高三轉(zhuǎn)學考試題及答案

一、單項選擇題（每題2分，共10題）1.設集合\(A=\{1,2,3\}\)，\(B=\{x|x^2-3x+2=0\}\)，則\(A\capB=(\)\)A.\(\{1\}\)B.\(\{2\}\)C.\(\{1,2\}\)D.\(\{1,2,3\}\)答案：C2.若\(\sin\alpha=\frac{1}{3}\)，且\(\alpha\in(0,\frac{\pi}{2})\)，則\(\cos\alpha=(\)\)A.\(\frac{2\sqrt{2}}{3}\)B.\(-\frac{2\sqrt{2}}{3}\)C.\(\frac{\sqrt{2}}{3}\)D.\(-\frac{\sqrt{2}}{3}\)答案：A3.函數(shù)\(y=\log_{2}(x+1)\)的定義域是\((\)\)A.\((-1,+\infty)\)B.\((0,+\infty)\)C.\([-1,+\infty)\)D.\((-\infty,-1)\)答案：A4.已知向量\(\vec{a}=(1,2)\)，\(\vec=(2,-1)\)，則\(\vec{a}\cdot\vec=(\)\)A.0B.3C.4D.-3答案：A5.在等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{1}=1\)，\(d=2\)，則\(a_{5}=(\)\)A.9B.10C.11D.12答案：A6.過點\((1,1)\)且斜率為\(2\)的直線方程是\((\)\)A.\(y-1=2(x-1)\)B.\(y+1=2(x+1)\)C.\(y-1=\frac{1}{2}(x-1)\)D.\(y+1=\frac{1}{2}(x+1)\)答案：A7.若\(f(x)=x^{2}+1\)，則\(f(f(1))=(\)\)A.3B.5C.2D.4答案：B8.雙曲線\(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\)的漸近線方程是\((\)\)A.\(y=\pm\frac{3}{4}x\)B.\(y=\pm\frac{4}{3}x\)C.\(y=\pm\frac{9}{16}x\)D.\(y=\pm\frac{16}{9}x\)答案：B9.一個正方體的棱長為\(a\)，則它的外接球的半徑為\((\)\)A.\(\frac{\sqrt{3}}{2}a\)B.\(\sqrt{3}a\)C.\(\frac{a}{2}\)D.\(\frac{\sqrt{2}}{2}a\)答案：A10.已知函數(shù)\(y=A\sin(\omegax+\varphi)(A>0,\omega>0)\)的最小正周期為\(\pi\)，且當\(x=\frac{\pi}{6}\)時，\(y=2\)，則\(A=(\)\)A.1B.2C.\(\sqrt{2}\)D.\(\sqrt{3}\)答案：B二、多項選擇題（每題2分，共10題）1.下列函數(shù)是奇函數(shù)的是\((\)\)A.\(y=x^{3}\)B.\(y=\sinx\)C.\(y=e^{x}\)D.\(y=\frac{1}{x}\)答案：ABD2.若\(a>b\)，則下列不等式一定成立的是\((\)\)A.\(a+c>b+c\)B.\(ac>bc\)C.\(\frac{a}{c^{2}}>\frac{c^{2}}(c\neq0)\)D.\(a^{2}>b^{2}\)答案：AC3.在\(\triangleABC\)中，若\(A=60^{\circ}\)，\(a=\sqrt{3}\)，\(b=1\)，則\(B=(\)\)A.\(30^{\circ}\)B.\(150^{\circ}\)C.\(45^{\circ}\)D.\(135^{\circ}\)答案：A4.下列向量組中，能作為平面內(nèi)所有向量基底的是\((\)\)A.\(\vec{e}_{1}=(0,0)\)，\(\vec{e}_{2}=(1,-2)\)B.\(\vec{e}_{1}=(-1,2)\)，\(\vec{e}_{2}=(5,7)\)C.\(\vec{e}_{1}=(3,5)\)，\(\vec{e}_{2}=(6,10)\)D.\(\vec{e}_{1}=(2,-3)\)，\(\vec{e}_{2}=(\frac{1}{2},-\frac{3}{4})\)答案：B5.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比\(q=2\)，\(a_{1}=1\)，則下列結(jié)論正確的是\((\)\)A.\(a_{n}=2^{n-1}\)B.\(a_{3}=4\)C.\(S_{n}=2^{n}-1\)D.\(S_{3}=7\)答案：ABCD6.直線\(l:y=kx+1\)與圓\(C:x^{2}+y^{2}=1\)的位置關系可能是\((\)\)A.相交B.相切C.相離D.無法確定答案：AB7.函數(shù)\(y=\cos^{2}x-\sin^{2}x\)的性質(zhì)有\(zhòng)((\)\)A.最小正周期為\(\pi\)B.是偶函數(shù)C.圖象關于點\((\frac{\pi}{4},0)\)對稱D.在\([0,\frac{\pi}{2}]\)上單調(diào)遞減答案：ABD8.對于二項式\((a+b)^{n}\)，下列說法正確的是\((\)\)A.二項式系數(shù)之和為\(2^{n}\)B.展開式中第\(k+1\)項的通項公式為\(T_{k+1}=C_{n}^{k}a^{n-k}b^{k}\)C.當\(n=6\)時，展開式中二項式系數(shù)最大的項是第\(4\)項D.當\(n=5\)時，展開式中所有項的系數(shù)之和為\(32\)答案：ABC9.下列關于函數(shù)\(y=\frac{1}{x}\)的說法正確的是\((\)\)A.在\((-\infty,0)\)和\((0,+\infty)\)上單調(diào)遞減B.圖象關于原點對稱C.當\(x>1\)時，\(y\in(0,1)\)D.