數(shù)學(xué)高考真題導(dǎo)數(shù)及答案

一、單項(xiàng)選擇題1.函數(shù)\(f(x)=x^3-3x^2+1\)的導(dǎo)數(shù)\(f^\prime(x)\)為（）A.\(3x^2-6x\)B.\(3x^2+6x\)C.\(x^2-3x\)D.\(x^2+3x\)答案：A2.曲線\(y=e^x\)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.\(y=-x-1\)答案：A3.已知函數(shù)\(f(x)\)的導(dǎo)函數(shù)為\(f^\prime(x)\)，且滿足\(f(x)=2xf^\prime(1)+\lnx\)，則\(f^\prime(1)\)等于（）A.\(-e\)B.\(-1\)C.\(1\)D.\(e\)答案：B4.函數(shù)\(f(x)=x^2-2\lnx\)的單調(diào)遞減區(qū)間是（）A.\((0,1)\)B.\((1,+\infty)\)C.\((-\infty,1)\)D.\((-1,1)\)答案：A5.若函數(shù)\(f(x)=ax^3+3x^2+x+b(a\gt0,b\inR)\)恰好有三個(gè)不同的單調(diào)區(qū)間，則實(shí)數(shù)\(a\)的取值范圍是（）A.\((0,3)\cup(3,+\infty)\)B.\([3,+\infty)\)C.\((0,3]\)D.\((0,3)\)答案：D6.函數(shù)\(f(x)=\frac{1}{3}x^3-4x+4\)在區(qū)間\([0,3]\)上的最大值為（）A.\(4\)B.\(1\)C.\(\frac{28}{3}\)D.\(\frac{4}{3}\)答案：A7.已知函數(shù)\(f(x)\)在\(R\)上可導(dǎo)，其導(dǎo)函數(shù)為\(f^\prime(x)\)，若\(f(x)\)滿足\((x-1)[f^\prime(x)-f(x)]\gt0\)，\(f(2-x)=f(x)e^{2-2x}\)，則下列判斷一定正確的是（）A.\(f(1)\ltf(0)\)B.\(f(2)\gte^2f(0)\)C.\(f(3)\gte^3f(0)\)D.\(f(4)\lte^4f(0)\)答案：C8.已知函數(shù)\(f(x)=x^3+ax^2+bx+a^2\)在\(x=1\)處有極值\(10\)，則\(f(2)\)等于（）A.\(11\)或\(18\)B.\(11\)C.\(18\)D.\(17\)或\(18\)答案：C9.設(shè)函數(shù)\(f(x)\)是定義在\(R\)上的奇函數(shù)，\(f^\prime(x)\)為其導(dǎo)函數(shù)，已知\(f(1)=0\)，當(dāng)\(x\gt0\)時(shí)，\(f(x)+xf^\prime(x)\lt0\)，則不等式\(x\cdotf(x)\gt0\)的解集為（）A.\((-1,0)\cup(0,1)\)B.\((-\infty,-1)\cup(1,+\infty)\)C.\((-1,0)\cup(1,+\infty)\)D.\((-\infty,-1)\cup(0,1)\)答案：A10.已知函數(shù)\(f(x)=e^x-\ln(x+m)\)，當(dāng)\(m\leqslant2\)時(shí)，下列結(jié)論正確的是（）A.\(f(x)\)有兩個(gè)零點(diǎn)B.\(f(x)\)有一個(gè)零點(diǎn)C.\(f(x)\)沒有零點(diǎn)D.\(f(x)\)有無數(shù)個(gè)零點(diǎn)答案：B二、多項(xiàng)選擇題1.下列求導(dǎo)運(yùn)算正確的是（）A.\((\sinx)^\prime=\cosx\)B.\((\log_2x)^\prime=\frac{1}{x\ln2}\)C.\((e^{-x})^\prime=e^{-x}\)D.\((\sqrt{x})^\prime=\frac{1}{2\sqrt{x}}\)答案：ABD2.已知函數(shù)\(f(x)=x^3-3x\)，則（）A.\(f(x)\)是奇函數(shù)B.\(f(x)\)在\(x=1\)處取得極大值C.\(f(x)\)在\((1,+\infty)\)上單調(diào)遞增D.\(f(x)\)的圖象關(guān)于點(diǎn)\((0,0)\)對(duì)稱答案：ABCD3.若函數(shù)\(f(x)=x^3-3ax^2+3bx\)在\(x=1\)處有極值，則下列說法正確的是（）A.\(a-b=1\)B.\(f(1)\)是極大值C.