關(guān)于高考數(shù)學(xué)試題和答案

一、單項(xiàng)選擇題(每題2分,共10題)1.集合\(A=\{1,2,3\}\),\(B=\{2,3,4\}\),則\(A\capB\)=（）A.\(\{1,2\}\)B.\(\{2,3\}\)C.\(\{3,4\}\)D.\(\{1,4\}\)2.函數(shù)\(y=\log_2(x+1)\)的定義域是()A.\((-1,+\infty)\)B.\([-1,+\infty)\)C.\((0,+\infty)\)D.\([0,+\infty)\)3.已知向量\(\vec{a}=(1,2)\),\(\vec=(-2,m)\),若\(\vec{a}\parallel\vec\),則\(m\)=()A.-4B.4C.-1D.14.等差數(shù)列\(zhòng)(\{a_n\}\)中,\(a_1=1\),\(a_3=5\),則\(a_5\)=()A.9B.8C.7D.65.函數(shù)\(y=\sin(2x+\frac{\pi}{3})\)的最小正周期是()A.\(\pi\)B.\(2\pi\)C.\(\frac{\pi}{2}\)D.\(\frac{\pi}{4}\)6.已知\(x\),\(y\)滿足約束條件\(\begin{cases}x+y\geq1\\x-y\leq1\\y\leq1\end{cases}\),則\(z=3x-y\)的最大值為()A.1B.2C.3D.47.曲線\(y=x^3-2x+1\)在點(diǎn)\((1,0)\)處的切線方程為()A.\(y=x-1\)B.\(y=-x+1\)C.\(y=2x-2\)D.\(y=-2x+2\)8.已知\(\sin\alpha=\frac{3}{5}\),\(\alpha\in(\frac{\pi}{2},\pi)\),則\(\cos\alpha\)=()A.\(\frac{4}{5}\)B.\(-\frac{4}{5}\)C.\(\frac{3}{4}\)D.\(-\frac{3}{4}\)9.拋物線\(y^2=4x\)的焦點(diǎn)坐標(biāo)是()A.\((0,1)\)B.\((0,-1)\)C.\((1,0)\)D.\((-1,0)\)10.若\(a=\log_3\pi\),\(b=\log_2\sqrt{3}\),\(c=\log_3\sqrt{2}\),則()A.\(a>b>c\)B.\(a>c>b\)C.\(b>a>c\)D.\(b>c>a\)二、多項(xiàng)選擇題(每題2分,共10題)1.下列函數(shù)中,是偶函數(shù)的有（）A.\(y=x^2\)B.\(y=\cosx\)C.\(y=\sinx\)D.\(y=|x|\)2.以下哪些是等比數(shù)列的性質(zhì)()A.\(a_n^2=a_{n-1}a_{n+1}(n\geq2)\)B.\(S_n,S_{2n}-S_n,S_{3n}-S_{2n}\)成等比數(shù)列C.\(a_m\cdota_n=a_p\cdota_q(m+n=p+q)\)D.\(a_n=a_1q^{n-1}\)3.已知直線\(l_1:ax+y+1=0\),\(l_2:x+ay+1=0\),若\(l_1\parallell_2\),則\(a\)的值可能為()A.1B.-1C.0D.24.對(duì)于雙曲線\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1(a>0,b>0)\),以下說(shuō)法正確的是()A.實(shí)軸長(zhǎng)為\(2a\)B.虛軸長(zhǎng)為\(2b\)C.離心率\(e=\frac{c}{a}(c^2=a^2+b^2)\)D.漸近線方程為\(y=\pm\frac{a}x\)5.以下函數(shù)在其定義域內(nèi)單調(diào)遞增的有()A.\(y=2^x\)B.\(y=\lnx\)C.\(y=x^3\)D.\(y=\tanx\)6.已知\(a,b\inR\),且\(ab>0\),則下列不等式成立的是()A.\(a+b\geq2\sqrt{ab}\)B.\(\frac{1}{a}+\frac{1}\geq\frac{2}{\sqrt{ab}}\)C.\(a^2+b^2\geq2ab\)D.\(\frac{a}+\frac{a}\geq2\)7.一個(gè)正方體的棱長(zhǎng)為\(a\),以下說(shuō)法正確的是()A.正方體的表面積為\(6a^2\)B.正方體的體積為\(a^3\)C.正方體的體對(duì)角線長(zhǎng)為\(\sqrt{3}a\)D.正方體的面對(duì)角線長(zhǎng)為\(\sqrt{2}a\)8.已知\(\vec{a},\vec\)為非零向量,以下說(shuō)法正確的是()A.\(\vec{a}\cdot\vec=|\vec{a}||\vec|\cos\theta\)(\(\theta\)為\(\vec{a},\vec\)夾角)B.若\(\vec{a}\cdot\vec=0\),則\(\vec{a}\perp\vec\)C.\(|\vec{a}+\vec|^2=|\vec{a}|^2+2\vec{a}\cdot\vec+|\vec|^2\)D.若\(\vec{a}\parallel\vec\),則\(\vec{a}\cdot\vec=|\vec{a}||\vec|\)9.已知函數(shù)\(y=f(x)\),其導(dǎo)數(shù)\(f^\prime(x)\)的圖象如下,關(guān)于\(y=f(x)\)說(shuō)法正確的是（）A.