高中高三考試題庫及答案

一、單項選擇題（每題2分，共10題）1.已知集合\(A=\{1,2,3\}\)，\(B=\{x|x^2-3x+2=0\}\)，則\(A\capB=(\)\()\)A.\(\{1\}\)B.\(\{2\}\)C.\(\{1,2\}\)D.\(\{1,2,3\}\)答案：C2.函數(shù)\(y=\sin(2x+\frac{\pi}{3})\)的最小正周期是（）A.\(\pi\)B.\(2\pi\)C.\(\frac{\pi}{2}\)D.\(\frac{2\pi}{3}\)答案：A3.若\(a=\log_{2}3\)，\(b=\log_{3}2\)，\(c=\log_{4}\frac{1}{3}\)，則（）A.\(a>b>c\)B.\(b>a>c\)C.\(a>c>b\)D.\(c>a>b\)答案：A4.已知向量\(\vec{a}=(1,2)\)，\(\vec=(x,1)\)，若\(\vec{a}\perp\vec\)，則\(x=(\)\()\)A.-2B.2C.\(-\frac{1}{2}\)D.\(\frac{1}{2}\)答案：A5.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，公比\(q=2\)，則\(a_{5}=(\)\()\)A.16B.32C.8D.64答案：A6.過點\((1,1)\)且與直線\(x-2y+3=0\)平行的直線方程是（）A.\(x-2y+1=0\)B.\(x-2y-1=0\)C.\(2x-y+1=0\)D.\(2x-y-1=0\)答案：A7.在\(\triangleABC\)中，\(a=3\)，\(b=4\)，\(C=60^{\circ}\)，則\(c=\)（）A.\(\sqrt{13}\)B.\(\sqrt{37}\)C.\(\sqrt{19}\)D.\(\sqrt{21}\)答案：A8.函數(shù)\(y=x^{3}-3x^{2}+1\)的單調(diào)遞減區(qū)間是（）A.\((2,+\infty)\)B.\((-\infty,0)\)C.\((0,2)\)D.\((-\infty,0)\cup(2,+\infty)\)答案：C9.已知橢圓\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)\)的離心率\(e=\frac{\sqrt{3}}{2}\)，短軸長為\(2\)，則橢圓的方程為（）A.\(\frac{x^{2}}{4}+y^{2}=1\)B.\(\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\)C.\(\frac{x^{2}}{4}+\frac{y^{2}}{16}=1\)D.\(x^{2}+\frac{y^{2}}{4}=1\)答案：A10.若\(\int_{0}^{a}(2x+1)dx=6\)，則\(a=(\)\()\)A.2B.3C.-2D.-3答案：A二、多項選擇題（每題2分，共10題）1.下列函數(shù)中，在\((0,+\infty)\)上單調(diào)遞增的是（）A.\(y=x^{2}\)B.\(y=\lnx\)C.\(y=2^{x}\)D.\(y=\frac{1}{x}\)答案：ABC2.已知\(\sin\alpha=\frac{3}{5}\)，\(\alpha\in(\frac{\pi}{2},\pi)\)，則\(\cos\alpha=(\)\()\)A.\(-\frac{4}{5}\)B.\(\frac{4}{5}\)C.\(-\frac{3}{5}\)D.\(\frac{3}{5}\)答案：A3.下面關(guān)于向量的說法正確的是（）A.若\(\vec{a}\parallel\vec\)，\(\vec\parallel\vec{c}\)，則\(\vec{a}\parallel\vec{c}\)B.若\(\vec{a}\cdot\vec=\vec{a}\cdot\vec{c}\)，則\(\vec=\vec{c}\)C.向量\(\vec{a}=(1,2)\)在向量\(\vec=(2,-1)\)方向上的投影為\(0\)D.\(\vert\vec{a}+\vec\vert\leqslant\vert\vec{a}\vert+\vert\vec\vert\)答案：CD4.下列命題正確的是（）A.若\(p\veeq\)為真命題，則\(p\)，\(q\)均為真命題B.命題“\(\existsx\inR\)，\(x^{2}-x-1<0\)”的否定是“\(\forallx\inR\)，\(x^{2}-x-1\geqslant0\)”C.“\(x=1\)”是“\(x^{2}-3x+2=0\)”的充分不必要條件D.若\(a>b\)，則\(ac^{2}>bc^{2}\)答案：BC5.若直線\(y=kx+1\)與圓\(x^{2}+y^{2}=1\)相交于\(A\)，\(B\)兩點，則弦長\(\vertAB\vert\)可能為（）A.\(\sqrt{2}\)B.\(\sqrt{3}\)C.\(2\)D.\(1\)答案：ABD6.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和為\(S_{n}\)，若\(a_{1}=1\)，\(d=2\)，則（）A.\(a_{n}=2n-1\)B.\(S_{n}=n^{2}\)C.數(shù)列\(zhòng)(\{\frac{S_{n}}{n}\}\)是等差數(shù)列D.若\(n>m\)，則\(S_{n}-S_{m}=n^{2}-m^{2}\)答案：ABC7.函數(shù)\(y=f(x)\)的圖象關(guān)于直線\(x=a\)對稱的充要條件是（）A.\(f(a+x)=f(a-x)\)B.