2025青海省成人高考專升本高等數(shù)學(xué)考試題庫及答案一、函數(shù)、極限與連續(xù)（一）選擇題1.函數(shù)\(y=\frac{1}{\ln(x-1)}\)的定義域是（）A.\((1,+\infty)\)B.\((0,1)\cup(1,+\infty)\)C.\((1,2)\cup(2,+\infty)\)D.\((0,2)\cup(2,+\infty)\)答案：C解析：要使函數(shù)\(y=\frac{1}{\ln(x-1)}\)有意義，則\(\begin{cases}x-1>0\\\ln(x-1)\neq0\end{cases}\)。由\(x-1>0\)得\(x>1\)；由\(\ln(x-1)\neq0\)，即\(x-1\neq1\)，得\(x\neq2\)。所以定義域為\((1,2)\cup(2,+\infty)\)。2.\(\lim\limits_{x\to0}\frac{\sin3x}{x}\)的值為（）A.0B.1C.2D.3答案：D解析：根據(jù)重要極限\(\lim\limits_{u\to0}\frac{\sinu}{u}=1\)，令\(u=3x\)，當(dāng)\(x\to0\)時，\(u\to0\)，則\(\lim\limits_{x\to0}\frac{\sin3x}{x}=\lim\limits_{x\to0}\frac{\sin3x}{x}\times\frac{3}{3}=3\lim\limits_{x\to0}\frac{\sin3x}{3x}=3\times1=3\)。（二）填空題1.已知\(f(x)=\frac{1}{1-x}\)，則\(f(f(x))=\)______。答案：\(\frac{x-1}{x}\)解析：將\(f(x)=\frac{1}{1-x}\)代入\(f(f(x))\)中，得到\(f(f(x))=\frac{1}{1-f(x)}=\frac{1}{1-\frac{1}{1-x}}=\frac{1}{\frac{1-x-1}{1-x}}=\frac{1-x}{-x}=\frac{x-1}{x}\)，\(x\neq0\)且\(x\neq1\)。2.\(\lim\limits_{x\to\infty}(1+\frac{2}{x})^{x}=\)______。答案：\(e^{2}\)解析：根據(jù)重要極限\(\lim\limits_{x\to\infty}(1+\frac{1}{x})^{x}=e\)，令\(t=\frac{x}{2}\)，則\(x=2t\)，當(dāng)\(x\to\infty\)時，\(t\to\infty\)。\(\lim\limits_{x\to\infty}(1+\frac{2}{x})^{x}=\lim\limits_{t\to\infty}(1+\frac{1}{t})^{2t}=\left[\lim\limits_{t\to\infty}(1+\frac{1}{t})^{t}\right]^{2}=e^{2}\)。（三）解答題1.求\(\lim\limits_{x\to1}\frac{x^{2}-1}{x-1}\)。解：對分子進(jìn)行因式分解，\(x^{2}-1=(x+1)(x-1)\)，則\(\lim\limits_{x\to1}\frac{x^{2}-1}{x-1}=\lim\limits_{x\to1}\frac{(x+1)(x-1)}{x-1}=\lim\limits_{x\to1}(x+1)\)，將\(x=1\)代入\(x+1\)得\(1+1=2\)。2.討論函數(shù)\(f(x)=\begin{cases}x+1,x<0\\0,x=0\\x-1,x>0\end{cases}\)在\(x=0\)處的連續(xù)性。解：首先求左極限\(\lim\limits_{x\to0^{-}}f(x)=\lim\limits_{x\to0^{-}}(x+1)=0+1=1\)；右極限\(\lim\limits_{x\to0^{+}}f(x)=\lim\limits_{x\to0^{+}}(x-1)=0-1=-1\)；函數(shù)值\(f(0)=0\)。因為\(\lim\limits_{x\to0^{-}}f(x)\neq\lim\limits_{x\to0^{+}}f(x)\)，所以\(\lim\limits_{x\to0}f(x)\)不存在，函數(shù)\(f(x)\)在\(x=0\)處不連續(xù)。二、一元函數(shù)微分學(xué)（一）選擇題1.設(shè)\(y=\sinx\)，則\(y'\)等于（）A.\(\cosx\)B.\(-\cosx\)C.\(\sinx\)D.\(-\sinx\)答案：A解析：根據(jù)求導(dǎo)公式\((\sinx)'=\cosx\)。2.曲線\(y=x^{3}\)在點\((1,1)\)處的切線方程為（）A.\(y=3x-2\)B.\(y=-3x+4\)C.\(y=3x+2\)D.