2025年高三理科期末試卷及答案

一、單項(xiàng)選擇題1.已知集合\(A=\{x|x^2-5x+6\leq0\}\)，\(B=\{x|2^x\gt4\}\)，則\(A\capB\)等于（）A.\((2,3]\)B.\([2,3]\)C.\((2,+\infty)\)D.\([3,+\infty)\)答案：A2.復(fù)數(shù)\(z=\frac{1+2i}{1-i}\)（\(i\)為虛數(shù)單位），則\(z\)的共軛復(fù)數(shù)\(\overline{z}\)為（）A.\(-\frac{1}{2}+\frac{3}{2}i\)B.\(-\frac{1}{2}-\frac{3}{2}i\)C.\(\frac{1}{2}+\frac{3}{2}i\)D.\(\frac{1}{2}-\frac{3}{2}i\)答案：B3.已知向量\(\overrightarrow{a}=(1,m)\)，\(\overrightarrow=(3,-2)\)，且\((\overrightarrow{a}+\overrightarrow)\perp\overrightarrow\)，則\(m\)的值為（）A.-8B.-6C.6D.8答案：D4.設(shè)\(a=\log_37\)，\(b=2^{1.1}\)，\(c=0.8^{3.1}\)，則（）A.\(b\lta\ltc\)B.\(c\lta\ltb\)C.\(c\ltb\lta\)D.\(a\ltc\ltb\)答案：B5.函數(shù)\(f(x)=\sin(2x+\frac{\pi}{3})\)的圖象向左平移\(\frac{\pi}{6}\)個(gè)單位長(zhǎng)度后得到\(g(x)\)的圖象，則\(g(x)\)的一個(gè)單調(diào)遞減區(qū)間是（）A.\([\frac{\pi}{12},\frac{7\pi}{12}]\)B.\([\frac{\pi}{6},\frac{2\pi}{3}]\)C.\([\frac{\pi}{3},\frac{5\pi}{6}]\)D.\([\frac{\pi}{2},\frac{3\pi}{4}]\)答案：C6.已知\(F_1,F_2\)是橢圓\(C:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a\gtb\gt0)\)的兩個(gè)焦點(diǎn)，\(P\)為橢圓\(C\)上一點(diǎn)，且\(\angleF_1PF_2=60^{\circ}\)，若\(\triangleF_1PF_2\)的面積為\(9\sqrt{3}\)，則\(b=\)（）A.3B.6C.9D.12答案：A7.執(zhí)行如圖所示的程序框圖，若輸入\(n=4\)，則輸出\(S\)的值為（）A.\(\frac{1}{2}\)B.\(\frac{5}{6}\)C.\(\frac{7}{12}\)D.\(\frac{11}{12}\)答案：D8.某幾何體的三視圖如圖所示，則該幾何體的體積為（）A.\(8\pi\)B.\(16\pi\)C.\(24\pi\)D.\(32\pi\)答案：B9.已知函數(shù)\(f(x)=\begin{cases}x^2+2x,x\geq0\\-x^2+2x,x\lt0\end{cases}\)，若\(f(a)\leq3\)，則\(a\)的取值范圍是（）A.\((-\infty,-3]\cup[1,+\infty)\)B.\([-3,1]\)C.\([-3,3]\)D.\((-\infty,-3]\cup[3,+\infty)\)答案：B10.已知\(f(x)\)是定義在\(R\)上的奇函數(shù)，當(dāng)\(x\gt0\)時(shí)，\(f(x)=x^2-4x\)，則不等式\(f(x)\gtx\)的解集用區(qū)間表示為（）A.\((-5,0)\cup(5,+\infty)\)B.\((-\infty,-5)\cup(0,5)\)C.\((-\infty,-5)\cup(5,+\infty)\)D.\((-5,0)\cup(0,5)\)答案：A二、多項(xiàng)選擇題1.下列說法正確的是（）A.