數(shù)學(xué)選修1測試題及答案

一、單項選擇題1.命題“若\(a\gtb\)，則\(a-1\gtb-1\)”的逆否命題是（）A.若\(a-1\ltb-1\)，則\(a\ltb\)B.若\(a-1\leqb-1\)，則\(a\leqb\)C.若\(a\leqb\)，則\(a-1\leqb-1\)D.若\(a-1\gtb-1\)，則\(a\gtb\)答案：B2.拋物線\(y=2x^{2}\)的焦點坐標(biāo)是（）A.\((0,\frac{1}{8})\)B.\((0,\frac{1}{2})\)C.\((\frac{1}{8},0)\)D.\((\frac{1}{2},0)\)答案：A3.橢圓\(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)的離心率為（）A.\(\frac{\sqrt{7}}{4}\)B.\(\frac{\sqrt{7}}{3}\)C.\(\frac{1}{4}\)D.\(\frac{1}{3}\)答案：A4.已知命題\(p\)：\(\existsx_{0}\inR\)，\(x_{0}^{2}+2x_{0}+2\leq0\)，則\(\negp\)為（）A.\(\forallx\inR\)，\(x^{2}+2x+2\gt0\)B.\(\forallx\inR\)，\(x^{2}+2x+2\leq0\)C.\(\existsx_{0}\inR\)，\(x_{0}^{2}+2x_{0}+2\gt0\)D.\(\existsx_{0}\inR\)，\(x_{0}^{2}+2x_{0}+2\geq0\)答案：A5.雙曲線\(\frac{x^{2}}{4}-\frac{y^{2}}{12}=1\)的漸近線方程為（）A.\(y=\pm\sqrt{3}x\)B.\(y=\pm\frac{\sqrt{3}}{3}x\)C.\(y=\pm2x\)D.\(y=\pm\frac{1}{2}x\)答案：A6.已知\(a\)，\(b\)，\(c\)成等比數(shù)列，\(a\)，\(x\)，\(b\)成等差數(shù)列，\(b\)，\(y\)，\(c\)也成等差數(shù)列，則\(\frac{a}{x}+\frac{c}{y}\)的值為（）A.\(\frac{1}{4}\)B.\(\frac{1}{2}\)C.\(2\)D.\(4\)答案：C7.若函數(shù)\(f(x)=x^{3}+ax^{2}+bx+c\)的圖象的對稱中心為\((0,1)\)，則\(b\)的值為（）A.\(-1\)B.\(1\)C.\(3\)D.\(-3\)答案：B8.函數(shù)\(f(x)=e^{x}\cosx\)在點\((0,f(0))\)處的切線方程為（）A.\(x-y+1=0\)B.\(x+y+1=0\)C.\(x+y-1=0\)D.\(x-y-1=0\)答案：A9.已知函數(shù)\(f(x)=x^{2}-2x+m\)，若\(f(x_{1})=f(x_{2})\)（\(x_{1}\neqx_{2}\)），則\(f(\frac{x_{1}+x_{2}}{2})\)的值為（）A.\(1\)B.\(2\)C.\(m-1\)D.\(m\)答案：C10.若函數(shù)\(f(x)=\frac{1}{3}x^{3}-ax^{2}+x-5\)無極值點，則實數(shù)\(a\)的取值范圍是（）A.\((-1,1)\)B.\([-1,1]\)C.\((-\infty,-1)\cup(1,+\infty)\)D.\((-\infty,-1]\cup[1,+\infty)\)答案：B二、多項選擇題1.下列說法正確的是（）A.“\(x\gt1\)”是“\(x^{2}\gt1\)”的充分不必要條件B.命題“\(\forallx\inR\)，\(x^{2}+x+1\lt0\)”的否定是“\(\existsx\inR\)，\(x^{2}+x+1\geq0\)”C.若\(p\)，\(q\)為兩個命題，則“\(p\)且\(q\)為真”是“\(p\)或\(q\)為真”的必要不充分條件D.若命題\(p\)：“\(\existsx\inR\)，\(x^{2}+x+1\lt0\)”，則\(\negp\)：“\(\forallx\inR\)，\(x^{2}+x+1\geq0\)”答案：ABD2.