2025成年人數(shù)學(xué)導(dǎo)數(shù)單調(diào)性應(yīng)用考試題及答案

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.\(0\)2.函數(shù)\(f(x)=x^{3}-3x\)的單調(diào)遞增區(qū)間是（）A.\((-\infty,-1)\)B.\((-1,1)\)C.\((1,+\infty)\)D.\((-\infty,-1)\)和\((1,+\infty)\)3.若函數(shù)\(f(x)\)在區(qū)間\((a,b)\)內(nèi)導(dǎo)數(shù)\(f^\prime(x)>0\)，則\(f(x)\)在\((a,b)\)上（）A.單調(diào)遞減B.單調(diào)遞增C.先增后減D.先減后增4.函數(shù)\(y=\lnx\)的導(dǎo)數(shù)是（）A.\(\frac{1}{x}\)B.\(x\)C.\(\lnx\)D.\(\frac{1}{x^{2}}\)5.函數(shù)\(f(x)=x^{2}+2x-1\)的極小值點(diǎn)是（）A.\(x=-1\)B.\(x=1\)C.\(x=0\)D.\(x=2\)6.函數(shù)\(y=e^{x}\)的導(dǎo)數(shù)是（）A.\(e^{x}\)B.\(xe^{x}\)C.\(e^{-x}\)D.\(\frac{1}{e^{x}}\)7.函數(shù)\(f(x)=x^{4}-2x^{2}+5\)的單調(diào)遞減區(qū)間是（）A.\((-\infty,-1)\)和\((0,1)\)B.\((-1,0)\)和\((1,+\infty)\)C.\((-\infty,-1)\)D.\((1,+\infty)\)8.若\(f^\prime(x_{0})=0\)，則\(x_{0}\)一定是\(f(x)\)的（）A.極大值點(diǎn)B.極小值點(diǎn)C.極值點(diǎn)D.可能是極值點(diǎn)9.函數(shù)\(y=\sinx\)的導(dǎo)數(shù)是（）A.\(\cosx\)B.\(-\cosx\)C.\(\sinx\)D.\(-\sinx\)10.函數(shù)\(f(x)=x^{3}+ax^{2}+bx+c\)，\(f^\prime(1)=0\)，\(f^\prime(2)=0\)，則\(a+b\)的值為（）A.\(-11\)B.\(11\)C.\(-9\)D.\(9\)二、多項(xiàng)選擇題（每題2分，共10題）1.以下函數(shù)導(dǎo)數(shù)正確的有（）A.\((x^{5})^\prime=5x^{4}\)B.\((\cosx)^\prime=\sinx\)C.\((e^{2x})^\prime=2e^{2x}\)D.\((\ln2x)^\prime=\frac{1}{x}\)2.函數(shù)\(f(x)=x^{3}-3x^{2}+2\)的單調(diào)區(qū)間說(shuō)法正確的是（）A.單調(diào)遞增區(qū)間是\((-\infty,0)\)B.單調(diào)遞增區(qū)間是\((2,+\infty)\)C.單調(diào)遞減區(qū)間是\((0,2)\)D.單調(diào)遞減區(qū)間是\((-\infty,2)\)3.若函數(shù)\(y=f(x)\)在區(qū)間\([a,b]\)上可導(dǎo)且單調(diào)遞增，則（）A.\(f^\prime(x)\geqslant0\)在\([a,b]\)上恒成立B.\(f^\prime(x)>0\)在\((a,b)\)內(nèi)恒成立C.\(f^\prime(a)\geqslant0\)D.\(f^\prime(b)\geqslant0\)4.函數(shù)\(f(x)\)在\(x=x_{0}\)處取得極值的必要條件有（）A.\(f^\prime(x_{0})=0\)B.\(f(x)\)在\(x=x_{0}\)處連續(xù)C.\(f^\prime(x)\)在\(x=x_{0}\)兩側(cè)異號(hào)D.\(f(x)\)在\(x=x_{0}\)處可導(dǎo)5.