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Question 1 · Free response · 中档
(5分)函数 $\mathbf { y } { = } \sqrt { \mathbf { x } \left( \mathbf { x } { - } 1 \right) } + \sqrt { \mathbf { x } }$ 的定义域为( )A $\{ \mathbf { x } | \mathbf { x } { \ge } 0 \}$ B $\{ \mathbf { x } | \mathbf { x } \ge 1 \}$ C $\{ \mathbf { \boldsymbol { x } } | \mathbf { \boldsymbol { x } } { \geq } 1 \} \cup \{ \mathbf { \boldsymbol { 0 } } \}$ D $\{ { \bf x } | 0 { \leq } { \bf x } { \leq } 1 \}$
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Question 3 · Free response · 中档
(5分)在 $\triangle A B C$ 中, $\overrightarrow { A B } = \overrightarrow { c } , \overrightarrow { A C } = \overrightarrow { b } .$ .若点D满足 $\overrightarrow { B D } = 2 \overrightarrow { D C }$ ,则 $\overrightarrow { \mathrm { A D } } =$ )A $\frac { 2 } { 3 } \overrightarrow { b } + \frac { 1 } { 3 } \overrightarrow { c }$ B $\frac { 5 } { 3 } \overrightarrow { \circ } - \frac { 2 } { 3 } \overrightarrow { \mathrm { b } }$ C $\frac { 2 } { 3 } \overrightarrow { b } - \frac { 1 } { 3 } \overrightarrow { c }$ D $\frac { 1 } { 3 } \overrightarrow { b } + \frac { 2 } { 3 } \overrightarrow { \mathrm { c } }$
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Question 4 · Single choice · 中档
(5分)设a∈R,且(a+i)<sup>2</sup>i为正实数,则$a=$( )
A. 2 B. 1 C. 0 D. ﹣1
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Question 5 · Single choice · 中档
(5分)已知等差数列 $\{ \mathsf { a } _ { \mathsf { n } } \}$ 满足 $a _ { 2 } + a _ { 4 } = 4 , ~ \mathsf { a } _ { 3 } + \mathsf { a } _ { 5 } = 1 0$ ,则它的前10项的和 $\mathsf { I S } _ { 1 0 } \mathrm { = }$ ( )
A. 138 B. 135 C. 95 D. 23
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Question 6 · Free response · 中档
(5分)若函数y=f(x)的图象与函数 $\scriptstyle \mathbf { y } = \ln { \sqrt { \mathbf { x } } } + 1$ 的图象关于直线y=x对称,则f(x)=( )A ${ \tt e } ^ { 2 { \tt x } \mathrm { ~ - ~ } 2 }$ B. $\mathsf { e } ^ { 2 \mathsf { x } }$ C ${ \mathsf { e } } ^ { 2 \mathsf { x } + 1 }$ D ${ \mathsf { e } } ^ { 2 \mathbf { x } + 2 }$
Answer key See explanation
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Question 9 · Free response · 中档
(5分)设奇函数f(x)在 $( 0 , \ + \infty )$ 上为增函数,且f(1)=0,则不等式$\frac { \underline { { \hat { \mathbf { f } } } } ( \mathbf { x } ) - \underline { { \hat { \mathbf { f } } } } ( - \mathbf { x } ) } { \mathbf { x } } < 0$ 的解集为( )
