高二導(dǎo)數(shù)和數(shù)列試題及答案

一、單項(xiàng)選擇題（每題2分，共10題）1.函數(shù)\(y=x^2\)的導(dǎo)數(shù)是（）A.\(2x\)B.\(x\)C.\(2\)D.\(1\)2.若數(shù)列\(zhòng)(\{a_n\}\)的通項(xiàng)公式\(a_n=2n+1\)，則\(a_3\)的值為（）A.\(5\)B.\(6\)C.\(7\)D.\(8\)3.函數(shù)\(f(x)=e^x\)在\(x=0\)處的切線斜率為（）A.\(0\)B.\(1\)C.\(e\)D.\(\frac{1}{e}\)4.已知等差數(shù)列\(zhòng)(\{a_n\}\)中，\(a_1=1\)，\(d=2\)，則\(a_5\)等于（）A.\(9\)B.\(11\)C.\(13\)D.\(15\)5.函數(shù)\(y=\sinx\)的導(dǎo)數(shù)是（）A.\(\cosx\)B.\(-\cosx\)C.\(\sinx\)D.\(-\sinx\)6.等比數(shù)列\(zhòng)(\{a_n\}\)中，\(a_1=1\)，\(q=2\)，則\(a_4\)為（）A.\(4\)B.\(8\)C.\(16\)D.\(32\)7.函數(shù)\(f(x)=x^3-3x\)的極值點(diǎn)為（）A.\(1\)B.\(-1\)C.\(1\)和\(-1\)D.\(0\)8.數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}=a_n+3\)，\(a_1=2\)，則\(a_4\)為（）A.\(11\)B.\(12\)C.\(13\)D.\(14\)9.函數(shù)\(y=\lnx\)在\(x=1\)處的切線方程為（）A.\(y=x-1\)B.\(y=-x+1\)C.\(y=x\)D.\(y=-x\)10.等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n=n^2\)，則\(a_2\)的值為（）A.\(1\)B.\(3\)C.\(5\)D.\(7\)二、多項(xiàng)選擇題（每題2分，共10題）1.以下函數(shù)求導(dǎo)正確的是（）A.\((x^3)^\prime=3x^2\)B.\((\cosx)^\prime=\sinx\)C.\((\lnx)^\prime=\frac{1}{x}\)D.\((e^{-x})^\prime=e^{-x}\)2.等差數(shù)列\(zhòng)(\{a_n\}\)的性質(zhì)正確的有（）A.\(a_n=a_1+(n-1)d\)B.若\(m+n=p+q\)，則\(a_m+a_n=a_p+a_q\)C.\(S_n=\frac{n(a_1+a_n)}{2}\)D.\(a_{n+1}-a_n=d\)（\(d\)為公差）3.等比數(shù)列\(zhòng)(\{a_n\}\)滿足（）A.\(a_n=a_1q^{n-1}\)B.若\(m+n=p+q\)，則\(a_m\cdota_n=a_p\cdota_q\)C.\(S_n=\frac{a_1(1-q^n)}{1-q}(q\neq1)\)D.\(a_{n+1}\diva_n=q\)（\(q\)為公比）4.函數(shù)\(f(x)=x^2-4x+3\)（）A.在\(x=2\)處取得最小值B.對(duì)稱軸為\(x=2\)C.單調(diào)遞減區(qū)間是\((-\infty,2)\)D.與\(x\)軸有兩個(gè)交點(diǎn)5.已知數(shù)列\(zhòng)(\{a_n\}\)是等差數(shù)列，公差\(d\gt0\)，則（）A.\(a_{n+1}\gta_n\)B.\(S_n\)單調(diào)遞增C.\(a_1\lta_2\)D.數(shù)列可能是常數(shù)列6.函數(shù)\(y=x^3-12x\)（）A.有極大值B.有極小值C.極大值點(diǎn)為\(x=-2\)D.極小值點(diǎn)為\(x=2\)7.對(duì)于數(shù)列\(zhòng)(\{a_n\}\)，以下說法正確的是（）A.若\(a_{n+1}-a_n=2\)，則\(\{a_n\}\)是等差數(shù)列B.若\(\frac{a_{n+1}}{a_n}=3\)，則\(\{a_n\}\)是等比數(shù)列C.常數(shù)列既是等差數(shù)列也是等比數(shù)列D.等差數(shù)列的通項(xiàng)公式是關(guān)于\(n\)的一次函數(shù)8.函數(shù)\(f(x)\)在某點(diǎn)\(x_0\)處導(dǎo)數(shù)存在，以下說法正確的是（）A.\(f(x)\)在\(x_0\)處連續(xù)B.\(f(x)\)在\(x_0\)處可導(dǎo)C.導(dǎo)數(shù)\(f^\prime(x_0)\)表示\(f(x)\)在\(x_0\)處切線的斜率D.\(f(x)\)在\(x_0\)處的切線一定存在9.已知等比數(shù)列\(zhòng)(\{a_n\}\)，\(a_1=1\)，\(q=2\)，則（）A.\(a_3=4\)B.\(S_3=7\)C.\(a_n=2^{n-1}\)D.\(S_n=2^n-1\)10.以下函數(shù)在\((0,+\infty)\)上單調(diào)遞增的是（）A.\(y=x^2\)B.\(y=\lnx\)C.\(y=e^x\)D.\(y=\frac{1}{x}\)三、判斷題（每題2分，共10題）1.