高中競(jìng)賽班考試題及答案

一、單項(xiàng)選擇題（每題2分，共10題）1.若函數(shù)\(y=f(x)\)的定義域是\([1,3]\)，則函數(shù)\(y=f(x+1)\)的定義域是（）A.[0,2]B.[1,3]C.[2,4]D.無(wú)法確定答案：A2.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=9\)，\(a_{6}=243\)，則\(\{a_{n}\}\)的通項(xiàng)公式為（）A.\(a_{n}=3^{n-1}\)B.\(a_{n}=3^{n}\)C.\(a_{n}=3^{n+1}\)D.\(a_{n}=3^{n-2}\)答案：A3.已知向量\(\vec{a}=(1,2)\)，\(\vec=(x,1)\)，若\(\vec{a}\perp\vec\)，則\(x=\)（）A.-2B.2C.\(-\frac{1}{2}\)D.\(\frac{1}{2}\)答案：A4.在\(\triangleABC\)中，\(A=60^{\circ}\)，\(a=\sqrt{3}\)，\(b=1\)，則\(B=\)（）A.\(30^{\circ}\)B.\(45^{\circ}\)C.\(60^{\circ}\)D.\(90^{\circ}\)答案：A5.若\(f(x)=\sin(x+\varphi)\)是偶函數(shù)，則\(\varphi=\)（）A.\(\frac{\pi}{2}+k\pi,k\inZ\)B.\(\pi+k\pi,k\inZ\)C.\(2k\pi,k\inZ\)D.\(\frac{\pi}{4}+k\pi,k\inZ\)答案：A6.函數(shù)\(y=x^{3}-3x^{2}+3x-1\)的極值點(diǎn)個(gè)數(shù)為（）A.0B.1C.2D.3答案：A7.若雙曲線\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a>0,b>0)\)的漸近線方程為\(y=\pm\frac{1}{2}x\)，則其離心率\(e=\)（）A.\(\frac{\sqrt{5}}{2}\)B.\(\frac{\sqrt{3}}{2}\)C.\(\sqrt{5}\)D.\(\sqrt{3}\)答案：A8.從\(5\)名男生和\(3\)名女生中選\(3\)人參加某項(xiàng)活動(dòng)，要求既有男生又有女生，則不同的選法有（）種。A.45B.30C.40D.35答案：A9.若\(z=1+i\)，則\(\frac{z^{2}-3z+6}{z+1}=\)（）A.\(1-i\)B.\(1+i\)C.\(-1-i\)D.\(-1+i\)答案：A10.若\(\log_{a}\frac{2}{3}<1\)，則\(a\)的取值范圍是（）A.\((0,\frac{2}{3})\cup(1,+\infty)\)B.\((\frac{2}{3},1)\)C.\((1,+\infty)\)D.\((0,1)\)答案：A二、多項(xiàng)選擇題（每題2分，共10題）1.下列函數(shù)中，在\((0,+\infty)\)上單調(diào)遞增的是（）A.\(y=x^{2}\)B.\(y=\log_{2}x\)C.\(y=2^{x}\)D.\(y=\frac{1}{x}\)答案：ABC2.已知\(\triangleABC\)的內(nèi)角\(A\)，\(B\)，\(C\)所對(duì)的邊分別為\(a\)，\(b\)，\(c\)，若\(a=2\)，\(b=3\)，\(c=\sqrt{7}\)，則（）A.\(A=60^{\circ}\)B.\(B=120^{\circ}\)C.\(\cosC=\frac{1}{2}\)D.\(\sinC=\frac{\sqrt{3}}{2}\)答案：ACD3.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：B4.若\(a,b\inR\)，則下列不等式一定成立的是（）A.\(a^{2}+b^{2}\geqslant2ab\)B.\(a+\frac{1}{a}\geqslant2\)C.\(\frac{a+b}{2}\geqslant\sqrt{ab}\)D.\(a^{2}+1>0\)答案：AD5.