\(f(1)\)是極小值D.\(f^\prime(1)=0\)答案：AD4.已知函數(shù)\(f(x)\)的導(dǎo)函數(shù)\(f^\prime(x)\)的圖象如圖所示，則（）A.函數(shù)\(f(x)\)在\((-\infty,-2)\)上單調(diào)遞增B.函數(shù)\(f(x)\)在\(x=-2\)處取得極大值C.函數(shù)\(f(x)\)在\((-2,2)\)上單調(diào)遞減D.函數(shù)\(f(x)\)在\(x=2\)處取得極小值答案：ABCD5.設(shè)函數(shù)\(f(x)=e^x-ax-1\)，則（）A.當(dāng)\(a=1\)時(shí)，\(f(x)\)在\((0,+\infty)\)上單調(diào)遞增B.當(dāng)\(a=0\)時(shí)，\(f(x)\)的最小值為\(0\)C.對(duì)任意\(a\lt0\)，\(f(x)\)在\(R\)上單調(diào)遞增D.存在\(a\gt0\)，使得\(f(x)\)有兩個(gè)零點(diǎn)答案：ACD6.已知函數(shù)\(f(x)=\frac{\lnx}{x}\)，則（）A.\(f(x)\)在\((0,e)\)上單調(diào)遞增B.\(f(x)\)在\((e,+\infty)\)上單調(diào)遞減C.\(f(x)\)的最大值為\(\frac{1}{e}\)D.\(f(x)\)在\((0,+\infty)\)上有兩個(gè)零點(diǎn)答案：ABC7.函數(shù)\(f(x)=x^2e^x\)的導(dǎo)數(shù)\(f^\prime(x)\)的零點(diǎn)有（）A.\(0\)B.\(-2\)C.\(2\)D.\(1\)答案：AB8.已知函數(shù)\(f(x)\)滿足\(f(x)=f^\prime(1)e^{x-1}-f(0)x+\frac{1}{2}x^2\)，則（）A.\(f(0)=1\)B.\(f^\prime(1)=e\)C.\(f(x)\)的單調(diào)遞增區(qū)間是\((0,+\infty)\)D.\(f(x)\)的單調(diào)遞減區(qū)間是\((-\infty,0)\)答案：ABD9.對(duì)于函數(shù)\(f(x)=x\lnx\)，下列說法正確的是（）A.\(f(x)\)在\((0,\frac{1}{e})\)上單調(diào)遞減B.\(f(x)\)在\((\frac{1}{e},+\infty)\)上單調(diào)遞增C.\(f(x)\)在\(x=\frac{1}{e}\)處取得極小值\(-\frac{1}{e}\)D.\(f(x)\)在\((0,+\infty)\)上無最大值答案：ABCD10.已知函數(shù)\(f(x)\)在\(R\)上可導(dǎo)，其導(dǎo)函數(shù)為\(f^\prime(x)\)，若\(f(x)\)滿足\(f^\prime(x)-f(x)\lt0\)，且\(f(0)=2\)，則（）A.\(f(1)\lt2e\)B.\(f(2)\lt2e^2\)C.不等式\(f(x)\gt2e^x\)的解集為\((-\infty,0)\)D.\(f(x)\)在\(R\)上單調(diào)遞減答案：ABC三、判斷題1.若函數(shù)\(y=f(x)\)在區(qū)間\((a,b)\)內(nèi)有\(zhòng)(f^\prime(x)\gt0\)，則\(y=f(x)\)在區(qū)間\((a,b)\)內(nèi)單調(diào)遞增。（）答案：對(duì)2.函數(shù)\(f(x)\)在\(x=x_0\)處的導(dǎo)數(shù)\(f^\prime(x_0)\)的幾何意義是曲線\(y=f(x)\)在點(diǎn)\((x_0,f(x_0))\)處的切線的斜率。（）答案：對(duì)3.函數(shù)\(f(x)=x^3\)在\(R\)上有極值。（）答案：錯(cuò)4.若函數(shù)\(f(x)\)在區(qū)間\([a,b]\)上的圖象是一條連續(xù)不斷的曲線，則\(f(x)\)在區(qū)間\([a,b]\)上一定有最值。（）答案：對(duì)5.函數(shù)\(y=\sinx\)的導(dǎo)數(shù)\(y^\prime=\cosx\)。（）答案：對(duì)6.若函數(shù)\(f(x)\)在\(x=x_0\)處可導(dǎo)，且\(f(x)\)在\(x=x_0\)處取得極值，則\(f^\prime(x_0)=0\)。（）答案：對(duì)7.函數(shù)\(f(x)=e^x\)的導(dǎo)數(shù)\(f^\prime(x)=e^x\)。（）答案：對(duì)8.