2025年四川統(tǒng)招數(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.\(\varnothing\)答案：B2.函數(shù)\(y=\log_2(x+1)\)的定義域是\((\)\)A.\((-1,+\infty)\)B.\([-1,+\infty)\)C.\((0,+\infty)\)D.\((-\infty,-1)\)答案：A3.已知向量\(\overrightarrow{a}=(1,2)\)，\(\overrightarrow=(x,-4)\)，若\(\overrightarrow{a}\parallel\overrightarrow\)，則\(x=(\)\)A.2B.-2C.8D.-8答案：B4.等差數(shù)列\(zhòng)(\{a_n\}\)中，\(a_1=1\)，\(a_3+a_5=14\)，其前\(n\)項(xiàng)和\(S_n=100\)，則\(n=(\)\)A.9B.10C.11D.12答案：B5.若\(\sin\alpha=\frac{3}{5}\)，\(\alpha\in(\frac{\pi}{2},\pi)\)，則\(\cos\alpha=(\)\)A.\(\frac{4}{5}\)B.\(-\frac{4}{5}\)C.\(\frac{3}{4}\)D.\(-\frac{3}{4}\)答案：B6.拋物線\(y^2=8x\)的焦點(diǎn)坐標(biāo)是\((\)\)A.\((2,0)\)B.\((-2,0)\)C.\((0,2)\)D.\((0,-2)\)答案：A7.函數(shù)\(f(x)=x^3-3x^2+2\)在區(qū)間\([-1,1]\)上的最大值是\((\)\)A.0B.2C.-2D.4答案：B8.已知\(x\)，\(y\)滿足約束條件\(\begin{cases}x+y\geq1\\x-y\leq1\\y\leq1\end{cases}\)，則\(z=2x-y\)的最大值為\((\)\)A.1B.2C.3D.4答案：C9.直線\(3x+4y-5=0\)與圓\(x^2+y^2=1\)的位置關(guān)系是\((\)\)A.相交B.相切C.相離D.無(wú)法確定答案：B10.從\(5\)名男生和\(3\)名女生中選\(3\)人參加活動(dòng)，至少有\(zhòng)(1\)名女生的選法有\(zhòng)((\)\)種A.56B.46C.36D.26答案：B二、多項(xiàng)選擇題1.下列函數(shù)中，在\((0,+\infty)\)上單調(diào)遞增的有\(zhòng)((\)\)A.\(y=x^2\)B.\(y=\frac{1}{x}\)C.\(y=\lgx\)D.\(y=2^x\)答案：ACD2.已知\(a\)，\(b\)，\(c\)為實(shí)數(shù)，則下列命題正確的有\(zhòng)((\)\)A.若\(a>b\)，則\(ac^2>bc^2\)B.若\(ac^2>bc^2\)，則\(a>b\)C.若\(a>b\)，\(ab>0\)，則\(\frac{1}{a}<\frac{1}\)D.若\(a>b\)，\(ab<0\)，則\(\frac{1}{a}>\frac{1}\)答案：BCD3.一個(gè)正方體的展開(kāi)圖如圖所示，\(A\)、\(B\)、\(C\)、\(D\)為原正方體的頂點(diǎn)，則在原來(lái)的正方體中（）A.\(AB\parallelCD\)B.\(AB\)與\(CD\)相交C.\(AB\perpCD\)D.\(AB\)與\(CD\)所成的角為\(60^{\circ}\)答案：D（注：這里題目有誤，應(yīng)該是讓選正確的選項(xiàng)組合，按正確理解選D）4.已知函數(shù)\(f(x)=\sin(2x+\varphi)\)，\(|\varphi|\lt\frac{\pi}{2}\)，若\(f(\frac{\pi}{6})=f(\frac{\pi}{3})\)，且\(f(x)\)在區(qū)間\((\frac{\pi}{6},\frac{\pi}{3})\)上有最小值，無(wú)最大值，則\(\varphi\)的值可能為\((\)\)A.\(-\frac{\pi}{6}\)B.\(\frac{\pi}{6}\)C.\(\frac{\pi}{3}\)D.\(-\frac{\pi}{3}\)答案：A5.