高二數(shù)學(xué)期末考試試卷及答案

一、單項(xiàng)選擇題1.已知命題\(p\)：\(\forallx\inR\)，\(x^{2}+1\gt0\)，則\(\negp\)是（）A.\(\forallx\inR\)，\(x^{2}+1\leq0\)B.\(\existsx\inR\)，\(x^{2}+1\lt0\)C.\(\existsx\inR\)，\(x^{2}+1\leq0\)D.\(\existsx\inR\)，\(x^{2}+1\gt0\)答案：C2.若雙曲線\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\)（\(a\gt0\)，\(b\gt0\)）的一條漸近線方程為\(y=\sqrt{3}x\)，則其離心率為（）A.\(\sqrt{2}\)B.\(\sqrt{3}\)C.\(2\)D.\(3\)答案：C3.已知\(\vec{a}=(1,-2)\)，\(\vec=(x,1)\)，若\(\vec{a}\perp\vec\)，則\(x\)的值為（）A.\(2\)B.\(-2\)C.\(1\)D.\(-1\)答案：A4.在等差數(shù)列\(zhòng)(\{a_{n}\}\)中，\(a_{3}+a_{5}=10\)，則\(a_{4}\)的值為（）A.\(5\)B.\(6\)C.\(8\)D.\(10\)答案：A5.拋物線\(y^{2}=8x\)的焦點(diǎn)坐標(biāo)是（）A.\((2,0)\)B.\((-2,0)\)C.\((0,2)\)D.\((0,-2)\)答案：A6.已知\(x\)，\(y\)滿足約束條件\(\begin{cases}x+y\geq1\\x-y\leq1\\y\leq1\end{cases}\)，則\(z=2x-y\)的最大值為（）A.\(1\)B.\(2\)C.\(3\)D.\(4\)答案：C7.橢圓\(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\)的焦距為（）A.\(2\sqrt{5}\)B.\(\sqrt{5}\)C.\(2\sqrt{13}\)D.\(\sqrt{13}\)答案：A8.已知\(\{a_{n}\}\)是等比數(shù)列，\(a_{2}=2\)，\(a_{5}=\frac{1}{4}\)，則公比\(q\)等于（）A.\(-\frac{1}{2}\)B.\(-2\)C.\(2\)D.\(\frac{1}{2}\)答案：D9.若直線\(l\)的方向向量為\(\vec{m}=(1,0,2)\)，平面\(\alpha\)的法向量為\(\vec{n}=(-2,0,-4)\)，則（）A.\(l\parallel\alpha\)B.\(l\perp\alpha\)C.\(l\subset\alpha\)D.\(l\)與\(\alpha\)斜交答案：B10.在\(\triangleABC\)中，\(a=3\)，\(b=\sqrt{6}\)，\(A=\frac{2\pi}{3}\)，則\(B\)等于（）A.\(\frac{\pi}{4}\)B.\(\frac{\pi}{6}\)C.\(\frac{\pi}{3}\)D.\(\frac{\pi}{12}\)答案：A二、多項(xiàng)選擇題1.下列說(shuō)法正確的是（）A.命題“若\(x^{2}=1\)，則\(x=1\)”的否命題為“若\(x^{2}=1\)，則\(x\neq1\)”B.命題“\(\existsx\inR\)，\(x^{2}+x+1\lt0\)”的否定是“\(\forallx\inR\)，\(x^{2}+x+1\geq0\)”C.命題“若\(x=y\)，則\(\sinx=\siny\)”的逆否命題為真命題D.“\(x=-1\)”是“\(x^{2}-5x-6=0\)”的必要不充分條件答案：BC2.已知橢圓\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a\gtb\gt0)\)的左、右焦點(diǎn)分別為\(F_{1}\)，\(F_{2}\)，\(P\)是橢圓上一點(diǎn)，且\(PF_{2}\perpx\)軸，若\(\anglePF_{1}F_{2}=30^{\circ}\)，則橢圓的（）A.離心率\(e=\frac{\sqrt{3}}{3}\)B.離心率\(e=\sqrt{3}-1\)C.