對(duì)于函數(shù)\(y=A\sin(\omegax+\varphi)(A>0,\omega>0)\)，其圖象相鄰的最高點(diǎn)與最低點(diǎn)的橫坐標(biāo)之差為\(\pi\)，且圖象過(guò)點(diǎn)\((0,\frac{1}{2})\)，則（）A.\(\omega=1\)B.\(\varphi=\frac{\pi}{6}\)C.\(A=1\)D.函數(shù)圖象關(guān)于\(x=\frac{\pi}{3}\)對(duì)稱答案：AC6.若\(z_{1},z_{2}\)為復(fù)數(shù)，則下列命題正確的是（）A.若\(\vertz_{1}-z_{2}\vert=0\)，則\(z_{1}=z_{2}\)B.若\(z_{1}=\overline{z_{2}}\)，則\(\vertz_{1}\vert=\vertz_{2}\vert\)C.若\(\vertz_{1}\vert=\vertz_{2}\vert\)，則\(z_{1}=\pmz_{2}\)D.若\(z_{1}^{2}+z_{2}^{2}=0\)，則\(z_{1}=z_{2}=0\)答案：AB7.在正方體\(ABCD-A_{1}B_{1}C_{1}D_{1}\)中，下列說(shuō)法正確的是（）A.\(AC_{1}\)與\(BD\)垂直B.\(A_{1}C_{1}\)與\(AB_{1}\)平行C.\(AD_{1}\)與\(A_{1}C_{1}\)異面D.\(A_{1}D\)與\(B_{1}C\)平行答案：ACD8.已知\(f(x)\)是定義在\(R\)上的奇函數(shù)，當(dāng)\(x>0\)時(shí)，\(f(x)=x^{2}-2x\)，則（）A.\(f(0)=0\)B.當(dāng)\(x<0\)時(shí)，\(f(x)=-x^{2}-2x\)C.\(f(x)\)在\((-\infty,-1)\)上單調(diào)遞增D.\(f(x)\)在\((-1,1)\)上單調(diào)遞減答案：ABC9.從\(1\)到\(9\)這\(9\)個(gè)數(shù)字中任取\(3\)個(gè)不同的數(shù)字組成一個(gè)三位數(shù)，則（）A.組成的三位數(shù)共有\(zhòng)(A_{9}^{3}\)個(gè)B.若這個(gè)三位數(shù)是偶數(shù)，則有\(zhòng)(C_{4}^{1}A_{8}^{2}\)個(gè)C.若這個(gè)三位數(shù)中數(shù)字\(1\)不在百位，則有\(zhòng)(A_{8}^{1}A_{8}^{2}\)個(gè)D.若這個(gè)三位數(shù)中數(shù)字\(1\)和\(2\)不能相鄰，則有\(zhòng)(A_{7}^{3}\)個(gè)答案：ABC10.若\(\{a_{n}\}\)為等差數(shù)列，\(S_{n}\)為其前\(n\)項(xiàng)和，\(a_{1}=1\)，\(S_{3}=9\)，則（）A.\(a_{n}=2n-1\)B.\(S_{n}=n^{2}\)C.\(a_{3}=5\)D.數(shù)列\(zhòng)(\{\frac{1}{a_{n}a_{n+1}}\}\)的前\(n\)項(xiàng)和\(T_{n}=\frac{n}{2n+1}\)答案：ABC三、判斷題（每題2分，共10題）1.若\(a>b\)，則\(ac^{2}>bc^{2}\)。（×）2.函數(shù)\(y=\sinx\)在\([0,2\pi]\)上的對(duì)稱軸為\(x=\frac{\pi}{2}\)和\(x=\frac{3\pi}{2}\)。（√）3.若\(\vec{a}\cdot\vec=0\)，則\(\vec{a}=\vec{0}\)或\(\vec=\vec{0}\)。（×）4.等比數(shù)列\(zhòng)(\{a_{n}\}\)中，若\(a_{1}<0\)，\(q>1\)，則數(shù)列\(zhòng)(\{a_{n}\}\)單調(diào)遞減。（×）5.函數(shù)\(y=f(x)\)在\(x=x_{0}\)處的導(dǎo)數(shù)\(f'(x_{0})\)就是曲線\(y=f(x)\)在點(diǎn)\((x_{0},f(x_{0}))\)處的切線斜率。（√）6.若\(z\inC\)，則\(\vertz\vert^{2}=z^{2}\)。（×）7.在\(\triangleABC\)中，\(A>B\)是\(\sinA>\sinB\)的充分不必要條件。