已知橢圓\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a\gtb\gt0)\)的左、右焦點分別為\(F_{1}\)，\(F_{2}\)，\(P\)是橢圓上一點，且\(\angleF_{1}PF_{2}=60^{\circ}\)，設(shè)\(\vertPF_{1}\vert=m\)，\(\vertPF_{2}\vert=n\)，則（）A.\(m+n=2a\)B.\(m^{2}+n^{2}=2b^{2}\)C.\(mn=\frac{4}{3}b^{2}\)D.\(\triangleF_{1}PF_{2}\)的面積為\(\frac{\sqrt{3}}{3}b^{2}\)答案：ACD3.對于雙曲線\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a\gt0,b\gt0)\)，下列說法正確的是（）A.實軸長為\(2a\)B.漸近線方程為\(y=\pm\frac{a}x\)C.離心率\(e=\frac{c}{a}\)（\(c^{2}=a^{2}+b^{2}\)）D.焦點到漸近線的距離為\(b\)答案：ABCD4.已知函數(shù)\(f(x)=x^{3}-3x\)，則（）A.\(f(x)\)是奇函數(shù)B.\(f(x)\)在\(x=1\)處取得極大值C.\(f(x)\)在\((-1,1)\)上單調(diào)遞增D.\(f(x)\)的圖象關(guān)于點\((0,0)\)對稱答案：ABCD5.已知拋物線\(y^{2}=2px(p\gt0)\)的焦點為\(F\)，過\(F\)的直線交拋物線于\(A\)，\(B\)兩點，\(O\)為坐標(biāo)原點，則（）A.\(\overrightarrow{OA}\cdot\overrightarrow{OB}=-\frac{3}{4}p^{2}\)B.若\(\vertAF\vert\cdot\vertBF\vert=4p^{2}\)，則直線\(AB\)的斜率為\(\pm\sqrt{3}\)C.以\(AF\)為直徑的圓與\(y\)軸相切D.設(shè)\(A(x_{1},y_{1})\)，\(B(x_{2},y_{2})\)，則\(\frac{1}{\vertAF\vert}+\frac{1}{\vertBF\vert}=\frac{2}{p}\)答案：ACD6.下列關(guān)于函數(shù)極值的說法正確的是（）A.函數(shù)\(f(x)\)在\(x=x_{0}\)處取得極值，則\(f^{\prime}(x_{0})=0\)B.若\(f^{\prime}(x_{0})=0\)，則\(x=x_{0}\)一定是函數(shù)\(f(x)\)的極值點C.函數(shù)\(f(x)\)在區(qū)間\((a,b)\)內(nèi)有極值，則\(f(x)\)在區(qū)間\((a,b)\)內(nèi)不是單調(diào)函數(shù)D.函數(shù)\(f(x)\)在\(x=x_{0}\)處取得極值，則\(f(x)\)在\(x=x_{0}\)處的導(dǎo)數(shù)存在且\(f^{\prime}(x_{0})=0\)答案：AC7.已知橢圓\(C_{1}:\frac{x^{2}}{a_{1}^{2}}+\frac{y^{2}}{b_{1}^{2}}=1(a_{1}\gtb_{1}\gt0)\)與雙曲線\(C_{2}:\frac{x^{2}}{a_{2}^{2}}-\frac{y^{2}}{b_{2}^{2}}=1(a_{2}\gt0,b_{2}\gt0)\)有相同的焦點\(F_{1}\)，\(F_{2}\)，\(P\)是\(C_{1}\)，\(C_{2}\)的一個公共點，且滿足\(\angleF_{1}PF_{2}=90^{\circ}\)，設(shè)\(\vertPF_{1}\vert=s\)，\(\vertPF_{2}\vert=t\)，則（）A.\(s^{2}+t^{2}=4c^{2}\)（\(c\)為半焦距）B.\(s+t=2a_{1}\)C.\(\verts-t\vert=2a_{2}\)D.\(\frac{1}{s^{2}}+\frac{1}{t^{2}}=\frac{2}{b_{1}^{2}}+\frac{2}{b_{2}^{2}}\)答案：ABCD8.已知函數(shù)\(f(x)=ax^{3}+bx^{2}+cx+d\)的圖象如圖所示，則（）A.\(a\gt0\)B.