行拟合更加符合 问题实际．·························································································4分 （2）令u x2，则先求出线性回归方程：yˆ mˆunˆ，·································5分 i i 1491625 0.81.11.52.43.7 ∵u =11，y 1.9，······················7分 5 5 n (u u)2 (111)2 (411)2 (911)2 (1611)2 (2511)2 =374，············9分 i i1 n (u u)(y y) i i 45.1 ∴mˆ  i1  0.121，···················································10分 n 374 (u u)2 i i1 理科数学参考答案 第 1 页 共 6 页由1.90.12111nˆ，得nˆ0.5690.57，·············································11分 即yˆ 0.12u0.57， ∴所求非线性回归方程为：yˆ 0.12x20.57．········································12分 8 2 19．解：设{b }的公比为q，则b b q2，所以  q2， n 3 1 27 3 2 所以q ，·····················································································2分 3 2 因为{b }的各项都为正，所以取q ，···················································3分 n 3 2 所以b ( )n．···················································································4分 n 3 若选①：由2a S 1(n1)，得2a S 1(n≥2) ，·····························5分 n n n1 n1 两式相减得：2a 2a a 0，整理得a 2a (n≥2)，························6分 n n1 n n n1 因为a 10，所以{a }是公比为2，首项为1的等比数列，····················· 7分 1 n ∴a 2n1，····················································································· 8分 n 1 4 ∴a b  ( )n，··············································································9分 n n 2 3 1 4 ∵y ( )x在R上为增函数，···························································10分 2 3 ∴数列{a b }单调递增，没有最大值，··················································11分 n n ∴不存在mN，使得对任意的nN，a b ≤a b 恒成立．··················· 12分 n n m m 若选择②：因为a a a 2(n 2)，且a 0，a 0， n1 n1 n 1 2 ∴{a }为等比数列，···········································································6分 n a 1 公比q 2  ，················································································7分 a 4 1 1 ∴a ( )n1，···················································································8分 n 4 1 2 2n 1 1 2 所以a b ( )n1( )n 4 4( )n≤4  ．·································10分 n n 4 3 3n 6 6 3 2 当且仅当n1时取得最大值 ，··························································11分 3 ∴存在m1，使得对任意的nN，a b ≤a b 恒成立．··························12分 n n m m 若选择③：因为a 1a ，所以a a 1(n≥2)，·····························5分 n n1 n n1 ∴{a }是以1为公差的等差数列，又a 1，············································6分 n 1 2 ∴a n，所以a b n( )n，···························································· 8分 n n n 3 理科数学参考答案 第 2 页 共 6 页2 2 3n 2 设c a b ，则c c n( )n(n1)( )n1 ( )n1，····················9分 n n n n n1 3 3 3 3 ∴当n3时，c c 0，所以c c ， n n1 n n1 当n3时，c c 0，所以c c ， n n1 n n1 当n3时，c c 0，所以c c ， n n1 n n1 则c c c c c ，································································11分 1 2 3 4 5 ∴存在m2或3，使得对任意的nN，a b ≤a b 恒成立．······················12分 n n m m 20．解：（1）设B(x，y)，C(x，y )， 1 1 2 2 1 直线BC的方程为：xmy4(m= )，···················································1分 k xmy4  联立x2 y2 ，消x整理得：(3m2 4)y2 24my360，·····················2分   1  4 3 24m 36 ∴y  y  ，y y  ······················································3分 1 2 3m2 4 1 2 3m2 4 y y y y 从而：k k  1  2  1 2 1 2 x 2 x 2 (my 6)(my 6) 1 2 1 2 y y  1 2 m2y y 6m(y y )36 1 2 1 2 36 3m2 4 1   36m2 144m2 4  36 3m2 4 3m2 4 1 ∴k k 为定值 .