职教组卷基于海量职教高考试题库建立的在线组卷及学习系统
职教组卷

选择知识点

  • 题型:解答题 题类:模拟题 难易度:较难

    年份:2020

    已知点\(P\)是抛物线\(C _{1}\):\(y ^{2} =4x\)的准线上任意一点,过点\(P\)作抛物线\(C _{1}\)的两条切线\(PA\),\(PB\),其中\(A\),\(B\)为切点.
    \((1)\)证明:直线\(AB\)过定点,并求出定点的坐标;
    \((2)\)若直线\(AB\)交椭圆\(C _{2}\):\( \dfrac {x^{2}}{4}+ \dfrac {y^{2}}{3} =1\)于\(C\),\(D\)两点,\(S _{1}\),\(S _{2}\)分别是\(\triangle PAB\),\(\triangle PCD\)的面积,求\( \dfrac {S_{1}}{S_{2}}\)的最小值.
  • 题型:填空题 题类:模拟题 难易度:较难

    年份:2020

    已知平面向量\( \overrightarrow {a}\),\( \overrightarrow {b}\),\( \overrightarrow {c}\),\( \overrightarrow {d}\)满足\(| \overrightarrow {a} |=| \overrightarrow {b} |=| \overrightarrow {c} |=1\),\( \overrightarrow {a} ⋅ \overrightarrow {c} = \overrightarrow {b} ⋅ \overrightarrow {c} = \dfrac {1- \overrightarrow {a}\cdot \overrightarrow {b}}{ \overrightarrow {a}\cdot \overrightarrow {d}} > 0\),\( \overrightarrow {c} ⋅ \overrightarrow {d} =0\),若平面向量\( \overrightarrow {s} =x \overrightarrow {a} +y \overrightarrow {b} (x , y > 0\)且\(xy=1)\),则\(| \overrightarrow {s} +2 \overrightarrow {c} |+| \overrightarrow {s} - \overrightarrow {d} |\)的最小值是______.
  • 题型:解答题 题类:模拟题 难易度:较难

    年份:2020

    数列\(\{a _{n} \}\),\(a _{1} =1\),\(a _{n+1} =2a _{n} -n ^{2} +3n(n∈N ^{*} ).\)
    \((\)Ⅰ\()\)是否存在常数\(λ\),\(μ\),使得数列\(\{a _{n} +λn ^{2} +μn\}\)是等比数列,若存在,求出\(λ\),\(μ\)的值,若不存在,说明理由.
    \((\)Ⅱ\()\)设\(b _{n} = \dfrac {1}{a_{n}+n-2^{n-1}},S_{n}=b_{1}+b_{2}+b_{3}+…+b_{n}\),证明:当\(n\geqslant 2\)时,\( \dfrac {n}{n+1} < S_{n} < \dfrac {5}{3}\).
  • 题型:解答题 题类:模拟题 难易度:较难

    年份:2020

    已知椭圆\(E\):\( \dfrac {x^{2}}{a^{2}}+ \dfrac {y^{2}}{b^{2}} =1(a > b > 0)\),过点\(A(2 , 1)\),且该椭圆的短轴端点与两焦点\(F _{1}\),\(F _{2}\)的张角为直角.
    \((\)Ⅰ\()\)求椭圆\(E\)的方程;
    \((\)Ⅱ\()\)过点\(B(0 , 3)\)且斜率大于\(0\)的直线与椭圆\(E\)相交于点\(P\),\(Q\),直线\(AP\),\(AQ\)与\(y\)轴相交于\(M\),\(N\)两点,求\(|BM|+|BN|\)的取值范围.
  • 题型:解答题 题类:模拟题 难易度:较难

    年份:2020

    已知数列\(a _{1}\),\(a _{2}\),\(…\),\(a _{10}\)满足:对任意的\(i\),\(j∈\{1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10\}\),若\(i\neq j\),则\(a _{i} \neq a _{j}\),且\(a _{i} ∈\{1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10\}\),设集合\(A=\{a _{i} +a _{i+1} +a _{i+2} |i=1 , 2 , 3 , 4 , 5 , 6 , 7 , 8\}\),集合\(A\)中元素最小值记为\(m(A)\),集合\(A\)中元素最大值记为\(n(A).\)如数列:\(7\),\(6\),\(2\),\(8\),\(3\),\(4\),\(9\),\(1\),\(5\),\(10\)时,\(A=\{13 , 14 , 15 , 16\}\),\(m(A)=13\),\(n(A)=16\).
    \((1)\)已知数列:\(10\),\(6\),\(1\),\(2\),\(7\),\(8\),\(3\),\(9\),\(5\),\(4\),写出集合\(A\)及\(m(A)\),\(n(A)\);
    \((2)\)求证:不存在 \(m(A)\geqslant 18\);
    \((3)\)求\(m(A)\)的最大值以及\(n(A)\)的最小值,并说明理由.
  • 题型:选择题 题类:模拟题 难易度:较难

