2025年二卷數(shù)學(xué)期中試卷及答案

一、單項(xiàng)選擇題1.已知集合\(A=\{x|x^2-3x+2=0\}\)，\(B=\{x|0<x<5,x\inN\}\)，則滿足條件\(A\subseteqC\subseteqB\)的集合\(C\)的個(gè)數(shù)為（）A.1B.2C.3D.4答案：D2.函數(shù)\(f(x)=\frac{1}{\sqrt{1-x}}+\lg(3x+1)\)的定義域是（）A.\((-\frac{1}{3},+\infty)\)B.\((-\infty,1)\)C.\((-\frac{1}{3},1)\)D.\((-\frac{1}{3},1]\)答案：C3.已知函數(shù)\(f(x)=a^{x-2}+1\)（\(a>0\)且\(a\neq1\)）的圖象恒過定點(diǎn)\(P\)，則點(diǎn)\(P\)的坐標(biāo)是（）A.\((2,1)\)B.\((2,2)\)C.\((1,1)\)D.\((1,2)\)答案：B4.已知\(\alpha\)是第二象限角，\(\sin\alpha=\frac{3}{5}\)，則\(\cos\alpha\)等于（）A.\(\frac{4}{5}\)B.\(-\frac{4}{5}\)C.\(\pm\frac{4}{5}\)D.\(-\frac{3}{5}\)答案：B5.已知\(\tan\alpha=2\)，則\(\frac{\sin\alpha-\cos\alpha}{\sin\alpha+\cos\alpha}\)的值為（）A.\(\frac{1}{3}\)B.\(-\frac{1}{3}\)C.3D.-3答案：A6.函數(shù)\(y=\sin(2x+\frac{\pi}{3})\)的最小正周期是（）A.\(\frac{\pi}{2}\)B.\(\pi\)C.\(2\pi\)D.\(4\pi\)答案：B7.已知向量\(\overrightarrow{a}=(1,2)\)，\(\overrightarrow=(m,4)\)，若\(\overrightarrow{a}\parallel\overrightarrow\)，則實(shí)數(shù)\(m\)的值為（）A.-2B.2C.-8D.8答案：B8.已知等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，若\(a_3=5\)，\(S_3=9\)，則\(a_5\)的值為（）A.7B.9C.11D.13答案：C9.在等比數(shù)列\(zhòng)(\{a_n\}\)中，\(a_2=2\)，\(a_5=16\)，則公比\(q\)等于（）A.-2B.2C.\(\frac{1}{2}\)D.4答案：B10.已知函數(shù)\(f(x)=x^3+ax^2+bx+c\)，若\(f(1)=f(2)=f(3)=0\)，則\(f(0)\)的值為（）A.6B.-6C.10D.-10答案：A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.\(y=\frac{1}{x^2}\)C.\(y=x+\frac{1}{x}\)D.\(y=|x|\)答案：ABD2.下列說法正確的是（）A.若\(a>b\)，則\(ac^2>bc^2\)B.若\(a>b\)，\(c>d\)，則\(a-c>b-d\)C.若\(a>b\)，\(c>d\)，則\(ac>bd\)D.若\(a>b>0\)，\(c<0\)，則\(\frac{c}{a}>\frac{c}\)答案：D3.已知向量\(\overrightarrow{a}=(1,1)\)，\(\overrightarrow=(-1,2)\)，則（）A.\(\overrightarrow{a}\cdot\overrightarrow=1\)B.\(|\overrightarrow{a}|=\sqrt{2}\)C.\(\overrightarrow{a}\)與\(\overrightarrow\)的夾角為\(\frac{\pi}{4}\)D.\(\overrightarrow{a}\)在\(\overrightarrow\)方向上的投影為\(\frac{\sqrt{5}}{5}\)答案：ABD4.已知數(shù)列\(zhòng)(\{a_n\}\)是等差數(shù)列，其前\(n\)項(xiàng)和為\(S_n\)，若（）A.