若\(p\)：\(\forallx\inR\)，\(x^2+x+1\lt0\)，則\(\negp\)：\(\existsx\inR\)，\(x^2+x+1\geq0\)B.若\(p\veeq\)為真命題，則\(p\wedgeq\)也為真命題C.命題“若\(x^2-3x+2=0\)，則\(x=1\)”的逆否命題為“若\(x\neq1\)，則\(x^2-3x+2\neq0\)”D.設(shè)\(a,b\inR\)，則“\(a\neq0\)”是“\(ab\neq0\)”的必要不充分條件答案：ACD2.已知函數(shù)\(f(x)=\cos(2x+\varphi)(0\lt\varphi\lt\pi)\)的圖象關(guān)于直線\(x=\frac{\pi}{6}\)對(duì)稱，則（）A.\(\varphi=\frac{\pi}{6}\)B.\(f(x)\)在\((0,\frac{\pi}{6})\)上單調(diào)遞減C.\(f(x)\)的圖象向左平移\(\frac{\pi}{6}\)個(gè)單位長(zhǎng)度后得到\(g(x)=\cos2x\)的圖象D.\(f(x)\)在\((0,\frac{\pi}{2})\)上有且僅有一個(gè)零點(diǎn)答案：BD3.已知雙曲線\(C:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a\gt0,b\gt0)\)的離心率為\(\sqrt{3}\)，實(shí)軸長(zhǎng)為\(2\)，則（）A.雙曲線\(C\)的漸近線方程為\(y=\pm\sqrt{2}x\)B.雙曲線\(C\)的方程為\(x^2-\frac{y^2}{2}=1\)C.雙曲線\(C\)的焦點(diǎn)到漸近線的距離為\(\sqrt{2}\)D.直線\(y=x+1\)與雙曲線\(C\)有兩個(gè)交點(diǎn)答案：ABC4.以下關(guān)于函數(shù)\(y=x^3-3x^2+2x\)的說法正確的是（）A.函數(shù)有三個(gè)零點(diǎn)B.函數(shù)的極大值點(diǎn)為\(x=1\)C.函數(shù)的極小值為\(0\)D.函數(shù)在區(qū)間\((-\infty,0)\)上單調(diào)遞增答案：AD5.已知\(a,b\gt0\)，且\(a+b=1\)，則（）A.\(a^2+b^2\geq\frac{1}{2}\)B.\(\frac{1}{a}+\frac{1}\geq4\)C.\(\sqrt{a}+\sqrt\leq\sqrt{2}\)D.\(ab\geq\frac{1}{4}\)答案：ABC6.已知\(\{a_n\}\)是等差數(shù)列，\(S_n\)為其前\(n\)項(xiàng)和，若\(a_1+a_3+a_5=3\)，\(S_4=4\)，則（）A.\(a_1=-1\)B.\(d=1\)C.\(a_2=0\)D.\(S_{10}=40\)答案：ABC7.已知\(f(x)\)是定義在\(R\)上的偶函數(shù)，且\(f(x+4)=f(x-2)\)，若當(dāng)\(x\in[-3,0]\)時(shí)，\(f(x)=6^{-x}\)，則（）A.\(f(2025)=6\)B.\(f(x)\)在\([0,6]\)上有\(zhòng)(3\)個(gè)零點(diǎn)C.\(f(x)\)的圖象關(guān)于直線\(x=3\)對(duì)稱D.\(f(x)\)在\([0,2]\)上單調(diào)遞減答案：ACD8.已知\(A,B,C\)是球\(O\)的球面上三點(diǎn)，\(AB=2\)，\(BC=4\)，\(\angleABC=60^{\circ}\)，且三棱錐\(O-ABC\)的體積為\(\frac{4\sqrt{3}}{3}\)，則球\(O\)的表面積為（）A.\(16\pi\)B.\(20\pi\)C.\(24\pi\)D.\(36\pi\)答案：BC9.已知\(f(x)=\sin(\omegax+\varphi)(\omega\gt0,|\varphi|\lt\frac{\pi}{2})\)的部分圖象如圖所示，則（）A.\(\omega=2\)B.\(\varphi=\frac{\pi}{6}\)C.\(f(x)\)的單調(diào)遞增區(qū)間為\([k\pi-\frac{\pi}{3},k\pi+\frac{\pi}{6}](k\inZ)\)D.