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Question 1 · Free response · 中档
集合A $ \{ x \vert - 1 \leq x \leq 2 \}$ $\mathrm { B } = \left\{ x | x < 1 \right\}$ ,则 $A \cap ( C _ { R } B )$ =【D】(A) $\{ x | x > 1 \}$ (B) $\{ x \mid x \geq 1 \}$ (C) $\left\{ x \vert 1 < x \leq 2 \right\}$ (D) $\left\{ x \vert 1 \leq x \leq 2 \right\}$
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Question 3 · Single choice · 中档
对于函数$f(x)=2sinxcosx$,下列选项中正确的是 【B】
A. f(x)在 $\frac { \pi } { 4 } , \frac { \pi } { 2 } )$ 上是递增的 B. f(x)的图象关于原点对称 C. f(x)的最小正周期为 $2 \pi$ D. f(x)的最大值为2
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Question 5 · Single choice · 中档
已知函数 若f $\mathrm { ~ ( ~ f ~ ~ ( ~ 0 ~ ) ~ ) ~ } { = } 4 \mathrm { a }$ ,则实数a等于【C】
A. $\frac { 1 } { 2 }$ B. $\frac { 4 } { 5 }$ C. 2 D. 9
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Question 8 · Single choice · 中档
已知抛物线 $y ^ { 2 } = 2 p x ( p > 0 )$ 的准线与圆 $x ^ { 2 } + y ^ { 2 } - 6 x - 7 = 0$ 相切,则p的值为【C】
A. $\frac { 1 } { 2 }$ B. 1 C. 2 D. 4
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Question 9 · Single choice · 中档
对于数列 $\left\{ a _ { _ n } \right\} , \mathrm { } ^ { \mathrm { } ^ { * } } a _ { _ { n + 1 } } > \left| a _ { _ n } \right| ( n = 1 , \ 2 , . . . ) ^ { , }$ 是 $\textstyle \left\{ a _ { n } \right\}$ 为递增数列”的【B】
A. 必要不充分条件 B. 充分不必要条件 C. 充要条件 D. 既不充分也不必要条件
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Question 11 · 中档
已知向量$a=$(2,-1),$b=$(-1,m),$c=$(-1,2),若(a+b)∥c,则$m=-1$
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Question 12 · Free response · 中档
观察下列等式: $1 ^ { 3 } + 2 ^ { 3 } = 3 ^ { 2 }$ $1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } = 6 ^ { 2 }$ $1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + 4 ^ { 3 } = 1 0 ^ { 2 }$ ,…,根据上述规律,第五个等式为 $1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + 4 ^ { 3 } + 5 ^ { 3 } + 6 ^ { 3 } = 2 1 ^ { 2 }$ 0
Answer key $1 ^ { 3 } + 2 ^ { 3 } + 3 ^ { 3 } + 4 ^ { 3 } + 5 ^ { 3 } + 6 ^ { 3 } = 2 1 ^ { 2 }$
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Question 15 · Fill in the blank · 中档
(考生注意:请在下列三题中任选一题作答,如果多做,则按所做的第一题评分) A.(不等式选做题)不等式$|x+3|-|x-2|\geq3$的解集为____
先写答案,再对照解析 等价写法不自动判错
直接输入 Formula Chemistry
π ² √ ≤ ≥ ∞ ∈ ∴
写好了,对答案
Answer key $\{x\,|\,x\geq1\}$
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Question 19 · Free response · 中档
(Ⅰ)样本中男生人数为40,由分层抽样比例为10%估计全校男生人数为400人。 (Ⅱ)由统计图知,样本中身高在170~185cm之间的学生有14+13+4+3+1=35人,样本容量为70,所以样本中学生身高在170~180cm之间的概率p=0.5 (Ⅲ)样本中女生身高在165~180cm之间的人数为10,身高在170~180cm之间的人数为4, 设A表示事件“从样本中身高在165~180cm之间的女生中任取2人,至少有1人身高在170~180cm之间”, 则 $P(A)=1-\dfrac{C_6^2}{C_{10}^2}=\dfrac{2}{3}$ (或 $P(A)=\dfrac{C_6^1 C_4^1 + C_4^2}{C_{10}^2}=\dfrac{2}{3}$)
Answer key 400;0.5;$\dfrac{2}{3}$
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