    年份:2020

    已知数列\(\{a _{n} \}\)满足:\(a _{n} > 0\),且\(a _{n} ^{2} =3a _{n+1} ^{2} -2a _{n+1} (n∈N ^{*} )\),下列说法正确的是\((\:\:\:\:)\)
    A.若\(a_{1}= \dfrac {1}{2}\),则\(a _{n} > a _{n+1}\) B.若\(a _{1} =2\),则\(a_{n}\geqslant 1+( \dfrac {3}{7})^{n-1}\) C.\(a _{1} +a _{5} \leqslant 2a _{3}\) D.\(|a_{n+2}-a_{n+1}|\geqslant \dfrac { \sqrt {3}}{3}|a_{n+1}-a_{n}|\)
  • 题型:解答题 题类:模拟题 难易度:较难

    年份:2020

    已知公差非零的等差数列\(\{a _{n} \}\)的前\(n\)项和为\(S _{n} (n∈N*)\),且\(a _{1}\),\(a _{2}\),\(a _{4}\)成等比数列,且\(S _{4} =10\),数列\(\{b _{n} \}\)满足\(b _{1} =2\),\(b_{n}-b_{n-1}=2^{n-1}(n\geqslant 2,n∈N^{*})\).
    \((1)\)求数列\(\{a _{n} \}\)和\(\{b _{n} \}\)的通项公式;
    \((2)\)设数列\(\{c _{n} \}\)满足\(c_{n}= \dfrac {\ln a_{n}}{b_{n}},(n∈N^{+})\),求证:\((1- \dfrac {1}{2^{n-1}} )\boldsymbol{⋅}\ln \sqrt {2} \leqslant c _{2} +…+c _{n} < \dfrac {3}{4}\),\((n∈N ^{*} , n\geqslant 2)\).
  • 题型:解答题 题类:模拟题 难易度:较难

    年份:2020

    已知圆\(C:(x- \sqrt {3})^{2}+y^{2}=16\),点\(G(- \sqrt {3},0)\),\(P\)是圆\(C\)上一动点,若线段\(PG\)的垂直平分线和\(CP\)相交于点\(M\).
    \((1)\)求点\(M\)的轨迹方程\(E\).
    \((2)\)已知直线\(l\):\(y=kx+m(m\neq 0)\)交曲线\(E\)于\(A\),\(B\)两点.
    ①若射线\(BO\)交椭圆\( \dfrac {x^{2}}{16}+ \dfrac {y^{2}}{4}=1\)于点\(Q\),求\(\triangle ABQ\)面积的最大值;
    ②若\(OA⊥OB\),\(OD\)垂直\(AB\)于点\(D\),求点\(D\)的轨迹方程.
  • 题型:选择题 题类:模拟题 难易度:较难

    年份:2020

    设\(a\),\(b∈R\),数列\(\{a _{n} \}\)满足\(a _{1} =a\),\(a _{n} =\ln a _{n+1} +b(n∈N*)\),则\((\:\:\:\:)\)
    A.若\(b=-2\),则\(a _{2020} > a\) B.若\(b=-2\),则\(a _{2020} < a\) C.若\(b=2\),则\(a _{2020} > a\) D.若\(b=2\),则\(a _{2020} < a\)
  • 题型:解答题 题类:模拟题 难易度:较难

    年份:2020

    已知数列\(\{a _{n} \}\)满足\(a _{1} =1\),\( \sqrt {a_{n}}- \sqrt {a_{n+1}}= \sqrt {a_{n}\cdot a_{n+1}}(n∈N^{*})\).
    \((\)Ⅰ\()\)求证:数列\(\{ \sqrt { \dfrac {1}{a_{n}}}\}\)为等差数列,并求\(a _{n}\);
    \((\)Ⅱ\()\)设\(b_{n}= \sqrt {1+2a_{n+1}}\),数列\(\{b _{n} \}\)的前\(n\)项和为\(S _{n}\),求证:\(S_{n} < n+1- \dfrac {1}{n+1}\).