理科數(shù)學(xué)試題和答案

一、單項(xiàng)選擇題(每題2分,共10題)1.函數(shù)\(y=\sinx\)的最小正周期是(）A.\(\frac{\pi}{2}\)B.\(\pi\)C.\(2\pi\)D.\(4\pi\)2.拋物線\(y^2=4x\)的焦點(diǎn)坐標(biāo)是()A.\((1,0)\)B.\((0,1)\)C.\((2,0)\)D.\((0,2)\)3.已知向量\(\vec{a}=(1,2)\),\(\vec=(-1,1)\),則\(\vec{a}+\vec\)等于()A.\((0,3)\)B.\((1,3)\)C.\((2,1)\)D.\((2,3)\)4.若\(\tan\alpha=2\),則\(\frac{\sin\alpha}{\cos\alpha}\)的值為()A.\(\frac{1}{2}\)B.\(2\)C.\(\pm2\)D.\(\frac{2}{3}\)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)\(f(x)=x^3-3x\)的極大值點(diǎn)是()A.\(1\)B.\(-1\)C.\(0\)D.\(2\)7.已知\(\log_{2}x=3\),則\(x\)的值為()A.\(6\)B.\(8\)C.\(16\)D.\(32\)8.從\(5\)名男生和\(3\)名女生中選\(2\)人參加比賽,選到一男一女的概率是()A.\(\frac{15}{28}\)B.\(\frac{5}{14}\)C.\(\frac{3}{14}\)D.\(\frac{1}{2}\)9.圓\((x-1)^2+(y+2)^2=9\)的圓心坐標(biāo)是()A.\((1,2)\)B.\((-1,2)\)C.\((1,-2)\)D.\((-1,-2)\)10.直線\(y=2x+1\)與直線\(y=2x-3\)的位置關(guān)系是()A.相交B.平行C.垂直D.重合二、多項(xiàng)選擇題(每題2分,共10題)1.下列函數(shù)中,是偶函數(shù)的有（）A.\(y=x^2\)B.\(y=\cosx\)C.\(y=\sinx\)D.\(y=|x|\)2.以下屬于基本不等式的有()A.\(a^2+b^2\geq2ab\)B.\(\frac{a+b}{2}\geq\sqrt{ab}(a\gt0,b\gt0)\)C.\(a+b\geq2\sqrt{ab}\)D.\(a^2+b^2+c^2\geqab+bc+ca\)3.一個(gè)正方體的棱長(zhǎng)為\(a\),以下正確的是()A.正方體的表面積為\(6a^2\)B.正方體的體積為\(a^3\)C.正方體的體對(duì)角線長(zhǎng)為\(\sqrt{3}a\)D.正方體的面對(duì)角線長(zhǎng)為\(\sqrt{2}a\)4.已知復(fù)數(shù)\(z=a+bi\)(\(a,b\inR\)),以下說(shuō)法正確的是()A.當(dāng)\(a=0\)時(shí),\(z\)是純虛數(shù)B.\(|z|=\sqrt{a^2+b^2}\)C.若\(z=\overline{z}\),則\(b=0\)D.\(z\cdot\overline{z}=a^2+b^2\)5.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(zhǎng)軸長(zhǎng)為\(2a\)B.短軸長(zhǎng)為\(2b\)C.焦距為\(2c\)(\(c^2=a^2-b^2\))D.離心率\(e=\frac{c}{a}\)且\(0\lte\lt1\)6.下列求導(dǎo)正確的是(）A.\((x^n)^\prime=nx^{n-1}\)B.\((\sinx)^\prime=\cosx\)C.\((\lnx)^\prime=\frac{1}{x}\)D.\((e^x)^\prime=e^x\)7.已知直線\(l_1:y=k_1x+b_1\),\(l_2:y=k_2x+b_2\),以下能判斷\(l_1\parallell_2\)的是()A.\(k_1=k_2\)且\(b_1\neqb_2\)B.\(k_1k_2=-1\)C.兩直線斜率都不存在且不重合D.\(k_1=k_2\)且\(b_1=b_2\)8.以下哪些是等比數(shù)列的性質(zhì)()A.\(a_{n}=a_{1}q^{n-1}\)B.\(a_{m}\cdota_{n}=a_{p}\cdota_{q}(m+n=p+q)\)C.\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}(q\neq1)\)D.\(S_{n}=na_{1}(q=1)\)9.空間中,下列說(shuō)法正確的是()A.垂直于同一條直線的兩條直線平行B.若直線\(a\)平行于平面\(\alpha\),直線\(b\subset\alpha\),則\(a\parallelb\)C.若兩個(gè)平面\(\alpha\)、\(\beta\)平行,直線\(a\subset\alpha\),則\(a\parallel\beta\)D.若一條直線垂直于一個(gè)平面內(nèi)兩條相交直線,則這條直線垂直于這個(gè)平面10.已知\(A(1,2)\),\(B(3,4)\),以下說(shuō)法正確的是（）A.直線\(AB\)的斜率為\(1\)B.\(AB\)中點(diǎn)坐標(biāo)為\((2,3)\)C.