2025年新高考導(dǎo)數(shù)考試題及答案

一、單項選擇題1.函數(shù)\(y=x^2+2x\)的導(dǎo)數(shù)為（）A.\(y'=2x\)B.\(y'=2x+2\)C.\(y'=x+2\)D.\(y'=2x^2+2\)答案：B2.若函數(shù)\(f(x)=e^x\)，則\(f'(0)\)的值為（）A.0B.1C.eD.\(\frac{1}{e}\)答案：B3.函數(shù)\(y=\sinx\)的導(dǎo)數(shù)是（）A.\(y'=\cosx\)B.\(y'=-\cosx\)C.\(y'=\sinx\)D.\(y'=-\sinx\)答案：A4.已知函數(shù)\(f(x)=x^3-3x^2+1\)，則\(f'(1)\)等于（）A.0B.-3C.3D.-6答案：B5.曲線\(y=\frac{1}{x}\)在點\((1,1)\)處的切線方程為（）A.\(x+y-2=0\)B.\(x-y-2=0\)C.\(x+y+2=0\)D.\(x-y+2=0\)答案：A6.函數(shù)\(f(x)=x\lnx\)的導(dǎo)數(shù)為（）A.\(f'(x)=\lnx+1\)B.\(f'(x)=\lnx\)C.\(f'(x)=x\lnx+1\)D.\(f'(x)=x\lnx\)答案：A7.若函數(shù)\(y=f(x)\)在點\((x_0,f(x_0))\)處的切線斜率為\(k=2\)，則\(f'(x_0)\)的值為（）A.0B.1C.2D.-2答案：C8.函數(shù)\(f(x)=\cos2x\)的導(dǎo)數(shù)是（）A.\(f'(x)=-2\sin2x\)B.\(f'(x)=2\sin2x\)C.\(f'(x)=-\sin2x\)D.\(f'(x)=\sin2x\)答案：A9.已知函數(shù)\(f(x)=\frac{1}{3}x^3-x\)，則\(f(x)\)的單調(diào)遞增區(qū)間是（）A.\((-\infty,-1)\)和\((1,+\infty)\)B.\((-1,1)\)C.\((-\infty,-\sqrt{3})\)和\((\sqrt{3},+\infty)\)D.\((-\sqrt{3},\sqrt{3})\)答案：A10.函數(shù)\(f(x)=e^x-x\)在區(qū)間\([-1,1]\)上的最小值是（）A.\(1+\frac{1}{e}\)B.1C.\(e-1\)D.0答案：B二、多項選擇題1.下列函數(shù)中，導(dǎo)數(shù)正確的是（）A.若\(y=x^5\)，則\(y'=5x^4\)B.若\(y=\frac{1}{x^2}\)，則\(y'=-\frac{2}{x^3}\)C.若\(y=\sqrt{x}\)，則\(y'=\frac{1}{2\sqrt{x}}\)D.若\(y=\cosx\)，則\(y'=\sinx\)答案：ABC2.曲線\(y=x^3-2x+1\)在點\((1,0)\)處的切線方程可能是（）A.\(x-y-1=0\)B.\(x+y-1=0\)C.\(2x-y-2=0\)D.\(2x+y-2=0\)答案：AC3.已知函數(shù)\(f(x)\)的導(dǎo)數(shù)\(f'(x)\)，且\(f'(x)>0\)的解集為\((a,b)\)，則下列說法正確的是（）A.\(f(x)\)在區(qū)間\((a,b)\)上單調(diào)遞增B.\(f(x)\)在區(qū)間\((-\infty,a)\)和\((b,+\infty)\)上單調(diào)遞減C.\(f(x)\)在\(x=a\)處取得極小值D.\(f(x)\)在\(x=b\)處取得極大值答案：AB4.函數(shù)\(f(x)=x^3-3x^2+2\)的極值點可能是（）A.\(x=0\)B.\(x=1\)C.\(x=2\)D.\(x=3\)答案：AC5.對于函數(shù)\(y=e^x\)，下列說法正確的是（）A.函數(shù)的導(dǎo)數(shù)\(y'=e^x\)B.函數(shù)在\(R\)上單調(diào)遞增C.函數(shù)圖象恒在\(x\)軸上方D.函數(shù)有最小值答案：ABC6.已知函數(shù)\(f(x)=\sinx+\cosx\)，則（）A.