若\(a_{10}=0\)，則\(S_{19}=0\)B.若\(S_{10}=S_{20}\)，則\(S_{30}=0\)C.若\(a_{10}>0\)，\(a_{11}<0\)，則\(S_{10}\)最大D.若\(S_{10}>0\)，\(S_{11}<0\)，則\(S_{5}\)最大答案：ABCD5.下列關(guān)于函數(shù)\(y=\sin(2x+\frac{\pi}{6})\)的說法正確的是（）A.圖象關(guān)于點(diǎn)\((\frac{\pi}{3},0)\)對稱B.圖象關(guān)于直線\(x=\frac{\pi}{6}\)對稱C.在區(qū)間\([-\frac{\pi}{6},\frac{\pi}{3}]\)上單調(diào)遞增D.在區(qū)間\([\frac{\pi}{6},\frac{2\pi}{3}]\)上單調(diào)遞減答案：BD6.已知函數(shù)\(f(x)\)是定義在\(R\)上的奇函數(shù)，當(dāng)\(x\geqslant0\)時(shí)，\(f(x)=x^2-2x\)，則（）A.\(f(-1)=1\)B.當(dāng)\(x<0\)時(shí)，\(f(x)=-x^2-2x\)C.函數(shù)\(f(x)\)有兩個(gè)零點(diǎn)D.函數(shù)\(f(x)\)在\([-1,1]\)上單調(diào)遞減答案：BD7.設(shè)\(a>0\)，\(b>0\)，且\(a+b=1\)，則（）A.\(ab\leqslant\frac{1}{4}\)B.\(\frac{1}{a}+\frac{1}\geqslant4\)C.\(a^2+b^2\geqslant\frac{1}{2}\)D.\(\sqrt{a}+\sqrt\leqslant\sqrt{2}\)答案：ABCD8.已知函數(shù)\(f(x)=\log_2(x^2-ax+3a)\)在\([2,+\infty)\)上是增函數(shù)，則實(shí)數(shù)\(a\)的取值范圍是（）A.\((-\infty,4]\)B.\((-4,4]\)C.\((0,4]\)D.\([4,+\infty)\)答案：B9.已知\(\alpha\)，\(\beta\)為銳角，\(\sin\alpha=\frac{3}{5}\)，\(\cos(\alpha+\beta)=-\frac{5}{13}\)，則（）A.\(\cos\alpha=\frac{4}{5}\)B.\(\sin(\alpha+\beta)=\frac{12}{13}\)C.\(\sin\beta=\frac{56}{65}\)D.\(\cos\beta=\frac{33}{65}\)答案：ABD10.已知函數(shù)\(f(x)\)滿足\(f(x+1)=f(x-1)\)，且當(dāng)\(x\in[-1,1]\)時(shí)，\(f(x)=x^2\)，則（）A.函數(shù)\(f(x)\)的周期為\(2\)B.\(f(2025)=1\)C.函數(shù)\(f(x)\)是偶函數(shù)D.函數(shù)\(f(x)\)在\([2,3]\)上單調(diào)遞增答案：ABCD三、判斷題1.空集是任何集合的真子集。（×）2.函數(shù)\(y=\frac{1}{x}\)在其定義域上是減函數(shù)。（×）3.若\(\sin\alpha=\sin\beta\)，則\(\alpha=\beta\)。（×）4.向量\(\overrightarrow{a}\)與向量\(\overrightarrow\)平行，則\(\overrightarrow{a}\)與\(\overrightarrow\)的方向相同或相反。（×）5.等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n=n^2+n\)，則\(a_n=2n\)。（√）6.函數(shù)\(y=A\sin(\omegax+\varphi)\)（\(A>0\)，\(\omega>0\)）的振幅是\(A\)，周期是\(T=\frac{2\pi}{\omega}\)。（√）7.若\(a>b>0\)，則\(\frac{1}{a}<\frac{1}\)。（√）8.