\(b\lt0\)C.\(c\gt0\)D.\(d\gt0\)答案：ABC9.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\(e\)，左、右焦點分別為\(F_{1}\)，\(F_{2}\)，\(P\)是雙曲線上一點，且\(\angleF_{1}PF_{2}=\theta\)，則（）A.當(dāng)\(\theta=90^{\circ}\)時，\(\triangleF_{1}PF_{2}\)的面積為\(b^{2}\)B.若\(\vertPF_{1}\vert=m\)，\(\vertPF_{2}\vert=n\)，則\(\vertm-n\vert=2a\)C.雙曲線的漸近線方程為\(y=\pm\frac{a}x\)D.\(e=\sqrt{1+\frac{b^{2}}{a^{2}}}\)答案：ABCD10.已知函數(shù)\(f(x)\)在\(R\)上可導(dǎo)，其導(dǎo)函數(shù)為\(f^{\prime}(x)\)，若\(f^{\prime}(x)\)滿足\(f^{\prime}(x)\gtf(x)\)，則（）A.\(e\cdotf(0)\gtf(1)\)B.\(f(2)\lte^{2}\cdotf(0)\)C.\(e^{3}\cdotf(-1)\gtf(2)\)D.\(f(2023)\gte^{2023}\cdotf(0)\)答案：CD三、判斷題1.命題“若\(x^{2}=1\)，則\(x=1\)”的否命題為“若\(x^{2}=1\)，則\(x\neq1\)”。（×）2.橢圓\(\frac{x^{2}}{m^{2}}+\frac{y^{2}}{n^{2}}=1(m\gt0,n\gt0)\)的焦點在\(x\)軸上。（×）3.雙曲線\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a\gt0,b\gt0)\)的漸近線方程為\(y=\pm\frac{a}x\)。（×）4.若函數(shù)\(f(x)\)在區(qū)間\((a,b)\)內(nèi)單調(diào)遞增，則\(f^{\prime}(x)\gt0\)在\((a,b)\)內(nèi)恒成立。（×）5.拋物線\(y^{2}=2px(p\gt0)\)的焦點坐標(biāo)是\((\frac{p}{2},0)\)。（√）6.命題“\(\forallx\inR\)，\(x^{2}+x+1\gt0\)”的否定是“\(\existsx\inR\)，\(x^{2}+x+1\leq0\)”。（√）7.若函數(shù)\(f(x)\)在\(x=x_{0}\)處可導(dǎo)，且\(f^{\prime}(x_{0})=0\)，則\(x=x_{0}\)是函數(shù)\(f(x)\)的極值點。（×）8.橢圓\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a\gtb\gt0)\)的離心率\(e=\frac{c}{a}\)，其中\(zhòng)(c=\sqrt{a^{2}-b^{2}}\)。（√）9.雙曲線\(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1(a\gt0,b\gt0)\)的實軸長為\(2b\)。（×）10.若函數(shù)\(y=f(x)\)在區(qū)間\([a,b]\)上的圖象是一條連續(xù)不斷的曲線，且\(f(a)\cdotf(b)\lt0\)，則函數(shù)\(y=f(x)\)在區(qū)間\((a,b)\)內(nèi)至少有一個零點。（√）四、簡答題1.求橢圓\(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)的長軸長、短軸長、焦點坐標(biāo)、離心率。答案：對于橢圓\(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)，\(a^{2}=25\)，\(b^{2}=16\)，則\(a=5\)，\(b=4\)。長軸長\(2a=10\)，短軸長\(2b=8\)。\(c=\sqrt{a^{2}-b^{2}}=\sqrt{25-16}=3\)，焦點坐標(biāo)為\((\pm3,0)\)。離心率\(e=\frac{c}{a}=\frac{3}{5}\)。2.