成考高數(shù)試題及答案

一、單項(xiàng)選擇題(每題2分,共10題)1.函數(shù)\(y=\sqrt{x-1}\)的定義域是(）A.\(x\geq0\)B.\(x\geq1\)C.\(x\gt1\)D.\(x\gt0\)2.\(\lim\limits_{x\to0}\frac{\sinx}{x}\)的值為()A.0B.1C.\(\infty\)D.不存在3.函數(shù)\(y=x^2\)的導(dǎo)數(shù)\(y^\prime\)是()A.\(2x\)B.\(x^2\)C.\(\frac{1}{2}x\)D.\(2\)4.\(\intxdx\)等于()A.\(\frac{1}{2}x^2+C\)B.\(x^2+C\)C.\(\frac{1}{2}x+C\)D.\(x+C\)5.函數(shù)\(y=\cosx\)的周期是()A.\(\pi\)B.\(2\pi\)C.\(3\pi\)D.\(4\pi\)6.直線\(y=2x+1\)的斜率是()A.1B.2C.\(\frac{1}{2}\)D.-27.已知\(f(x)=3x+1\),則\(f(2)\)的值為()A.5B.6C.7D.88.若\(a\gtb\),則()A.\(a^2\gtb^2\)B.\(ac\gtbc\)C.\(a+c\gtb+c\)D.\(\frac{a}{c}\gt\frac{c}\)9.函數(shù)\(y=e^x\)在點(diǎn)\((0,1)\)處的切線方程是()A.\(y=x+1\)B.\(y=x-1\)C.\(y=-x+1\)D.\(y=-x-1\)10.方程\(x^2-5x+6=0\)的根為()A.\(x=2\),\(x=3\)B.\(x=-2\),\(x=-3\)C.\(x=1\),\(x=6\)D.\(x=-1\),\(x=-6\)二、多項(xiàng)選擇題(每題2分,共10題)1.下列函數(shù)中,是奇函數(shù)的有（）A.\(y=x^3\)B.\(y=\sinx\)C.\(y=\cosx\)D.\(y=e^x\)2.下列極限存在的有()A.\(\lim\limits_{x\to0}\frac{1}{x}\)B.\(\lim\limits_{x\to0}x\sin\frac{1}{x}\)C.\(\lim\limits_{x\to\infty}\frac{\sinx}{x}\)D.\(\lim\limits_{x\to\infty}x\)3.下列求導(dǎo)正確的有()A.\((x^n)^\prime=nx^{n-1}\)B.\((\sinx)^\prime=\cosx\)C.\((\lnx)^\prime=\frac{1}{x}\)D.\((e^x)^\prime=e^x\)4.下列積分計(jì)算正確的有()A.\(\int1dx=x+C\)B.\(\intx^2dx=\frac{1}{3}x^3+C\)C.\(\int\frac{1}{x}dx=\ln|x|+C\)D.\(\int\cosxdx=\sinx+C\)5.以下哪些是二次函數(shù)的形式()A.\(y=ax^2+bx+c\)(\(a\neq0\))B.\(y=a(x-h)^2+k\)(\(a\neq0\))C.\(y=ax+b\)D.\(y=\frac{a}{x}\)6.下列不等式中,正確的有（）A.\(x^2\geq0\)B.\(|x|\geq0\)C.\(x^2+1\gt0\)D.\(-x^2\leq0\)7.已知直線\(l_1:y=k_1x+b_1\),\(l_2:y=k_2x+b_2\),若\(l_1\parallell_2\),則()A.\(k_1=k_2\)B.\(b_1=b_2\)C.\(k_1k_2=-1\)D.兩直線無(wú)交點(diǎn)8.函數(shù)\(y=\log_ax\)(\(a\gt0\)且\(a\neq1\))的性質(zhì)有()A.定義域?yàn)閈((0,+\infty)\)B.當(dāng)\(a\gt1\)時(shí),函數(shù)單調(diào)遞增C.過(guò)點(diǎn)\((1,0)\)D.值域?yàn)閈(R\)9.以下屬于基本初等函數(shù)的有(）A.冪函數(shù)B.指數(shù)函數(shù)C.對(duì)數(shù)函數(shù)D.三角函數(shù)10.方程\(ax^2+bx+c=0\)(\(a\neq0\)),其判別式\(\Delta=b^2-4ac\),當(dāng)()時(shí),方程有兩個(gè)不同實(shí)根。A.\(\Delta\gt0\)B.\(\Delta=0\)C.\(\Delta\lt0\)D.\(a\)、\(b\)、\(c\)同號(hào)三、判斷題(每題2分,共10題)1.