ti杯高二數(shù)學(xué)競賽試題和答案

一、單項選擇題(每題2分,共10題)1.函數(shù)\(y=\sinx\)的最小正周期是(）A.\(\pi\)B.\(2\pi\)C.\(3\pi\)D.\(4\pi\)2.直線\(y=2x+1\)的斜率是()A.\(1\)B.\(2\)C.\(-1\)D.\(-2\)3.拋物線\(y^2=4x\)的焦點坐標(biāo)是()A.\((1,0)\)B.\((0,1)\)C.\((-1,0)\)D.\((0,-1)\)4.若\(\vec{a}=(1,2)\),\(\vec=(3,4)\),則\(\vec{a}\cdot\vec\)等于()A.\(5\)B.\(11\)C.\(10\)D.\(13\)5.等差數(shù)列\(zhòng)(\{a_{n}\}\)中,\(a_{1}=1\),\(a_{3}=5\),則公差\(d\)為()A.\(1\)B.\(2\)C.\(3\)D.\(4\)6.函數(shù)\(f(x)=x^3\)的導(dǎo)數(shù)\(f^\prime(x)\)是()A.\(3x^2\)B.\(2x^3\)C.\(x^2\)D.\(3x\)7.已知\(\cos\alpha=\frac{1}{2}\),且\(0\lt\alpha\lt2\pi\),則\(\alpha\)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.雙曲線\(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\)的漸近線方程是()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.從\(5\)名男生和\(3\)名女生中選\(2\)人參加活動,至少有\(zhòng)(1\)名女生的選法有()種A.\(18\)B.\(28\)C.\(22\)D.\(36\)10.若\(x\gt0\),\(y\gt0\)且\(x+y=1\),則\(xy\)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、多項選擇題(每題2分,共10題)1.以下哪些是偶函數(shù)(）A.\(y=x^2\)B.\(y=\cosx\)C.\(y=x^3\)D.\(y=\sinx\)2.下列屬于基本不等式的是()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.橢圓\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a\gtb\gt0)\)的性質(zhì)正確的有()A.長軸長為\(2a\)B.短軸長為\(2b\)C.離心率\(e=\frac{c}{a}\)(\(c^2=a^2-b^2\))D.焦點在\(y\)軸4.以下哪些是等比數(shù)列的性質(zhì)()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\(\vec{a}=(x_1,y_1)\),\(\vec=(x_2,y_2)\),則（）A.\(\vec{a}+\vec=(x_1+x_2,y_1+y_2)\)B.\(\vec{a}-\vec=(x_1-x_2,y_1-y_2)\)C.\(\vec{a}\cdot\vec=x_1x_2+y_1y_2\)D.若\(\vec{a}\parallel\vec\),則\(x_1y_2-x_2y_1=0\)6.對于函數(shù)\(y=\tanx\),正確的是（）A.定義域為\(x\neqk\pi+\frac{\pi}{2},k\inZ\)B.周期為\(\pi\)C.是奇函數(shù)D.值域為\(R\)7.以下哪些點在直線\(2x+y-1=0\)上()A.\((0,1)\)B.\((1,-1)\)C.\((-1,3)\)D.\((2,-3)\)8.以下哪些是直線的方程形式()A.點斜式\(y-y_0=k(x-x_0)\)B.斜截式\(y=kx+b\)C.兩點式\(\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}\)D.一般式\(Ax+By+C=0\)(\(A^2+B^2\neq0\))9.已知集合\(A=\{1,2,3\}\),集合\(B=\{2,3,4\}\),則（）A.\(A\capB=\{2,3\}\)B.\(A\cupB=\{1,2,3,4\}\)C.\(A-B=\{1\}\)D.\(B-A=\{4\}\)10.以下哪些是導(dǎo)數(shù)的運算法則()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、判斷題(每題2分,共10題)1.函數(shù)\(y=\lgx\)的定義域是\((0,+\infty)\)。()2.圓\(x^{2}+y^{2}=r^{2}\)的圓心是\((0,0)\),半徑是\(r\)。(）3.若\(a\gtb\),則\(a^2\gtb^2\)。（)4.向量\(\vec{0}\)與任意向量平行。()5.數(shù)列\(zhòng)(1,-1,1,-1,\cdots\)是等比數(shù)列。()6.函數(shù)\(y=\sinx\)在\([0,\pi]\)上單調(diào)遞增。()7.直線\(x=1\)的斜率不存在。()8.若\(f(x)\)是奇函數(shù),則\(f(0)=0\)。（)9.橢圓的離心率\(e\)的范圍是\((0,1)\)。()10.二項式\((a+b)^n\)展開式的通項公式是\(T_{r+1}=C_{n}^{r}a^{n-r}b^{r}\)。(）答案:1.√2.√3.×4.√5.√6.×7.√8.×9.√10.√四、簡答題(每題5分,共4題)1.求函數(shù)\(y=x^2-2x+3\)的對稱軸和頂點坐標(biāo)。答案:對于二次函數(shù)\(y=ax^2+bx+c\),對稱軸公式為\(x=-\frac{2a}\)。此函數(shù)\(a=1\),\(b=-2\),則對稱軸\(x=1\)。把\(x=1\)代入函數(shù)得\(y=2\),所以頂點坐標(biāo)為\((1,2)\)。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：因為\(\sin^{2}\alpha+\cos^{2}\alpha=1\),\(\sin\alpha=\frac{3}{5}\),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.求過點\((1,2)\)且斜率為\(3\)的直線方程。答案:根據(jù)直線的點斜式方程\(y-y_0=k(x-x_0)\),已知點\((1,2)\),斜率\(k=3\),則直線方程為\(y-2=3(x-1)\),整理得\(y=3x-1\)。4.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)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、討論題(每題5分,共4題)1.討論函數(shù)\(y=\frac{1}{x}\)的單調(diào)性。答案:函數(shù)\(y=\frac{1}{x}\)定義域為\(x\neq0\)。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\((-\infty,0)\)單調(diào)遞減;在\((0,+\infty)\)同理可證也單調(diào)遞減。2.探討直線與圓的位置關(guān)系有哪些判斷方法。答案:一是幾何法,通過比較圓心到直線的距離\(d\)與圓半徑\(r\)的大小,\(d\gtr\)時相離,\(d=r\)時相切,\(d\ltr\)時相交;二是代數(shù)法,聯(lián)立直線與圓的方程,消元后得一元二次方程,根據(jù)判別式\(\Delta\)判斷,\(\Delta\lt0\)相離,\(\Delta=0\)相切,\(\Delta\gt0\)相交。3.分析等比數(shù)列前\(n\)項和公式推導(dǎo)過程中運用的思想方法。答案:推導(dǎo)等比數(shù)列前\(n\)項和公式運用了錯位相減法。設(shè)等比數(shù)列\(zhòng)(\{a_{n}\}\),\(S_{n}=a_{1}+a_{1}q+a_{1}q^{2}+\cdots+a_{1}q^{n