2025年建湖高一數(shù)學(xué)試卷及答案

一、單項(xiàng)選擇題1.已知集合\(A=\{x|x^2-3x+2=0\}\)，\(B=\{x|0<x<6,x\inN\}\)，則滿足\(A\subseteqC\subseteqB\)的集合\(C\)的個(gè)數(shù)為（）A.4B.8C.7D.16答案：B2.函數(shù)\(y=\sqrt{3-2x-x^2}\)的定義域是（）A.\([-3,1]\)B.\((-\infty,-3]\cup[1,+\infty)\)C.\([-1,3]\)D.\((-\infty,-1]\cup[3,+\infty)\)答案：A3.已知冪函數(shù)\(y=f(x)\)的圖象過點(diǎn)\((4,2)\)，則\(f\left(\dfrac{1}{2}\right)\)的值為（）A.\(\sqrt{2}\)B.\(\dfrac{\sqrt{2}}{2}\)C.\(\dfrac{1}{4}\)D.\(4\)答案：B4.若\(a=0.3^{0.3}\)，\(b=0.3^2\)，\(c=\log_{0.3}2\)，則\(a\)，\(b\)，\(c\)的大小關(guān)系為（）A.\(a<b<c\)B.\(b<a<c\)C.\(c<a<b\)D.\(c<b<a\)答案：D5.已知\(\sin\alpha=\dfrac{3}{5}\)，\(\alpha\in\left(\dfrac{\pi}{2},\pi\right)\)，則\(\tan\alpha\)的值為（）A.\(\dfrac{3}{4}\)B.\(-\dfrac{3}{4}\)C.\(\dfrac{4}{3}\)D.\(-\dfrac{4}{3}\)答案：B6.函數(shù)\(y=\sin\left(2x+\dfrac{\pi}{3}\right)\)的最小正周期是（）A.\(\dfrac{\pi}{2}\)B.\(\pi\)C.\(2\pi\)D.\(4\pi\)答案：B7.已知向量\(\overrightarrow{a}=(1,2)\)，\(\overrightarrow=(m,-1)\)，若\(\overrightarrow{a}\parallel(\overrightarrow{a}+\overrightarrow)\)，則\(m\)的值為（）A.\(\dfrac{1}{2}\)B.\(-\dfrac{1}{2}\)C.3D.\(-3\)答案：B8.在\(\triangleABC\)中，\(a=3\)，\(b=5\)，\(\sinA=\dfrac{1}{3}\)，則\(\sinB\)等于（）A.\(\dfrac{1}{5}\)B.\(\dfrac{5}{9}\)C.\(\dfrac{\sqrt{5}}{3}\)D.1答案：B9.已知等差數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和為\(S_n\)，若\(a_3+a_5=18\)，則\(S_7\)的值為（）A.63B.49C.36D.25答案：A10.已知\(x>0\)，\(y>0\)，且\(x+y=1\)，則\(\dfrac{1}{x}+\dfrac{4}{y}\)的最小值為（）A.9B.8C.7D.6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.\(y=\dfrac{1}{x^3}\)C.\(y=1+\sinx\)D.\(y=2^{|x|}\)答案：AD2.以下關(guān)于函數(shù)\(y=\log_2x\)的說法正確的是（）A.在\((0,+\infty)\)上單調(diào)遞增B.過定點(diǎn)\((1,0)\)C.是奇函數(shù)D.定義域?yàn)閈(R\)答案：AB3.已知\(\alpha\)是第二象限角，\(\sin\alpha=\dfrac{3}{5}\)，則以下正確的有（）A.\(\cos\alpha=-\dfrac{4}{5}\)B.\(\tan\alpha=-\dfrac{3}{4}\)C.\(\sin\left(\alpha+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{10}\)D.\(\cos\left(\alpha-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{10}\)答案：ABC4.下列向量中，與向量\(\overrightarrow{a}=(1,-2)\)垂直的有（）A.\(\overrightarrow=(2,1)\)B.\(\overrightarrow{c}=(-1,2)\)C.\(\overrightarrowuci6cq0=(4,2)\)D.\(\overrightarrow{e}=(2,-1)\)答案：AC5.對(duì)于\(\triangleABC\)，以下說法正確的是（）A.若\(a^2+b^2<c^2\)，則\(\triangleABC\)是鈍角三角形B.若\(A>B\)，則\(\sinA>\sinB\)C.若\(a=8\)，\(c=10\)，\(B=60^{\circ}\)，則符合條件的\(\triangleABC\)有兩個(gè)D.若\(\sin^2A+\sin^2B<\sin^2C\)，則\(\triangleABC\)是鈍角三角形答案：ABD6.等差數(shù)列\(zhòng)(\{a_n\}\)的公差\(d\neq0\)，首項(xiàng)\(a_1=1\)，若\(a_1\)，\(a_2\)，\(a_5\)成等比數(shù)列，則（）A.\(a_n=2n-1\)B.\(S_n=n^2\)C.\(d=2\)D.數(shù)列\(zhòng)(\{2^{a_n}\}\)的前\(n\)項(xiàng)和為\(\dfrac{2(4^n-1)}{3}\)答案：ABCD7.已知\(x>0\)，\(y>0\)，且\(x+2y=1\)，則（）A.\(xy\leqslant\dfrac{1}{8}\)B.\(\dfrac{1}{x}+\dfrac{2}{y}\geqslant9\)C.\(x^2+4y^2\geqslant\dfrac{1}{2}\)D.\(\log_2x+\log_2y\leqslant-3\)答案：ABCD8.