函數(shù)\(y=\frac{1}{x}\)在定義域內(nèi)是單調(diào)遞減函數(shù)。()2.\(\lim\limits_{x\to\infty}e^x=0\)。()3.函數(shù)\(y=x^3+1\)的導(dǎo)數(shù)是\(y^\prime=3x^2+1\)。()4.\(\int_{-1}^{1}x^3dx=0\)。()5.若\(f(x)\)在\(x=a\)處連續(xù),則\(\lim\limits_{x\toa}f(x)=f(a)\)。（)6.直線\(y=3x-2\)與直線\(y=-\frac{1}{3}x+1\)垂直。()7.函數(shù)\(y=\sqrt{x+2}\)的定義域是\(x\geq-2\)。()8.二次函數(shù)\(y=x^2-2x+3\)的圖象開(kāi)口向下。()9.若\(a\gtb\),\(c\gt0\),則\(ac\gtbc\)。（)10.函數(shù)\(y=\sin^2x+\cos^2x=1\)恒成立。(）四、簡(jiǎn)答題(每題5分,共4題)1.求函數(shù)\(y=x^3-3x^2+1\)的導(dǎo)數(shù)。答案:根據(jù)求導(dǎo)公式\((x^n)^\prime=nx^{n-1}\),\(y^\prime=(x^3-3x^2+1)^\prime=3x^2-6x\)。2.計(jì)算\(\int(2x+3)dx\)。答案:根據(jù)積分公式\(\intx^ndx=\frac{1}{n+1}x^{n+1}+C\)(\(n\neq-1\)),\(\int(2x+3)dx=2\intxdx+3\int1dx=2\times\frac{1}{2}x^2+3x+C=x^2+3x+C\)。3.求函數(shù)\(y=\frac{1}{x-2}\)的定義域。答案:要使函數(shù)有意義,則分母不為零,即\(x-2\neq0\),解得\(x\neq2\),所以定義域?yàn)閈(x\in(-\infty,2)\cup(2,+\infty)\)。4.已知直線過(guò)點(diǎn)\((1,2)\)且斜率為\(3\),求直線方程。答案：根據(jù)直線點(diǎn)斜式方程\(y-y_0=k(x-x_0)\)（其中\(zhòng)((x_0,y_0)\)為直線上一點(diǎn),\(k\)為斜率）,可得直線方程為\(y-2=3(x-1)\),整理得\(y=3x-1\)。五、討論題(每題5分,共4題)1.討論函數(shù)\(y=x^2-4x+3\)的單調(diào)性。答案:先求導(dǎo)數(shù)\(y^\prime=2x-4\),令\(y^\prime=0\),得\(x=2\)。當(dāng)\(x\lt2\)時(shí),\(y^\prime\lt0\),函數(shù)單調(diào)遞減;當(dāng)\(x\gt2\)時(shí),\(y^\prime\gt0\),函數(shù)單調(diào)遞增。2.討論極限\(\lim\limits_{x\to0}\frac{\sin2x}{x}\)的求解方法。答案:可利用重要極限\(\lim\limits_{x\to0}\frac{\sinx}{x}=1\),將\(\lim\limits_{x\to0}\frac{\sin2x}{x}\)變形為\(2\lim\limits_{x\to0}\frac{\sin2x}{2x}\),令\(t=2x\),當(dāng)\(x\to0\)時(shí),\(t\to0\),則\(2\lim\limits_{x\to0}\frac{\sin2x}{2x}=2\lim\limits_{t\to0}\frac{\sint}{t}=2\)。3.討論如何判斷直線\(Ax+By+C=0\)與圓\((x-a)^2+(y-b)^2=r^2\)的位置關(guān)系。答案:計(jì)算圓心\((a,b)\)到直線的距離\(d=\frac{|Aa+Bb+C|}{\sqrt{A^2+B^2}}\)。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。4.討論二次函數(shù)\(y=ax^2+bx+c\)(\(a\neq0\))的最值情況。答案：當(dāng)\(a\gt0\)時(shí),函數(shù)圖象開(kāi)口向上,有最小值,\(x=-\frac{2a}\)時(shí),\(y_{min}=\frac{4ac-b^2}{4a}\)；當(dāng)\(a\lt0\)時(shí),函數(shù)圖象開(kāi)口向下,有最大值,\(x=-\frac{2a}\)時(shí),\(y_{max}=\frac{4ac-