對數(shù)函數(shù)\(y=\log_ax\)（\(a>0\)且\(a\neq1\)）的圖象恒過點(diǎn)\((1,0)\)。（√）9.等比數(shù)列\(zhòng)(\{a_n\}\)中，若\(a_1=1\)，\(q=2\)，則\(a_3=4\)。（√）10.若函數(shù)\(f(x)\)在區(qū)間\([a,b]\)上的圖象是連續(xù)不斷的，且\(f(a)\cdotf(b)<0\)，則函數(shù)\(f(x)\)在區(qū)間\((a,b)\)內(nèi)至少有一個(gè)零點(diǎn)。（√）四、簡答題1.已知集合\(A=\{x|x^2-5x+6=0\}\)，\(B=\{x|mx-1=0\}\)，且\(B\subseteqA\)，求實(shí)數(shù)\(m\)的值。答案：先解方程\(x^2-5x+6=0\)，即\((x-2)(x-3)=0\)，得\(x=2\)或\(x=3\)，所以\(A=\{2,3\}\)。因?yàn)閈(B\subseteqA\)，當(dāng)\(B=\varnothing\)時(shí)，\(mx-1=0\)無解，\(m=0\)；當(dāng)\(B\neq\varnothing\)時(shí)，\(x=\frac{1}{m}\)，若\(\frac{1}{m}=2\)，則\(m=\frac{1}{2}\)；若\(\frac{1}{m}=3\)，則\(m=\frac{1}{3}\)。綜上，\(m=0\)或\(\frac{1}{2}\)或\(\frac{1}{3}\)。2.已知函數(shù)\(f(x)=\sin(2x+\frac{\pi}{6})+\frac{1}{2}\)，求函數(shù)\(f(x)\)的單調(diào)遞增區(qū)間。答案：根據(jù)正弦函數(shù)單調(diào)性，令\(2k\pi-\frac{\pi}{2}\leqslant2x+\frac{\pi}{6}\leqslant2k\pi+\frac{\pi}{2}\)，\(k\inZ\)。先解不等式\(2k\pi-\frac{\pi}{2}\leqslant2x+\frac{\pi}{6}\)，得\(2k\pi-\frac{2\pi}{3}\leqslant2x\)，即\(k\pi-\frac{\pi}{3}\leqslantx\)；再解\(2x+\frac{\pi}{6}\leqslant2k\pi+\frac{\pi}{2}\)，得\(2x\leqslant2k\pi+\frac{\pi}{3}\)，即\(x\leqslantk\pi+\frac{\pi}{6}\)。所以\(f(x)\)的單調(diào)遞增區(qū)間是\([k\pi-\frac{\pi}{3},k\pi+\frac{\pi}{6}]\)，\(k\inZ\)。3.已知向量\(\overrightarrow{a}=(1,2)\)，\(\overrightarrow=(-2,1)\)，求\((\overrightarrow{a}+2\overrightarrow)\cdot\overrightarrow{a}\)的值。答案：先求\(\overrightarrow{a}+2\overrightarrow\)，\(2\overrightarrow=2(-2,1)=(-4,2)\)，則\(\overrightarrow{a}+2\overrightarrow=(1-4,2+2)=(-3,4)\)。再求\((\overrightarrow{a}+2\overrightarrow)\cdot\overrightarrow{a}\)，根據(jù)向量數(shù)量積坐標(biāo)運(yùn)算公式，\((\overrightarrow{a}+2\overrightarrow)\cdot\overrightarrow{a}=-3\times1+4\times2=-3+8=5\)。4.已知等差數(shù)列\(zhòng)(\{a_n\}\)中，\(a_1=1\)，\(a_3=5\)，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\(a_n=a_1+(n-1)d\)。已知\(a_1=1\)，\(a_3=5\)，由\(a_3=a_1+2d\)，可得\(5=1+2d\)，解得\(d=2\)。所以數(shù)列\(zhòng)(\{a_n\}\)的通項(xiàng)公式\(a_n=1