2027版高三《入学检测》（湖南省名校联合体）数学答案_2027版《高三上学期开学摸底模拟卷》全国多省（数学）

高三数学一参考答案
一、选择题（本大题共 8小题，每小题 5分，共 40分）
1．D 2．A 3．C 4．A 5．A 6．B 7．D 8．C
二、选择题（本大题共 3小题，每小题 6分，共 18分）
9．BD 10．CD 11．ACD
三、填空题（本大题共 3小题，每小题 5分，共 15分）
2
12． 13． 5+1 14．16
2
四、解答题（本大题共 5小题，共 77分．解答应写出文字说明、证明过程或演算步骤）
15．【解析】（1）由0.002×20+0.004×20+0.009×20+a×20+0.012×20+0.003×20=1，解得
a=0.02． ················································································································· （2分）
x =0.002×20×10+0.004×20×30+0.009×20×50+0.02×20×70+0.012×20×90+
0.003×20×110=68． ································································································ （4分）
（2）由频率分布直方图可知，在[
60,80
)，[
80,100
)的用户数之比为5∶3，
所以用分层抽样抽取的8人中，有3人是忠实粉丝， ··························································· （5分）
从8人中任取2人，X 取0，1，2，
C2 5 C1C1 15 C2 3
P ( X =0 )= 5 = ，P ( X =1 )= 5 3 = ，P ( X =2 )= 3 = ，·· （11分，算对一个得两分）
C2 14 C2 28 C2 28
8 8 8
所以X 的分布列为
X 0 1 2
5 15 3
P
14 28 28
3
所以X 的期望E ( X )= ． ························································································ （13分）
4
16．【解析】（1）证明：因为平面PAD⊥平面ABCD，交线为AD，AB⊂平面ABCD且
AB ⊥ AD，
所以AB⊥平面PAD． ································································································ （3分）
又DM ⊂平面PAD，所以DM ⊥ AB，
因为△PAD为等边三角形，且M 为PA的中点，所以DM ⊥ PA，
因为ABPA= A，且AB，PA⊂平面PAB，
所以DM ⊥平面PAB． ······························································································· （6分）
（2）取AD的中点O，连接PO，CO，由等边三角形性质知PO⊥ AD，
学科网（北京）股份有限公司由面面垂直性质可得PO⊥平面ABCD．
1
因为AO= AD=2= BC且AO//BC，
2
所以四边形ABCO为平行四边形，得OC//AB，即OC ⊥ AD． ·········································· （8分）
以O为原点，分别以OC，OD，OP为x， y，z 轴建立空间直角坐标系，
则各点坐标为：P ( 0,0,2 3 ) ，B ( 2,−2,0 )，C ( 2,0,0 )，D ( 0,2,0 )，A ( 0,−2,0 )． ············· （9分）
( )
因为M 为PA的中点，所以M 0,−1, 3 ，
  
( ) ( )
可得向量：PB= 2,−2,−2 3 ，DC =( 2,−2,0 )，DM = 0,−3, 3 ．

设平面MCD的法向量为n =( x,y,z )，


 n⋅DM =0, −3y+ 3z =0,
由

 得
n⋅DC =0,  2x−2y =0,
 ( )
解得x= y，z = 3y，取y =1，则n = 1,1, 3 ． ························································ （12分）
设直线PB与平面MCD所成角为θ，