以下函數(shù)中，在\((0,+\infty)\)上單調(diào)遞增的有（）A.\(y=x^{3}\)B.\(y=\lnx\)C.\(y=e^{x}\)D.\(y=\frac{1}{x}\)6.函數(shù)\(f(x)=x^{2}-4x+3\)的極值情況正確的是（）A.極小值為\(-1\)B.極小值點(diǎn)為\(x=2\)C.極大值為\(3\)D.不存在極大值7.對(duì)于函數(shù)\(y=f(x)\)，導(dǎo)數(shù)\(f^\prime(x)\)與單調(diào)性的關(guān)系，下列說(shuō)法正確的是（）A.\(f^\prime(x)>0\)的區(qū)間是\(f(x)\)的單調(diào)遞增區(qū)間B.\(f^\prime(x)<0\)的區(qū)間是\(f(x)\)的單調(diào)遞減區(qū)間C.\(f(x)\)單調(diào)遞增則\(f^\prime(x)>0\)D.\(f(x)\)單調(diào)遞減則\(f^\prime(x)<0\)8.已知函數(shù)\(f(x)\)的導(dǎo)數(shù)\(f^\prime(x)=3x^{2}-6x\)，則（）A.\(f(x)\)的單調(diào)遞增區(qū)間是\((-\infty,0)\)和\((2,+\infty)\)B.\(f(x)\)的單調(diào)遞減區(qū)間是\((0,2)\)C.\(x=0\)是極大值點(diǎn)D.\(x=2\)是極小值點(diǎn)9.下列函數(shù)中導(dǎo)數(shù)不為\(1\)的有（）A.\(y=x+1\)B.\(y=x^{2}\)C.\(y=\lnx\)D.\(y=e^{x}\)10.函數(shù)\(f(x)\)在區(qū)間\((a,b)\)內(nèi)可導(dǎo)，\(x_{1},x_{2}\in(a,b)\)且\(x_{1}<x_{2}\)，若\(f(x)\)單調(diào)遞增，則（）A.\(f(x_{1})<f(x_{2})\)B.\(f^\prime(x_{1})\leqslantf^\prime(x_{2})\)C.\(f(x_{1})\leqslantf(x_{2})\)D.\(f^\prime(x_{1})<f^\prime(x_{2})\)三、判斷題（每題2分，共10題）1.函數(shù)\(y=5\)的導(dǎo)數(shù)是\(5\)。（）2.若\(f^\prime(x)>0\)在區(qū)間\((a,b)\)內(nèi)成立，則\(f(x)\)在\([a,b]\)上單調(diào)遞增。（）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.函數(shù)\(y=x^{3}\)的單調(diào)遞增區(qū)間是\((-\infty,+\infty)\)。（）5.函數(shù)\(f(x)\)的導(dǎo)數(shù)\(f^\prime(x)\)在區(qū)間\((a,b)\)內(nèi)小于\(0\)，則\(f(x)\)在\((a,b)\)內(nèi)單調(diào)遞減。（）6.函數(shù)\(y=\cosx\)的導(dǎo)數(shù)是\(\sinx\)。（）7.若函數(shù)\(f(x)\)在區(qū)間\((a,b)\)上單調(diào)遞增，則\(f^\prime(x)\)在\((a,b)\)內(nèi)恒大于\(0\)。（）8.函數(shù)\(f(x)=x^{2}\)在\(x=0\)處取得極小值。（）9.函數(shù)\(y=\ln(x+1)\)的導(dǎo)數(shù)是\(\frac{1}{x+1}\)。（）10.函數(shù)\(f(x)\)在某點(diǎn)處導(dǎo)數(shù)不存在，則該點(diǎn)一定不是極值點(diǎn)。（）四、簡(jiǎn)答題（每題5分，共4題）1.簡(jiǎn)述利用導(dǎo)數(shù)判斷函數(shù)單調(diào)性的步驟。答案：先求函數(shù)定義域，再對(duì)函數(shù)求導(dǎo)，接著令導(dǎo)數(shù)大于\(0\)，解不等式得單調(diào)遞增區(qū)間；令導(dǎo)數(shù)小于\(0\)，解不等式得單調(diào)遞減區(qū)間。