················································································5分 1 2 4 y （2）直线AB的方程为：y 1 (x2)，················································6分 x 2 1 6y 6y 令x4，得到y  1  1 ，······················································7分 M x 2 my 6 1 1 6y 同理：y  2 ．··········································································8分 N my 6 2 6y 6y 从而 |MN|| y y || 1  2 | M N my 6 my 6 1 2 36| y y |  1 2 ······················································9分 |m2y y 6m(y y )36| 1 2 1 2 12 m2 4 又 | y y | (y  y )24y y  ， 1 2 1 2 1 2 3m2 4 理科数学参考答案 第 3 页 共 6 页144 |m2y y 6m(y  y )36| ，·····················································10分 1 2 1 2 3m2 4 所以|MN|3 m2 4 ，·······································································11分 1 因为：m [3，4]，所以|MN|[3 5，6 3]， k 即线段MN长度的取值范围为[3 5，6 3]．·············································12分 21．解：（1）由a=1时， f(x)(x1)(ex 1)，·············································1分 由 f(x)0解得：x>0或x<−1；由 f(x)0解得：−10，即h(a)在区间( ，e2)上单调递增， 2 2 1 又h(e2)e2 0，则 a(lna)2≤a2e2恒成立．····································· 11分 2 4 综上：0≤a≤ e2．···········································································12分 3 3 22．解：（1）①当B在线段AO上时，由|OA|‧|OB|=4，则B（2，）或（2， ）； 2 ②当B不在线段AO上时，设B（ρ，θ），且满足|OA|‧|OB|=4， 4 ∴A（ ，），·············································································1分  4 4 又∵A在曲线l上，则 cos() sin()2 ，··························· 3分   ∴2sin2cos，······································································4分 3  又∵≤≤ ，即0≤≤ . 2 2 综上所述，曲线C的极坐标方程为：  3 2sin2cos（0≤≤ ），或2(=或= )．·························5分 2 2 3 （2）①若曲线C为：2(=或= )，此时P，Q重合，不符合题意； 2   ②若曲线C为：2sin2cos（0≤≤ ），设l ：（0≤≤ ）， 1 2 2 ， 又l 1 与曲线C交于点P，联立 2sin2cos， 得： 2sin2cos，····································································6分 P ， 又l 1 与曲线l交于点Q，联立 sincos2， 2 得：  ，·······································································7分 Q sincos 又∵M是P，Q的中点，   1    P Q sincos (0≤≤ ) ，·······························8分 M 2 sincos 2  令sincost ，则t  2sin( )， 4    3 又∵0≤≤ ，则 ≤ ≤ ，且1≤t≤ 2 ， 2 4 4 4 理科数学参考答案 第 5 页 共 6 页1 1 ∴ t (1≤t≤ 2)，且 t 在1，2上是增函数，······················9分 M t M t   1 2    ∴0≤ ≤ 2  ，且当  时，即 时等号成立． M 2 2 4 2 4 2 ∴ OM 的最大值为 ．····································································10分 2 23．解：（1）由 f(x)≤3的解集为[n，1]，可知，1是方程 f(x)=3的根， ∴ f(1)=3+|m+1|=3，则m=−1，······························································ 1分 ∴ f(x)=|2x+1|+|x−1|， 1 1 ①当x≤ 时， f(x)=−3x≤3，即x≥−1，解得：−1≤x≤ ，··················2分 2 2 1 1 ②当 x1时， f(x)=x+2≤3，解得：  x1，·································3分 2 2 ③当x≥1时， f(x)=3x≤3，解得：x=1．················································4分 综上所述： f(x)的解集为[−1，1]，所以m=−1，n=−1．······························5分 1 2 （2）由（1）可知m=−1，则  2．······················································6分 2a b 1 2 1 2 令 x，  y，则2a ，b ， 2a b x y 又a，b 均为正数，则x y2（x0，y0）， 由基本不等式得，2x y≥2 xy，·······················································7分 ∴xy≤1，当且仅当x=y=1时，等号成立． 1 所以有 ≥1，当且仅当x=y=1时，等号成立．········································8分 xy 4 4 又16a2 b2 4(2a)2 b2   x2 y2 4 4 8 ≥2   (当且仅当x=y时，等号成立)．·······9分 x2 y2 xy 1 ∴16a2 b2≥8成立，(当且仅当，a ，b2时等号成立)．·····················10分 2 理科数学参考答案 第 6 页 共 6 页