2O2O陜西高考數(shù)學(xué)理科試題及答案VIP

2O2O陜西高考數(shù)學(xué)理科試題及答案

一、單項(xiàng)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.\(\{1,2,3\}\)2.復(fù)數(shù)\(z=\frac{1+i}{1-i}\)，則\(|z|=(\)\)A.\(0\)B.\(1\)C.\(\sqrt{2}\)D.\(2\)3.已知向量\(\overrightarrow{a}=(1,m)\)，\(\overrightarrow=(3,-2)\)，且\((\overrightarrow{a}+\overrightarrow)\perp\overrightarrow\)，則\(m=(\)\)A.-8B.-6C.6D.84.已知等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，若\(a_3=3\)，\(S_6=21\)，則數(shù)列\(zhòng)(\{a_n\}\)的公差\(d=(\)\)A.1B.-1C.2D.-25.函數(shù)\(f(x)=\sin(2x+\frac{\pi}{3})\)的最小正周期為\((\)\)A.\(4\pi\)B.\(2\pi\)C.\(\pi\)D.\(\frac{\pi}{2}\)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.\(y=\pm\frac{2\sqrt{5}}{5}x\)C.\(y=\pm\frac{5}{4}x\)D.\(y=\pm\frac{4}{5}x\)7.已知\(a=\log_20.2\)，\(b=2^{0.2}\)，\(c=0.2^{0.3}\)，則\((\)\)A.\(a\ltb\ltc\)B.\(a\ltc\ltb\)C.\(c\lta\ltb\)D.\(b\ltc\lta\)8.執(zhí)行如圖所示的程序框圖，若輸入\(n\)的值為\(3\)，則輸出\(s\)的值是\((\)\)A.1B.2C.4D.79.某圓柱的高為\(2\)，底面周長(zhǎng)為\(16\)，其三視圖如右圖。圓柱表面上的點(diǎn)\(M\)在正視圖上的對(duì)應(yīng)點(diǎn)為\(A\)，圓柱表面上的點(diǎn)\(N\)在左視圖上的對(duì)應(yīng)點(diǎn)為\(B\)，則在此圓柱側(cè)面上，從\(M\)到\(N\)的路徑中，最短路徑的長(zhǎng)度為\((\)\)A.\(2\sqrt{17}\)B.\(2\sqrt{5}\)C.\(3\)D.\(2\)10.已知函數(shù)\(f(x)=\begin{cases}x^2+2x,x\geq0\\2x-x^2,x\lt0\end{cases}\)，若\(f(a)+f(a^2-2)\lt0\)，則實(shí)數(shù)\(a\)的取值范圍是\((\)\)A.\((-\infty,-1)\cup(2,+\infty)\)B.\((-1,2)\)C.\((-2,1)\)D.\((-\infty,-2)\cup(1,+\infty)\)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.\(y=\cosx\)B.\(y=x^3\)C.\(y=\ln(x^2+1)\)D.\(y=e^x\)2.已知直線\(l\)過點(diǎn)\((1,0)\)且垂直于\(x\)軸，若\(l\)被拋物線\(y^2=4ax\)截得的線段長(zhǎng)為\(4\)，則拋物線的焦點(diǎn)坐標(biāo)為\((\)\)A.\((1,0)\)B.\((2,0)\)C.\((0,1)\)D.\((0,2)\)3.下列關(guān)于向量的運(yùn)算，正確的是\((\)\)A.\(\overrightarrow{a}\cdot\overrightarrow=\overrightarrow\cdot\overrightarrow{a}\)B.\((\overrightarrow{a}+\overrightarrow)\cdot\overrightarrow{c}=\overrightarrow{a}\cdot\overrightarrow{c}+\overrightarrow\cdot\overrightarrow{c}\)C.