2024届四川省绵阳市高三上学期第一次诊断性考试 理数答案
绵阳市高中 2021 级第一次诊断性考试
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.
BCDAC ADBBD CC
二、填空题:本大题共 4小题,每小题 5分,共 20 分.
13.7 14. 15.9
16.1
三、解答题:本大题共 6小题,共 70 分.
17.解:(1)由 a1,a2,a4成等比数列,则 ,································2 分
∴,
可解得 ,··················································································3 分
∴数列{an}的前 项和 ;·······························5 分
(2) ①,···············································6 分
当 时, ,可得 ,·······················································7 分
可得 ②,·····································································8 分
由②式-①式,得 ,············································9 分
∴
······································································11 分
.························································································12 分
1
8.解:(1)∵ ,则 ,·····················································1 分
又
,··························································2 分
∴
,·······················································································4 分
∴
;·······································································5 分
(2)
由题意, ,···················································6 分
∵
·································································7 分
∴
·····································8 分
∴
Z,·····························································9 分
∴
,····················································10 分
∴的最小值为 .·········································································12 分
19.解:(1)∵ 为奇函数,
∴,解得:m=2.··························································5 分
(2)当 m>0 时,2x2+m>0 ,
∴函数 不可能有两个零点.·······························6 分
当m <0 时,由 ,解得: 或 m2,·································7 分
要使得 f(x)仅有两个零点,则 ,·········································8 分
即 ,此方程无解.
故m=0,即 ,······························································9 分
令 ,则 ,
,解得: 或 , 解得: ,
故 在 , 上递增,在 上递减,···························10 分
又 ,
故函数 仅有一个零点.························································12 分
20.解:(1)∵cos(CB)sinA=cos(CA)sinB
∴
(cosCcosB+sinCsinB)sinA=(cosCcosA+sinCsinA)sinB································2 分
∴cosCcosBsinA=cosCcosAsinB······························································3 分
又∵△ABC 为斜三角形,则 cosC≠0,
cos∴BsinA=cosAsinB,········································································5 分
sin(∴AB)=0,又 A,B为△ABC 的内角,
∴A=B;··························································································6 分
(2)
由△ABC 的面积 S= ,
∴S= absinC= ,则 bsinC=1,即 =sinC,·········································7 分
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