下列函數(shù)中，值域是\((0,+\infty)\)的有（）A.\(y=\dfrac{1}{x^2}\)B.\(y=2x+1(x>0)\)C.\(y=\sqrt{x^2+1}\)D.\(y=\left(\dfrac{1}{2}\right)^{x}\)答案：AD9.已知函數(shù)\(y=A\sin(\omegax+\varphi)\)（\(A>0\)，\(\omega>0\)，\(|\varphi|<\dfrac{\pi}{2}\)）的部分圖象，則（）A.\(A=2\)B.\(\omega=2\)C.\(\varphi=\dfrac{\pi}{6}\)D.函數(shù)的單調(diào)遞增區(qū)間為\(\left[k\pi-\dfrac{\pi}{3},k\pi+\dfrac{\pi}{6}\right](k\inZ)\)答案：ABCD10.設(shè)\(S_n\)為等比數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和，滿足\(a_1=1\)，且\(S_5=3S_4+1\)，則（）A.\(q=2\)B.\(a_n=2^{n-1}\)C.\(S_n=2^{n}-1\)D.數(shù)列\(zhòng)(\{a_n\}\)是遞增數(shù)列答案：ABCD三、判斷題1.空集是任何集合的真子集。（）答案：×2.函數(shù)\(y=\dfrac{1}{\sqrt{x-1}}\)的定義域是\((1,+\infty)\)。（）答案：√3.若\(a>b\)，則\(a^2>b^2\)。（）答案：×4.函數(shù)\(y=\sinx\)的圖象關(guān)于原點(diǎn)對(duì)稱。（）答案：√5.向量\(\overrightarrow{a}=(1,2)\)與向量\(\overrightarrow=(2,4)\)共線。（）答案：√6.在\(\triangleABC\)中，若\(a=2\)，\(b=3\)，\(\sinA=\dfrac{1}{3}\)，則\(\sinB=\dfrac{1}{2}\)。（）答案：×7.等差數(shù)列\(zhòng)(\{a_n\}\)中，若\(a_3+a_7=10\)，則\(a_5=5\)。（）答案：√8.函數(shù)\(y=\log_2(x^2+1)\)的值域是\([0,+\infty)\)。（）答案：√9.若\(x>0\)，\(y>0\)，且\(x+y=2\)，則\(xy\)的最大值為\(1\)。（）答案：√10.函數(shù)\(y=\cos\left(2x+\dfrac{\pi}{3}\right)\)的最小正周期是\(\pi\)。（）答案：√四、簡(jiǎn)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：解方程\(x^2-5x+6=0\)，即\((x-2)(x-3)=0\)，得\(x=2\)或\(x=3\)，所以\(A=\{2,3\}\)。當(dāng)\(B=\varnothing\)時(shí)，\(mx-1=0\)無解，此時(shí)\(m=0\)。當(dāng)\(B\neq\varnothing\)時(shí)，\(x=\dfrac{1}{m}\)。若\(\dfrac{1}{m}=2\)，則\(m=\dfrac{1}{2}\)；若\(\dfrac{1}{m}=3\)，則\(m=\dfrac{1}{3}\)。綜上，\(m=0\)或\(m=\dfrac{1}{2}\)或\(m=\dfrac{1}{3}\)。2.已知函數(shù)\(f(x)=x^2+2ax+1\)，\(x\in[-1,2]\)，求\(f(x)\)的最小值\(g(a)\)。答案：函數(shù)\(f(x)=x^2+2ax+1\)的圖象開口向上，對(duì)稱軸為\(x=-a\)。當(dāng)\(-a\leqslant-1\)即\(a\geqslant1\)時(shí)，\(f(x)\)在\([-1,2]\)上單調(diào)遞增，\(g(a)=f(-1)=2-2a\)。當(dāng)\(-1<-a<2\)即\(-2<a<1\)時(shí)，\(g(a)=f(-a)=1-a^2\)。當(dāng)\(-a\geqslant2\)即\(a\leqslant-2\)時(shí)，\(f(x)\)在\([-1,2]\)上單調(diào)遞減，\(g(a)=f(2)=5+4a\)。所以\(g(a)=\begin{cases}5+4a,a\leqslant-2\\1-a^2,-2<a<1\\2-2a,a\geqslant1\end{cases}\)。3.已知\(\sin\alpha+\cos\alpha=\dfrac{1}{5}\)，\(\alpha\in(0,\pi)\)，求\(\tan\alpha\)的值。答案：將\(\sin\alpha+\cos\alpha=\dfrac{1}{5}\)兩邊平方得\((\sin\alpha+\cos\alpha)^2=\dfrac{1}{25}\)，即\(1+2\sin\alpha\cos\alpha=\dfrac{1}{25}\)，所以\(2\sin\alpha\cos\alpha=-\dfrac{24}{25}\)。因?yàn)閈(\alpha\in(0,\pi)\)，\(2\sin\alpha\cos\alpha<0\)，所以\(\sin\alpha>0\)，\(\cos\alpha<0\)。\((\sin\alpha-\cos\alpha)^2=1-2\sin\alpha\cos\alpha=1+\dfrac{24}{25}=\dfrac{49}{25}\)，則\(\sin\alpha-\cos\alpha=\dfrac{7}{5}\)。聯(lián)立\(\begin{cases}\sin\alpha+\cos\alpha=\dfrac{1}{5}\\\sin\alpha-\cos\alpha=\dfrac{7}{5}\end{cases}\)，解得\(\begin{cases}\sin\alpha=\dfrac{4}{5}\\\cos\alpha=-\dfrac{3}{5}\end{cases}\)，所以\(\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=-\dfra