2025年江蘇高考數(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\(A\cupB\)等于（）A.\(\{1,2\}\)B.\(\{1,2,3,4\}\)C.\(\{0,1,2,3,4\}\)D.\(\varnothing\)答案：B2.若復(fù)數(shù)\(z\)滿足\((1+i)z=2i\)，則\(z\)的共軛復(fù)數(shù)\(\overline{z}\)等于（）A.\(1+i\)B.\(1-i\)C.\(-1+i\)D.\(-1-i\)答案：B3.已知向量\(\overrightarrow{a}=(1,m)\)，\(\overrightarrow=(3,-2)\)，且\((\overrightarrow{a}+\overrightarrow)\perp\overrightarrow\)，則\(m\)等于（）A.-8B.-6C.6D.8答案：D4.函數(shù)\(f(x)=\frac{1}{\sqrt{1-x^2}}+\ln(2x+1)\)的定義域?yàn)椋ǎ〢.\((-\frac{1}{2},1)\)B.\((-\frac{1}{2},1]\)C.\((-\frac{1}{2},+\infty)\)D.\((1,+\infty)\)答案：A5.等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，若\(a_3+a_5+a_7=15\)，則\(S_9\)的值為（）A.60B.45C.36D.27答案：B6.已知\(\sin\alpha=\frac{3}{5}\)，\(\alpha\in(\frac{\pi}{2},\pi)\)，則\(\tan(\alpha+\frac{\pi}{4})\)等于（）A.\(\frac{1}{7}\)B.7C.-\(\frac{1}{7}\)D.-7答案：C7.拋物線\(y^2=4x\)的焦點(diǎn)為\(F\)，點(diǎn)\(A\)在拋物線上且\(|AF|=3\)，則點(diǎn)\(A\)的橫坐標(biāo)為（）A.1B.2C.\(\frac{5}{2}\)D.3答案：B8.函數(shù)\(y=\sin(2x+\frac{\pi}{3})\)的圖象可由函數(shù)\(y=\sin2x\)的圖象（）A.向左平移\(\frac{\pi}{3}\)個(gè)單位長(zhǎng)度得到B.向右平移\(\frac{\pi}{3}\)個(gè)單位長(zhǎng)度得到C.向左平移\(\frac{\pi}{6}\)個(gè)單位長(zhǎng)度得到D.向右平移\(\frac{\pi}{6}\)個(gè)單位長(zhǎng)度得到答案：C9.若函數(shù)\(f(x)=x^3-3x+a\)有\(zhòng)(3\)個(gè)不同的零點(diǎn)，則實(shí)數(shù)\(a\)的取值范圍是（）A.\((-2,2)\)B.\([-2,2]\)C.\((-\infty,-2)\)D.\((2,+\infty)\)答案：A10.已知\(x>0\)，\(y>0\)，且\(\frac{1}{x}+\frac{9}{y}=1\)，則\(x+y\)的最小值為（）A.12B.16C.20D.24答案：B二、多項(xiàng)選擇題1.下列函數(shù)中，是偶函數(shù)的有（）A.\(y=x^2+1\)B.\(y=\cosx\)C.\(y=\ln|x|\)D.\(y=2^x\)答案：ABC2.已知直線\(l_1\)：\(ax+2y+6=0\)，直線\(l_2\)：\(x+(a-1)y+a^2-1=0\)，則下列說法正確的是（）A.若\(l_1\parallell_2\)，則\(a=-1\)或\(a=2\)B.若\(l_1\perpl_2\)，則\(a=\frac{2}{3}\)C.當(dāng)\(a=2\)時(shí)，\(l_1\)與\(l_2\)重合D.當(dāng)\(a=-1\)時(shí)，\(l_1\parallell_2\)答案：BD3.對(duì)于函數(shù)\(f(x)=\sin(2x+\frac{\pi}{6})\)，下列說法正確的是（）A.函數(shù)\(f(x)\)的最小正周期為\(\pi\)B.函數(shù)\(f(x)\)圖象的一條對(duì)稱軸方程為\(x=\frac{\pi}{6}\)C.函數(shù)\(f(x)\)在區(qū)間\([-\frac{\pi}{6},\frac{\pi}{3}]\)上單調(diào)遞增D.