寒假某cg的数学卷子....

2.4

1~12 BDABDDABC@A@B#D#
13. [5,+\(\infty\)]
14. \(\frac{x^2}{8}\)+\(\frac{y^2}{4} = 1\)
15. 0
16. 2 1

解:\((1).\) \(f^{'}(x)\)=\(\frac{1}{4}-\frac{a}{x^2}-\frac{1}{x}\)
令x=1 \(f^{'}(1)={-\frac{3}{4}}-a=-2\)
解得 \(a={\frac{5}{4}}\)
(2).
\(x>0\)
\(f^{'}(x)\)=\(\frac{1}{4}-\frac{5}{4\times x^2}-\frac{1}{x}\)=\(\frac{x^2-5-4x}{4\times x^2}\)
令 \(x^2-5-4x>0\) 解得 \(x>5/x<-1\)
$\therefore f(x)增区间为(5,+\infty)\quad f(x)减区间为(0,5) \( 最小值为\)f(5)=-\frac{1}{2}-ln^5$
无最大值

解:\((1).\) \(\because\quad AD=DC\quad B为AC中点\)
\(\therefore BD\bot AC \quad-> DC=2\quad DB=1 \quad BC=\sqrt{3}\)

$ \because\left{
\begin{aligned}
DBEF为矩形\
\quad平面BDEF ⊥平面ACD
\end{aligned}
\right.
$
-> \(ED\bot平面ACD\)

\(\therefore V_{B-DEC}=\frac{\sqrt{3}}{2}\times2\times\frac{1}{3}=\frac{\sqrt{3}}{3}\)
(2).

如图以B为原点建立如图所示的直角坐标系
$D=(0,1,0) \quad C=(\sqrt{3},0,0)\quad E=(0,1,2) \quad M=(0,0,t) \quad A(-\sqrt{3},0,0)\
BM=\frac{1}{2} \( \)\vec{DM}=(0,-1,t)\( \)\vec{AC}=(2\sqrt{3},0,0)\( \)\vec{AE}=(\sqrt{3},1,2)\( \)设面ACE的法向量为\vec{n}=(x,y,z)\( \)\because\left{
\begin{aligned}
\vec{n}\cdot\vec{AC}=0\
\vec{n}\cdot\vec{AE}=0
\end{aligned}
\right.
\( ->\)\vec{n}=(0,-4\sqrt{3},2\sqrt{3})\( 又\)\because\vec{DM}=(0,-1,t)\( \)\vec{DM}=\vec{n}\( \)\therefore t=\frac{1}{2}\( \)\therefore BM=\frac{1}{2}$

(1).由题可知\(\quad a=1\quad b=\sqrt{3}\)
\(\therefore 方程为\frac{y^2}{1}-\frac{x^2}{3}=1\)
(2).

\(A(0,1) \quad B(0,-1)\\ Q(x1,y1) \quad P(-x1,y1)\)
\(k_AP=\frac{y1-1}{-x1}\\ k_BQ=\frac{y1+1}{x1}\\ y_AP = k_AP*x+1 \\ y_AP = k_BQ*x-1\\ 联立得 x=\frac{x1}{y1}\quad y=\frac{1}{y1}\\ 又\because \frac{y1^2}{1}-\frac{x1^2}{3}=1\\ \therefore x^2+3y^2=3\)

(1)
\(\because\left\{ \begin{aligned} PD⊥底面ABCD\\ 运算可得DB\bot BC \end{aligned} \right.\\ \)
\(\because\left\{ \begin{aligned} BC\bot DB\\ BC\bot PD \\ PD\cap DB=D \end{aligned} \right. \)
->
BC⊥平面 PBD
\(\because BC\subset 面PBC\)
->平面 PBD ⊥平面 PBC ;
(2)
\(C作CK\bot PB 于K\)

\(\because\left\{ \begin{aligned} 平面 PBD ⊥平面 PBC \\ CK\subset 面PBC\\ CK\bot PB\\ PB为面PBD和面PBC的交棱 \end{aligned} \right. \)面
->\(CK\bot面PBD\)
设PD=t
-> \(\frac{\sqrt{6}}{2}=\frac{\sqrt{t^2+1}}{\sqrt{2}}\)
->\(t=\sqrt{2}\)
\(H(0,\frac{2}{3},0) P(0,0,\sqrt{2})\\ B(1,1,0) C(0,2,0)\)
\(\vec{PC}=(0,2,-\sqrt{2})\)
\(\vec{PB}=(1,1,-\sqrt{2})\)
\(\vec{PH}=(0,\frac{2}{3},-\sqrt{2})\)
\(设面PBH的法向量为\vec{n_1}=(x,y,z)\)
\(\because\left\{ \begin{aligned} \vec{n}\cdot\vec{PB}=0\\ \vec{n}\cdot\vec{PH}=0 \end{aligned} \right. \)
->\(\vec{n_1}=(-\frac{\sqrt{2}}{3},\sqrt{2},\frac{2}{3})\)
\(设面PBC的法向量为\vec{n_2}=(x,y,z)\)
\(\because\left\{ \begin{aligned} \vec{n}\cdot\vec{PB}=0\\ \vec{n}\cdot\vec{PC}=0 \end{aligned} \right. \)
->\(\vec{n_2}=(\sqrt{2},\sqrt{2},2)\)
\(cos<\vec{n_1},\vec{n_2}>=\frac{\vec{n_1}\cdot{\vec{n_2}}}{|n_1|\times|n_2|}=\frac{\sqrt{3}}{3}\)
\(二面角H−PB−C的余弦值即为cos<\vec{n_1},\vec{n_2}>=\frac{\sqrt{3}}{3}\)
21.
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