2025年三卷高考數(shù)學(xué)試卷及答案

一、單項(xiàng)選擇題1.已知集合\(A=\{x|x^2-3x+2=0\}\)，\(B=\{x|0<x<5,x\inN\}\)，則\(A\cupB=\)（）A.\(\{1,2\}\)B.\(\{1,2,3,4\}\)C.\(\{0,1,2,3,4\}\)D.\(\{-1,0,1,2,3,4\}\)答案：B2.復(fù)數(shù)\(z=\frac{2-i}{1+2i}\)（\(i\)為虛數(shù)單位）的共軛復(fù)數(shù)\(\overline{z}\)的虛部為（）A.\(1\)B.\(-1\)C.\(i\)D.\(-i\)答案：A3.已知向量\(\vec{a}=(1,m)\)，\(\vec=(3,-2)\)，且\((\vec{a}+\vec)\perp\vec\)，則\(m=\)（）A.\(-8\)B.\(-6\)C.\(6\)D.\(8\)答案：D4.函數(shù)\(f(x)=2\sin(\omegax+\varphi)(\omega>0,|\varphi|<\frac{\pi}{2})\)的部分圖象如圖所示，則\(\omega\)，\(\varphi\)的值分別是（）A.\(2\)，\(-\frac{\pi}{3}\)B.\(2\)，\(-\frac{\pi}{6}\)C.\(4\)，\(-\frac{\pi}{6}\)D.\(4\)，\(\frac{\pi}{3}\)答案：A5.若雙曲線\(C:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a>0,b>0)\)的一條漸近線被圓\((x-2)^2+y^2=4\)所截得的弦長(zhǎng)為\(2\)，則\(C\)的離心率為（）A.\(2\)B.\(\sqrt{3}\)C.\(\sqrt{2}\)D.\(\frac{2\sqrt{3}}{3}\)答案：A6.執(zhí)行如圖所示的程序框圖，如果輸入的\(x=0\)，\(y=1\)，\(n=1\)，則輸出\(x\)，\(y\)的值滿足（）A.\(y=2x\)B.\(y=3x\)C.\(y=4x\)D.\(y=5x\)答案：C7.已知\(a=\log_20.2\)，\(b=2^{0.2}\)，\(c=0.2^{0.3}\)，則（）A.\(a<b<c\)B.\(a<c<b\)C.\(c<a<b\)D.\(b<c<a\)答案：B8.設(shè)\(F_1,F_2\)是橢圓\(C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)\)的左、右焦點(diǎn)，過(guò)\(F_1\)的直線\(l\)與\(C\)交于\(A\)，\(B\)兩點(diǎn)，若\(AB\perpAF_2\)，\(|AB|:|AF_2|=3:4\)，則\(C\)的離心率為（）A.\(\frac{\sqrt{2}}{2}\)B.\(\frac{\sqrt{3}}{2}\)C.\(\frac{5}{7}\)D.\(\frac{\sqrt{17}}{5}\)答案：C9.已知函數(shù)\(f(x)\)是定義在\(R\)上的奇函數(shù)，當(dāng)\(x\geqslant0\)時(shí)，\(f(x)=x(1+x)\)，則當(dāng)\(x<0\)時(shí)，\(f(x)=\)（）A.\(x(1+x)\)B.\(-x(1+x)\)C.\(x(1-x)\)D.\(-x(1-x)\)答案：C10.已知\(f(x)\)是定義域?yàn)閈((-\infty,+\infty)\)的奇函數(shù)，滿足\(f(1-x)=f(1+x)\)．若\(f(1)=2\)，則\(f(1)+f(2)+f(3)+\cdots+f(50)=\)（）A.\(-50\)B.\(0\)C.\(2\)D.\(50\)答案：C二、多項(xiàng)選擇題1.下列函數(shù)中，在區(qū)間\((0,+\infty)\)上單調(diào)遞增的是（）A.\(y=x^{\frac{1}{2}}\)B.\(y=2^{-x}\)C.\(y=\log_{\frac{1}{2}}x\)D.\(y=\frac{1}{1-x}\)答案：AD2.已知向量\(\vec{a}=(1,-2)\)，\(\vec=(m,4)\)，且\(\vec{a}\parallel\vec\)，則（）A.