嘉祥高考三模數(shù)學(xué)試題及答案

一、單項(xiàng)選擇題（每題2分，共10題）1.集合\(A=\{1,2,3\}\)，\(B=\{2,3,4\}\)，則\(A\capB=(\)\)A.\(\{1,2,3,4\}\)B.\(\{2,3\}\)C.\(\{1,3\}\)D.\(\{1,4\}\)2.已知\(i\)為虛數(shù)單位，復(fù)數(shù)\(z=1+2i\)，則\(\vertz\vert=(\)\)A.\(\sqrt{5}\)B.\(\sqrt{3}\)C.\(\sqrt{2}\)D.\(5\)3.函數(shù)\(y=\sinx\)的最小正周期是\((\)\)A.\(\pi\)B.\(2\pi\)C.\(4\pi\)D.\(\frac{\pi}{2}\)4.已知向量\(\overrightarrow{a}=(1,2)\)，\(\overrightarrow=(-1,m)\)，若\(\overrightarrow{a}\parallel\overrightarrow\)，則\(m=(\)\)A.\(2\)B.\(-2\)C.\(\frac{1}{2}\)D.\(-\frac{1}{2}\)5.等差數(shù)列\(zhòng)(\{a_n\}\)中，\(a_1=1\)，\(a_3=5\)，則\(a_5=(\)\)A.\(9\)B.\(10\)C.\(11\)D.\(12\)6.雙曲線\(\frac{x^{2}}{4}-\frac{y^{2}}{5}=1\)的漸近線方程為\((\)\)A.\(y=\pm\frac{\sqrt{5}}{2}x\)B.\(y=\pm\frac{2}{\sqrt{5}}x\)C.\(y=\pm\frac{5}{4}x\)D.\(y=\pm\frac{4}{5}x\)7.若\(\log_{2}x=3\)，則\(x=(\)\)A.\(8\)B.\(6\)C.\(3\)D.\(2\)8.函數(shù)\(f(x)=x^{3}\)在點(diǎn)\((1,1)\)處的切線斜率為\((\)\)A.\(1\)B.\(2\)C.\(3\)D.\(4\)9.從\(5\)名同學(xué)中選\(2\)名參加比賽，不同選法有\(zhòng)((\)\)A.\(10\)種B.\(20\)種C.\(60\)種D.\(120\)種10.已知\(a=0.3^{2}\)，\(b=2^{0.3}\)，\(c=\log_{2}0.3\)，則\(a\)，\(b\)，\(c\)大小關(guān)系為\((\)\)A.\(a\ltb\ltc\)B.\(c\lta\ltb\)C.\(c\ltb\lta\)D.\(a\ltc\ltb\)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\(zhòng)((\)\)A.\(y=x^{2}\)B.\(y=\cosx\)C.\(y=\sinx\)D.\(y=x^{3}\)2.一個(gè)正方體的棱長為\(2\)，則以下正確的有\(zhòng)((\)\)A.表面積為\(24\)B.體積為\(8\)C.對角線長為\(2\sqrt{3}\)D.棱長總和為\(24\)3.已知直線\(l_1\)：\(y=k_1x+b_1\)，\(l_2\)：\(y=k_2x+b_2\)，若\(l_1\perpl_2\)，則\((\)\)A.\(k_1k_2=-1\)B.\(k_1=k_2\)C.\(k_1+k_2=0\)D.兩直線斜率之積為\(-1\)（前提兩直線斜率都存在）4.以下屬于基本初等函數(shù)的有\(zhòng)((\)\)A.冪函數(shù)B.指數(shù)函數(shù)C.對數(shù)函數(shù)D.三角函數(shù)5.橢圓\(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)的性質(zhì)正確的有\(zhòng)((\)\)A.焦點(diǎn)在\(x\)軸上B.\(a=3\)C.\(b=2\)D.\(c=\sqrt{5}\)6.已知\(\sin\alpha=\frac{1}{2}\)，\(\alpha\in(0,\pi)\)，則\(\alpha\)可能的值為\((\)\)A.\(\frac{\pi}{6}\)B.\(\frac{5\pi}{6}\)C.\(\frac{\pi}{3}\)D.\(\frac{2\pi}{3}\)7.以下說法正確的是\((\)\)A.空集是任何集合的子集B.任何一個(gè)集合至少有兩個(gè)子集C.若\(A\subseteqB\)且\(B\subseteqA\)，則\(A=B\)D.集合\(\{x\midx^{2}-1=0\}\)的元素為\(1\)和\(-1\)8.設(shè)等比數(shù)列\(zhòng)(\{a_n\}\)的公比為\(q\)，則\((\)\)A.