廣州高考二模試題答案

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.\(2\pi\)B.\(\pi\)C.\(\frac{\pi}{2}\)D.\(4\pi\)答案：A2.已知集合\(A=\{1,2,3\}\),\(B=\{2,3,4\}\),則\(A\capB\)等于()A.\(\{1,2\}\)B.\(\{2,3\}\)C.\(\{3,4\}\)D.\(\{1,4\}\)答案：B3.復(fù)數(shù)\(z=1+i\),則\(\vertz\vert\)等于()A.\(1\)B.\(\sqrt{2}\)C.\(2\)D.\(2\sqrt{2}\)答案：B4.直線\(y=2x+1\)的斜率是()A.\(\frac{1}{2}\)B.\(-\frac{1}{2}\)C.\(2\)D.\(-2\)答案：C5.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.\(4\)答案：B6.已知向量\(\vec{a}=(1,2)\),\(\vec=(-1,m)\),若\(\vec{a}\parallel\vec\),則\(m\)的值為()A.\(2\)B.\(-2\)C.\(\frac{1}{2}\)D.\(-\frac{1}{2}\)答案：B7.函數(shù)\(y=\log_2x\)的定義域是()A.\((0,+\infty)\)B.\([0,+\infty)\)C.\((-\infty,0)\)D.\((-\infty,0]\)答案：A8.已知\(\cos\alpha=\frac{1}{2}\),且\(0\lt\alpha\lt\frac{\pi}{2}\),則\(\sin\alpha\)的值為()A.\(\frac{1}{2}\)B.\(\frac{\sqrt{3}}{2}\)C.\(-\frac{1}{2}\)D.\(-\frac{\sqrt{3}}{2}\)答案：B9.拋物線\(y^2=4x\)的焦點(diǎn)坐標(biāo)是()A.\((0,1)\)B.\((0,-1)\)C.\((1,0)\)D.\((-1,0)\)答案：C10.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：A二、多項(xiàng)選擇題(每題2分,共10題)1.以下哪些是奇函數(shù)(）A.\(y=x^3\)B.\(y=\sinx\)C.\(y=\cosx\)D.\(y=x\)答案:ABD2.下列屬于基本不等式的是(）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.\(a^2+b^2\leq2ab\)答案:AB3.已知直線\(l_1:y=k_1x+b_1\),\(l_2:y=k_2x+b_2\),則\(l_1\parallell_2\)的條件有（）A.\(k_1=k_2\)B.\(k_1=k_2\)且\(b_1\neqb_2\)C.\(k_1k_2=-1\)D.\(k_1\neqk_2\)答案：B4.以下哪些是等比數(shù)列()A.\(1,2,4,8,\cdots\)B.\(1,-1,1,-1,\cdots\)C.\(1,1,1,1,\cdots\)D.\(1,0,1,0,\cdots\)答案:ABC5.下列函數(shù)在\((0,+\infty)\)上單調(diào)遞增的有(）A.\(y=x^2\)B.\(y=2^x\)C.\(y=\log_2x\)D.\(y=\frac{1}{x}\)答案:ABC6.已知\(\alpha\)是第二象限角,\(\sin\alpha=\frac{3}{5}\),則以下正確的是（）A.\(\cos\alpha=-\frac{4}{5}\)B.\(\tan\alpha=-\frac{3}{4}\)C.\(\cos\alpha=\frac{4}{5}\)D.\(\tan\alpha=\frac{3}{4}\)答案:AB7.橢圓\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a\gtb\gt0)\)的性質(zhì)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\(e=\frac{c}{a}\)答案:ABCD8.以下哪些向量運(yùn)算正確(）A.\(\vec{a}+\vec=\vec+\vec{a}\)B.\((\vec{a}+\vec)+\vec{c}=\vec{a}+(\vec+\vec{c})\)C.\(\lambda(\vec{a}+\vec)=\lambda\vec{a}+\lambda\vec\)D.\(\vec{a}-\vec=\vec-\vec{a}\)答案:ABC9.已知函數(shù)\(f(x)\)的定義域?yàn)閈([-1,2]\),則函數(shù)\(f(2x-1)\)的定義域可能是（）A.