ti杯高二数学竞赛试题和答案

ti杯高二数学竞赛试题和答案

一、单项选择题(每题2分,共10题)1.函数\(y=\sinx\)的最小正周期是(）A.\(\pi\)B.\(2\pi\)C.\(3\pi\)D.\(4\pi\)2.直线\(y=2x+1\)的斜率是()A.\(1\)B.\(2\)C.\(-1\)D.\(-2\)3.抛物线\(y^2=4x\)的焦点坐标是()A.\((1,0)\)B.\((0,1)\)C.\((-1,0)\)D.\((0,-1)\)4.若\(\vec{a}=(1,2)\),\(\vec{b}=(3,4)\),则\(\vec{a}\cdot\vec{b}\)等于()A.\(5\)B.\(11\)C.\(10\)D.\(13\)5.等差数列\(\{a_{n}\}\)中,\(a_{1}=1\),\(a_{3}=5\),则公差\(d\)为()A.\(1\)B.\(2\)C.\(3\)D.\(4\)6.函数\(f(x)=x^3\)的导数\(f^\prime(x)\)是()A.\(3x^2\)B.\(2x^3\)C.\(x^2\)D.\(3x\)7.已知\(\cos\alpha=\frac{1}{2}\),且\(0\lt\alpha\lt2\pi\),则\(\alpha\)等于()A.\(\frac{\pi}{3}\)B.\(\frac{5\pi}{3}\)C.\(\frac{\pi}{3}\)或\(\frac{5\pi}{3}\)D.\(\frac{2\pi}{3}\)8.双曲线\(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\)的渐近线方程是()A.\(y=\pm\frac{3}{4}x\)B.\(y=\pm\frac{4}{3}x\)C.\(y=\pm\frac{2}{3}x\)D.\(y=\pm\frac{3}{2}x\)9.从\(5\)名男生和\(3\)名女生中选\(2\)人参加活动,至少有\(1\)名女生的选法有()种A.\(18\)B.\(28\)C.\(22\)D.\(36\)10.若\(x\gt0\),\(y\gt0\)且\(x+y=1\),则\(xy\)的最大值是()A.\(\frac{1}{4}\)B.\(\frac{1}{2}\)C.\(1\)D.\(2\)答案:1.B2.B3.A4.D5.B6.A7.C8.B9.C10.A二、多项选择题(每题2分,共10题)1.以下哪些是偶函数(）A.\(y=x^2\)B.\(y=\cosx\)C.\(y=x^3\)D.\(y=\sinx\)2.下列属于基本不等式的是()A.\(a^2+b^2\geq2ab\)B.\(a+b\geq2\sqrt{ab}\)(\(a\gt0,b\gt0\)）C.\(a^2-b^2=(a+b)(a-b)\)D.\((a+b)^2=a^2+2ab+b^2\)3.椭圆\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a\gtb\gt0)\)的性质正确的有()A.长轴长为\(2a\)B.短轴长为\(2b\)C.离心率\(e=\frac{c}{a}\)(\(c^2=a^2-b^2\))D.焦点在\(y\)轴4.以下哪些是等比数列的性质()A.\(a_{n}=a_{1}q^{n-1}\)B.\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)(\(q\neq1\))C.\(a_{m}\cdota_{n}=a_{p}\cdota_{q}\)(\(m+n=p+q\))D.\(a_{n+1}-a_{n}=d\)5.已知向量\(\vec{a}=(x_1,y_1)\),\(\vec{b}=(x_2,y_2)\),则（）A.\(\vec{a}+\vec{b}=(x_1+x_2,y_1+y_2)\)B.\(\vec{a}-\vec{b}=(x_1-x_2,y_1-y_2)\)C.\(\vec{a}\cdot\vec{b}=x_1x_2+y_1y_2\)D.若\(\vec{a}\parallel\vec{b}\),则\(x_1y_2-x_2y_1=0\)6.对于函数\(y=\tanx\),正确的是（）A.定义域为\(x\neqk\pi+\frac{\pi}{2},k\inZ\)B.周期为\(\pi\)C.是奇函数D.值域为\(R\)7.以下哪些点在直线\(2x+y-1=0\)上()A.\((0,1)\)B.\((1,-1)\)C.\((-1,3)\)D.\((2,-3)\)8.以下哪些是直线的方程形式()A.点斜式\(y-y_0=k(x-x_0)\)B.斜截式\(y=kx+b\)C.两点式\(\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\)D.