已知雙曲線\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a\gt0,b\gt0)\)的左、右焦點(diǎn)分別為\(F_1,F_2\)，過(guò)\(F_2\)的直線與雙曲線右支交于\(A\)，\(B\)兩點(diǎn)，若\(\triangleABF_1\)是等腰三角形，且\(\angleBAF_1=90^{\circ}\)，則雙曲線的離心率為\((\)\)A.\(\sqrt{2}\)B.\(\sqrt{2}+1\)C.\(\sqrt{3}\)D.\(\sqrt{3}+1\)答案：B6.下列關(guān)于概率的說(shuō)法正確的有\(zhòng)((\)\)A.事件\(A\)發(fā)生的概率\(P(A)\)滿足\(0\leqP(A)\leq1\)B.若\(P(A)=0\)，則\(A\)為不可能事件C.若\(P(A)=1\)，則\(A\)為必然事件D.互斥事件一定是對(duì)立事件，對(duì)立事件不一定是互斥事件答案：ABC7.已知\(a\)，\(b\)為正實(shí)數(shù)，且\(a+b=1\)，則下列結(jié)論正確的有\(zhòng)((\)\)A.\(ab\leq\frac{1}{4}\)B.\(\frac{1}{a}+\frac{1}\geq4\)C.\(a^2+b^2\geq\frac{1}{2}\)D.\(\sqrt{a}+\sqrt\leq\sqrt{2}\)答案：ABCD8.已知函數(shù)\(f(x)=x^3-3x\)，則下列說(shuō)法正確的有\(zhòng)((\)\)A.\(f(x)\)是奇函數(shù)B.\(f(x)\)的極大值為\(2\)，極小值為\(-2\)C.\(f(x)\)在\((-\infty,-1)\)上單調(diào)遞增D.\(f(x)\)在\((1,+\infty)\)上單調(diào)遞減答案：ABC9.已知圓\(C_1:x^2+y^2-2x-4y+4=0\)與圓\(C_2:x^2+y^2-4x-2y+1=0\)，則兩圓的位置關(guān)系是\((\)\)A.相交B.外切C.內(nèi)切D.相離答案：A10.已知數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n=n^2\)，則下列說(shuō)法正確的有\(zhòng)((\)\)A.\(a_1=1\)B.\(a_n=2n-1\)C.數(shù)列\(zhòng)(\{a_n\}\)是等差數(shù)列D.數(shù)列\(zhòng)(\{a_n\}\)是等比數(shù)列答案：ABC三、判斷題1.空集是任何集合的子集。（√）2.函數(shù)\(y=\sinx\)的最小正周期是\(2\pi\)。（√）3.若直線\(l_1:y=k_1x+b_1\)，\(l_2:y=k_2x+b_2\)，且\(k_1=k_2\)，則\(l_1\parallell_2\)。（×，當(dāng)\(b_1=b_2\)時(shí)兩直線重合）4.若\(a\)，\(b\)為實(shí)數(shù)，且\(a\cdotb=0\)，則\(a=0\)或\(b=0\)。（√）5.橢圓\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a\gtb\gt0)\)的離心率\(e=\frac{c}{a}\)，其中\(zhòng)(c^2=a^2-b^2\)。（√）6.若向量\(\overrightarrow{a}\cdot\overrightarrow=0\)，則\(\overrightarrow{a}\perp\overrightarrow\)。（×，當(dāng)\(\overrightarrow{a}=\overrightarrow{0}\)或\(\overrightarrow=\overrightarrow{0}\)時(shí)不成立）7.函數(shù)\(y=\log_ax\)（\(a>0\)且\(a\neq1\)）在\((0,+\infty)\)上一定是單調(diào)函數(shù)。（√）8.等比數(shù)列\(zhòng)(\{a_n\}\)中，若\(m+n=p+q\)，則\(a_m\cdota_n=a_p\cdota_q\)。（√）9.若直線\(l\)垂直于平面\(\alpha\)內(nèi)的無(wú)數(shù)條直線，則\(l\perp\alpha\)。