函數(shù)\(y=x^4\)的導(dǎo)數(shù)是\(y^\prime=4x^3\)。（）2.常數(shù)列\(zhòng)(2,2,2,\cdots\)是等比數(shù)列。（）3.函數(shù)\(f(x)\)在\(x=x_0\)處導(dǎo)數(shù)為\(0\)，則\(x_0\)一定是極值點(diǎn)。（）4.等差數(shù)列\(zhòng)(\{a_n\}\)中，若\(a_3=5\)，\(a_5=9\)，則公差\(d=2\)。（）5.函數(shù)\(y=\cosx\)在\([0,\pi]\)上單調(diào)遞減。（）6.等比數(shù)列\(zhòng)(\{a_n\}\)中，\(a_1=1\)，\(q=-1\)，則\(a_{2023}=1\)。（）7.函數(shù)\(f(x)=x\lnx\)的導(dǎo)數(shù)是\(f^\prime(x)=\lnx+1\)。（）8.若數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n=n^2+1\)，則\(a_n=2n-1\)。（）9.函數(shù)\(y=e^{-x}\)的導(dǎo)數(shù)是\(y^\prime=-e^{-x}\)。（）10.等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n\)不可能是關(guān)于\(n\)的二次函數(shù)。（）四、簡(jiǎn)答題（每題5分，共4題）1.求函數(shù)\(y=x^3-2x^2+1\)的導(dǎo)數(shù)。答案：根據(jù)求導(dǎo)公式\((x^n)^\prime=nx^{n-1}\)，\(y^\prime=3x^2-4x\)。2.已知等差數(shù)列\(zhòng)(\{a_n\}\)中，\(a_1=3\)，\(d=2\)，求\(a_5\)和\(S_5\)。答案：\(a_n=a_1+(n-1)d\)，則\(a_5=3+(5-1)\times2=11\)；\(S_n=\frac{n(a_1+a_n)}{2}\)，\(S_5=\frac{5\times(3+11)}{2}=35\)。3.等比數(shù)列\(zhòng)(\{a_n\}\)中，\(a_1=2\)，\(q=3\)，求\(a_4\)和\(S_4\)。答案：\(a_n=a_1q^{n-1}\)，\(a_4=2\times3^{4-1}=54\)；\(S_n=\frac{a_1(1-q^n)}{1-q}(q\neq1)\)，\(S_4=\frac{2\times(1-3^4)}{1-3}=80\)。4.求函數(shù)\(f(x)=x^2-6x+8\)的單調(diào)區(qū)間。答案：\(f^\prime(x)=2x-6\)，令\(f^\prime(x)\gt0\)，得\(x\gt3\)，單調(diào)遞增區(qū)間為\((3,+\infty)\)；令\(f^\prime(x)\lt0\)，得\(x\lt3\)，單調(diào)遞減區(qū)間為\((-\infty,3)\)。五、討論題（每題5分，共4題）1.討論函數(shù)\(y=x^3-3x^2+2\)的極值情況。答案：\(y^\prime=3x^2-6x=3x(x-2)\)。令\(y^\prime=0\)，得\(x=0\)或\(x=2\)。當(dāng)\(x\lt0\)，\(y^\prime\gt0\)；\(0\ltx\lt2\)，\(y^\prime\lt0\)；\(x\gt2\)，\(y^\prime\gt0\)。所以\(x=0\)是極大值點(diǎn)，\(y(0)=2\)；\(x=2\)是極小值點(diǎn)，\(y(2)=-2\)。2.已知數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n=n^2-n\)，討論數(shù)列\(zhòng)(\{a_n\}\)的性質(zhì)。答案：當(dāng)\(n=1\)時(shí)，\(a_1=S_1=0\)；當(dāng)\(n\geq2\)時(shí)，\(a_n=S_n-S_{n-1}=n^2-n-[(n-1)^2-(n-1)]=2n-2\)，\(n=1\)時(shí)也滿足。\(a_{n+1}-a_n=2\)，所以\(\{a_n\}\)是首項(xiàng)為\(0\)，公差為\(2\)的等差數(shù)列。3.討論函數(shù)\(f(x)=e^x-x\)的單調(diào)性。答案：\(f^\prime(x)=e^x-1\)。令\(f^\prime(x)=0\)，得\(x=0\)。當(dāng)\(x\lt0\)時(shí)，\(f^\prime(x)\lt0\)，\(f(x)\)單調(diào)遞減；當(dāng)\(x\gt0\)時(shí)，\(f^\prime(x)\gt0\)，\(f(x)\)單調(diào)遞增。4.對(duì)于等比數(shù)列\(zhòng)(\{a_n\}\)，公比\(q\gt1\)時(shí)，討論其前\(n\)項(xiàng)和\(S_n\)的變化情況。答案：\(S_n=\frac{a_1(1-q^n)}{1-q}(q\neq1)\)。當(dāng)\(a_1\gt0\)時(shí)，\(q^n\)隨\(n\)增大而增大，\(S_n\)單調(diào)遞增；當(dāng)\(a_1\lt0\)時(shí)，\(q^n\)隨\(n\)增大而增大，\(S_