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            • 1. 在同一平面直角坐标系中,经过伸缩变换\(\begin{cases} x′= \dfrac{1}{2}x, \\ y′= \dfrac{1}{3}y \end{cases}\)后,曲线 \(C\)\(x\)\({\,\!}^{2}+\) \(y\)\({\,\!}^{2}=36\)变为何种曲线,并求曲线的焦点坐标.
            • 2.

              已知曲线\(C\)\({\,\!}_{1}\)的极坐标方程是\(ρ=1\),在以极点\(O\)为坐标原点,极轴为\(x\)轴的正半轴的平面直角坐标系中,将曲线\(C\)\({\,\!}_{1}\)所有点的横坐标伸长为原来的\(3\)倍,得到曲线\(C\)\({\,\!}_{2}\)
              \((1)\)求曲线\(C\)\({\,\!}_{2}\)的参数方程;

              \((2)\)直线\(l\)过点\(M(1,0)\),倾斜角为\( \dfrac{π}{4}\),与曲线\(C\)\({\,\!}_{2}\)交于\(A\),\(B\)两点,求\(|MA|·|MB|\)的值.

            • 3.

              已知圆\(O\)的半径为\(1\),\(PA\)\(PB\)为该圆的两条切线,\(A\)\(B\)为两切点,那么\(\overrightarrow{PA}\cdot \overrightarrow{PB}\)的最小值为      

            • 4. 如图,半径为\(1\),圆心角为\( \dfrac {3π}{2}\)的圆弧\( \hat AB\)上有一点\(C\).
              \((1)\)若\(C\)为圆弧\(AB\)的中点,点\(D\)在线段\(OA\)上运动,求\(| \overrightarrow{OC}+ \overrightarrow{OD}|\)的最小值;
              \((2)\)若\(D\),\(E\)分别为线段\(OA\),\(OB\)的中点,当\(C\)在圆弧\( \hat AB\)上运动时,求\( \overrightarrow{CE}⋅ \overrightarrow{CD}\)的取值范围.
            • 5.

              I.已知数列\(\{a_{n}\}\)的前\(n\)项和为\(S_{n}\),把满足条件\(a_{n+1}\leqslant S_{n}(n∈N^{*})\)的所有数列\(\{a_{n}\}\)构成的集合记为\(M\).

              \((1)\)若数列\(\{a_{n}\}\)通项为\(a_{n}= \dfrac{1}{2^{n}}\),求证:\(\{a_{n}\}∈M\);

              \((2)\)若数列\(\{a_{n}\}\)是等差数列,且\(\{a_{n}+n\}∈M\),求公差\(d\)的值;

              \((3)\)若数列\(\{a\)\({\,\!}_{n}\)\(\}\)的各项均为正数,且\(\{a\)\({\,\!}_{n}\)\(\}∈M\),数列\(\{\)\( \dfrac{4^{n}}{a_{n}}\)\(\}\)中是否存在无穷多项依次成等差数列,若存在,给出一个数列\(\{a\)\({\,\!}_{n}\)\(\}\)的通项;若不存在,说明理由.


              \(II.\)已知矩阵\(A=\left[ \begin{matrix} 1 & a \\ -1 & b \\ \end{matrix} \right]\)的一个特征值为\(2\),其对应的一个特征向量为\(\alpha =\left[ \begin{matrix} 2 \\ 1 \\ \end{matrix} \right].\)矩阵\({{B}^{-1}}=\left[ \begin{matrix} 1 & a \\ 0 & b \\ \end{matrix} \right]\),求\({{(AB)}^{-1}}\).


              \(III.\)在极坐标系中,设直线\(\theta =\dfrac{{ }\!\!\pi\!\!{ }}{{3}}\)与曲线\({{\rho }^{2}}-10\rho \cos \theta +4=0\)相交于\(A\),\(B\)两点,求线段\(AB\) 中点的极坐标\(.\)  

              \(IV.\)、如图,四棱锥\(PABCD\)的底面\(ABCD\)是菱形,\(AC\)与\(BD\)交于点\(O\),\(OP⊥\)底面\(ABCD\),\(M\)为\(PC\)的中点,\(AC=4\),\(BD=2\),\(OP=4\).


              \((1)\) 求直线\(AP\)与\(BM\)所成角的余弦值;

              \((2)\) 求平面\(ABM\)与平面\(PAC\)所成锐二面角的余弦值.


              \(V.\)已知\(F_{n}(x)=\sum_{k=0}^{n}[(-1)^{k}C\rlap{_{n}}{^{k}}f_{k}(x)](n∈N^{*}).\)

              \((1)\)若\(f_{k}(x)=x^{k}\),求\(F_{2\;015}(2)\)的值;

              \((2)\)若\(f_{k}(x)= \dfrac{x}{x+k}(x∉\{0,-1,…,-n\})\),求证:\(F_{n}(x)= \dfrac{n!}{(x+1)(x+2)…(x+n)}\).

            • 6.
              如图,\(⊙O\)中\( \hat AB\)的中点为\(P\),弦\(PC\),\(PD\)分别交\(AB\)于\(E\),\(F\)两点.
              \((1)\)若\(∠PFB=2∠PCD\),求\(∠PCD\)的大小;
              \((2)\)若\(EC\)的垂直平分线与\(FD\)的垂直平分线交于点\(G\),证明:\(OG⊥CD\).
            • 7.
              如图,\(\triangle ABC\)是直角三角形,\(∠ABC=90^{\circ}\),以\(AB\)为直径的圆\(O\)交\(AC\)于点\(E\),点\(D\)是\(BC\)边的中点,连接\(OD\)交圆\(O\)于点\(M\).
              \((1)\)求证:\(O\)、\(B\)、\(D\)、\(E\)四点共圆;
              \((2)\)求证:\(2DE^{2}=DM⋅AC+DM⋅AB\).
            • 8.

              选修\(4—1\):几何证明选讲

              如图,\(\triangle ABC\)是直角三角形,\(∠ABC=90^{\circ}\),以\(AB\)为直径的圆\(O\)交\(AC\)于点\(E\),点\(D\)是\(BC\)边的中点,\(OD\)交圆\(O\)于点\(M\).

              \((\)Ⅰ\()\)求证:\(O\)、\(B\)、\(D\)、\(E\)四点共圆;

              \((\)Ⅱ\()\)求证:\(AB+AC=\dfrac{2D{{E}^{2}}}{DM}\).

            • 9.

              如图,\(⊙O_{1}\)与\(⊙O_{2}\)相交于\(A\)、\(B\)两点,\(AB\)是\(⊙O_{2}\)的直径,过\(A\)点作\(⊙O_{1}\)的切线交\(⊙O_{2}\)于点\(E\),并与\(BO_{1}\)的延长线交于点\(P\),\(PB\)分别与\(⊙O_{1}\),\(⊙O_{2}\)交于\(C\),\(D\)两点.

              \((1)\)求证:\(PA·PD=PE·PC\);

              \((2)\)求证:\(AD=AE\).

            • 10.
              如图,\(\triangle OAB\)是等腰三角形,\(∠AOB=120^{\circ}.\)以\(O\)为圆心,\( \dfrac {1}{2}OA\)为半径作圆.
              \((\)Ⅰ\()\)证明:直线\(AB\)与\(⊙O\)相切;
              \((\)Ⅱ\()\)点\(C\),\(D\)在\(⊙O\)上,且\(A\),\(B\),\(C\),\(D\)四点共圆,证明:\(AB/\!/CD\).
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