\(|AB|=2\sqrt{2}\)D.直線\(AB\)的方程為\(y-2=x-1\)三、判斷題(每題2分,共10題)1.\(y=\log_{a}x\)(\(a\gt0\)且\(a\neq1\))的定義域是\((0,+\infty)\)。()2.若\(a\gtb\),則\(a^2\gtb^2\)。（)3.向量\(\vec{a}\cdot\vec=|\vec{a}|\times|\vec|\sin\theta\)(\(\theta\)為\(\vec{a}\)與\(\vec\)的夾角)。()4.函數(shù)\(y=\cos(x+\frac{\pi}{2})\)是奇函數(shù)。()5.雙曲線\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)的漸近線方程是\(y=\pm\frac{a}x\)。()6.若\(f^\prime(x_0)=0\),則\(x_0\)一定是函數(shù)\(f(x)\)的極值點(diǎn)。(）7.兩個(gè)平面垂直,則一個(gè)平面內(nèi)垂直于交線的直線垂直于另一個(gè)平面。（)8.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和\(S_{n}=na_{1}+\frac{n(n-1)}{2}d\)。()9.復(fù)數(shù)\(z=3+4i\)的共軛復(fù)數(shù)是\(3-4i\)。()10.對(duì)于任意實(shí)數(shù)\(a,b\),\((a+b)^2=a^2+2ab+b^2\)恒成立。（）四、簡(jiǎn)答題(每題5分,共4題)1.求函數(shù)\(y=x^2-2x+3\)的最小值及對(duì)稱軸。答案:將函數(shù)化為頂點(diǎn)式\(y=(x-1)^2+2\),所以對(duì)稱軸為\(x=1\),當(dāng)\(x=1\)時(shí),\(y\)有最小值\(2\)。2.已知\(\sin\alpha=\frac{3}{5}\),\(\alpha\)是第二象限角,求\(\cos\alpha\)的值。答案：因?yàn)閈(\sin^2\alpha+\cos^2\alpha=1\),\(\sin\alpha=\frac{3}{5}\),所以\(\cos^2\alpha=1-(\frac{3}{5})^2=\frac{16}{25}\)。又\(\alpha\)是第二象限角,\(\cos\alpha\lt0\),所以\(\cos\alpha=-\frac{4}{5}\)。3.求過(guò)點(diǎn)\((1,2)\)且斜率為\(3\)的直線方程。答案:由直線的點(diǎn)斜式方程\(y-y_1=k(x-x_1)\)(其中\(zhòng)((x_1,y_1)=(1,2)\),\(k=3\)),可得\(y-2=3(x-1)\),整理得\(3x-y-1=0\)。4.計(jì)算\(\int_{0}^{1}(x^2+1)dx\)。答案:\(\int_{0}^{1}(x^2+1)dx=(\frac{1}{3}x^3+x)\big|_{0}^{1}=(\frac{1}{3}\times1^3+1)-(\frac{1}{3}\times0^3+0)=\frac{4}{3}\)。五、討論題(每題5分,共4題)1.討論函數(shù)\(y=\frac{1}{x}\)在\((0,+\infty)\)上的單調(diào)性,并說(shuō)明理由。答案:函數(shù)\(y=\frac{1}{x}\)在\((0,+\infty)\)上單調(diào)遞減。設(shè)\(0\ltx_1\ltx_2\),則\(f(x_1)-f(x_2)=\frac{1}{x_1}-\frac{1}{x_2}=\frac{x_2-x_1}{x_1x_2}\gt0\),即\(f(x_1)\gtf(x_2)\),所以單調(diào)遞減。2.已知直線\(l\)與圓\(C\),討論直線與圓的位置關(guān)系有幾種判定方法。答案:兩種方法。一是幾何法,比較圓心到直線的距離\(d\)與圓半徑\(r\)的大小,\(d\gtr\)時(shí)相離,\(d=r\)時(shí)相切,\(d\ltr\)時(shí)相交;二是代數(shù)法,聯(lián)立直線與圓的方程得方程組,消元后看判別式\(\Delta\),\(\Delta\gt0\)相交,\(\Delta=0\)相切,\(\Delta\lt0\)相離。3.討論在等比數(shù)列中,若\(a_{n}\gt0\),\(a_{1}a_{9}=64\),\(a_{4}=4\),求公比\(q\)的值。答案：由等比數(shù)列性質(zhì)\(a_{1}a_{9}=a_{5}^2=64\),又\(a_{n}\gt0\),所以\(a_{5}=8\)。則\(q=\frac{a_{5}}{a_{4}}=\frac{8}{4}=2\)。4.討論如何利用導(dǎo)數(shù)判斷函數(shù)的單調(diào)性。答案:先求函數(shù)\(f(x)\)的導(dǎo)數(shù)\(f^\prime(x)\)。若在某區(qū)間內(nèi)\(f^\prime(x)\gt0\),則\(f(x)\)在該區(qū)間單調(diào)遞增;若\(f^\prime(x)\lt0\),則\(f(x)\)在該區(qū)間單調(diào)遞減;若\(f^\prime(x)=0