（×，需垂直于平面內(nèi)兩條相交直線才行）10.對(duì)于函數(shù)\(y=f(x)\)，若\(f(a+x)=f(a-x)\)，則函數(shù)\(y=f(x)\)的圖象關(guān)于直線\(x=a\)對(duì)稱。（√）四、簡(jiǎn)答題1.已知函數(shù)\(f(x)=x^2-2x+3\)，求\(f(x)\)在區(qū)間\([0,3]\)上的最大值和最小值。答案：對(duì)\(f(x)=x^2-2x+3\)進(jìn)行配方得\(f(x)=(x-1)^2+2\)。函數(shù)對(duì)稱軸為\(x=1\)，開(kāi)口向上。在區(qū)間\([0,3]\)上，當(dāng)\(x=1\)時(shí)，\(f(x)\)取得最小值\(f(1)=2\)；當(dāng)\(x=3\)時(shí)，\(f(x)\)取得最大值\(f(3)=3^2-2\times3+3=6\)。2.已知等差數(shù)列\(zhòng)(\{a_n\}\)中，\(a_3=5\)，\(a_5=9\)，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\(d\)，則\(a_5-a_3=2d\)，由\(a_3=5\)，\(a_5=9\)可得\(2d=9-5=4\)，即\(d=2\)。又\(a_3=a_1+2d=5\)，把\(d=2\)代入得\(a_1+2\times2=5\)，解得\(a_1=1\)。所以通項(xiàng)公式\(a_n=a_1+(n-1)d=1+2(n-1)=2n-1\)。3.已知\(\sin\alpha=\frac{1}{3}\)，\(\alpha\in(\frac{\pi}{2},\pi)\)，求\(\cos(\alpha+\frac{\pi}{4})\)的值。答案：因?yàn)閈(\sin\alpha=\frac{1}{3}\)，\(\alpha\in(\frac{\pi}{2},\pi)\)，根據(jù)\(\sin^2\alpha+\cos^2\alpha=1\)，可得\(\cos\alpha=-\sqrt{1-\sin^2\alpha}=-\sqrt{1-(\frac{1}{3})^2}=-\frac{2\sqrt{2}}{3}\)。則\(\cos(\alpha+\frac{\pi}{4})=\cos\alpha\cos\frac{\pi}{4}-\sin\alpha\sin\frac{\pi}{4}=(-\frac{2\sqrt{2}}{3})\times\frac{\sqrt{2}}{2}-\frac{1}{3}\times\frac{\sqrt{2}}{2}=-\frac{4+\sqrt{2}}{6}\)。4.求過(guò)點(diǎn)\(P(2,3)\)且與直線\(2x-y+1=0\)平行的直線方程。答案：已知直線\(2x-y+1=0\)的斜率為\(2\)，因?yàn)樗笾本€與之平行，所以斜率也為\(2\)。由點(diǎn)斜式\(y-y_0=k(x-x_0)\)（其中\(zhòng)((x_0,y_0)\)為直線上一點(diǎn)，\(k\)為斜率），過(guò)點(diǎn)\(P(2,3)\)，斜率\(k=2\)，則直線方程為\(y-3=2(x-2)\)，整理得\(2x-y-1=0\)。五、討論題1.討論函數(shù)\(f(x)=\frac{1}{x}\)在\((0,+\infty)\)和\((-\infty,0)\)上的單調(diào)性，并證明。答案：\(f(x)=\frac{1}{x}\)在\((0,+\infty)\)上單調(diào)遞減，在\((-\infty,0)\)上單調(diào)遞減。證明：在\((0,+\infty)\)上任取\(x_1\)，\(x_2\)，且\(x_1\ltx_2\)，則\(f(x_1)-f(x_2)=\frac{1}{x_1}-\frac{1}{x_2}=\frac{x_2-x_1}{x_1x_2}\)。因?yàn)閈(0\ltx_1\ltx_2\)，所以\(x_2-x_1\gt0\)，\(x_1x_2\gt0\)，即\(f(x_1)-f(x_2)\gt0\)，\(f(x_1)\gtf(x_2)\)，所以\(f(x)\)在\((0,+\infty)\)上單調(diào)遞減。同理可證在\((-\infty,0)\)上也單調(diào)遞減。2.已知圓\(C\)的方程為\(x