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Question 1 · Free response · 中档
(5分)已知集合 $\mathbf { \delta A } { = } \{ { \mathbf { x } } | { \mathbf { x } } ^ { 2 } - { \mathbf { x } } - 2 { < } 0 \}$ $\mathsf { B } = \{ \mathsf { x } | - 1 { < } \mathsf { x } { < } 1 \}$ ,则( )A $\mathsf { A } \subsetneq \mathsf { B }$ B $\mathsf { B } \subsetneq \mathsf { A }$ C.A=B D.A∩B=∅
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Question 2 · Single choice · 中档
(5分)复数 $z = \frac { - 3 + i } { 2 + i }$ 的共轭复数是( )
A. 2+i B. 2﹣i C. ﹣1+i D. ﹣1﹣i
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Question 4 · Free response · 中档
(5分)设 ${ \sf F } _ { 1 } , \ { \sf F } _ { 2 }$ 是椭圆E: $\frac { x ^ { 2 } } { a ^ { 2 } } + \frac { y ^ { 2 } } { b ^ { 2 } } = 1 ( a > b > 0 )$ 的左、右焦点,P为直线 $\pmb { \mathrm { x } } \pmb { = }$ $\frac { 3 \tt a } { 2 }$ 上一点, $\triangle \mathsf { F } _ { 2 } \mathsf { P } \mathsf { F } _ { 1 }$ 是底角为 $3 0 ^ { \circ }$ 的等腰三角形,则E的离心率为(A $\frac { 1 } { 2 }$ B $\frac { 2 } { 3 }$ C $\frac { 3 } { 4 }$ D $\frac { 4 } { 5 }$
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Question 5 · Single choice · 中档
(5分)已知正三角形ABC的顶点A(1,1),B(1,3),顶点C在第一象限,若点 $( { \bf { x } } , { \bf { \Lambda } } { \bf { y } } )$ 在 $\triangle A B C$ 内部,则 ${ \sf z } = - { \sf x } ^ { + } { \sf y }$ 的取值范围是( )
A. ( $\left( 1 - { \sqrt { 3 } } , \ 2 \right)$ B. (0,2) C. ( $( { \sqrt { 3 } } - 1 , ~ 2 )$ D. (0, $\scriptstyle 1 + { \sqrt { 3 } } )$
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Question 6 · Single choice · 中档
(5分)如果执行右边的程序框图,输入正整数N(N≥2)和实数 $\mathsf { a } _ { 1 }$ , $\mathtt { a } _ { 2 }$ ,…, $\mathsf { a } _ { \mathsf { n } } ,$ ,输出A,B,则( )
A. $\mathsf { A } ^ { + } \mathsf { B }$ 为 $\mathsf { I } _ { 1 }$ , $\mathsf { a } _ { 2 } , \ldots , \mathsf { a } _ { \mathsf { n } }$ 的和 B. $\textstyle { \frac { A + B } { 2 } }$ 为 $\mathsf { a } _ { 1 }$ , $\mathsf { a } _ { 2 } , \ldots , \mathsf { a } _ { \mathsf { n } }$ 的算术平均数 C. A和B分别是 $\mathsf { a } _ { 1 } , ~ \mathsf { a } _ { 2 } , ~ . . . , ~ \mathsf { a } _ { \mathsf { n } }$ 中最大的数和最小的数 D. A和B分别是 $\mathsf { a } _ { 1 }$ , $\mathsf { a } _ { 2 } , \ldots , \mathsf { a } _ { \mathsf { n } }$ 中最小的数和最大的数
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Question 7 · Single choice · 中档
(5分)如图,网格纸上小正方形的边长为1,粗线画出的是某几何体的三视图,则此几何体的体积为( )
<table><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td colspan="2"></td><td></td><td></td><td colspan="2"></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr><tr><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td><td></td></tr></table>
A. 6 B. 9 C. 12 D. 18
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Question 8 · Free response · 中档
(5分)平面α截球O的球面所得圆的半径为1,球心O到平面α的距离为 ${ \sqrt { 2 } } ,$ 则此球的体积为( )A $\sqrt { 6 } \pi$ B $4 \sqrt { 3 } \pi$ C $4 \sqrt { 6 } \pi$ D $6 \sqrt { 3 } \pi$
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Question 9 · Free response · 中档
(5分)已知 $\scriptstyle 1 \omega > 0 , 0 < \phi < \pi$ ,直线 $x = \frac { \pi } { 4 }$ 和 $\scriptstyle x = { \frac { 5 \pi } { 4 } }$ 是函数f $( \mathbf { x } ) = \sin ( \omega \mathbf { x } + \phi )$ 图象的两条相邻的对称轴,则φ=(A $\frac { \pi } { 4 }$ B $\frac { \pi } { 3 }$ C $\frac { \pi } { 2 }$ D $\frac { 3 \pi } { 4 }$
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Question 10 · Free response · 中档