PB⋅n
则sinθ=|cos〈PB,n〉|= 

，
PB ⋅ n
( )
2×1+(−2 )×1+ −2 3 × 3 −6 6 3
代入得sinθ= = = = ， ················ （14分）
22 +(−2)2 +(−2 3)2 × 12 +12 +( 3)2 2 5× 5 10 5
3
所以tanθ= ．
4
3
故直线PB与平面MCD所成角的正切值为 ． ······························································· （15分）
4
y−1
17．【解析】（1）设点H ( x,y )．由题意，x≠0，HA，HB的斜率分别为k = ，
HA x
y+1
k = ，
HB x
学科网（北京）股份有限公司1 y−1 y+1 1 y2 −1 1
代入k ⋅k = ，得 ⋅ = ⇒ = ， ······················································ （2分）
HA HB 4 x x 4 x2 4
( )
x2
整理得4 y2 −1 = x2 ⇒4y2 −4= x2 ⇒ y2 − =1．
4
这是焦点在y轴上的双曲线，其顶点为( 0,±1 )，即点A，B．
x2
因此曲线Γ的方程为y2 − =1． ·················································································· （5分）
4
（2）显然y >0，
Q
法一：若切线斜率存在，设切线CD的方程为y =kx+m，由CD与圆O相切得m2 =k2 +1． ··· （6分）
 y −m   y −m 
设点Q Q ,y ，D D ,y  ．
 k Q   k D 
y −m y −m
由OQ⊥OD可得 Q ⋅ D + y y =0，
k k Q D
y −m  −ky y −m 
解得y = Q ，故D Q , Q ． ································································ （8分）
D my −1  my −1 my −1 
 
Q Q Q
2 2
x2  y −m  1  −ky 
代入双曲线y2 − =1可得 Q  −  Q  =1， ············································ （11分）
4  my −1  4  my −1 
   
Q Q
5 2 5
展开化简得− y2 +1=0⇒ y = ． ······································································ （13分）
4 Q Q 5
 5  5
若CD斜率不存在，此时易得CD:x=−1或x=1，与双曲线联立解得D−1,− 或1,− ，
   
 2   2 
 
2 5
又由OQ⋅OD=0，点Q与点D的横坐标相同，解得y = ． ······································ （14分）
Q 5
 
( ) ( )
所以OQ⋅OF = x ,y ⋅ 0, 5 = 5y =2为定值． ····················································· （15分）
Q Q Q
法二：设点Q ( x ,y )，因为OQ⊥OD，则可令点D ( ky ,−kx )．
0 0 0 0
因为切线CD与单位圆相切，故O到直线CD的距离为1，
1 1
在Rt△QOD中，面积S = DO ⋅ OQ = DQ ×1，
2 2
1 1
即 k x2 + y2 ⋅ x2 + y2 = k2 +1⋅ x2 + y2 ， ·························································· （8分）
2 0 0 0 0 2 0 0
( )
化简得k2 x2 + y2 =k2 +1．① ····················································································· （9分）
0 0
学科网（北京）股份有限公司 y2 
又D ( ky ,−kx )在双曲线上，可得k2 x2 − 0 =1，② ··················································· （11分）
0 0  0 4 
5 2 5
①−②，得 y2 =1⇒ y = ， ·············································································· （13分）
4 0 0 5
 
2 5
所以OQ⋅OF = 5y = 5× =2为定值． ······························································· （15分）
0 5
法三：设点D ( x ,y )，C ( x ,y )，若切线斜率存在，设直线CD的方程为y =kx+m，
0 0 1 1
由CD与圆O相切得且m2 =k2 +1． ·············································································· （6分）
y =kx+m,
  1
联立 x2 得 k2 − x2 +2kmx+k2 =0，
y2 − =1,  4
 4
−8km 4k2
由韦达定理，得x +x = ，x x = ， ·························································· （8分）
0 1 4k2 −1 0 1 4k2 −1
 y −m 
由OQ⊥OD且点Q在直线y =kx+m上，设Q Q ,y ，
 k Q 
x x
代入垂直条件并结合y =kx +m，推导得y = 0 1 ， ··········································· （10分）
0 0 Q k
( x −x )
2 1 0
4(x x )2
平方得y2 = 0 1 ，
Q k2 ( x +x )2 −4x x 
 1 0 0 1
4 2 5
代入韦达定理及m2 =k2 +1化简可得y2 = ⇒ y = ， ············································ （13分）
Q 5 Q 5
 5  5
若CD斜率不存在，此时易得CD:x=−1或x=1，与双曲线联立解得D−1,− 或D1,− ，
   
2 2
   
 
2 5
又由OQ⋅OD=0，点Q与点D的横坐标相同，解得y = ． ······································ （14分）
Q 5
 
