高效備考：2025年高考數(shù)學(xué)應(yīng)用題模擬試卷（進(jìn)階版）一、選擇題（本大題共12小題，每小題5分，共60分）1.已知函數(shù)$f(x)=x^3-3x^2+4x+1$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減2.已知函數(shù)$f(x)=2^x-3^x$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減3.已知函數(shù)$f(x)=x^3+3x^2-9x+5$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減4.已知函數(shù)$f(x)=\sqrt{4x^2+1}$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減5.已知函數(shù)$f(x)=\frac{1}{x^2+1}$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減6.已知函數(shù)$f(x)=\ln(x+1)$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減7.已知函數(shù)$f(x)=\frac{x^2+1}{x+1}$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減8.已知函數(shù)$f(x)=\frac{1}{x^2}-\frac{1}{x}$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減9.已知函數(shù)$f(x)=\frac{x^2-1}{x^2+1}$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減10.已知函數(shù)$f(x)=\frac{x^2}{x^2+1}$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減11.已知函數(shù)$f(x)=\frac{1}{x^2}-\frac{1}{x}$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減12.已知函數(shù)$f(x)=\frac{x^2-1}{x^2+1}$，則下列說(shuō)法正確的是（）A.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增B.函數(shù)在$(-\infty,0)$上單調(diào)遞減C.函數(shù)在$(0,+\infty)$上單調(diào)遞增D.函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減二、填空題（本大題共12小題，每小題5分，共60分）13.若函數(shù)$f(x)=x^3-3x^2+4x+1$在$(-\infty,+\infty)$上單調(diào)遞增，則下列說(shuō)法正確的是（）A.函數(shù)的導(dǎo)數(shù)$f'(x)\geq0$B.函數(shù)的導(dǎo)數(shù)$f'(x)\leq0$C.函數(shù)的導(dǎo)數(shù)$f'(x)<0$D.函數(shù)的導(dǎo)數(shù)$f'(x)>0$14.若函數(shù)$f(x)=2^x-3^x$在$(-\infty,+\infty)$上單調(diào)遞減，則下列說(shuō)法正確的是（）A.函數(shù)的導(dǎo)數(shù)$f'(x)\geq0$B.函數(shù)的導(dǎo)數(shù)$f'(x)\leq0$C.函數(shù)的導(dǎo)數(shù)$f'(x)<0$D.函數(shù)的導(dǎo)數(shù)$f'(x)>0$15.若函數(shù)$f(x)=x^3+3x^2-9x+5$在$(-\infty,+\infty)$上單調(diào)遞增，則下列說(shuō)法正確的是（）A.函數(shù)的導(dǎo)數(shù)$f'(x)\geq0$B.函數(shù)的導(dǎo)數(shù)$f'(x)\leq0$C.函數(shù)的導(dǎo)數(shù)$f'(x)<0$D.函數(shù)的導(dǎo)數(shù)$f'(x)>0$16.若函數(shù)$f(x)=\sqrt{4x^2+1}$在$(-\infty,+\infty)$上單調(diào)遞增，則下列說(shuō)法正確的是（）A.函數(shù)的導(dǎo)數(shù)$f'(x)\geq0$B.函數(shù)的導(dǎo)數(shù)$f'(x)\leq0$C.函數(shù)的導(dǎo)數(shù)$f'(x)<0$D.函數(shù)的導(dǎo)數(shù)$f'(x)>0$17.若函數(shù)$f(x)=\frac{1}{x^2+1}$在$(-\infty,+\infty)$上單調(diào)遞增，則下列說(shuō)法正確的是（）A.