\(m=-2\)B.\(m=2\)C.\(\vec{a}\cdot\vec=-10\)D.\(\vec{a}\cdot\vec=10\)答案：AC3.對(duì)于函數(shù)\(f(x)=\sinx+\sqrt{3}\cosx\)，下列說(shuō)法正確的是（）A.函數(shù)\(f(x)\)的圖象關(guān)于點(diǎn)\((\frac{\pi}{3},0)\)對(duì)稱B.存在\(\alpha\in(0,\frac{\pi}{3})\)，使\(f(\alpha)=1\)C.存在\(\alpha\in(0,\frac{\pi}{3})\)，使函數(shù)\(f(x+\alpha)\)的圖象關(guān)于\(y\)軸對(duì)稱D.函數(shù)\(f(x)\)在\([0,\frac{\pi}{6}]\)上的值域?yàn)閈([1,2]\)答案：CD4.已知雙曲線\(C:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a>0,b>0)\)的左、右焦點(diǎn)分別為\(F_1,F_2\)，過(guò)\(F_2\)的直線\(l\)與雙曲線\(C\)的右支交于\(A\)，\(B\)兩點(diǎn)，若\(|AF_1|=|AB|\)，\(\angleF_1AB=90^{\circ}\)，則（）A.雙曲線\(C\)的離心率\(e=\sqrt{5-2\sqrt{2}}\)B.雙曲線\(C\)的離心率\(e=\sqrt{5+2\sqrt{2}}\)C.若\(a=1\)，則雙曲線\(C\)的標(biāo)準(zhǔn)方程為\(x^2-\frac{y^2}{\sqrt{2}+1}=1\)D.若\(a=1\)，則雙曲線\(C\)的標(biāo)準(zhǔn)方程為\(x^2-\frac{y^2}{\sqrt{2}-1}=1\)答案：AC5.已知函數(shù)\(f(x)=x^3-3x^2+2\)，則（）A.函數(shù)\(f(x)\)有兩個(gè)極值點(diǎn)B.\(f(x)\)在區(qū)間\((-\infty,0)\)上單調(diào)遞增C.\(f(x)\)在區(qū)間\((0,2)\)上單調(diào)遞減D.\(f(x)\)在區(qū)間\((2,+\infty)\)上單調(diào)遞增答案：ABCD6.已知\(a\)，\(b\)，\(c\)滿足\(c<b<a\)，且\(ac<0\)，那么下列選項(xiàng)中一定成立的是（）A.\(ab>ac\)B.\(c(b-a)<0\)C.\(cb^2<ab^2\)D.\(ac(a-c)>0\)答案：A7.已知橢圓\(C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)\)的左、右焦點(diǎn)分別為\(F_1,F_2\)，\(P\)為橢圓\(C\)上一點(diǎn)，且\(\overrightarrow{PF_1}\cdot\overrightarrow{PF_2}=0\)，若\(\trianglePF_1F_2\)的面積為\(9\)，則（）A.\(b=3\)B.\(a-c=3\)C.\(\trianglePF_1F_2\)的周長(zhǎng)為\(2a+2c\)D.\(a^2-c^2=9\)答案：AC8.已知函數(shù)\(y=f(x)\)的定義域?yàn)閈(R\)，且\(f(x+1)\)是偶函數(shù)，\(f(x-1)\)是奇函數(shù)，則（）A.\(f(0)=0\)B.\(f(x+4)=f(x)\)C.\(f(3)=0\)D.\(f(x)\)的圖象關(guān)于點(diǎn)\((-1,0)\)對(duì)稱答案：ABD9.已知\(x\)，\(y\)滿足約束條件\(\begin{cases}x-y+1\geqslant0\\x+y-3\leqslant0\\y\geqslant1\end{cases}\)，則（）A.\(z=x+y\)的最大值為\(4\)B.\(z=x-y\)的最大值為\(1\)C.\(z=2x+y\)的最小值為\(3\)D.\(z=x+2y\)的最小值為\(5\)答案：AB10.