所以OQ⋅OF = 5y =2为定值． ··············································································· （15分）
Q
法四：显然切线CD的斜率不为0，设切线CD:x=my+b，
由CD与圆O相切得b2 =m2 +1． ·················································································· （6分）
设点Q ( my +b,y ) ，D ( my +b,y )，
Q Q D D
由OQ⊥OD可得 ( my +b )( my +b )+ y y =0，
Q D Q D
学科网（北京）股份有限公司− ( my +b )  y − ( my +b )
解得y = Q ，故D Q , Q ． ···················································· （8分）
D by +m by +m by +m 
Q  Q Q 
x2 − ( my +b ) 2 1  y  2
代入双曲线y2 − =1可得 Q  −  Q  =1， ······································ （11分）
4  by +m  4  by +m 
 Q   Q 
2 5
展开化简得−5y2 +4=0，解得y = （显然y >0，负值舍去）． ····························· （13分）
Q Q 5 Q
 
( ) ( )
所以OQ⋅OF = x ,y ⋅ 0, 5 = 5y =2为定值． ····················································· （15分）
Q Q Q
1
18．【解析】（1）当a =1时， f ( x )= xex − x3+x， f ( 0 )=0×e0 −0+0=0，故切点为( 0,0 )．
3
f′( x )=ex +xex −x2 +1，故 f′( 0 )=e0 +0×e0 −0+1=1+1=2．
切线斜率为2，得切线方程：y−0=2 ( x−0 )，
所以 f ( x )的图象在 ( 0, f ( 0 )) 处的切线方程为y =2x． ····················································· （4分）
（2）（ⅰ） f′( x )=( x+1 ) ex −ax2 +a=( x+1 )( ex −ax+a ) ，设g ( x )=ex −ax+a，
若a≤0，则g ( x )单调递增，即其至多有一个零点，即 f ( x )至多有两个极值点，不可能有三个极值点，
故a>0． ·················································································································· （5分）
当x≤0时，g ( x )>0恒成立，故x∈(−∞,−1 )时 f′( x )<0；
当x∈(−1,0 )时 f′( x )>0，即−1为 f ( x )的一个极小值点． ··············································· （7分）
又g′( x )=ex −a，则g ( x )在(−∞,lna )上单调递减，在( lna,+∞)上单调递增，
即g ( x )有最小值g ( lna )=2a−alna．
 lna>0,
f ( x )有三个极值点，即g ( x )有两个变号零点x ，x ，即 解得a∈ ( e2,+∞ ) ． ·· （9分）
2 3 2a−alna<0,
ex 2 −ax +a=0, ex 2 =a ( x −1 ) ,
（ⅱ）由（ⅰ）知x =−1，0< x <lna< x ，且 2 即 2 ······· （10分）
1 2 3 ex 3 −ax +a=0, ex 3 =a ( x −1 ) .
3 3
1 1 1 1 1
要证 + + >0，即证 + −1>0， ································································· （11分）
x x x x x
1 2 3 2 3
即证x +x −x x >0，即证x x −x −x +1<1，即( x −1 )( x −1 )<1． ························· （13分）
2 3 2 3 2 3 2 3 2 3
即证ex 2 +x 3 <a2，即证x +x <2lna． ·········································································· （15分）
2 3
设G ( x )= g ( x )−g ( 2lna−x )，
学科网（北京）股份有限公司a2
则G′( x )= g′( x )+g′( 2lna−x )=ex −a+e2lna−x −a=ex + −2a≥2a−2a=0． ·············· （16分）
ex
则G ( x )单调递增，而G ( lna )=0，则G ( x )<0，即g ( x )= g ( x )< g ( 2lna−x )，x <2lna−x ，
2 3 2 2 3 2
1 1 1
即x +x <2lna，即 + + >0证毕． ································································· （17分）
2 3 x x x
1 2 3
19．【解析】（1）在原递推式中以a 作主元，可因式分解得( a −2a )( a −a −2 )=0．
n+1 n+1 n n+1 n
因此可得a =2a ，或a =a +2．（亦可直接用求根公式） ·········································· （2分）
n+1 n n+1 n
已知a =2，则a =2a =4或a +2=4，即a ∈{ 4 }；
1 2 1 1 2
则a 可能为2a =8或a +2=6，即a ∈{ 6,8 }；
3 2 2 3
则a 的可能取值为：若a =6，则a ∈{ 8,12 }；
4 3 4
若a =8，则a ∈{ 10,16 }，
3 4
所以a ∈{ 8,10,12,16 }．
4
又因为i =3 ( a =2,a =4,a =8 )，故k的所有可能取值为10，12，16． ··························· （4分）
8 1 2 3
（2）一方面，构造数列{
a
}如下：
r
 2r,1≤r ≤n− j,
a =
r 2r−1+2r−n+j,n− j+1≤r ≤n,
显然符合原式且此时a =2n−1+2j，即i ≤n． ································································ （6分）
i k
k
另一方面，若i ≤n−1，则k ≤2n−1，但k =2n−1+2j，显然k >2n−1，矛盾．
k
因此只能有i =n． ···································································· （8分，用其他方法证明亦可．）
k
构造方法亦可写成：取a =2a ( l ≠n− j )，a =a +2 ( l =n− j )．
l+1 l l+1 l
（3）易知当k为奇数时，i =0．以下讨论k为偶数时的情形．
k
n−1
设2n−1 ≤k <2n ( n∈N* ) ，且记k =∑b ⋅2i ，其中b =1，b ∈{ 0,1 }( i =1,2,,n−2 )．
i n−1 i
i=1
n−1
记ω( k )=∑b ．下面尝试证明：i =n+ω( k )−2． ······················································ （10分）
i k
i=1
a
考虑c = n ，则c =1，且满足：c =2c 或c =c +1．
n 2 1 n+1 n n+1 n
k k k
则显然i 是满足c = 的i的最小值，此时 ∈N*，且2n−2 ≤ <2n−1，
k i 2 2 2
学科网（北京）股份有限公司k
由上述分解，可将 写为唯一的二进制格式，即(b b b) ，
2 n−1 n−2 1 2
下考虑将该二进制数从左往右写出，从1开始（显然b =1），每写一个数字即对应一个新的二进制数，
n−1
k
直到得到 的二进制数．
2
每多写一个0，对应十进制中即为给上一个数乘2，
每多写一个1，对应十进制中即为给上一个数乘2再加1．
则从(1) 到(b b b) 的过程即可等价于c 到c 的过程（显然这一路径具有最小性），
2 n−1 n−2 1 2 1 i
k
写0时，相当于进行一步递推；写1时，相当于进行两步递推，故总递推步数为：
k 
“该二进制数的位数”−1+“二进制数中1的个数”−1，即n−3+ω
 