\(a_{n}=a_{1}q^{n-1}\)B.若\(m+n=p+s\)，則\(a_m\cdota_n=a_p\cdota_s\)C.\(S_n=\frac{a_1(1-q^{n})}{1-q}\)（\(q\neq1\)）D.\(S_n=na_1\)（\(q=1\)）9.已知函數(shù)\(f(x)\)在\(x=x_0\)處可導(dǎo)，則\((\)\)A.\(f^\prime(x_0)=\lim\limits_{\Deltax\to0}\frac{f(x_0+\Deltax)-f(x_0)}{\Deltax}\)B.函數(shù)在\(x=x_0\)處的切線方程為\(y-f(x_0)=f^\prime(x_0)(x-x_0)\)C.\(f^\prime(x_0)\)是函數(shù)\(f(x)\)在\(x=x_0\)處的切線斜率D.若\(f^\prime(x_0)=0\)，則\(x=x_0\)是函數(shù)\(f(x)\)的極值點(diǎn)10.從\(1\)，\(2\)，\(3\)，\(4\)，\(5\)中任取\(3\)個(gè)數(shù)，組成無重復(fù)數(shù)字的三位數(shù)，則以下說法正確的有\(zhòng)((\)\)A.可以組成\(60\)個(gè)三位數(shù)B.其中偶數(shù)有\(zhòng)(24\)個(gè)C.其中奇數(shù)有\(zhòng)(36\)個(gè)D.能被\(5\)整除的有\(zhòng)(12\)個(gè)三、判斷題（每題2分，共10題）1.直線\(x=1\)的斜率不存在。（）2.若\(a\gtb\)，則\(a^{2}\gtb^{2}\)。（）3.函數(shù)\(y=2x+1\)是增函數(shù)。（）4.拋物線\(y^{2}=4x\)的焦點(diǎn)坐標(biāo)是\((1,0)\)。（）5.若\(\overrightarrow{a}\cdot\overrightarrow=0\)，則\(\overrightarrow{a}\perp\overrightarrow\)。（）6.數(shù)列\(zhòng)(1\)，\(2\)，\(3\)，\(4\)，\(5\)是等差數(shù)列也是等比數(shù)列。（）7.函數(shù)\(y=\frac{1}{x}\)的定義域是\(x\neq0\)。（）8.若\(\sin\alpha\gt0\)，則\(\alpha\)是第一象限角。（）9.空間中垂直于同一條直線的兩條直線平行。（）10.二項(xiàng)式\((a+b)^n\)展開式的通項(xiàng)公式為\(T_{r+1}=C_{n}^{r}a^{n-r}b^{r}\)。（）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(diǎn)坐標(biāo)。答案：對于二次函數(shù)\(y=ax^{2}+bx+c\)，對稱軸\(x=-\frac{2a}\)，此函數(shù)\(a=1\)，\(b=-2\)，則對稱軸\(x=1\)。把\(x=1\)代入得\(y=2\)，頂點(diǎn)坐標(biāo)為\((1,2)\)。2.已知\(\tan\alpha=2\)，求\(\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}\)的值。答案：分子分母同時(shí)除以\(\cos\alpha\)，則\(\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}=\frac{\tan\alpha+1}{\tan\alpha-1}\)，把\(\tan\alpha=2\)代入得\(\frac{2+1}{2-1}=3\)。3.求過點(diǎn)\((1,2)\)且與直線\(2x-y+1=0\)平行的直線方程。答案：兩直線平行斜率相等，已知直線斜率為\(2\)，設(shè)所求直線方程為\(y-2=2(x-1)\)，整理得\(2x-y=0\)。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)\(y=\log_{a}x\)（\(a\gt0\)且\(a\neq1\)）的單調(diào)性。答案：當(dāng)\(a\gt1\)時(shí)，函數(shù)\(y=\log_{a}x\)在\((0,+\infty)\)上單調(diào)遞增；當(dāng)\(0\lta\lt1\)時(shí)，函數(shù)\(y=\log_{a}x\)在\((0,+\infty)\)上單調(diào)遞減。2.如何判斷直線與圓的位置關(guān)系？答案：可通過比較圓心到直線的距離\(d\)與圓半徑\(r\)的大小判斷。若\(d\gtr\)，直線與圓相離；若\(d=r\)，直線與圓相切；若\(d\ltr\)，直線與圓相交。3