“元三维大联考”2023级高三第一次诊断考试数学答案_251104四川省元三维大联考2023级高三第一次诊断考试（全科）_“元三维大联考”2023级高三第一次诊断考试数学

高中 2023 级第一次诊断性考试 数学参考答案及评分标准 一、选择题：本题共8小题，每小题5分，共40分． 1．C 2．D 3．B 4．D 5．A 6．B 7．A 8．C 二、选择题：本大题共3小题，每小题6分，共18分．全部选对的得6分，选对但不 全的得部分分，有选错的得0分． 9．BCD 10．ACD 11．AC 三、填空题：本题共3个小题，每小题5分，共15分． 11 12．1； 13． ； 14．(1，0)(0，1) 2 四、解答题：本题共5小题，第15题13分，第16、17小题15分，第18、19小题17 分，共77分．解答应写出文字说明、证明过程或演算步骤． 15．解：（1）∵ f(x)sin(x)的最小正周期为， ∴2，·······················································································2分  3 3 3 又 f( ) ，即sin() ，则sin ，···························4分 2 2 2 2  2 又( ，)，则 ，······························································5分 2 3 2 ∴ f(x)sin(2x )；····································································6分 3   2  （2）由题知：g(x) f(x )sin[2(x ) ] sin(2x )，···········8分 6 6 3 3    4 由x[0， ]，则 ≤2x ≤ ，··················································10分 2 3 3 3 3  ∴ ≤sin(2x )≤1，································································12分 2 3 3 故g(x)的值域为[ ，1]．·····························································13分 2 数学参考答案与评分标准 第1页，共6页x(x2)，x≥2， 16．解：（1）当m2时， f(x)x|x2| ·····················3分 x(x2)，x2， 故 f(x)的单调递增区间为：(，2]，[1，)，单调递减区间为：[−2，−1]； ·····································································································6分 （2）由题知x∈[1，2]， f(x)≤2， 即对任意x∈[1，2]，x xm≤2恒成立，············································ 7分 2 2 2 2 2 ∴ xm≤ ，即 ≤xm≤ ，则 x≤m≤ x， x x x x x 2 2 ∴x ≤m≤x ，·····································································9分 x x 2 令g(x)x ，易知g(x)在x[1，2]单调递增，故g(x) g(2)1， x max ∴m≥1，则m≤1，····································································11分 2 令h(x)x ，易知g(x)在x[1， 2]单调递减，在x[ 2，2]单调递增， x 故h(x) h( 2)2 2 ，·······························································13分 min ∴m≤2 2，即m≥2 2 ，···························································14分 综上：2 2≤m≤1．·································································· 15分 17．解：（1）因为：当n≥2时，a 2a n2， n n1 所以：a n2[a (n1)](n2)，·················································3分 n n1 因为：a 12，············································································4分 1 所以数列{a n}是以2为首项，2为公比的等比数列，···························5分 n 所以：k 1，b0，a 2n n；·················································7分 n （2）由（1）可知：b a 2na  4nn2n ，·····································9分 n n n 设T 2222 323 n2n ① n 则2T 22 223324 n2n1 ②·············································10分 n 数学参考答案与评分标准 第2页，共6页由②−①整理得：T (222 23 2n)n2n1 (n1)2n12，···· 13分 n 4n1 2 ∴S 442434nT  (n1)2n1 ．··························15分 n n 3 3 2 4318b 18．解：（1）f(x)定义域为R，且图象关于点( ， )对称， 3 27 2 4318b ∴ f( ) ， 3 27 解得：c=1；····················································································3分 （2） f(x)3x24xbb(3x24x) ，3x24x[0，7]，······················4分 ①b≥7时， f(x)b(3x24x)≥0，此时f(x)在[−1，0]单调递增， ∴ f(x)  f(0)1；········································································5分 max ②b≤0时， f(x)b(3x24x)≤0，此时f(x)在[−1，0]单调递减， ∴ f(x)  f(1)4b ；··································································6分 max ③0b7时，存在x (1，0)，使得 f(x )b(3x 24x )=0 ， 0 0 0 0 且当x[1，x )时， f(x)0；当x(x，0]时， f(x)0， 0 0 即f(x)在区间[1，x )单调递减，在(x，0]单调递增， 0 0 4b,0b≤3， 此时 