函數(shù)\(f(x)\)的導(dǎo)數(shù)\(f^\prime(x)\)在區(qū)間\((a,b)\)上大于零，則\(f(x)\)在區(qū)間\((a,b)\)上的圖象是下凸的。（）答案：錯(cuò)9.若函數(shù)\(f(x)\)在區(qū)間\((a,b)\)內(nèi)有\(zhòng)(f^\prime(x)\lt0\)，則\(f(x)\)在區(qū)間\((a,b)\)內(nèi)單調(diào)遞減。（）答案：對(duì)10.函數(shù)\(f(x)=\lnx\)的導(dǎo)數(shù)\(f^\prime(x)=\frac{1}{x}\)。（）答案：對(duì)四、簡答題1.求函數(shù)\(f(x)=x^3-3x^2+2\)的單調(diào)區(qū)間。答案：先對(duì)函數(shù)求導(dǎo)，\(f^\prime(x)=3x^2-6x=3x(x-2)\)。令\(f^\prime(x)\gt0\)，即\(3x(x-2)\gt0\)，解得\(x\lt0\)或\(x\gt2\)，所以\(f(x)\)的單調(diào)遞增區(qū)間為\((-\infty,0)\)和\((2,+\infty)\)。令\(f^\prime(x)\lt0\)，即\(3x(x-2)\lt0\)，解得\(0\ltx\lt2\)，所以\(f(x)\)的單調(diào)遞減區(qū)間為\((0,2)\)。2.求曲線\(y=x^2e^x\)在點(diǎn)\((1,e)\)處的切線方程。答案：首先求導(dǎo)，根據(jù)求導(dǎo)公式\((uv)^\prime=u^\primev+uv^\prime\)，\(y^\prime=(x^2)^\primee^x+x^2(e^x)^\prime=2xe^x+x^2e^x\)。把\(x=1\)代入導(dǎo)函數(shù)得\(y^\prime|_{x=1}=2e+e=3e\)，即切線的斜率為\(3e\)。由點(diǎn)斜式可得切線方程為\(y-e=3e(x-1)\)，整理得\(3ex-y-2e=0\)。3.已知函數(shù)\(f(x)=ax^3+bx^2+cx\)在\(x=1\)處取得極值，且\(f^\prime(0)=3\)，求\(a\)，\(b\)，\(c\)的值。答案：對(duì)\(f(x)\)求導(dǎo)得\(f^\prime(x)=3ax^2+2bx+c\)。因?yàn)閈(f(x)\)在\(x=1\)處取得極值，所以\(f^\prime(1)=3a+2b+c=0\)。又因?yàn)閈(f^\prime(0)=3\)，即\(c=3\)。把\(c=3\)代入\(3a+2b+c=0\)得\(3a+2b+3=0\)。還需其他條件才能確定\(a\)，\(b\)的值，僅現(xiàn)有條件可表示\(b=-\frac{3a+3}{2}\)。4.求函數(shù)\(f(x)=\frac{\lnx}{x}\)在區(qū)間\([1,e]\)上的最大值和最小值。答案：對(duì)\(f(x)\)求導(dǎo)，\(f^\prime(x)=\frac{\frac{1}{x}\cdotx-\lnx}{x^2}=\frac{1-\lnx}{x^2}\)。在區(qū)間\([1,e]\)上，令\(f^\prime(x)=0\)，即\(1-\lnx=0\)，解得\(x=e\)。當(dāng)\(x\in[1,e)\)時(shí)，\(f^\prime(x)\gt0\)，\(f(x)\)單調(diào)遞增；當(dāng)\(x=e\)時(shí)，\(f^\prime(x)=0\)。所以\(f(x)\)在\(x=e\)處取得最大值\(f(e)=\frac{1}{e}\)，在\(x=1\)處取得最小值\(f(1)=0\)。五、討論題1.討論函數(shù)\(f(x)=x^3-3ax+1\)的單調(diào)性與極值情況。答案：對(duì)\(f(x)\)求導(dǎo)得\(f^\prime(x)=3x^2-3a=3(x^2-a)\)。當(dāng)\(a\leqslant0\)時(shí)，\(f^\prime(x)\geqslant0\)恒成立，\(f(x)\)在\(R\)上單調(diào)遞增，無極值。當(dāng)\(a\gt0\)時(shí)，令\(f^\prime(x)=0\)，得\(x=\pm\sqrt{a}\)。當(dāng)\(x\lt-\sqrt{a}\)或\(x\gt\sqrt{a}\)時(shí)，\(f^\prime(x)\gt0\)，\(f(x)\)單調(diào)遞增；當(dāng)\(-\sqrt{a}\ltx\lt\sqr