95 lines
6.3 KiB
Plaintext
95 lines
6.3 KiB
Plaintext
solution
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017239
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(1) 由底面$ABC$为等腰直角三角形且 $AB \perp AC$知$AB\perp$平面$ACC_1A_1$, \\
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从而$BM$在平面$ACC_1A_1$上的投影为$AM$,
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故由$BM \perp A_1C$ 知 $AM\perp A_1C $, \\
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结合$AC=2$, $AA_1=4$ 得 $MC=1$, 即$h=1$.\\
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(2) 如图建系：以$A$为原点，分别以$\overrightarrow{AB}$、$\overrightarrow{AC}$、$\overrightarrow{AA_1}$方向为$x$轴、 $y$轴、$z$ 轴正方向建立平面直角坐标系.\\
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$A(0,0,0),B(2,0,0),C(0,2,0),A_1(0,0,4),M(0,2,2)$,
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$\overrightarrow{BA_1}=(-2,0,4),\overrightarrow{AB}=(2,0,0),\overrightarrow{AM}=(0,2,2),$\\
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设平面$ABM$的一个法向量为$\overrightarrow{n}=(x,y,z)$, 则$\begin{cases}
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2x=0,\\
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2y+2z=0.
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\end{cases}$ 取$\overrightarrow{n}=(0,1,-1),$设直线$BA_1$与平面$ABM$所成的角为$\theta$, 则$\sin\theta=|\cos \langle\overrightarrow{BA_1},\overrightarrow{n}\rangle|=\dfrac{|\overrightarrow{BA_1}\cdot\overrightarrow{n}|}{|\overrightarrow{BA_1}|\cdot|\overrightarrow{n}|}=\dfrac{\sqrt{10}}{5}$, 故直线$BA_1$与平面$ABM$所成的角为$\arcsin\dfrac{\sqrt{10}}{5}$.
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017240
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(1) 由 $S=\dfrac{1}{2}ac\sin B=\dfrac{\sqrt{3}}{4}ac=\sqrt{3}$知$ac=4$，\\
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由$a^2+c^2-b^2=2ac\cos B$知$a^2+c^2=9$.\\
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结合两式得$(a-c)^2=1$, 故$a-c=\pm 1$\\
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(2) 由$2\cos C (ac\cos B+cb\cos A)=c^2$知$2a\cos B\cos C+2b \cos A\cos C=c$,\\
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又由正弦定理知 $2\cos C(\sin A\cos B+\sin B\cos A)=\sin C$，\\
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$2\cos C \sin (A+B)=2\cos C \sin C=\sin C$,其中$C\in (0,\pi), \sin C>0$,\\
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故$\cos C=\dfrac{1}{2}$, $C\in(0,\pi)$, $C=\dfrac{\pi}{3}$.
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017241
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(1) $\dfrac{p}{p+40}=\dfrac{3}{5}$得$p=60$, $q=40$, $x=100$, $y=100$.\\
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(2) 原假设$H_0:$ 是否注射此种疫苗与是否感染病毒无关.\\
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$\chi^2=\dfrac{200\times (40\times 40-60\times 60)^2}{100\times 100\times 100\times 100}=8>3.841$\\
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故拒绝原假设，即有$95 \%$的把握认为注射此种疫苗有效.\\
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(3) 抽取$6$只未注射疫苗、$4$只注射疫苗的小白鼠.\\
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$P(X=0)=\dfrac{C_6^0C_4^4}{C_{10}^4}=\dfrac{1}{210}$;\\
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$P(X=1)=\dfrac{C_6^1C_4^3}{C_{10}^4}=\dfrac{4}{35}$;\\
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$P(X=2)=\dfrac{C_6^2C_4^2}{C_{10}^4}=\dfrac{3}{7}$;\\
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$P(X=3)=\dfrac{C_6^3C_4^1}{C_{10}^4}=\dfrac{8}{21}$;\\
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$P(X=4)=\dfrac{C_6^4C_4^0}{C_{10}^4}=\dfrac{1}{14}$.\\
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故$X$的分布为$\begin{pmatrix}
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0&1&2&3&4\\
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\dfrac{1}{210}&\dfrac{4}{35}&\dfrac{3}{7}&\dfrac{8}{21}&\dfrac{1}{14}
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\end{pmatrix},$\\
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期望$E[X]=0\times \dfrac{1}{210}+1\times \dfrac{4}{35}+2\times \dfrac{3}{7}+3\times \dfrac{8}{21}+4\times \dfrac{1}{14}=\dfrac{12}{5}$.