已知函數(shù)\(f(x)=\begin{cases}2^x-1,x\leqslant0\\f(x-1)+1,x>0\end{cases}\)，把函數(shù)\(g(x)=f(x)-x\)的零點(diǎn)按從小到大的順序排列成一個(gè)數(shù)列\(zhòng)(\{a_n\}\)，則（）A.\(a_1=0\)B.\(a_2=1\)C.\(a_n=n-1\)D.數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n=\frac{n(n-1)}{2}\)答案：ABCD三、判斷題1.若\(a>b\)，則\(ac^2>bc^2\)。（×）2.函數(shù)\(y=\sinx\)在區(qū)間\([0,2\pi]\)上的圖象關(guān)于直線\(x=\pi\)對(duì)稱。（×）3.向量\(\vec{a}=(1,2)\)與向量\(\vec=(2,4)\)共線。（√）4.雙曲線\(\frac{x^2}{4}-\frac{y^2}{9}=1\)的漸近線方程為\(y=\pm\frac{3}{2}x\)。（√）5.若\(A\)，\(B\)為互斥事件，則\(P(A)+P(B)=1\)。（×）6.函數(shù)\(y=\log_2(x^2+1)\)的值域是\([0,+\infty)\)。（√）7.直線\(x+y+1=0\)與圓\((x-1)^2+y^2=1\)相切。（×）8.若\(a\)，\(b\)，\(c\)成等比數(shù)列，則\(b^2=ac\)。（√）9.函數(shù)\(f(x)=x^3\)是奇函數(shù)。（√）10.已知\(a=\log_32\)，\(b=\log_23\)，則\(a<b\)。（√）四、簡(jiǎn)答題1.已知等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，\(a_3=5\)，\(S_6=36\)。求數(shù)列\(zhòng)(\{a_n\}\)的通項(xiàng)公式。答案：設(shè)等差數(shù)列\(zhòng)(\{a_n\}\)的公差為\(d\)，首項(xiàng)為\(a_1\)。由\(a_3=5\)可得\(a_1+2d=5\)；由\(S_6=36\)可得\(6a_1+\frac{6\times5}{2}d=36\)，即\(6a_1+15d=36\)。聯(lián)立方程組\(\begin{cases}a_1+2d=5\\6a_1+15d=36\end{cases}\)，解方程組得\(a_1=1\)，\(d=2\)。所以\(a_n=a_1+(n-1)d=1+2(n-1)=2n-1\)。2.已知函數(shù)\(f(x)=x^3-3x^2+2\)，求函數(shù)\(f(x)\)的單調(diào)區(qū)間。答案：對(duì)\(f(x)=x^3-3x^2+2\)求導(dǎo)得\(f^\prime(x)=3x^2-6x=3x(x-2)\)。令\(f^\prime(x)>0\)，即\(3x(x-2)>0\)，解得\(x<0\)或\(x>2\)，所以\(f(x)\)的單調(diào)遞增區(qū)間為\((-\infty,0)\)和\((2,+\infty)\)；令\(f^\prime(x)<0\)，即\(3x(x-2)<0\)，解得\(0<x<2\)，所以\(f(x)\)的單調(diào)遞減區(qū)間為\((0,2)\)。3.已知橢圓\(C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)\)的離心率為\(\frac{\sqrt{2}}{2}\)，且過(guò)點(diǎn)\((\sqrt{2},1)\)。求橢圓\(C\)的標(biāo)準(zhǔn)方程。答案：因?yàn)闄E圓離心率\(e=\frac{c}{a}=\frac{\sqrt{2}}{2}\)，即\(c=\frac{\sqrt{2}}{2}a\)，又\(a^2=b^2+c^2\)，所以\(a^2=b^2+\frac{1}{2}a^2\)，即\(a^2=2b^2\)。橢圓過(guò)點(diǎn)\((\sqrt{2},1)\)，則\(\frac{(\sqrt{2})^2}{a^2}+\frac{1^2}{b^2}=1\)，將\(a^2=2b^2\)代入可得\(\frac{2}{2b^2}+\frac{1}{b^2}=1\)，解得\(b^2=2\)，則\(a^2=4\)。所以橢圓\(C\)