南京市2020-2021学年度第一学期期末调研测试高一数学答案(终稿)

南京市2020－2021学年度第一学期期末学情调研 高一数学参考答案 2021.01
一、单项选择题
1．C 2．B 3．A 4．D 5．B 6．C 7．D 8．A
二、多项选择题
9．AB 10．BD 11．ACD 12．AB
三、填空题
13．12
14．4 15．119 16．32 四、解答题
17．（本小题满分10分）
解：（1）由 2x ＋1x －1＜1，得 x ＋2x －1
＜0，所以A ＝{x |－2＜x ＜1}． B ＝{x |2x 2＋(m －2)x －m ＜0}＝{x |(x －1)(2x ＋m )＜0}．
当m ＝1时，B ＝{x |－12
＜x ＜1}． ·························································· 3分 所以A ∪B ＝{x |－2＜x ＜1}． ······························································· 4分
（2）因为“x ∈A ”是“x ∈B ”的必要条件，所以B －⊂A ． ······························· 6分
若－m 2
＞1，不符合题意； ···································································· 7分 若－m 2
＝1即m ＝－2时，B ＝∅，符合题意； ··········································· 8分 若－m 2＜1，则B ＝{x |－m 2
＜x ＜1}， 所以－2≤－m 2
＜1，解得－2＜m ≤4．····················································· 9分 综上，m ∈[－2，4]． ········································································· 10分
18．（本小题满分12分）
解：（1）因为sin (π＋α)cos (π－α)＝sin αcos α，且sin (π＋α)cos (π－α)＝18
， 所以sin αcos α＝18
． ············································································· 2分 故 (cos α－sin α)2＝cos 2α－2sin αcos α＋sin 2α
＝1－2sin αcos α＝1－2×18＝34
． ········································ 4分 又因为0＜α＜π4
，所以cos α＞sin α，即cos α－sin α＞0， 所以cos α－sin α＝32
． 所以cos α＋cos(π2
＋α)＝cos α－sin α＝32． ················································ 6分
（2）法一：由（1）知sin αcos α＝18
，又因为sin 2α＋cos 2α＝1， 所以 sin αcos αsin 2α＋cos 2α＝18． 因为0＜α＜π4
，cos α≠0， 所以tan α tan 2α＋1＝18
，即tan 2α－8tan α＋1＝0， ············································ 9分 解得tan α＝4－15或tan α＝4＋15． ··················································· 10分
因为0＜α＜π4
，所以0＜tan α＜1， 所以tan α＝4－15． ········································································· 12分
法二： 由（1）知⎩
⎨⎧cos α－sin α＝32，sin αcos α＝18
． 因为0＜α＜π4，所以cos α＞sin α＞0， 故⎩⎪⎨⎪⎧ cos α＝3＋5 4，sin α＝－3＋54， ······························································ 10分
所以tan α＝sin αcos α
＝4－15． ································································· 12分 19．（本小题满分12分）
解：（1）原式＝5＋[(2)－3]－23
＋log 3( 3)4
＝5＋4＋4
＝13． ·
·············································································· 4分 （2）法一：因为y ＝log 0.4x 在(0，＋∞)上递减，y ＝log 4x 在(0，＋∞)上递增，
所以a ＝log 0.43＜log 0.41＝0，b ＝log 43＞log 41＝0，
故ab ＜0． ························································································ 6分
因为1a ＋1b
＝log 30.4＋log 34＝log 3(0.4×4)＝log 31.6， 且y ＝log 3x 在(0，＋∞)递增，
所以0＝log 31＜log 31.6＜log 33＝1，即0＜1a ＋1b
＜1． ································· 10分 所以0＞ab (1a ＋1b
)＞ab ，即ab ＜a ＋b ＜0． ·············································· 12分 法二：因为a ＝log 0.43，b ＝log 43，
所以a ＋b ＝log 0.43＋log 43＝lg3lg0.4＋lg3lg4＝lg3×lg4＋lg0.4lg0.4·lg4＝lg3×lg1.6lg0.4·lg4
， 因为lg3＞0，lg4＞0，lg1.6＞0，lg0.4＜0，
所以a ＋b ＜0． ·················································································· 6分
(a ＋b )－ab ＝lg3×lg1.6lg0.4·lg4－lg3lg0.4×lg3lg4＝lg3×lg1.6－lg3lg0.4·lg4 ＝lg3×lg 1.63lg0.4·lg4＝lg3×lg 815lg0.4·lg4