，
2
k 
则i =n−3+ω   +1=n+ω( k )−2． ······································································· （13分）
k 2
（用其他方法完成该结论的证明亦可，若提出该结论而未证明，扣除证明过程的相应分数）
当2n−1 ≤k <2n时，满足ω( k )=s的k的个数为Cs−1 ( 1≤s≤n−1 )，
n−2
这是因为最高位b =1固定，其余n−2位中恰有s−1个1． ············································· （14分）
n−1
2n−1 n−1 n−1 n−1
于是 ∑ i =∑Cs−1 ( n+s−2 )=( n−1 )∑Cs−1 +∑( s−1 ) Cs−1
k n−2 n−2 n−2
k=2n−1 s=1 s=1 s=1
=( n−1 ) 2n−2 +( n−2 ) 2n−3 =2n−3( 3n−4 )． ·································································· （15分）
2n n
故∑i
k
=∑

2t−3( 3t−4 )

+n．
k=1 t=2
n
记A=∑

2s−3( 3s−4 )

，
s=2
n
则2A=∑2s−2( 3s−4 )，
s=2
 n 
相减得−A=2−3

8+3∑2s −2n+1( 3n−4 )

=−2−2n−2( 3n−7 )，
 
s=3
故A=2n−2( 3n−7 )+2， ··························································································· （16分）
2n
因此∑i =2n−2( 3n−7 )+2+n． ················································································ （17分）
k
k=1
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