f(x) max{f(1)，f(0)}max{1，4b}  ·················· 7分 max 1，3b7， 4b，b≤3， 综上：h(b) ···································································8分 1，b3， 当b(，3]时，h(b)单调递减，此时h(b)的最小值为h(3)=1； 当b(3，)时，h(b)1，································································9分 综上所述：h(b)的最小值为1；··························································10分 （3）g(x) f(x)ax2m=x3(a2)x2bx1m ， a，x ，x 是函数g(x) f(x)ax2m的三个互异零点，即g(a)g(x )g(x ) ， 1 2 1 2 也即g(x)=g(a)的三个根是a，x ，x ，················································· 11分 1 2 代入得：x3(a2)x2bx1m=a3(a2)a2ba1m ， 数学参考答案与评分标准 第3页，共6页整理得：x3a3(a2)(x2a2)b(xa)0 ， ∴(xa)(x2axa2)(a2)(xa)(xa)b(xa)0 ， 即：(xa)[x2(2a2)x2a22ab]0 的三根是a，x ，x ， 1 2 所以x ，x 必然为方程x2(2a2)x2a22ab0的两个相异实根，·······13分 1 2 则(2a2)24(2a22ab)4(b1a2)0 ， 所以ba21，则b1，································································15分 又方程x2(2a2)x2a22ab0两个根都不同于a，则b5a24a， ∴对于b1，a( b1， b1) ，且b5a24a，使得a，x 1 ，x 2 是函数 g(x) f(x)ax2m 的三个互异零点，····················································16分 ∴b的取值范围为(1，)．····························································17分 19．解：（1） f(x)不为偶函数， 理由如下：若 f(x)为偶函数，则只需要 f(x) f(x)， 1 1 即ex  x3kx2 x1ex  x3kx2 x1恒成立，···························1分 6 6 1 即ex ex  x32x0恒成立， 3 而该等式显然对任意实数不恒成立， 故 f(x)不为偶函数；········································································3分 x2 （2）∵ f(x)ex 12kx ， 2 ∴ f(0)0且 f(x)ex 2k x ，则 f(x)ex 1， 又x0，故 f(x)0，∴ f(x)在(0，)上单调递增，······················4分 1 ①当k≤ 时， f(x) f(0)12k≥0，对x(0，)恒成立， 2 ∴ f(x)在(0，)上单调递增， ∴ f(x) f(0)0， 数学参考答案与评分标准 第4页，共6页∴ f(x)在(0，)上单调递增， f(x)0， ∴ f(x)在(0，)上无极值点，也没有零点，不满足题意；·····················5分 1 ②当k  时，f(0)12k 0，又 f(x)在(0，)上单调递增，且当x， 2 f(x)，因此x 0，使 f(x )0， 0 0 ∴当x(0，x )时，f(x)0，f(x)单调递减，当x(x ,)时，f(x)0，f(x) 0 0 单调递增，····························································································6分 ∴ f(x ) f(0)0又x时， f(x)， 0 ∴由零点存在性定理知：x (x ,)，使 f(x )0， 1 0 1 ∴当x(0，x )时，f(x)0，f(x)单调递减，当x(x，)时，f(x)0，f(x) 1 1 单调递增， ∴ f(x)在(0，)有唯一的极值点x ，················································7分 1 又 f(x ) f(0)0且，当x时， f(x)， 1 由零点存在性定理知：x (x，)，使 f(x )0， 2 1 2 ∴ f(x)在(0，)有唯一的零点x ，·················································· 8分 2 1 综上所述：k  ，满足题意；···························································9分 2 ex1 1 （ii）要证： f(2x x ) (x x )30，由 f(x )0， 1 2 3 2 1 2 ex1 1 即证： f(2x x ) f(x ) (x x )30 ， 1 2 2 3 2 1 ex1 1 即f[x (x x )] f[x (x x )] (x x )30 ，······················10分 1 2 1 1 2 1 3 2 1 令t  x x ，由（i）知t 0， 2 1 ex1 1 即证当t 0时， f(x t) f(x t) t30恒成立，······················11分 1 1 3 数学参考答案与评分标准 第5页，共6页ex1 1 令h(t) f(x t) f(x t) t3(t 0) ， 1 1 3 即证：h(t)0在t(0，)恒成立，注意到h(0)0， ∵h(t)f(x t) f(x t)(ex11)t2， 1 1 ∴h(t)ex1 t ex1 t 4kx 2x 2t2(ex1 1)t2 ，且h(0)0，···········12分 1 1 x2 又由 f(x )0，知ex1 12kx  1 0，·········································13分 1 1 2 x2 ∴ex1 2kx 1 1 ，∴h(t)ex1(et et 2t2)，且h(0)0， 1 2 h(t)ex1(et et 2t)，且h(0)0，············································ 14分 h(t)ex1(et et 2)≤e（x1 2 etet 2）0（*），······················15分 ∴从而（*）知h(t)在(0，)单调递减， 故h(t)＜h(0)0，······································································16分 ∴h(t)在(0，)单调递减，故h(t)＜h(0)0； h(t)在(0，)单调递减，故h(t)＜h(0)0， 故原不等式成立．···········································································17分 数学参考答案与评分标准 第6页，共6页