\(f'(x)=\cosx-\sinx\)B.\(f(x)\)的最大值為\(\sqrt{2}\)C.\(f(x)\)的最小正周期為\(2\pi\)D.\(f(x)\)在\((0,\frac{\pi}{4})\)上單調(diào)遞增答案：ABCD7.若函數(shù)\(f(x)=ax^2+bx+c\)（\(a\neq0\)）的導(dǎo)數(shù)為\(f'(x)\)，且\(f'(x)\)是一次函數(shù)，則（）A.\(f'(x)=2ax+b\)B.當(dāng)\(a>0\)時，\(f(x)\)的圖象開口向上C.\(f(x)\)的對稱軸為\(x=-\frac{2a}\)D.\(f(x)\)在對稱軸處取得最值答案：ABCD8.函數(shù)\(f(x)=\lnx\)的性質(zhì)有（）A.定義域為\((0,+\infty)\)B.導(dǎo)數(shù)\(f'(x)=\frac{1}{x}\)C.函數(shù)在定義域上單調(diào)遞增D.函數(shù)圖象過點\((1,0)\)答案：ABCD9.曲線\(y=x^4\)與直線\(y=kx\)相切，則（）A.切點坐標(biāo)為\((0,0)\)時，\(k=0\)B.切點坐標(biāo)為\((1,1)\)時，\(k=4\)C.切點坐標(biāo)為\((-1,1)\)時，\(k=-4\)D.切點坐標(biāo)為\((2,16)\)時，\(k=8\)答案：ABC10.已知函數(shù)\(f(x)\)滿足\(f'(x)=x^2-4\)，則（）A.\(f(x)=\frac{1}{3}x^3-4x+C\)（\(C\)為常數(shù)）B.\(f(x)\)的單調(diào)遞增區(qū)間是\((-\infty,-2)\)和\((2,+\infty)\)C.\(f(x)\)的單調(diào)遞減區(qū)間是\((-2,2)\)D.\(f(x)\)可能有極值答案：ABCD三、判斷題1.函數(shù)\(y=5\)的導(dǎo)數(shù)是\(y'=0\)。（）答案：對2.若函數(shù)\(f(x)\)在\(x=x_0\)處可導(dǎo)，則\(f(x)\)在\(x=x_0\)處一定連續(xù)。（）答案：對3.曲線\(y=f(x)\)在點\((x_0,f(x_0))\)處的切線與曲線\(y=f(x)\)只有一個交點。（）答案：錯4.函數(shù)\(f(x)=x^3\)的單調(diào)遞增區(qū)間是\((-\infty,+\infty)\)。（）答案：對5.若函數(shù)\(f(x)\)的導(dǎo)數(shù)\(f'(x)\)在區(qū)間\((a,b)\)上恒小于\(0\)，則\(f(x)\)在區(qū)間\((a,b)\)上單調(diào)遞減。（）答案：對6.函數(shù)\(y=\cos^2x\)的導(dǎo)數(shù)是\(y'=-2\cosx\sinx\)。（）答案：對7.函數(shù)\(f(x)=e^{-x}\)的導(dǎo)數(shù)是\(f'(x)=e^{-x}\)。（）答案：錯8.函數(shù)\(y=\lnx^2\)與\(y=2\lnx\)的導(dǎo)數(shù)相同。（）答案：錯9.若函數(shù)\(f(x)\)在區(qū)間\([a,b]\)上有最大值和最小值，則\(f(x)\)在區(qū)間\([a,b]\)上一定有極值。（）答案：錯10.曲線\(y=x^2\)在點\((-1,1)\)處的切線方程為\(2x+y+1=0\)。（）答案：對四、簡答題1.求函數(shù)\(y=x^3-4x^2+5x-1\)的導(dǎo)數(shù)。答案：根據(jù)求導(dǎo)公式\((x^n)^\prime=nx^{n-1}\)，對函數(shù)\(y=x^3-4x^2+5x-1\)求導(dǎo)。\(y^\prime=(x^3)^\prime-(4x^2)^\prime+(5x)^\prime-(1)^\prime=3x^2-8x+5\)。2.已知函數(shù)\(f(x)=x^2+2ax+1\)在\(x=1\)處的切線斜率為\(2\)，求\(a\)的值。答案：先對\(f(x)=x^2+2ax+1\)求導(dǎo)，\(f^\prime(x)=2x+2a\)。