2.已知函數(shù)\(f(x)=x^{3}-3x\)，求其極值。答案：對(duì)\(f(x)\)求導(dǎo)得\(f^\prime(x)=3x^{2}-3\)，令\(f^\prime(x)=0\)，解得\(x=\pm1\)。當(dāng)\(x=-1\)時(shí)，\(f(-1)=2\)為極大值；當(dāng)\(x=1\)時(shí)，\(f(1)=-2\)為極小值。3.函數(shù)\(y=x^{2}+bx+c\)在\(x=1\)處取得極值\(-1\)，求\(b\)，\(c\)的值。答案：求導(dǎo)得\(y^\prime=2x+b\)，因?yàn)樵赲(x=1\)處取得極值，所以\(y^\prime(1)=2+b=0\)，得\(b=-2\)。又\(f(1)=1+b+c=-1\)，將\(b=-2\)代入得\(c=0\)。4.說(shuō)明函數(shù)極值與單調(diào)性的關(guān)系。答案：函數(shù)在單調(diào)遞增區(qū)間與單調(diào)遞減區(qū)間的轉(zhuǎn)折點(diǎn)處可能取得極值。在極值點(diǎn)兩側(cè)函數(shù)單調(diào)性改變，左側(cè)遞增右側(cè)遞減為極大值點(diǎn)，左側(cè)遞減右側(cè)遞增為極小值點(diǎn)。五、討論題（每題5分，共4題）1.討論函數(shù)\(f(x)=x^{3}-ax^{2}\)的單調(diào)性與\(a\)的取值關(guān)系。答案：求導(dǎo)得\(f^\prime(x)=3x^{2}-2ax=x(3x-2a)\)。當(dāng)\(a=0\)時(shí)，\(f^\prime(x)\geqslant0\)，\(f(x)\)在\(R\)上遞增；當(dāng)\(a>0\)時(shí)，\(f(x)\)在\((-\infty,0)\)和\((\frac{2a}{3},+\infty)\)遞增，在\((0,\frac{2a}{3})\)遞減；當(dāng)\(a<0\)時(shí)，\(f(x)\)在\((-\infty,\frac{2a}{3})\)和\((0,+\infty)\)遞增，在\((\frac{2a}{3},0)\)遞減。2.已知函數(shù)\(f(x)\)在區(qū)間\([a,b]\)上連續(xù)且可導(dǎo)，若\(f^\prime(x)\geqslant0\)，且\(f^\prime(x)\)在\((a,b)\)內(nèi)只有有限個(gè)零點(diǎn)，討論\(f(x)\)的單調(diào)性。答案：因?yàn)閈(f^\prime(x)\geqslant0\)，且\(f^\prime(x)\)在\((a,b)\)內(nèi)只有有限個(gè)零點(diǎn)。在非零點(diǎn)處\(f^\prime(x)>0\)，函數(shù)單調(diào)遞增，有限個(gè)零點(diǎn)不影響整體趨勢(shì)，所以\(f(x)\)在\([a,b]\)上單調(diào)遞增。3.討論函數(shù)\(y=\frac{1}{x}+x\)的單調(diào)性。答案：求導(dǎo)得\(y^\prime=1-\frac{1}{x^{2}}=\frac{x^{2}-1}{x^{2}}\)。令\(y^\prime>0\)，解得\(x>1\)或\(x<-1\)，此時(shí)函數(shù)遞增；令\(y^\prime<0\)，解得\(-1<x<0\)或\(0<x<1\)，此時(shí)函數(shù)遞減，定義域?yàn)閈(x\neq0\)。4.若函數(shù)\(f(x)\)與\(g(x)\)在區(qū)間\((a,b)\)內(nèi)導(dǎo)數(shù)相等，討論\(f(x)\)與\(g(x)\)的關(guān)系。答案：因?yàn)閈(f(x)\)與\(g(x)\)在區(qū)間\((a,b)\)內(nèi)導(dǎo)數(shù)相等，根據(jù)導(dǎo)數(shù)性質(zhì)，\(f(x)\)與\(g(x)\)相差一個(gè)常數(shù)。即\(