當\(x<-1\)時，\(y\in(-1,0)\)答案：ABC10.已知\(\alpha\)，\(\beta\)是空間中兩個不同的平面，\(m\)，\(n\)是平面\(\alpha\)，\(\beta\)外兩條不同的直線，則下列說法正確的是\((\)\)A.若\(m\paralleln\)，\(m\parallel\alpha\)，則\(n\parallel\alpha\)B.若\(m\perpn\)，\(m\perp\alpha\)，則\(n\parallel\alpha\)C.若\(m\parallel\alpha\)，\(\alpha\parallel\beta\)，則\(m\parallel\beta\)D.若\(m\perp\alpha\)，\(\alpha\parallel\beta\)，則\(m\perp\beta\)答案：D三、判斷題（每題2分，共10題）1.空集是任何集合的子集。（）答案：對2.若\(a\inR\)，則\(a^{2}\geqslant0\)。（）答案：對3.函數(shù)\(y=3^{x}\)與\(y=\log_{3}x\)的圖象關于直線\(y=x\)對稱。（）答案：對4.若\(\vec{a}\)與\(\vec\)垂直，則\(\vec{a}\cdot\(\vec=0\)。（）答案：對5.等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項和\(S_{n}=na_{1}+\frac{n(n-1)}{2}d\)。（）答案：對6.直線\(Ax+By+C=0(A,B\)不同時為\(0)\)的斜率\(k=-\frac{A}{B}\)。（）答案：錯7.對于函數(shù)\(y=f(x)\)，若\(f(a)=f(b)\)，則\(a=b\)。（）答案：錯8.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\(zhòng)(c^{2}=a^{2}-b^{2}\)。（）答案：對9.兩個向量的夾角范圍是\([0,\pi]\)。（）答案：對10.若\(p\)則\(q\)的逆命題是若\(q\)則\(p\)。（）答案：對四、簡答題（每題5分，共4題）1.求函數(shù)\(y=\frac{1}{x-1}\)在\(x=3\)處的導數(shù)。答案：首先對\(y=\frac{1}{x-1}=(x-1)^{-1}\)求導，根據(jù)求導公式\((X^{n})^\prime=nX^{n-1}\)，可得\(y^\prime=-(x-1)^{-2}=-\frac{1}{(x-1)^{2}}\)，當\(x=3\)時，\(y^\prime=-\frac{1}{4}\)。2.已知\(\triangleABC\)中，\(AB=3\)，\(AC=4\)，\(\angleA=60^{\circ}\)，求\(BC\)的長。答案：根據(jù)余弦定理\(BC^{2}=AB^{2}+AC^{2}-2AB\cdotAC\cdot\cosA\)，代入\(AB=3\)，\(AC=4\)，\(\angleA=60^{\circ}\)，可得\(BC^{2}=3^{2}+4^{2}-2\times3\times4\times\frac{1}{2}=25-12=13\)，所以\(BC=\sqrt{13}\)。3.求過點\((1,2)\)且與直線\(y=2x-1\)平行的直線方程。答案：因為所求直線與\(y=2x-1\)平行，所以斜率\(k=2\)，又過點\((1,2)\)，根據(jù)點斜式方程\(y-y_{1}=k(x-x_{1})\)，可得直線方程為\(y-2=2(x-1)\)，即\(y=2x\)。4.已知等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比\(q=3\)，\(a_{1}=1\)，求其前\(n\)項和\(S_{n}\)。答案：根據(jù)等比數(shù)列前\(n\)項和公式\(S_{n}=\frac{a_{1}(1-q^{n})}{1-q}\)，將\(q=3\)，\(a_{1}=1\)代入可得\(S_{n}=\frac{1\times(1-3^{n})}{1-3}=\frac{3^{n}-1}{2}\)。五、討論題（每題5分，共4題）1.討論函數(shù)\(y=x^{3}-3x\)的單調(diào)性。答案：對\(y=x^{3}-3x\)求導得\(y^\prime=3x^{2}-3=3(x+1)(x-1)\)。令\(y^\prime>0\)，解得\(x>1\)或\(x<-1\)；令\(y^\prime<0\)，解得\(-1<x<1\)。所以函數(shù)在\((-\infty,-1)\)和\((1,+\infty)\)上單調(diào)遞增，在\((-1,1)\)上單調(diào)遞減。2.在\(\triangleABC\)中，已知\(a=5\)，\(b=7\)，\(c=8\)，討論\(\triangleABC\)的形狀。答案：根據(jù)余弦定理\(\cosA=\frac{b^{2}+c^{2}-a^{2}}{2bc}=\frac{49+64-25}{2\times7\times8}=\frac{11}{14}>0\)，\(\cosB=\frac{a^{2}+c^{2}-b^{2}}{2ac}=\frac{25+64-49}{2\times5\times8}=\frac{1}{2}>0\)，\(\cosC=\frac{a^{2}+b^{2}-c^{2}}{2ab}=\frac{25+49-64}{2\times5\times7}=\frac{1}{7}>0\)，三個角都是銳角，所以\(\triangleABC\)是銳角三角形。3.討論直線\(y=kx+b\)與橢圓\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)\)的位置關系。答案：將\(y=kx+b\)代入橢圓方程得\(\frac{x^{2}}{a^{2}