短軸長(zhǎng)與長(zhǎng)軸長(zhǎng)之比為\(\frac{\sqrt{2}}{2}\)D.短軸長(zhǎng)與長(zhǎng)軸長(zhǎng)之比為\(\frac{\sqrt{6}}{3}\)答案：AD3.設(shè)\(\{a_{n}\}\)是等差數(shù)列，\(S_{n}\)是其前\(n\)項(xiàng)和，且\(S_{5}\ltS_{6}\)，\(S_{6}=S_{7}\gtS_{8}\)，則下列結(jié)論正確的是（）A.\(d\lt0\)B.\(a_{7}=0\)C.\(S_{9}\gtS_{5}\)D.\(S_{6}\)與\(S_{7}\)均為\(S_{n}\)的最大值答案：ABD4.已知向量\(\vec{a}=(1,-1)\)，\(\vec=(2,x)\)，若\(\vec{a}\cdot\vec=1\)，則（）A.\(x=1\)B.\(|\vec|=\sqrt{5}\)C.\(\vec{a}\)與\(\vec\)的夾角為\(45^{\circ}\)D.\(\vec{a}\parallel\vec\)答案：ABC5.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\(\sqrt{3}\)，則（）A.雙曲線\(C\)的漸近線方程為\(y=\pm\sqrt{2}x\)B.\(\frac{b^{2}}{a^{2}}=2\)C.雙曲線\(C\)的漸近線與圓\((x-2)^{2}+y^{2}=1\)相離D.雙曲線\(C\)的漸近線與圓\((x-2)^{2}+y^{2}=1\)相交答案：ABC6.設(shè)等比數(shù)列\(zhòng)(\{a_{n}\}\)的公比為\(q\)，其前\(n\)項(xiàng)和為\(S_{n}\)，前\(n\)項(xiàng)積為\(T_{n}\)，并且滿足條件\(a_{1}\gt1\)，\(a_{7}a_{8}\gt1\)，\(\frac{a_{7}-1}{a_{8}-1}\lt0\)，則下列結(jié)論正確的是（）A.\(0\ltq\lt1\)B.\(a_{7}a_{9}\gt1\)C.\(T_{n}\)的最大值為\(T_{7}\)D.\(S_{n}\)的最大值為\(S_{7}\)答案：AC7.已知拋物線\(y^{2}=2px(p\gt0)\)，過(guò)其焦點(diǎn)且斜率為\(1\)的直線交拋物線于\(A\)，\(B\)兩點(diǎn)，若線段\(AB\)的中點(diǎn)的縱坐標(biāo)為\(2\)，則（）A.\(p=1\)B.\(p=2\)C.\(|AB|=8\)D.\(|AB|=10\)答案：BC8.已知\(\vec{a}=(1,2)\)，\(\vec=(-3,4)\)，\(\vec{c}=\vec{a}+\lambda\vec(\lambda\inR)\)，則（）A.當(dāng)\(\lambda=-\frac{1}{5}\)時(shí)，\(|\vec{c}|\)最小B.當(dāng)\(|\vec{c}|\)最小時(shí)，\(\vec\perp\vec{c}\)C.當(dāng)\(\lambda=1\)時(shí)，\(\vec{a}\)與\(\vec{c}\)夾角的余弦值為\(\frac{2\sqrt{5}}{5}\)D.當(dāng)\(\lambda=1\)時(shí)，\(\vec{a}\)與\(\vec{c}\)夾角的余弦值為\(\frac{\sqrt{5}}{5}\)答案：ABC9.已知橢圓\(C\)：\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a\gtb\gt0)\)的左、右頂點(diǎn)分別為\(A_{1}\)，\(A_{2}\)，點(diǎn)\(P\)是橢圓上異于\(A_{1}\)，\(A_{2}\)的點(diǎn)，若直線\(PA_{1}\)，\(PA_{2}\)的斜率之積為\(-\frac{1}{2}\)，則（）A.橢圓\(C\)的離心率為\(\frac{\sqrt{2}}{2}\)B.橢圓\(C\)的離心率為\(\frac{1}{2}\)C.橢圓\(C\)的短軸長(zhǎng)是長(zhǎng)軸長(zhǎng)的\(\sqrt{2}\)倍D.橢圓\(C\)的短軸長(zhǎng)是長(zhǎng)軸長(zhǎng)的\(\frac{\sqrt{2}}{2}\)倍答案：AD10.