\((\overrightarrow{a}\cdot\overrightarrow)\overrightarrow{c}=\overrightarrow{a}(\overrightarrow\cdot\overrightarrow{c})\)D.\(|\overrightarrow{a}\cdot\overrightarrow|\leq|\overrightarrow{a}||\overrightarrow|\)4.已知\(a\gt0\)，\(b\gt0\)，且\(a+b=1\)，則下列結(jié)論正確的有\(zhòng)((\)\)A.\(\frac{1}{a}+\frac{1}\geq4\)B.\(a^2+b^2\geq\frac{1}{2}\)C.\(\sqrt{a}+\sqrt\leq\sqrt{2}\)D.\(\frac{1}{a^2}+\frac{1}{b^2}\geq8\)5.一個(gè)正方體的展開圖如圖所示，\(A\)、\(B\)、\(C\)、\(D\)為原正方體的頂點(diǎn)，則在原來的正方體中\(zhòng)((\)\)A.\(AB\parallelCD\)B.\(AB\)與\(CD\)相交C.\(AB\perpCD\)D.\(AB\)與\(CD\)所成的角為\(60^{\circ}\)6.已知函數(shù)\(f(x)=\sin(\omegax+\varphi)(\omega\gt0,|\varphi|\lt\frac{\pi}{2})\)的部分圖象如圖所示，則\((\)\)A.\(\omega=2\)B.\(\varphi=-\frac{\pi}{6}\)C.\(f(x)\)的單調(diào)遞增區(qū)間為\([k\pi-\frac{\pi}{6},k\pi+\frac{\pi}{3}],k\inZ\)D.\(f(x)\)的圖象關(guān)于點(diǎn)\((\frac{\pi}{3},0)\)對(duì)稱7.已知橢圓\(C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a\gtb\gt0)\)的左、右焦點(diǎn)分別為\(F_1,F_2\)，離心率為\(\frac{\sqrt{3}}{3}\)，過\(F_2\)的直線\(l\)交\(C\)于\(A,B\)兩點(diǎn)，若\(\triangleAF_1B\)的周長(zhǎng)為\(4\sqrt{3}\)，則下列說法正確的有\(zhòng)((\)\)A.\(a=\sqrt{3}\)B.\(b=\sqrt{2}\)C.橢圓\(C\)的方程為\(\frac{x^2}{3}+\frac{y^2}{2}=1\)D.\(\triangleAF_1B\)的面積最大值為\(2\sqrt{3}\)8.已知函數(shù)\(f(x)=x^3-3x^2+2\)，則\((\)\)A.\(f(x)\)有兩個(gè)極值點(diǎn)B.\(f(x)\)在\(x=1\)處取得極大值C.\(f(x)\)在\(x=2\)處取得極小值D.\(f(x)\)的圖象關(guān)于點(diǎn)\((1,0)\)對(duì)稱9.下列說法正確的是\((\)\)A.命題“\(\forallx\inR,x^2+1\gt0\)”的否定是“\(\existsx\inR,x^2+1\leq0\)”B.若\(p\)：\(a\inA\)，\(q\)：\(a\inB\)，且\(A\subseteqB\)，則\(p\)是\(q\)的充分不必要條件C.若復(fù)合命題\(p\wedgeq\)為假命題，則\(p\)，\(q\)都是假命題D.“若\(x^2=1\)，則\(x=1\)”的逆否命題為“若\(x\neq1\)，則\(x^2\neq1\)”10.已知\(f(x)\)是定義在\(R\)上的奇函數(shù)，當(dāng)\(x\geq0\)時(shí)，\(f(x)=x^2-2x\)，則\((\)\)A.\(f(-1)=1\)B.當(dāng)\(x\lt0\)時(shí)，\(f(x)=-x^2-2x\)C.\(f(x)\)的單調(diào)遞增區(qū)間為\((-\infty,-1)\)和\((1,+\infty)\)D.\(f(x)\)的值域?yàn)閈(R\)三、判斷題（每題2分，共10題）1.空集是任何集合的子集。（）2.函數(shù)\(y=\frac{1}{x}\)在其定義域上是減函數(shù)。（）3.若\(a\gtb\)，則\(a^2\gtb^2\)。（）4.直線\(Ax+By+C=0\)（\(A\)、\(B\)不同時(shí)為\(0\)）的斜率為\(-\frac{A}{B}\)。