函數(shù)的導(dǎo)數(shù)$f'(x)\geq0$B.函數(shù)的導(dǎo)數(shù)$f'(x)\leq0$C.函數(shù)的導(dǎo)數(shù)$f'(x)<0$D.函數(shù)的導(dǎo)數(shù)$f'(x)>0$18.若函數(shù)$f(x)=\ln(x+1)$在$(-\infty,+\infty)$上單調(diào)遞增，則下列說(shuō)法正確的是（）A.函數(shù)的導(dǎo)數(shù)$f'(x)\geq0$B.函數(shù)的導(dǎo)數(shù)$f'(x)\leq0$C.函數(shù)的導(dǎo)數(shù)$f'(x)<0$D.函數(shù)的導(dǎo)數(shù)$f'(x)>0$19.若函數(shù)$f(x)=\frac{x^2+1}{x+1}$在$(-\infty,+\infty)$上單調(diào)遞增，則下列說(shuō)法正確的是（）A.函數(shù)的導(dǎo)數(shù)$f'(x)\geq0$B.函數(shù)的導(dǎo)數(shù)$f'(x)\leq0$C.函數(shù)的導(dǎo)數(shù)$f'(x)<0$D.函數(shù)的導(dǎo)數(shù)$f'(x)>0$20.若函數(shù)$f(x)=\frac{1}{x^2}-\frac{1}{x}$在$(-\infty,+\infty)$上單調(diào)遞增，則下列說(shuō)法正確的是（）A.函數(shù)的導(dǎo)數(shù)$f'(x)\geq0$B.函數(shù)的導(dǎo)數(shù)$f'(x)\leq0$C.函數(shù)的導(dǎo)數(shù)$f'(x)<0$D.函數(shù)的導(dǎo)數(shù)$f'(x)>0$三、解答題（本大題共4小題，每小題15分，共60分）21.已知函數(shù)$f(x)=x^3-3x^2+4x+1$，求$f(x)$的單調(diào)區(qū)間。22.已知函數(shù)$f(x)=2^x-3^x$，求$f(x)$的單調(diào)區(qū)間。23.已知函數(shù)$f(x)=\frac{x^2}{x^2+1}$，求$f(x)$的單調(diào)區(qū)間。24.已知函數(shù)$f(x)=\frac{1}{x^2}-\frac{1}{x}$，求$f(x)$的單調(diào)區(qū)間。四、證明題要求：證明下列命題，并給出證明過(guò)程。25.證明：若函數(shù)$f(x)=ax^2+bx+c$（$a\neq0$）在$x=1$處取得極小值，則$a>0$且$b=2a$。五、計(jì)算題要求：計(jì)算下列表達(dá)式的值，并化簡(jiǎn)結(jié)果。26.計(jì)算下列極限：\[\lim_{x\to0}\frac{\sinx-x}{x^3}\]27.計(jì)算下列定積分：\[\int_0^1(x^2-2x+1)\,dx\]六、應(yīng)用題要求：根據(jù)下列條件，解答實(shí)際問(wèn)題。28.一輛汽車以60公里/小時(shí)的速度行駛，當(dāng)速度減為30公里/小時(shí)時(shí)，行駛了10分鐘。求汽車在減速過(guò)程中行駛的總路程。本次試卷答案如下：一、選擇題1.D解析：函數(shù)$f(x)=x^3-3x^2+4x+1$的導(dǎo)數(shù)為$f'(x)=3x^2-6x+4$，令$f'(x)=0$，解得$x=1$或$x=\frac{2}{3}$。當(dāng)$x<\frac{2}{3}$或$x>1$時(shí)，$f'(x)>0$，函數(shù)單調(diào)遞增；當(dāng)$\frac{2}{3}<x<1$時(shí)，$f'(x)<0$，函數(shù)單調(diào)遞減。因此，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減。2.B解析：函數(shù)$f(x)=2^x-3^x$的導(dǎo)數(shù)為$f'(x)=2^x\ln2-3^x\ln3$，由于$\ln2>0$和$\ln3>0$，當(dāng)$x<0$時(shí)，$2^x>0$且$3^x>0$，因此$f'(x)<0$，函數(shù)單調(diào)遞減；當(dāng)$x>0$時(shí)，$2^x>0$且$3^x>0$，因此$f'(x)>0$，函數(shù)單調(diào)遞增。3.A解析：函數(shù)$f(x)=x^3+3x^2-9x+5$的導(dǎo)數(shù)為$f'(x)=3x^2+6x-9$，令$f'(x)=0$，解得$x=-3$或$x=1$。當(dāng)$x<-3$或$x>1$時(shí)，$f'(x)>0$，函數(shù)單調(diào)遞增；當(dāng)$-3<x<1$時(shí)，$f'(x)<0$，函數(shù)單調(diào)遞減。4.A解析：函數(shù)$f(x)=\sqrt{4x^2+1}$的導(dǎo)數(shù)為$f'(x)=\frac{8x}{\sqrt{4x^2+1}}$，由于$x^2\geq0$，$4x^2+1>0$，因此$f'(x)>0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增。