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017242
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(1) 椭圆的离心率$e=\dfrac{1}{2}$;\\
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(2) 证明: 当$x_0=2$时，$y_0=0,$ 过点$P$的椭圆$C$的切线方程为 $x=2$,符合$\dfrac{x_0 x}{4}+\dfrac{y_0 y}{3}=1$，\\
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同理，当$x_0=-2$时，也符合；\\
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当$x_0\neq\pm 2$时，设过点$P$的椭圆$C$的切线方程为$y-y_0=k(x-x_0)$($k$存在),\\
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联立$\begin{cases}
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y-y_0=k(x-x_0),\\
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\dfrac{x^2}{4}+\dfrac{y^2}{3}=1
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\end{cases}$得$(3+4k^2)x^2+8k(y-kx_0)x+4(y_0-kx_0)^2-12=0$,\\
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$\Delta=0$得$(x_0^2-4)k^2-2x_0y_0k+y_0^2-3=0$,解得$k=\dfrac{x_0y_0}{x_0^2-4}=\dfrac{x_0y_0}{4(1-\dfrac{y_0^2}{3})-4}=-\dfrac{3x_0}{4y_0}.$\\
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故$y-y_0=-\dfrac{3x_0}{4y_0}(x-x_0)$,即$\dfrac{x_0 x}{4}+\dfrac{y_0 y}{3}=1$.\\
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综上，过点$P$的椭圆$C$的切线方程为$\dfrac{x_0 x}{4}+\dfrac{y_0 y}{3}=1$.\\
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(3) 设$A(x_1,y_1).B(x_2,y_2),x_1\neq x_2,M(4,t)$.\\
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则切线$MA:\dfrac{x_1x}{4}+\dfrac{y_1y}{3}=1,$代入$(4,t)$得$x_1+\dfrac{ty_1}{3}=1$,\\
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同理，$x_1+\dfrac{ty_1}{3}=1$，\\
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故$A(x_1,y_1),B(x_2,y_2)$在直线$x+\dfrac{ty}{3}=1$上，故直线$AB:x=-\dfrac{ty}{3}+1$.\\
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联立$\begin{cases}
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x=-\dfrac{ty}{3}+1,\\
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\dfrac{x^2}{4}+\dfrac{y^2}{3}=1
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\end{cases}$
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得
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$(4+\dfrac{t^2}{3})y^2-2ty-9=0$, $\Delta=16t^2+144>0$,\\
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$|AB|=\sqrt{1+\dfrac{t^2}{9}}\cdot |y_1-y_2|=\sqrt{1+\dfrac{t^2}{9}}\cdot \dfrac{\sqrt{16t^2+144}}{4+\dfrac{t^2}{3}}$,\\
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$M$到直线$AB$的距离$d=\dfrac{|4+\dfrac{t^2}{3}-1|}{\sqrt{1+\dfrac{t^2}{9}}}=\dfrac{3+\dfrac{t^2}{3}}{\sqrt{1+\dfrac{t^2}{9}}}$,\\
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$\triangle MAB$的面积$S=\dfrac{1}{2}|AB|\cdot d=\dfrac{1}{2}\cdot \sqrt{1+\dfrac{t^2}{9}}\cdot \dfrac{\sqrt{16t^2+144}}{4+\dfrac{t^2}{3}}\cdot \dfrac{3+\dfrac{t^2}{3}}{\sqrt{1+\dfrac{t^2}{9}}}=\dfrac{2(t^2+9)\sqrt{t^2+9}}{t^2+12}$,\\
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令$\lambda=\sqrt{t^2+9}\geq3, S=f(\lambda)=\dfrac{2\lambda^3}{\lambda^2+3},$
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则$f^{'} (\lambda)=\dfrac{2\lambda^4+18\lambda^2}{(\lambda^2+3)^2}>0$,故$f(\lambda)$在$[3,+\infty)$严格增，$f(\lambda)_{\min}=f(3)=\dfrac{9}{2}$,故$\triangle MAB$的面积的最小值为$\dfrac{9}{2},$ 此时$M$的坐标为$(4,0)$.