（）5.若向量\(\overrightarrow{a}\)與\(\overrightarrow\)共線，則存在實(shí)數(shù)\(\lambda\)，使得\(\overrightarrow{a}=\lambda\overrightarrow\)。（）6.若\(p\)：\(x\gt1\)，\(q\)：\(x\gt2\)，則\(p\)是\(q\)的必要不充分條件。（）7.橢圓\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a\gtb\gt0)\)的長(zhǎng)軸長(zhǎng)為\(2a\)。（）8.函數(shù)\(y=\sinx\)的圖象向左平移\(\frac{\pi}{2}\)個(gè)單位得到\(y=\cosx\)的圖象。（）9.若\(a\)，\(b\)，\(c\)成等比數(shù)列，則\(b^2=ac\)。（）10.函數(shù)\(y=\log_2(x^2+1)\)的值域是\([0,+\infty)\)。（）四、簡(jiǎn)答題（每題5分，共4題）1.求函數(shù)\(y=\sin(2x+\frac{\pi}{4})\)的單調(diào)遞增區(qū)間。答案：令\(2k\pi-\frac{\pi}{2}\leq2x+\frac{\pi}{4}\leq2k\pi+\frac{\pi}{2},k\inZ\)，解得\(k\pi-\frac{3\pi}{8}\leqx\leqk\pi+\frac{\pi}{8},k\inZ\)，所以單調(diào)遞增區(qū)間是\([k\pi-\frac{3\pi}{8},k\pi+\frac{\pi}{8}],k\inZ\)。2.已知等差數(shù)列\(zhòng)(\{a_n\}\)中，\(a_1=1\)，\(a_3=5\)，求其前\(n\)項(xiàng)和\(S_n\)。答案：設(shè)公差為\(d\)，\(a_3=a_1+2d=1+2d=5\)，解得\(d=2\)。\(S_n=na_1+\frac{n(n-1)}{2}d=n\times1+\frac{n(n-1)}{2}\times2=n^2\)。3.已知向量\(\overrightarrow{a}=(1,2)\)，\(\overrightarrow=(-3,4)\)，求\(\overrightarrow{a}\cdot\overrightarrow\)以及\(\overrightarrow{a}\)在\(\overrightarrow\)方向上的投影。答案：\(\overrightarrow{a}\cdot\overrightarrow=1\times(-3)+2\times4=5\)。\(\vert\overrightarrow\vert=\sqrt{(-3)^2+4^2}=5\)，\(\overrightarrow{a}\)在\(\overrightarrow\)方向上的投影為\(\frac{\overrightarrow{a}\cdot\overrightarrow}{\vert\overrightarrow\vert}=\frac{5}{5}=1\)。4.已知橢圓\(\frac{x^2}{9}+\frac{y^2}{4}=1\)，求其焦點(diǎn)坐標(biāo)和離心率。答案：由橢圓方程\(\frac{x^2}{9}+\frac{y^2}{4}=1\)，得\(a^2=9\)，\(b^2=4\)，則\(c=\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt{5}\)。焦點(diǎn)坐標(biāo)為\((\pm\sqrt{5},0)\)，離心率\(e=\frac{c}{a}=\frac{\sqrt{5}}{3}\)。五、討論題（每題5分，共4題）1.討論函數(shù)\(f(x)=x^3-3x\)的單調(diào)性與極值情況。答案：\(f^\prime(x)=3x^2-3=3(x+1)(x-1)\)。令\(f^\prime(x)\gt0\)，得\(x\lt-1\)或\(x\gt1\)，\(f(x)\)在\((-\infty,-1)\)和\((1,+\infty)\)遞增；令\(f^\prime(x)\lt0\)，得\(-1\ltx\lt1\)，\(f(x)\)在\((-1,1)\)遞減。\(x=-1\)時(shí)取極大值\(f(-1)=2\)；\(x=1\)時(shí)取極小值\(f(1)=-2\)。2.已知直線\(l\)與拋物線\(y^2=4x\)相