(5分)等轴双曲线C的中心在原点,焦点在x轴上,C与抛物线 $\mathsf { y } ^ { 2 } \mathop { = } \mathtt { 1 } 6 \mathsf { x }$ 的准线交于点A和点B, $\vert A B \vert = 4 \sqrt { 3 }$ ,则C的实轴长为( )A $\sqrt { 2 }$ B $2 \sqrt { 2 }$ C.4 D.8
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Question 12 · Single choice · 中档
(5分)数列 $\{ \mathsf { a } _ { \mathsf { n } } \}$ 满足 ${ \sf a } _ { \sf n + 1 } + \mathrm { ~ ~ \left( ~ - ~ \right)} 1 { \sf n } _ { { \sf a } _ { \sf n } = 2 \sf n } - 1$ ,则 $\{ \mathsf { a } _ { \mathsf { n } } \}$ 的前60项和为()
A. 3690 B. 3660 C. 1845 D. 1830
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Question 16 · Free response · 中档
(5分)设函数f $( \mathbf { x } ) = \frac { ( \mathbf { x } + 1 ) \mathbf { \Sigma } ^ { 2 } + \mathsf { s i n } \mathbf { x } } { \mathbf { \Sigma } _ { \mathbf { x } } \mathbf { \Sigma } ^ { 2 } + 1 }$ 的最大值为M,最小值为m,则 $\mathsf { M } + \mathsf { m }$ $= 2$
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Question 18 · Free response · 中档
(12分)某花店每天以每枝5元的价格从农场购进若干枝玫瑰花,然后以每枝10元的价格出售.如果当天卖不完,剩下的玫瑰花做垃圾处理
(Ⅰ)若花店一天购进17枝玫瑰花,求当天的利润y(单位:元)关于当天需求量n(单位:枝,n∈N)的函数解析式
(Ⅱ)花店记录了100天玫瑰花的日需求量(单位:枝),整理得如表:
<table><tr><td rowspan=1 colspan=1>日需求量n</td><td rowspan=1 colspan=1>14</td><td rowspan=1 colspan=1>15</td><td rowspan=1 colspan=1>16</td><td rowspan=1 colspan=1>17</td><td rowspan=1 colspan=1>18</td><td rowspan=1 colspan=1>19</td><td rowspan=1 colspan=1>20</td></tr><tr><td rowspan=1 colspan=1>频数</td><td rowspan=1 colspan=1>10</td><td rowspan=1 colspan=1>20</td><td rowspan=1 colspan=1>16</td><td rowspan=1 colspan=1>16</td><td rowspan=1 colspan=1>15</td><td rowspan=1 colspan=1>13</td><td rowspan=1 colspan=1>10</td></tr></table>
(i)假设花店在这100天内每天购进17枝玫瑰花,求这100天的日利润(单位:元)的平均数;
(ii)若花店一天购进17枝玫瑰花,以100天记录的各需求量的频率作为各需求量发生的概率,求当天的利润不少于75元的概率
Answer key See explanation
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Question 19 · Free response · 中档
(12分)如图,三棱柱 $\mathsf { A B C } - \mathsf { A } _ { 1 } \mathsf { B } _ { 1 } \mathsf { C } _ { 1 }$ 中,侧棱垂直底面, $\angle A C B = 9 0 ^ { \circ }$ $A C = B C$ $\mathsf { \Lambda } = \frac { 1 } { 2 } \mathsf { A } \mathsf { A } _ { 1 }$ ,D是棱 $\mathsf { A A } _ { 1 }$ 的中点
(Ⅰ)证明:平面 ${ \mathsf { B D C } } _ { 1 } \bot$ 平面BDC
(Ⅱ)平面 $\mathsf { B D C } _ { 1 }$ 分此棱柱为两部分,求这两部分体积的比
Answer key See explanation
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Question 22 · Free response · 中档
(10分)如图,D,E分别为△ABC边AB,AC的中点,直线DE交△ABC的外接圆于F,G两点,若CF∥AB,证明:
①CD=BC;
② $\triangle B C D \sim \triangle G B D$
Answer key (1) 证明:①∵D,E分别为 $\triangle A B C$ 边AB,AC的中点 ∴DF∥BC,AD=DB ∵AB∥CF,∴四边形BDFC是平行四边形 ∴CF∥BD,CF=BD ∴CF∥AD,CF=AD ∴四边形ADCF是平行四边形 ∴AF=CD ∵ ,∴BC=AF,∴CD=BC; (2) ②由①知 $\widehat { B C } = \widehat { A F }$ ,所以 $\widehat { B F } = \widehat { A C }$ 所以 $\angle B G D = \angle D B C$ 因为GF∥BC,所以 $\angle B D G = \angle A D F = \angle D B C = \angle B D C$ 所以 $\Delta B C D \sim \Delta G B D$
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Question 23 · Free response · 中档
已知曲线$C_1$的参数方程是$\begin{cases}x=2\cos\phi\\y=3\sin\phi\end{cases}$($\phi$为参数),以坐标原点为极点,$x$轴的正半轴为极轴建立坐标系,曲线$C_2$的坐标系方程是$\rho=2$,正方形$ABCD$的顶点都在$C_2$上,且$A$,$B$,$C$,$D$依逆时针次序排列,点$A$的极坐标为$\left(2,\dfrac{\pi}{3}\right)$. (1)求点$A$,$B$,$C$,$D$的直角坐标; (2)设$P$为$C_1$上任意一点,求$|PA|^2+|PB|^2+|PC|^2+|PD|^2$的取值范围.
Answer key (1)$(1,\sqrt{3})$,$(-\sqrt{3},1)$,$(-1,-\sqrt{3})$,$(\sqrt{3},-1)$;(2)$[32,52]$
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