\([0,\frac{3}{2}]\)B.\([-1,2]\)C.\([-3,3]\)D.\([-\frac{1}{2},\frac{3}{2}]\)答案：A10.以下哪些是三角函數(shù)的誘導(dǎo)公式()A.\(\sin(\alpha+2k\pi)=\sin\alpha\)(\(k\inZ\))B.\(\cos(\alpha+\pi)=-\cos\alpha\)C.\(\tan(\alpha+\frac{\pi}{2})=-\frac{1}{\tan\alpha}\)D.\(\sin(\frac{\pi}{2}-\alpha)=\cos\alpha\)答案:ABD三、判斷題(每題2分,共10題)1.空集是任何集合的子集。(）答案：對(duì)2.若\(a\gtb\),則\(a^2\gtb^2\)。()答案：錯(cuò)3.函數(shù)\(y=x^0\)的定義域是\(x\neq0\)。()答案：對(duì)4.等差數(shù)列的通項(xiàng)公式一定是關(guān)于\(n\)的一次函數(shù)。()答案：錯(cuò)5.向量\(\vec{a}\cdot\vec=\vert\vec{a}\vert\vert\vec\vert\cos\theta\)(\(\theta\)為\(\vec{a}\)與\(\vec\)的夾角)。()答案：對(duì)6.若\(z=a+bi\)(\(a,b\inR\)),則\(\overline{z}=a-bi\)。()答案：對(duì)7.橢圓\(\frac{x^2}{4}+\frac{y^2}{9}=1\)的焦點(diǎn)在\(x\)軸上。()答案：錯(cuò)8.\(y=\sinx\)與\(y=\cosx\)的最小正周期相同。()答案：對(duì)9.函數(shù)\(y=2x+1\)與\(y=2x-1\)平行。()答案：對(duì)10.若\(a\gt0\),\(b\gt0\),且\(ab=1\),則\(a+b\geq2\)。()答案：對(duì)四、簡(jiǎn)答題(每題5分,共4題)1.求函數(shù)\(y=x^2-2x+3\)的對(duì)稱(chēng)軸和頂點(diǎn)坐標(biāo)。答案:對(duì)于二次函數(shù)\(y=ax^2+bx+c\),對(duì)稱(chēng)軸\(x=-\frac{2a}\),此函數(shù)\(a=1\),\(b=-2\),對(duì)稱(chēng)軸\(x=1\)。把\(x=1\)代入函數(shù)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\(\frac{\sin\alpha+\cos\alpha}{\sin\alpha-\cos\alpha}=\frac{\tan\alpha+1}{\tan\alpha-1}\),將\(\tan\alpha=2\)代入得\(\frac{2+1}{2-1}=3\)。3.寫(xiě)出直線\(2x+3y-6=0\)的斜截式方程。答案:將直線方程\(2x+3y-6=0\)移項(xiàng)得\(3y=-2x+6\),兩邊同時(shí)除以\(3\),斜截式方程為\(y=-\frac{2}{3}x+2\)。4.已知等比數(shù)列\(zhòng)(\{a_n\}\)中,\(a_1=1\),\(q=2\),求\(a_5\)的值。答案：等比數(shù)列通項(xiàng)公式\(a_n=a_1q^{n-1}\),將\(a_1=1\),\(q=2\),\(n=5\)代入得\(a_5=1\times2^{5-1}=16\)。五、討論題(每題5分,共4題)1.討論函數(shù)\(y=\frac{1}{x}\)在\((-\infty,0)\)和\((0,+\infty)\)上的單調(diào)性。答案:在\((-\infty,0)\)和\((0,+\infty)\)上分別任取\(x_1\ltx_2\),計(jì)算\(f(x_1)-f(x_2)=\frac{1}{x_1}-\frac{1}{x_2}=\frac{x_2-x_1}{x_1x_2}\)。在這兩個(gè)區(qū)間內(nèi)\(x_2-x_1\gt0\),\(x_1x_2\gt0\),所以\(f(x_1)-f(x_2)\gt0\),即函數(shù)在這兩個(gè)區(qū)間上均單調(diào)遞減。2.討論直線與圓的位置關(guān)系有哪些判定方法。答案:一是幾何法,通過(guò)圓心到直線的距離\(d\)與圓半徑\(r\)比較,\(d\gtr\)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,消元后根據(jù)判別式\(\Delta\)判斷,\(\Delta\gt0\)相交,\(\Delta=0\)相切,\(\Delta\lt0\)相離。