函數(shù)\(f(x)\)的圖象可由函數(shù)\(y=\sin2x\)的圖象向左平移\(\frac{\pi}{12}\)個(gè)單位長(zhǎng)度得到答案：ABD4.已知\(a\)，\(b\)，\(c\)為實(shí)數(shù)，且\(a>b>0\)，則下列不等式正確的是（）A.\(\frac{1}{a}<\frac{1}\)B.\(ac^2>bc^2\)C.\(a^2>ab>b^2\)D.\(\lg(a-b)>0\)答案：AC5.設(shè)等比數(shù)列\(zhòng)(\{a_n\}\)的公比為\(q\)，其前\(n\)項(xiàng)和為\(S_n\)，前\(n\)項(xiàng)積為\(T_n\)，并且滿足條件\(a_1>1\)，\(a_{7}a_{8}>1\)，\(\frac{a_{7}-1}{a_{8}-1}<0\)，則下列結(jié)論正確的是（）A.\(0<q<1\)B.\(a_{7}a_{9}>1\)C.\(S_n\)的最大值為\(S_8\)D.\(T_n\)的最大值為\(T_7\)答案：AD6.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\(F_1,F_2\)，點(diǎn)\(P\)在橢圓上，\(\trianglePF_1F_2\)是以\(PF_1\)為底邊的等腰三角形，且\(60^{\circ}<\anglePF_2F_1<120^{\circ}\)，則該橢圓的離心率\(e\)的取值范圍是（）A.\((\frac{\sqrt{3}-1}{2},\frac{1}{2})\)B.\((\frac{1}{2},\frac{\sqrt{2}}{2})\)C.\((\frac{\sqrt{2}}{2},\frac{\sqrt{3}}{2})\)D.\((\frac{\sqrt{3}}{2},1)\)答案：A7.已知函數(shù)\(f(x)\)是定義在\(R\)上的奇函數(shù)，當(dāng)\(x\geqslant0\)時(shí)，\(f(x)=x(1+x)\)，則下列結(jié)論正確的是（）A.當(dāng)\(x<0\)時(shí)，\(f(x)=x(1-x)\)B.函數(shù)\(f(x)\)的圖象關(guān)于原點(diǎn)對(duì)稱C.函數(shù)\(f(x)\)的單調(diào)遞增區(qū)間是\((-\infty,+\infty)\)D.對(duì)任意的\(x_1,x_2\inR\)，都有\(zhòng)(f(x_1)-f(x_2)\leqslant2\)答案：ABCD8.已知\(a\)，\(b\)為正實(shí)數(shù)，且\(a+b=1\)，則下列說法正確的是（）A.\(ab\leqslant\frac{1}{4}\)B.\(\frac{1}{a}+\frac{1}\geqslant4\)C.\(\sqrt{a}+\sqrt\leqslant\sqrt{2}\)D.\(a^2+b^2\geqslant\frac{1}{2}\)答案：ABCD9.已知圓\(C:(x-1)^2+(y-2)^2=25\)，直線\(l:(2m+1)x+(m+1)y-7m-4=0\)，則下列說法正確的是（）A.直線\(l\)恒過定點(diǎn)\((3,1)\)B.直線\(l\)與圓\(C\)可能相離C.直線\(l\)被圓\(C\)截得的弦長(zhǎng)最短時(shí)，直線\(l\)的方程為\(2x-y-5=0\)D.直線\(l\)被圓\(C\)截得的弦長(zhǎng)最短時(shí)，弦長(zhǎng)為\(4\sqrt{5}\)答案：ACD10.已知函數(shù)\(f(x)=\begin{cases}x^2+2x,x\geqslant0\\-x^2+2x,x<0\end{cases}\)，則下列說法正確的是（）A.函數(shù)\(f(x)\)是奇函數(shù)B.函數(shù)\(f(x)\)在\(R\)上單調(diào)遞增C.若\(f(a-2)+f(a^2)\leqslant0\)，則\(a\in[-2,1]\)D.函數(shù)\(y=f(x)-\frac{1}{2}\)有\(zhòng)(3\)個(gè)零點(diǎn)答案：ABC三、判斷題1.若\(a\)，\(b\)，\(c\)為實(shí)數(shù)，且\(a>b\)，則\(ac^2>bc^2\)。（×）2.函數(shù)\(y=\log_2(x^2+1)\)的值域是\([0,+\infty)\)。（√）3.直線\(x=1\)的傾斜角是\(90^{\circ}\)。