一般式\(Ax+By+C=0\)(\(A^2+B^2\neq0\))9.已知集合\(A=\{1,2,3\}\),集合\(B=\{2,3,4\}\),则（）A.\(A\capB=\{2,3\}\)B.\(A\cupB=\{1,2,3,4\}\)C.\(A-B=\{1\}\)D.\(B-A=\{4\}\)10.以下哪些是导数的运算法则()A.\((u+v)^\prime=u^\prime+v^\prime\)B.\((uv)^\prime=u^\primev+uv^\prime\)C.\((\frac{u}{v})^\prime=\frac{u^\primev-uv^\prime}{v^2}\)(\(v\neq0\))D.\((x^n)^\prime=nx^{n-1}\)答案:1.AB2.AB3.ABC4.ABC5.ABCD6.ABCD7.ABCD8.ABCD9.ABCD10.ABCD三、判断题(每题2分,共10题)1.函数\(y=\lgx\)的定义域是\((0,+\infty)\)。()2.圆\(x^{2}+y^{2}=r^{2}\)的圆心是\((0,0)\),半径是\(r\)。(）3.若\(a\gtb\),则\(a^2\gtb^2\)。（)4.向量\(\vec{0}\)与任意向量平行。()5.数列\(1,-1,1,-1,\cdots\)是等比数列。()6.函数\(y=\sinx\)在\([0,\pi]\)上单调递增。()7.直线\(x=1\)的斜率不存在。()8.若\(f(x)\)是奇函数,则\(f(0)=0\)。（)9.椭圆的离心率\(e\)的范围是\((0,1)\)。()10.二项式\((a+b)^n\)展开式的通项公式是\(T_{r+1}=C_{n}^{r}a^{n-r}b^{r}\)。(）答案:1.√2.√3.×4.√5.√6.×7.√8.×9.√10.√四、简答题(每题5分,共4题)1.求函数\(y=x^2-2x+3\)的对称轴和顶点坐标。答案:对于二次函数\(y=ax^2+bx+c\),对称轴公式为\(x=-\frac{b}{2a}\)。此函数\(a=1\),\(b=-2\),则对称轴\(x=1\)。把\(x=1\)代入函数得\(y=2\),所以顶点坐标为\((1,2)\)。2.已知\(\sin\alpha=\frac{3}{5}\),\(\alpha\)是第二象限角,求\(\cos\alpha\)和\(\tan\alpha\)的值。答案：因为\(\sin^{2}\alpha+\cos^{2}\alpha=1\),\(\sin\alpha=\frac{3}{5}\),所以\(\cos\alpha=-\sqrt{1-(\frac{3}{5})^2}=-\frac{4}{5}\)。\(\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{\frac{3}{5}}{-\frac{4}{5}}=-\frac{3}{4}\)。3.求过点\((1,2)\)且斜率为\(3\)的直线方程。答案:根据直线的点斜式方程\(y-y_0=k(x-x_0)\),已知点\((1,2)\),斜率\(k=3\),则直线方程为\(y-2=3(x-1)\),整理得\(y=3x-1\)。4.已知等差数列\(\{a_{n}\}\)中,\(a_{1}=1\),\(a_{3}=5\),求\(a_{5}\)的值。答案：先求公差\(d\),\(a_{3}=a_{1}+2d\),即\(5=1+2d\),解得\(d=2\)。\(a_{5}=a_{1}+4d=1+4\times2=9\)。五、讨论题(每题5分,共4题)1.讨论函数\(y=\frac{1}{x}\)的单调性。答案:函数\(y=\frac{1}{x}\)定义域为\(x\neq0\)。在\((-\infty,0)\)上,任取\(x_1\ltx_2\lt0\),\(y_1-y_2=\frac{1}{x_1}-\frac{1}{x_2}=\frac{x_2-x_1}{x_1x_2}\gt0\),即\(y_1\gty_2\),所以在\((-\infty,0)\)单调递减;在\((0,+\infty)\)同理可证也单调递减。2.探讨直线与圆的位置关系有哪些判断方法。答案:一是几何法,通过比较圆心到直线的距离\(d\)与圆半径\(r\)的大小,\(d\gtr\)时相离,\(d=r\)时相切,\(d\ltr\)时相交;二是代数法,联立直线与圆的方程,消元后得一元二次方程,根据判别式\(\Delta\)判断,\(\Delta\lt0\)相离,\(\Delta=0\)相切,\(\Delta\gt0\)相交。3.分析等比数列前\(n\)项和公式推导过程中运用的思想方法。答案:推导等比数列前\(n\)项和公式运用了错位相减法。设等比数列\(\{a_{n}\}\),\(S_{n}=a_{1}+a_{1}q+a_{1}q^{2}+\cdots+a_{1}q^{n