（×）8.若\(a,b\inR\)，則\((a+b)^{2}\geqslant4ab\)恒成立。（×）9.從\(n\)個(gè)不同元素中取出\(m(m\leqslantn)\)個(gè)元素的組合數(shù)\(C_{n}^{m}\)與順序有關(guān)。（×）10.若函數(shù)\(y=f(x)\)是偶函數(shù)，則\(y=f(x)\)的圖象關(guān)于\(y\)軸對(duì)稱。（√）四、簡(jiǎn)答題（每題5分，共4題）1.求函數(shù)\(y=\frac{1}{x-1}+x(x>1)\)的最小值。答案：\(y=\frac{1}{x-1}+x=\frac{1}{x-1}+(x-1)+1\)，因?yàn)閈(x>1\)，所以\(x-1>0\)，根據(jù)基本不等式\(a+b\geqslant2\sqrt{ab}\)，這里\(a=\frac{1}{x-1}\)，\(b=x-1\)，則\(y\geqslant2\sqrt{\frac{1}{x-1}\times(x-1)}+1=3\)，當(dāng)且僅當(dāng)\(\frac{1}{x-1}=x-1\)即\(x=2\)時(shí)取等號(hào)，最小值為3。2.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}=5\)，\(a_{5}=9\)，求\(a_{n}\)。答案：設(shè)等差數(shù)列\(zhòng)(\{a_{n}\}\)的公差為\(d\)，則\(a_{5}-a_{3}=2d\)，\(2d=9-5=4\)，\(d=2\)，又\(a_{3}=a_{1}+2d\)，\(5=a_{1}+4\)，\(a_{1}=1\)，所以\(a_{n}=a_{1}+(n-1)d=1+(n-1)\times2=2n-1\)。3.若\(a,b\inR\)，且\(a+b=1\)，求證\(a^{2}+b^{2}\geqslant\frac{1}{2}\)。答案：因?yàn)閈((a+b)^{2}=a^{2}+b^{2}+2ab\)，\(a+b=1\)，所以\(1=a^{2}+b^{2}+2ab\leqslanta^{2}+b^{2}+(a^{2}+b^{2})=2(a^{2}+b^{2})\)，則\(a^{2}+b^{2}\geqslant\frac{1}{2}\)。4.求直線\(y=x+1\)被圓\(x^{2}+y^{2}=4\)截得的弦長(zhǎng)。答案：圓\(x^{2}+y^{2}=4\)的圓心\((0,0)\)，半徑\(r=2\)，圓心到直線\(y=x+1\)即\(x-y+1=0\)的距離\(d=\frac{\vert0-0+1\vert}{\sqrt{1^{2}+(-1)^{2}}}=\frac{\sqrt{2}}{2}\)，根據(jù)弦長(zhǎng)公式\(l=2\sqrt{r^{2}-d^{2}}\)，弦長(zhǎng)\(l=2\sqrt{4-\frac{1}{2}}=\sqrt{14}\)。五、討論題（每題5分，共4題）1.討論函數(shù)\(y=\frac{ax+1}{x+2}(a\neq\frac{1}{2})\)的單調(diào)性。答案：\(y=\frac{ax+1}{x+2}=\frac{a(x+2)+1-2a}{x+2}=a+\frac{1-2a}{x+2}\)，當(dāng)\(1-2a>0\)即\(a<\frac{1}{2}\)時(shí)，\(y=\frac{1-2a}{x+2}+a\)在\((-\infty,-2)\)和\((-2,+\infty)\)上單調(diào)遞減；當(dāng)\(1-2a<0\)即\(a>\frac{1}{2}\)時(shí)，\(y=\frac{1-2a}{x+2}+a\)在\((-\infty,-2)\)和\((-2,+\infty)\)上單調(diào)遞增。2.討論等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比\(q\)對(duì)數(shù)列單調(diào)性的影響。答案：當(dāng)\(a_{1}>0\)，\(q>1\)或\(a_{1}<0\)，\(0<q<1\)時(shí)，數(shù)列\(zhòng)(\{a_{n}\}\)單調(diào)遞增；當(dāng)\(a_{1}>0\)