\(f(x)\)在\((-\infty,x_1)\)單調(diào)遞增B.\(f(x)\)在\((x_1,x_2)\)單調(diào)遞減C.\(f(x)\)在\((x_2,+\infty)\)單調(diào)遞增D.\(x=x_1\)為\(f(x)\)的極大值點(diǎn)10.以下哪些是概率的基本性質(zhì)()A.\(0\leqP(A)\leq1\)B.\(P(\Omega)=1\)(\(\Omega\)為樣本空間)C.\(P(\varnothing)=0\)D.若\(A\),\(B\)互斥,則\(P(A\cupB)=P(A)+P(B)\)三、判斷題(每題2分,共10題)1.空集是任何集合的子集。()2.若\(a>b\),則\(a^2>b^2\)。（)3.函數(shù)\(y=\frac{1}{x}\)在定義域內(nèi)是減函數(shù)。()4.直線\(Ax+By+C=0\)（\(A\),\(B\)不同時(shí)為\(0\))的斜率為\(-\frac{A}{B}\)。（)5.橢圓\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)\)的離心率\(e\)滿足\(0<e<1\)。()6.若\(\sin\alpha=\sin\beta\),則\(\alpha=\beta\)。（)7.向量\(\vec{a}\)與\(\vec\)的數(shù)量積\(\vec{a}\cdot\vec\)是一個(gè)向量。()8.數(shù)列\(zhòng)(\{a_n\}\)的前\(n\)項(xiàng)和\(S_n=n^2\),則\(a_n=2n-1\)。（)9.函數(shù)\(y=\cosx\)的圖象關(guān)于\(y\)軸對(duì)稱(chēng)。()10.若\(a,b,c\)成等差數(shù)列,則\(2b=a+c\)。（）四、簡(jiǎn)答題(每題5分,共4題)1.求函數(shù)\(y=x^2-2x+3\)在區(qū)間\([0,3]\)上的最值。答案:\(y=x^2-2x+3=(x-1)^2+2\),對(duì)稱(chēng)軸\(x=1\)。當(dāng)\(x=1\)時(shí),\(y_{min}=2\);當(dāng)\(x=3\)時(shí),\(y_{max}=6\)。2.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\(a_n\)。答案：設(shè)公差為\(d\),由\(S_5=5a_3=25\)得\(a_3=5\)。又\(a_3=a_1+2d=5\),\(S_5=5a_1+\frac{5\times4}{2}d=25\),解得\(a_1=1\),\(d=2\),所以\(a_n=1+2(n-1)=2n-1\)。3.求過(guò)點(diǎn)\((1,2)\)且與直線\(2x-y+1=0\)平行的直線方程。答案:因?yàn)樗笾本€與\(2x-y+1=0\)平行,所以斜率\(k=2\)。由點(diǎn)斜式得\(y-2=2(x-1)\),即\(2x-y=0\)。4.已知\(\sin\alpha=\frac{4}{5}\),\(\alpha\in(\frac{\pi}{2},\pi)\),求\(\sin2\alpha\)的值。答案：因?yàn)閈(\alpha\in(\frac{\pi}{2},\pi)\),\(\sin\alpha=\frac{4}{5}\),則\(\cos\alpha=-\frac{3}{5}\)。\(\sin2\alpha=2\sin\alpha\cos\alpha=2\times\frac{4}{5}\times(-\frac{3}{5})=-\frac{24}{25}\)。五、討論題(每題5分,共4題)1.討論函數(shù)\(y=\frac{1}{x^2-1}\)的定義域、值域、單調(diào)性。答案:定義域?yàn)閈(x^2-1\neq0\),即\(x\neq\pm1\)。令\(t=x^2-1\geq-1\)且\(t\neq0\),則\(y=\frac{1}{t}\),值域?yàn)閈((-\infty,-1]\cup(0,+\infty)\)。在\((-\infty,-1)\)和\((-1,0)\)上單調(diào)遞增,在\((0,1)\)和\((1,+\infty)\)上單調(diào)遞減。2.已知橢圓\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)\)和直線\(y=kx+m\),討論直線與橢圓的位置關(guān)系判定方法。答案:將\(y=kx+m\)代入橢圓方程得一個(gè)關(guān)于\(x\)的一元二次方程,通過(guò)判別式\(\Delta\)判斷。\(\Delta>0\)時(shí),直線與橢圓相交;\(\Delta=0\)時(shí),直線與橢圓相切;\(\Delta<0\)時(shí),直線與橢圓相離。3.討論在立體幾何中,如何證明線面垂直。答案：可利用定義,證明直線與平面內(nèi)任意一條直線垂直；也可用判定定理,證明直線與平面內(nèi)兩條相交直線都垂直；還可利用面面垂直性質(zhì),若兩平面垂直,在一個(gè)平面內(nèi)垂直交線的直線垂直另一平面。4.討論概率在實(shí)際生活中的應(yīng)用及意義。答案:在保險(xiǎn)、風(fēng)險(xiǎn)評(píng)估、抽獎(jiǎng)、質(zhì)量控制等方面應(yīng)用廣泛。能幫助人們量化不確定