A $\left( ~ - ~ 1 , ~ 0 \right) \cup \left( ~ 1 , ~ + \infty \right)$ B $( \mathbf { \Sigma } - \infty , \mathbf { \Sigma } - \mathbf { \Sigma } _ { 1 } ) \cup \mathbf { \Sigma } ( 0 , \mathbf { \Sigma } _ { 1 } )$ C $\left( \mathbf { \Sigma } - \infty , \mathbf { \Sigma } - \mathbf { 1 } \right) \cup \left( \mathbf { 1 } , \mathbf { \Sigma } + \infty \right)$ D $( \mathbf { \Sigma } - \mathbf { \Sigma } 1 , \mathbf { \Sigma } 0 ) \cup \mathbf { \Sigma } ( 0 , \mathbf { \Sigma } 1 )$
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Question 10 · Free response · 中档
(5分)若直线 $\frac { x } { a } + \frac { y } { b } = 1$ 与圆 $\mathsf { x } ^ { 2 } + \mathsf { y } ^ { 2 } = 1$ 有公共点,则(A $a ^ { 2 } + b ^ { 2 } \leq 1$ B $a ^ { 2 } + b ^ { 2 } \geq 1$ C $\frac { 1 } { { \mathsf { a } } ^ { 2 } } + \frac { 1 } { { \mathsf { b } } ^ { 2 } } \lneq 1$ D $\frac { 1 } { { \mathsf { a } } ^ { 2 } } + \frac { 1 } { { \mathsf { b } } ^ { 2 } } \gg 1$
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Question 11 · Free response · 中档
(5分)已知三棱柱 $\mathsf { A B C } - \mathsf { A } _ { 1 } \mathsf { B } _ { 1 } \mathsf { C } _ { 1 }$ 的侧棱与底面边长都相等, $\mathsf { A } _ { 1 }$ 在底面ABC内的射影为 $\triangle A B C$ 的中心,则 $\mathsf { A B } _ { 1 }$ 与底面ABC所成角的正弦值等于( )
A $\frac 1 3$ B $\frac { \sqrt { 2 } } { 3 }$ C $\frac { \sqrt { 3 } } { 3 }$ D $\frac { 2 } { 3 }$
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Question 12 · Single choice · 中档
(5分)如图,一环形花坛分成A,B,C,D四块,现有4种不同的花供选种,要求在每块里种1种花,且相邻的2块种不同的花,则不同的种法总数为()
A. 96 B. 84 C. 60 D. 48
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Question 14 · Free response · 中档
(5分)已知抛物线 $\mathsf { y } = \mathsf { a } \mathsf { x } ^ { 2 } - 1$ 的焦点是坐标原点,则以抛物线与两坐标轴的三个交点为顶点的三角形面积为 2
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Question 16 · Free response · 中档
(5分)等边三角形ABC与正方形ABDE有一公共边AB,二面角C﹣AB﹣D的 余弦值为 $\frac { \sqrt { 3 } } { 3 }$ ,M,N分别是AC,BC的中点,则EM,AN所成角的余弦值等于 $\frac { 1 } { 6 } -$
Answer key $\frac { 1 } { 6 }$
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Question 18 · Free response · 中档
(12分)四棱锥 $\mathsf { A } ^ { \mathrm { ~ - ~ } } \mathsf { B C D E }$ 中,底面BCDE为矩形,侧面ABC⊥底面BCDE,BC=2$C D = { \sqrt { 2 } } , A B = A C$
(Ⅰ)证明: $\mathsf { A D \bot C E }$
(Ⅱ)设CE与平面ABE所成的角为45°,求二面角C﹣AD﹣E的大小
Answer key See explanation
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Question 20 · Free response · 中档
(12分)已知5只动物中有1只患有某种疾病,需要通过化验血液来确定患病的动物.血液化验结果呈阳性的即为患病动物,呈阴性即没患病.下面是两种化验方法:
方案甲:逐个化验,直到能确定患病动物为止
方案乙:先任取3只,将它们的血液混在一起化验.若结果呈阳性则表明患病动物为这3只中的1只,然后再逐个化验,直到能确定患病动物为止;若结果呈阴性则在另外2只中任取1只化验
(Ⅰ)求依方案甲所需化验次数不少于依方案乙所需化验次数的概率;
(Ⅱ)ξ表示依方案乙所需化验次数,求ξ的期望
Answer key (1) $\frac { 1 8 } { 2 5 }$; (2) 2 . 4
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