． ········································ 10分 因为lg3＞0，lg4＞0，lg 815
＜0，lg0.4＜0， 所以(a ＋b )－ab ＞0，即a ＋b ＞ab ，
综上，ab ＜a ＋b ＜0． ········································································ 12分
解：（1）因为函数f (x )＝x |x －a |为R 上的奇函数，
所以f (－x )＝－f (x ) 对任意x ∈R 成立，
即(－x )·|－x －a |＝－x ·|x －a |对任意x ∈R 成立， ···································· 2分 所以|－x －a |＝|x －a |，所以a ＝0． ····················································· 4分
（2）由f (sin 2x )＋f (t －2cos x )≥0得f (sin 2x )≥－f (t －2cos x )，
因为函数f (x )为R 上的奇函数， 所以f (sin 2x )≥f (2cos x －t )． ························ 6分
由（1）得，f (x )＝x |x |＝⎩⎨⎧x 2，x ≥0，－x 2，x ＜0，
是R 上的单调增函数， 故sin 2x ≥2cos x －t 对任意x ∈[π3，7π6
]恒成立． ·········································· 8分 所以t ≥2cos x －sin 2x 对任意x ∈[π3，7π6
]恒成立． 因为2cos x －sin 2x ＝cos 2x ＋2cos x －1＝(cos x ＋1)2－2，
令m ＝cos x ，由x ∈[π3，7π6]，得cos x ∈[－1，12]，即m ∈[－1，12
]． ·············· 10分 所以y ＝(m ＋1)2－2的最大值为14，故t ≥14
， 即t 的最小值为14
． ············································································ 12分 21．（本小题满分12分）
解：（1）因为小球振动过程中最高点与最低点的距离为10 cm ，所以A ＝102
＝5． ·· 2分 因为在一次振动中，小球从最高点运动至最低点所用时间为1 s ，所以周期为2，
即T ＝2＝2πω
，所以ω＝π． ··································································· 4分 所以h ＝5sin(πt ＋π4
)，t ≥0． ································································· 5分 （2）由题意，当t ＝14
时，小球第一次到达最高点， 以后每隔一个周期都出现一次最高点， ··················································· 7分 因为小球在t 0 s 内经过最高点的次数恰为50次，
所以14＋49T ≤t 0＜14
＋50T ． ··································································· 9分 因为T ＝2，所以9814≤t ＜10014
， 所以t 0的取值范围为[9814，10014
)． ······················································· 12分 （注：t 0的取值范围不考虑开闭）
解：（1）当a ＝－2时，f (x )＝－2x 2＋1．
方程f (x )＝x 可化为2x 2＋x －1＝0，解得x ＝－1或x ＝12
， 所以f (x )的不动点为－1和 12
． ······························································ 2分 （2）①因为函数f (x )有两个不动点x 1，x 2，
所以方程f (x )＝x ，即ax 2－x ＋1＝0的两个实数根为x 1，x 2，
记p (x )＝ax 2－x ＋1，则p (x )的零点为x 1和x 2，
因为x 1＜2＜x 2，所以a ·p (2)＜0，
即a (4a －1)＜0，解得0＜a ＜14
． 所以实数a 的取值范围为(0，14
)． ·························································· 6分 ②因为g (x )＝log a [f (x )－x ]＝log a (ax 2－x ＋1)．
方程g (x )＝x 可化为log a (ax 2－x ＋1)＝x ，即⎩⎨⎧a x ＝ax 2－x ＋1，
ax 2－x ＋1＞0．
因为0＜a ＜14
，△＝1－4a ＞0，所以p (x )＝0有两个不相等的实数根． 设p (x )＝ax 2－x ＋1＝0的两个实数根为m ，n ，不妨设m ＜n ．
因为函数p (x )＝ax 2－x ＋1图象的对称轴为直线x ＝12a ，p (1)＝a ＞0，12a ＞1，p (1a
)＝1＞0， 所以1＜m ＜12a ＜n ＜1a
． 记h (x )＝a x －(ax 2－x ＋1)，
因为h (1)＝0，且p (1)＝a ＞0，所以x ＝1是方程g (x )＝x 的实数根，
所以1是g (x )的一个不动点． ······························································· 8分 h (n )＝a n －(an 2－n ＋1)＝a n ＞0，
因为0＜a ＜14，所以1a ＞4，h (1a
)＝a 1a －1＜a 4－1＜0， 且h (x )的图象在[n ，1a
]上的图象是不间断曲线， 所以∃x 0∈(n ，1a
)，使得h (x 0)＝0， ························································· 10分 又因为p (x )在(n ，1a
)上单调递增，所以p (x 0)＞p (n )＝0， 所以x 0是g (x )的一个不动点，
综上，g (x )在(a ，＋∞)上至少有两个不动点． ········································· 12分。