5.D解析：函數(shù)$f(x)=\frac{1}{x^2+1}$的導(dǎo)數(shù)為$f'(x)=-\frac{2x}{(x^2+1)^2}$，當(dāng)$x>0$或$x<0$時(shí)，$f'(x)<0$，函數(shù)單調(diào)遞減；當(dāng)$x=0$時(shí)，$f'(x)=0$，函數(shù)在$x=0$處取得極值。二、填空題13.A解析：函數(shù)$f(x)=x^3-3x^2+4x+1$的導(dǎo)數(shù)為$f'(x)=3x^2-6x+4$，由于$f'(x)\geq0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增。14.C解析：函數(shù)$f(x)=2^x-3^x$的導(dǎo)數(shù)為$f'(x)=2^x\ln2-3^x\ln3$，由于$f'(x)<0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞減。15.A解析：函數(shù)$f(x)=x^3+3x^2-9x+5$的導(dǎo)數(shù)為$f'(x)=3x^2+6x-9$，由于$f'(x)\geq0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增。16.A解析：函數(shù)$f(x)=\sqrt{4x^2+1}$的導(dǎo)數(shù)為$f'(x)=\frac{8x}{\sqrt{4x^2+1}}$，由于$f'(x)\geq0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增。17.A解析：函數(shù)$f(x)=\frac{1}{x^2+1}$的導(dǎo)數(shù)為$f'(x)=-\frac{2x}{(x^2+1)^2}$，由于$f'(x)\geq0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增。18.A解析：函數(shù)$f(x)=\ln(x+1)$的導(dǎo)數(shù)為$f'(x)=\frac{1}{x+1}$，由于$f'(x)\geq0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增。19.A解析：函數(shù)$f(x)=\frac{x^2+1}{x+1}$的導(dǎo)數(shù)為$f'(x)=\frac{x^2+2x}{(x+1)^2}$，由于$f'(x)\geq0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增。20.A解析：函數(shù)$f(x)=\frac{1}{x^2}-\frac{1}{x}$的導(dǎo)數(shù)為$f'(x)=-\frac{2x}{x^3}$，由于$f'(x)\geq0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增。三、解答題21.解析：已知函數(shù)$f(x)=x^3-3x^2+4x+1$，導(dǎo)數(shù)$f'(x)=3x^2-6x+4$，令$f'(x)=0$，解得$x=1$或$x=\frac{2}{3}$。當(dāng)$x<\frac{2}{3}$或$x>1$時(shí)，$f'(x)>0$，函數(shù)單調(diào)遞增；當(dāng)$\frac{2}{3}<x<1$時(shí)，$f'(x)<0$，函數(shù)單調(diào)遞減。因此，函數(shù)在$(-\infty,\frac{2}{3})$和$(1,+\infty)$上單調(diào)遞增，在$(\frac{2}{3},1)$上單調(diào)遞減。22.解析：已知函數(shù)$f(x)=2^x-3^x$，導(dǎo)數(shù)$f'(x)=2^x\ln2-3^x\ln3$，令$f'(x)=0$，解得$x=\frac{\ln(\ln3/\ln2)}{\ln2}$。當(dāng)$x<\frac{\ln(\ln3/\ln2)}{\ln2}$時(shí)，$f'(x)<0$，函數(shù)單調(diào)遞減；當(dāng)$x>\frac{\ln(\ln3/\ln2)}{\ln2}$時(shí)，$f'(x)>0$，函數(shù)單調(diào)遞增。23.解析：已知函數(shù)$f(x)=\frac{x^2}{x^2+1}$，導(dǎo)數(shù)$f'(x)=\frac{2x(x^2+1)-x^2\cdot2x}{(x^2+1)^2}=\frac{2x}{(x^2+1)^2}$，由于$f'(x)>0$，函數(shù)在$(-\infty,+\infty)$上單調(diào)遞增。24.解析：已知函數(shù)$f(x)=\frac{1}{x^2}-\frac{1}{x}$，導(dǎo)數(shù)$f'(x)=-\frac{2x}{x^3}$，令$f'(x)=0$，解得$x=0$。當(dāng)$x