\(y=-3x-4\)答案：A解析：首先對\(y=x^{3}\)求導(dǎo)，\(y'=3x^{2}\)，將\(x=1\)代入導(dǎo)數(shù)\(y'\)中，得到切線的斜率\(k=y'\vert_{x=1}=3\times1^{2}=3\)。根據(jù)點斜式方程\(y-y_{0}=k(x-x_{0})\)（其中\(zhòng)((x_{0},y_{0})=(1,1)\)，\(k=3\)），可得切線方程為\(y-1=3(x-1)\)，即\(y=3x-2\)。（二）填空題1.設(shè)\(y=e^{2x}\)，則\(y''=\)______。答案：\(4e^{2x}\)解析：先求一階導(dǎo)數(shù)，根據(jù)復(fù)合函數(shù)求導(dǎo)法則，若\(y=e^{u}\)，\(u=2x\)，則\(y'=\frac{dy}{du}\cdot\frac{du}{dx}=e^{u}\cdot2=2e^{2x}\)。再求二階導(dǎo)數(shù)\(y''=(2e^{2x})'=2\times2e^{2x}=4e^{2x}\)。2.函數(shù)\(y=x^{2}-2x+3\)的單調(diào)遞減區(qū)間是______。答案：\((-\infty,1)\)解析：對函數(shù)\(y=x^{2}-2x+3\)求導(dǎo)，\(y'=2x-2\)。令\(y'<0\)，即\(2x-2<0\)，解得\(x<1\)，所以函數(shù)的單調(diào)遞減區(qū)間是\((-\infty,1)\)。（三）解答題1.設(shè)\(y=x\lnx\)，求\(y'\)。解：根據(jù)乘積的求導(dǎo)法則\((uv)'=u'v+uv'\)，這里\(u=x\)，\(v=\lnx\)。\(u'=(x)'=1\)，\(v'=(\lnx)'=\frac{1}{x}\)，則\(y'=(x\lnx)'=1\times\lnx+x\times\frac{1}{x}=\lnx+1\)。2.求函數(shù)\(f(x)=x^{3}-3x^{2}+1\)的極值。解：首先求導(dǎo)\(f'(x)=3x^{2}-6x=3x(x-2)\)。令\(f'(x)=0\)，即\(3x(x-2)=0\)，解得\(x=0\)或\(x=2\)。然后求二階導(dǎo)數(shù)\(f''(x)=6x-6\)。當(dāng)\(x=0\)時，\(f''(0)=6\times0-6=-6<0\)，所以\(f(x)\)在\(x=0\)處取得極大值，\(f(0)=0^{3}-3\times0^{2}+1=1\)。當(dāng)\(x=2\)時，\(f''(2)=6\times2-6=6>0\)，所以\(f(x)\)在\(x=2\)處取得極小值，\(f(2)=2^{3}-3\times2^{2}+1=8-12+1=-3\)。三、一元函數(shù)積分學(xué)（一）選擇題1.\(\intx^{2}dx\)等于（）A.\(\frac{1}{3}x^{3}+C\)B.\(\frac{1}{2}x^{2}+C\)C.\(x^{3}+C\)D.\(2x+C\)答案：A解析：根據(jù)不定積分公式\(\intx^{n}dx=\frac{1}{n+1}x^{n+1}+C\)（\(n\neq-1\)），當(dāng)\(n=2\)時，\(\intx^{2}dx=\frac{1}{3}x^{3}+C\)。2.\(\int_{0}^{1}2xdx\)的值為（）A.0B.1C.2D.3答案：B解析：先求\(\int2xdx=x^{2}+C\)，再根據(jù)牛頓-萊布尼茨公式\(\int_{a}^F'(x)dx=F(b)-F(a)\)，這里\(F(x)=x^{2}\)，\(a=0\)，\(b=1\)，則\(\int_{0}^{1}2xdx=x^{2}\vert_{0}^{1}=1^{2}-0^{2}=1\)。（二）填空題1.若\(\intf(x)dx=F(x)+C\)，則\(\intf(2x)dx=\)______。答案：\(\frac{1}{2}F(2x)+C\)解析：令\(u=2x\)，則\(du=2dx\)，\(dx=\frac{1}{2}du\)。\(\intf(2x)dx=\frac{1}{2}\intf(u)du=\frac{1}{2}F(u)+C=\frac{1}{2}F(2x)+C\)。2.由曲線\(y=x^{2}\)與直線\(x=0\)，\(x=1\)，\(y=0\)所圍成的平面圖形的面積\(S=\)______。答案：\(\frac{1}{3}\)解析：根據(jù)定積分的幾何意義，\(S=\int_{0}^{1}x^{2}dx=\frac{1}{3}x^{3}\vert_{0}^{1}=\frac{1}{3}(1^{3}-0^{3})=\frac{1}{3}\)。（三）解答題1.求\(\int\frac{1}{x^{2}+4}dx\)。