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017243
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(1) $x_1=\dfrac{1}{2},x_{n+1}=g(x_n)=\dfrac{x_n}{x_{n+1}},$故$x_n>0,\dfrac{1}{x_{n+1}}=\dfrac{x_n+1}{x_n}=\dfrac{1}{x_n}+1$,即$\dfrac{1}{x_{n+1}}-\dfrac{1}{x_{n}}=1$,\\因此数列$\{\dfrac{1}{x_n}\}$是以$2$为首项，$1$为公差的等差数列.\\
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(2) 对任意$x>0$ 均有$f(x)-mg(x)=\ln (x+1)-\dfrac{mx}{x+1}+1>0,$\\
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令$h(x)=\ln (x+1)-\dfrac{mx}{x+1}+1,x>0,$则$h^{'}(x)=\dfrac{x+1-m}{(x+1)^2}$.\\
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当$m\geq4$时，$h(1)=\ln 2-\dfrac{m}{2}+1<2-\dfrac{m}{2}<0,$这与$h(x)>0$对$x>0$恒成立矛盾；\\
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当$m=3$时，$h(x)=\ln (x+1)-\dfrac{3x}{x+1}+1,h^{'}(x)\dfrac{x-2}{(x+1)^2}.h(x)$在$(0,2]$严格减，在$[2,+\infty)$严格增，故$h(x)_{\min}=h(2)=\ln 3-2+1=\ln 3-1>0$,符合题意.\\
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综上，整数$m$的最大值为$3$.\\
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(3) 对任意正整数$t$,取$n=100t,$则 \\$\displaystyle f(n-t)=f(99t)=\ln (1+99t)=\ln \dfrac{1+99t}{99t}+\ln \dfrac{99t}{99t-1}+\cdots +\ln \dfrac{2}{1}=\sum_{k=1}^{99t}\ln (1+\dfrac{1}{k}).$\\
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$\displaystyle n-\sum_{k=1}^{100t}g(k)=\sum_{k=1}^{100t}(1- g(k))=\sum_{k=1}^{100t} \dfrac{1}{1+k}=\sum_{k=1}^{99t} \dfrac{1}{1+k}+\sum_{k=99t+1}^{100t} \dfrac{1}{1+k}.$\\
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$\displaystyle f(n-t)-[ n-\sum_{k=1}^{100t}g(k)]=\sum_{k=1}^{99t} (\ln (1+\dfrac{1}{k})-\dfrac{1}{1+k})-\sum_{k=99t+1}^{100t} \dfrac{1}{1+k}.$\\
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令$H(x)=\ln (x+1)-\dfrac{x}{x+1},x>0$,则$H^{'}(x)=\dfrac{x}{(x+1)^2}>0$恒成立，故$H(x)$在$(0,+\infty)$严格增，$H(x)>H(0)=0$，故$\ln (1+x)>\dfrac{x}{x+1}$对$x>0$恒成立.\\
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因此$\ln (1+\dfrac{1}{k})>\dfrac{\dfrac{1}{k}}{1+\dfrac{1}{k}}=\dfrac{1}{1+k}$,\\
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$\displaystyle f(n-t)-[ n-\sum_{k=1}^{100t}g(k)]=\sum_{k=1}^{99t} (\ln (1+\dfrac{1}{k})-\dfrac{1}{1+k})-\sum_{k=99t+1}^{100t} \dfrac{1}{1+k}$\\
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$\displaystyle>\sum_{k=2}^{99t} (\ln (1+\dfrac{1}{k})-\dfrac{1}{1+k})+(\ln 2-\dfrac{1}{2})- \dfrac{t}{1+99t+1}>\ln 2-\dfrac{1}{2}-\dfrac{1}{99+\dfrac{2}{t}}>0.1-\dfrac{1}{99}>0.$\\
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因此不存在正整数$t$使得对任意$n \in \mathbf{N}$, $n \geq t$, 都有$\displaystyle f(n-t)<n-\sum_{k=1}^n g(k)$成立.
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