（√）4.若向量\(\overrightarrow{a}\)，\(\overrightarrow\)滿足\(\overrightarrow{a}\cdot\overrightarrow=0\)，則\(\overrightarrow{a}=\overrightarrow{0}\)或\(\overrightarrow=\overrightarrow{0}\)。（×）5.若數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}=2a_n\)，則\(\{a_n\}\)是等比數(shù)列。（×）6.函數(shù)\(y=\sinx\)的圖象與\(y=\cosx\)的圖象的對(duì)稱軸相同。（×）7.橢圓\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)\)的離心率\(e=\frac{c}{a}\)，其中\(zhòng)(c^2=a^2-b^2\)。（√）8.若\(a\)，\(b\)為正實(shí)數(shù)，且\(a+b=1\)，則\(\frac{1}{a}+\frac{1}\geqslant4\)。（√）9.函數(shù)\(f(x)=x^3\)在\(R\)上是增函數(shù)。（√）10.若\(z\)是復(fù)數(shù)，則\(|z|^2=z\cdot\overline{z}\)。（√）四、簡(jiǎn)答題1.已知函數(shù)\(f(x)=2\sinx\cosx+2\sqrt{3}\cos^2x-\sqrt{3}\)。（1）求函數(shù)\(f(x)\)的最小正周期；（2）求函數(shù)\(f(x)\)的單調(diào)遞增區(qū)間。答案：（1）化簡(jiǎn)\(f(x)=\sin2x+\sqrt{3}(2\cos^2x-1)=\sin2x+\sqrt{3}\cos2x=2\sin(2x+\frac{\pi}{3})\)，最小正周期\(T=\frac{2\pi}{2}=\pi\)。（2）令\(2k\pi-\frac{\pi}{2}\leqslant2x+\frac{\pi}{3}\leqslant2k\pi+\frac{\pi}{2},k\inZ\)，解得\(k\pi-\frac{5\pi}{12}\leqslantx\leqslantk\pi+\frac{\pi}{12},k\inZ\)，所以單調(diào)遞增區(qū)間是\([k\pi-\frac{5\pi}{12},k\pi+\frac{\pi}{12}],k\inZ\)。2.已知等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，\(a_1=1\)，\(S_5=25\)。（1）求數(shù)列\(zhòng)(\{a_n\}\)的通項(xiàng)公式；（2）設(shè)\(b_n=2^{a_n}\)，求數(shù)列\(zhòng)(\{b_n\}\)的前\(n\)項(xiàng)和\(T_n\)。答案：（1）由\(S_5=5a_1+\frac{5\times4}{2}d=5+10d=25\)，解得\(d=2\)，則\(a_n=a_1+(n-1)d=1+2(n-1)=2n-1\)。（2）\(b_n=2^{2n-1}\)，\(\{b_n\}\)是首項(xiàng)\(b_1=2\)，公比\(q=4\)的等比數(shù)列，\(T_n=\frac{2(1-4^n)}{1-4}=\frac{2(4^n-1)}{3}\)。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\(\frac{\sqrt{2}}{2}\)，且過點(diǎn)\((\sqrt{2},1)\)。（1）求橢圓\(C\)的方程；（2）設(shè)直線\(l\)與橢圓\(C\)交于\(A\)，\(B\)兩點(diǎn)，\(O\)為坐標(biāo)原點(diǎn)，若\(\overrightarrow{OA}\cdot\overrightarrow{OB}=0\)，求\(O\)到直線\(l\)距離的取值范圍。答案：（1）由離心率\(e=\frac{c}{a}=\frac{\sqrt{2}}{2}\)，\(a^2=b^2+c^2\)，且橢圓過點(diǎn)\((\sqrt{2},1)\)，可得\(\frac{2}{a^2}+\frac{1}{b^2}=1\)，聯(lián)立解得\(a^2=4\)，\(b^2=2\)，橢圓方程為\(\frac{x^2}{4}+\f