解：將原式變形為\(\int\frac{1}{x^{2}+4}dx=\int\frac{1}{4(\frac{x^{2}}{4}+1)}dx=\frac{1}{4}\int\frac{1}{(\frac{x}{2})^{2}+1}dx\)。令\(u=\frac{x}{2}\)，則\(du=\frac{1}{2}dx\)，\(dx=2du\)。\(\frac{1}{4}\int\frac{1}{(\frac{x}{2})^{2}+1}dx=\frac{1}{4}\times2\int\frac{1}{u^{2}+1}du=\frac{1}{2}\arctanu+C=\frac{1}{2}\arctan\frac{x}{2}+C\)。2.計算\(\int_{0}^{2}(x^{2}+1)dx\)。解：先求\(\int(x^{2}+1)dx=\intx^{2}dx+\int1dx=\frac{1}{3}x^{3}+x+C\)。再根據(jù)牛頓-萊布尼茨公式\(\int_{0}^{2}(x^{2}+1)dx=(\frac{1}{3}x^{3}+x)\vert_{0}^{2}\)\(=(\frac{1}{3}\times2^{3}+2)-(\frac{1}{3}\times0^{3}+0)=\frac{8}{3}+2=\frac{14}{3}\)。四、多元函數(shù)微積分學(xué)（一）選擇題1.設(shè)\(z=x^{2}y\)，則\(\frac{\partialz}{\partialx}\)等于（）A.\(2xy\)B.\(x^{2}\)C.\(2x\)D.\(y\)答案：A解析：求\(\frac{\partialz}{\partialx}\)時，將\(y\)看作常數(shù)，對\(x\)求導(dǎo)，根據(jù)求導(dǎo)公式\((x^{n})'=nx^{n-1}\)，則\(\frac{\partialz}{\partialx}=2xy\)。2.設(shè)\(z=e^{xy}\)，則\(dz\)等于（）A.\(e^{xy}(xdx+ydy)\)B.\(e^{xy}(ydx+xdy)\)C.\(xe^{xy}dx\)D.\(ye^{xy}dy\)答案：B解析：先求\(\frac{\partialz}{\partialx}=ye^{xy}\)，\(\frac{\partialz}{\partialy}=xe^{xy}\)。根據(jù)全微分公式\(dz=\frac{\partialz}{\partialx}dx+\frac{\partialz}{\partialy}dy\)，可得\(dz=e^{xy}(ydx+xdy)\)。（二）填空題1.設(shè)\(z=\ln(x+y)\)，則\(\frac{\partial^{2}z}{\partialx\partialy}=\)______。答案：\(-\frac{1}{(x+y)^{2}}\)解析：先求\(\frac{\partialz}{\partialx}=\frac{1}{x+y}\)，再對\(\frac{\partialz}{\partialx}\)關(guān)于\(y\)求偏導(dǎo)，\(\frac{\partial^{2}z}{\partialx\partialy}=-\frac{1}{(x+y)^{2}}\)。2.設(shè)\(D\)是由\(x=0\)，\(y=0\)，\(x+y=1\)所圍成的區(qū)域，則\(\iint_{D}dxdy=\)______。答案：\(\frac{1}{2}\)解析：根據(jù)二重積分的幾何意義，\(\iint_{D}dxdy\)表示區(qū)域\(D\)的面積，區(qū)域\(D\)是一個直角三角形，兩直角邊分別為\(1\)，所以面積\(S=\frac{1}{2}\times1\times1=\frac{1}{2}\)。（三）解答題1.設(shè)\(z=x^{2}+y^{2}\)，求在點\((1,1)\)處沿\(l=(1,1)\)方向的方向?qū)?shù)。解：首先求\(\frac{\partialz}{\partialx}=2x\)，\(\frac{\partialz}{\partialy}=2y\)。在點\((1,1)\)處，\(\frac{\partialz}{\partialx}\vert_{(1,1)}=2\)，\(\frac{\partialz}{\partialy}\vert_{(1,1)}=2\)。方向\(l=(1,1)\)的單位向量\(\vec{e}=\frac{l}{\vertl\vert}=\frac{(1,1)}{\sqrt{1^{2}+1^{2}}}=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\)。根據(jù)方向?qū)?shù)公式\(\frac{\partialz}{\partiall}=\frac{\partialz}{\partialx}\cos\alpha+\frac{\parti