因為函數(shù)在\(x=1\)處切線斜率為\(2\)，即\(f^\prime(1)=2\)，將\(x=1\)代入\(f^\prime(x)\)得\(2\times1+2a=2\)，\(2+2a=2\)，\(2a=0\)，解得\(a=0\)。3.求函數(shù)\(f(x)=x^3-3x\)的極值。答案：對\(f(x)=x^3-3x\)求導(dǎo)得\(f^\prime(x)=3x^2-3\)，令\(f^\prime(x)=0\)，即\(3x^2-3=0\)，\(x^2=1\)，解得\(x=\pm1\)。當(dāng)\(x\lt-1\)時，\(f^\prime(x)>0\)；當(dāng)\(-1\ltx\lt1\)時，\(f^\prime(x)\lt0\)；當(dāng)\(x>1\)時，\(f^\prime(x)>0\)。所以\(x=-1\)時取極大值\(f(-1)=(-1)^3-3\times(-1)=2\)；\(x=1\)時取極小值\(f(1)=1^3-3\times1=-2\)。4.求曲線\(y=e^x\)在點\((0,1)\)處的切線方程。答案：先對\(y=e^x\)求導(dǎo)，\(y^\prime=e^x\)。曲線在點\((0,1)\)處的切線斜率\(k=y^\prime|_{x=0}=e^0=1\)。根據(jù)點斜式方程\(y-y_0=k(x-x_0)\)（其中\(zhòng)((x_0,y_0)=(0,1)\)，\(k=1\)），可得切線方程為\(y-1=1\times(x-0)\)，即\(y=x+1\)。五、討論題1.討論函數(shù)\(f(x)=x^3-3x^2+3x\)的單調(diào)性與極值情況。答案：對\(f(x)=x^3-3x^2+3x\)求導(dǎo)得\(f^\prime(x)=3x^2-6x+3=3(x^2-2x+1)=3(x-1)^2\)。因為\((x-1)^2\geq0\)恒成立，所以\(f^\prime(x)\geq0\)恒成立，僅當(dāng)\(x=1\)時\(f^\prime(x)=0\)。這說明函數(shù)\(f(x)\)在\((-\infty,+\infty)\)上單調(diào)遞增，無極值。2.已知函數(shù)\(f(x)=ax^3+bx^2+cx\)（\(a\neq0\)），其導(dǎo)數(shù)\(f^\prime(x)\)的圖象是開口向下的拋物線且與\(x\)軸有兩個交點，討論\(f(x)\)的單調(diào)性與極值情況。答案：\(f^\prime(x)=3ax^2+2bx+c\)，因為\(f^\prime(x)\)圖象開口向下，所以\(a\lt0\)，且與\(x\)軸有兩個交點設(shè)為\(x_1,x_2\)（\(x_1\ltx_2\)）。當(dāng)\(x\ltx_1\)或\(x>x_2\)時，\(f^\prime(x)\lt0\)，\(f(x)\)單調(diào)遞減；當(dāng)\(x_1\ltx\ltx_2\)時，\(f^\prime(x)>0\)，\(f(x)\)單調(diào)遞增。所以\(f(x)\)在\(x=x_1\)處取得極小值，在\(x=x_2\)處取得極大值。3.討論函數(shù)\(y=\lnx-x\)在區(qū)間\((0,+\infty)\)上的單調(diào)性、極值及最值情況。答案：對\(y=\lnx-x\)求導(dǎo)得\(y^\prime=\frac{1}{x}-1=\frac{1-x}{x}\)。令\(y^\prime=0\)，得\(x=1\)。當(dāng)\(0\ltx\lt1\)時，\(y^\prime>0\)，函數(shù)單調(diào)遞增；當(dāng)\(x>1\)時，\(y^\prime\lt0\)，函數(shù)單調(diào)遞減。所以\(x=1\)處取極大值\(y(1)=\ln1-1=-1\)，此即為最大值，函數(shù)在\((0,+\infty)\)上無最小值。4.已知函數(shù)\(f(x)=\sinx+\cosx\)，討論其在區(qū)間\([0,2\pi]\)上的單調(diào)性與極值情況。答案：\(f(x)=\sinx+\c