\(f(x)\)的圖象關(guān)于點(diǎn)\((\frac{k\pi}{2}+\frac{\pi}{6},0)(k\inZ)\)對(duì)稱答案：ABC10.已知\(a,b\inR\)，且\(a\gtb\gt0\)，則（）A.\(a^2\gtab\gtb^2\)B.\(\frac{a}\gt\frac{a+1}{b+1}\)C.\(a+\frac{1}\gtb+\frac{1}{a}\)D.\(2^a+2^{-b}\gt2^b+2^{-a}\)答案：ACD三、判斷題1.命題“若\(x^2=1\)，則\(x=1\)”的否命題為“若\(x^2=1\)，則\(x\neq1\)”。（×）2.若\(\sin\alpha=\frac{1}{2}\)，則\(\alpha=\frac{\pi}{6}\)。（×）3.函數(shù)\(y=\log_2(x^2+1)\)的值域是\([0,+\infty)\)。（√）4.已知直線\(l_1:ax+3y-1=0\)，\(l_2:x+by+1=0\)，若\(l_1\perpl_2\)，則\(ab=-3\)。（×）5.若向量\(\overrightarrow{a}=(1,2)\)，\(\overrightarrow=(3,4)\)，則\(\overrightarrow{a}\cdot\overrightarrow=11\)。（√）6.橢圓\(\frac{x^2}{16}+\frac{y^2}{9}=1\)的長(zhǎng)軸長(zhǎng)為\(8\)。（√）7.函數(shù)\(f(x)=x^3\)在\(R\)上是增函數(shù)。（√）8.若數(shù)列\(zhòng)(\{a_n\}\)滿足\(a_{n+1}=2a_n\)，則\(\{a_n\}\)是等比數(shù)列。（×）9.已知\(a\gt0\)，\(b\gt0\)，且\(a+b=1\)，則\(\frac{1}{a}+\frac{1}\geq4\)。（√）10.函數(shù)\(y=\sinx+\cosx\)的最大值為\(\sqrt{2}\)。（√）四、簡(jiǎn)答題1.已知函數(shù)\(f(x)=\sin^2x+\sqrt{3}\sinx\cosx\)。（1）求\(f(x)\)的最小正周期；（2）求\(f(x)\)在區(qū)間\([0,\frac{\pi}{2}]\)上的最大值和最小值。答案：（1）化簡(jiǎn)\(f(x)=\frac{1-\cos2x}{2}+\frac{\sqrt{3}}{2}\sin2x=\sin(2x-\frac{\pi}{6})+\frac{1}{2}\)，根據(jù)周期公式\(T=\frac{2\pi}{\omega}\)（\(\omega=2\)），得最小正周期\(T=\pi\)。（2）\(x\in[0,\frac{\pi}{2}]\)時(shí)，\(2x-\frac{\pi}{6}\in[-\frac{\pi}{6},\frac{5\pi}{6}]\)，\(\sin(2x-\frac{\pi}{6})\)在\(2x-\frac{\pi}{6}=\frac{\pi}{2}\)即\(x=\frac{\pi}{3}\)時(shí)取最大值\(1\)，\(f(x)\)最大值為\(\frac{3}{2}\)；在\(2x-\frac{\pi}{6}=-\frac{\pi}{6}\)即\(x=0\)時(shí)取最小值\(-\frac{1}{2}\)，\(f(x)\)最小值為\(0\)。2.已知等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，\(a_3=5\)，\(S_6=36\)。（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）設(shè)等差數(shù)列\(zhòng)(\{a_n\}\)公差為\(d\)，由\(a_3=5\)得\(a_1+2d=5\)，\(S_6=6a_1+\frac{6\times5}{2}d=36\)即\(6a_1+15d=36\)，聯(lián)立解得\(a_1=1\)，\(d=2\)，則\(a_n=1+2(n-1)=2n-1\)。（2