已知\(a\)，\(b\)，\(c\)分別為\(\triangleABC\)三個(gè)內(nèi)角\(A\)，\(B\)，\(C\)的對(duì)邊，且\((a+b)(\sinA-\sinB)=(c-b)\sinC\)，則（）A.\(A=\frac{\pi}{6}\)B.\(A=\frac{\pi}{3}\)C.若\(a=2\)，則\(\triangleABC\)面積的最大值為\(\sqrt{3}\)D.若\(a=2\)，則\(\triangleABC\)面積的最大值為\(2\sqrt{3}\)答案：BC三、判斷題1.命題“若\(a\gtb\)，則\(a^{2}\gtb^{2}\)”是真命題。（×）2.橢圓\(\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m\gt0,n\gt0)\)的焦點(diǎn)坐標(biāo)是\((\pm\sqrt{m^{2}-n^{2}},0)\)。（×）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，\(a_{1}=1\)，\(q=2\)，則\(S_{5}=31\)。（√）5.拋物線\(y^{2}=4x\)的準(zhǔn)線方程是\(x=-1\)。（√）6.若\(\{a_{n}\}\)是等差數(shù)列，則\(S_{n}\)，\(S_{2n}-S_{n}\)，\(S_{3n}-S_{2n}\)仍成等差數(shù)列。（√）7.雙曲線\(\frac{x^{2}}{9}-\frac{y^{2}}{16}=1\)的漸近線方程是\(y=\pm\frac{4}{3}x\)。（√）8.已知向量\(\vec{a}=(1,2)\)，\(\vec=(-2,m)\)，若\(\vec{a}\parallel\vec\)，則\(m=-4\)。（√）9.在\(\triangleABC\)中，\(a=3\)，\(b=5\)，\(\sinA=\frac{1}{3}\)，則滿足條件的\(\triangleABC\)有兩個(gè)。（√）10.若直線\(l\)的方向向量與平面\(\alpha\)的法向量垂直，則\(l\parallel\alpha\)。（×）四、簡(jiǎn)答題1.已知等差數(shù)列\(zhòng)(\{a_{n}\}\)的前\(n\)項(xiàng)和為\(S_{n}\)，\(a_{3}=5\)，\(S_{5}=25\)。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\(d\)。由\(a_{3}=5\)可得\(a_{1}+2d=5\)；由\(S_{5}=25\)，根據(jù)等差數(shù)列求和公式\(S_{n}=na_{1}+\frac{n(n-1)d}{2}\)，有\(zhòng)(5a_{1}+\frac{5\times4d}{2}=25\)，即\(5a_{1}+10d=25\)，化簡(jiǎn)得\(a_{1}+2d=5\)。聯(lián)立解得\(a_{1}=1\)，\(d=2\)。所以\(a_{n}=a_{1}+(n-1)d=1+2(n-1)=2n-1\)。2.已知橢圓\(C\)：\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a\gtb\gt0)\)的離心率為\(\frac{\sqrt{2}}{2}\)，且過(guò)點(diǎn)\((\sqrt{2},1)\)。求橢圓\(C\)的方程。答案：因?yàn)殡x心率\(e=\frac{c}{a}=\frac{\sqrt{2}}{2}\)，即\(c=\frac{\sqrt{2}}{2}a\)，又\(a^{2}=b^{2}+c^{2}\)，所以\(a^{2}=b^{2}+\frac{1}{2}a^{2}\)，即\(a^{2}=2b^{2}\)。橢圓過(guò)點(diǎn)\((\sqrt{2},1)\)，將其代入橢圓方程得\(\frac{(\sqrt{2})^{2}}{a^{2}}+\frac{1^{2}}{b^{2}}=1\)，把\(a^{2}=2b^{2}\)代入此式，可得\(\frac{2}{2b^{2}}+\frac{1}{b^{2}}=1\)，解得\(b^{2}=2\)，則\(a^{2}=4\)。所以橢圓\(C\)的方程為\(\frac{x^{2}}{4}+\frac{y^{2}}{2}=1\)。3.已