\(f(x)=f(2a-x)\)C.\(f(x-a)=f(x+a)\)D.\(f(-x)=f(2a+x)\)答案：AB8.在\(\triangleABC\)中，\(\sinA\sinB<\cosA\cosB\)，則\(\triangleABC\)（）A.是銳角三角形B.是直角三角形C.是鈍角三角形D.形狀不確定答案：C9.已知\(f(x)\)是定義在\(R\)上的奇函數(shù)，當\(x>0\)時，\(f(x)=x^{2}+x\)，則（）A.\(f(0)=0\)B.\(f(-1)=-2\)C.\(f(x)\)在\((-\infty,0)\)上單調(diào)遞減D.\(f(x)\)在\(R\)上是增函數(shù)答案：AB10.已知\(a>0\)，\(b>0\)，\(a+b=1\)，則（）A.\(ab\leqslant\frac{1}{4}\)B.\(\frac{1}{a}+\frac{1}\geqslant4\)C.\(\sqrt{a}+\sqrt\leqslant\sqrt{2}\)D.\(a^{2}+b^{2}\geqslant\frac{1}{2}\)答案：ABCD三、判斷題（每題2分，共10題）1.若\(a>b\)，則\(a^{2}>b^{2}\)。（）答案：錯誤2.函數(shù)\(y=\cosx\)的圖象關(guān)于\(y\)軸對稱。（）答案：正確3.若直線\(l_{1}\)：\(y=k_{1}x+b_{1}\)與直線\(l_{2}\)：\(y=k_{2}x+b_{2}\)垂直，則\(k_{1}\cdotk_{2}=-1\)。（）答案：錯誤4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}=1\)，\(a_{3}=4\)，則\(a_{2}=2\)。（）答案：錯誤5.若\(f(x)\)是偶函數(shù)，則\(f(x)=f(-x)\)對于定義域內(nèi)任意\(x\)都成立。（）答案：正確6.向量\(\vec{a}=(1,0)\)，\(\vec=(0,1)\)，則\(\vec{a}\cdot\vec=0\)。（）答案：正確7.函數(shù)\(y=\frac{1}{x}\)在\((-\infty,0)\cup(0,+\infty)\)上是減函數(shù)。（）答案：錯誤8.在\(\triangleABC\)中，\(A+B+C=\pi\)。（）答案：正確9.若\(\lim\limits_{x\rightarrow1}\frac{x^{2}-1}{x-1}=2\)。（）答案：正確10.若\(z=a+bi(a,b\inR)\)，則\(\vertz\vert=\sqrt{a^{2}+b^{2}}\)。（）答案：正確四、簡答題（每題5分，共4題）1.求函數(shù)\(y=\frac{1}{x-1}+\sqrt{x+2}\)的定義域。答案：要使函數(shù)有意義，則\(x-1\neq0\)且\(x+2\geqslant0\)，解得\(x\geqslant-2\)且\(x\neq1\)。2.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{7}=13\)，求\(a_{n}\)。答案：設(shè)等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)，則\(a_{7}-a_{3}=4d=8\)，\(d=2\)，\(a_{1}=a_{3}-2d=1\)，\(a_{n}=a_{1}+(n-1)d=2n-1\)。3.求曲線\(y=x^{3}-3x^{2}+2x\)在點\((1,0)\)處的切線方程。答案：\(y'=3x^{2}-6x+2\)，當\(x=1\)時，\(y'=-1\)，切線方程為\(y-0=-1\times(x-1)\)，即\(x+y-1=0\)。4.已知向量\(\vec{a}=(1,2)\)，\(\vec=(3,-1)\)，求\(\vec{a}\)與\(\vec\)的夾角\(\theta\)。答案：\(\vec{a}\cdot\vec=1\times3+2\times(-1)=1\)，\(\vert\vec{a}\vert=\sqrt{1^{2}+2^{2}}=\sqrt{5}\)，\(\vert\vec\vert=\sqrt{3^{2}+(-1)^{2}}=\sqrt{10}\)，\(\cos\theta=\frac{\vec{a}\cdot\vec}{\vert\vec{a}\vert\vert\vec\vert}=\frac{1}{\sqrt{50}}=\frac{\sqrt{2}}{10}\)，\(\theta=\arccos\frac{\sqrt{2}}{10}\)。五、討論題（每題5分，共4題）1.討論函數(shù)\(y=ax^{2}+bx+c(a\neq0)\)的單調(diào)性。答案：函數(shù)\(y=ax^{2}+bx+c(a\neq0)\)的對稱軸為\(x=-\frac{2a}\)。當\(a>0\)時，在\((-\infty,-\frac{2a})\)上單調(diào)遞減，在\((-\frac{2a},+\infty)\)上單調(diào)遞增；當\(a<0\)時，在\((-\infty,-\frac{2a})\)上單調(diào)遞增，在\((-\frac{2a},+\infty)\)上單調(diào)遞減。2.請闡述等差數(shù)列與等比數(shù)列的區(qū)別與聯(lián)系。答案：區(qū)別：等差數(shù)列是后一項與前一項的差為定值，等比數(shù)列是后一項與前一項的比為定值。聯(lián)系：當?shù)缺葦?shù)列公比\(q=1\)時類似等差數(shù)列；部分求和公式推導思路有相似性。3.在\(\triang