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2022年7月29日 星期五

92年大學學測-數學詳解

大學入學考試中心
九十二學年度學科能力測驗試題

第 一 部 分 : 選 擇 題  

壹 、 單 一 選 擇 題 

解答:$${1\over n} +{2\over n} +\cdots +{10\over n} ={55\over n} \in \mathbb{Z} \Rightarrow n=1,5,11,55 \Rightarrow 共四個正整數n,故選\bbox[red, 2pt]{(4)}$$
解答:$$f(x)=x^3-2x^2-x+5 =(x-2)(x^2-1)+3 \Rightarrow f(2)=3\\ 而 g(x)=f(f(x))= f(x)^3-2f(x)^2-f(x)+5 \Rightarrow g(2)=3^3-2\cdot 3^2-3+5=11 ,故選\bbox[red, 2pt]{(5)}$$
解答:$$(4+3i)(\cos \theta +i\sin \theta)= 5(\cos \alpha+i\sin \alpha)(\cos \theta +i\sin \theta), 其中\cases{\cos \alpha=4/5\\ \sin \alpha=3/5}\\ =5(\cos(\alpha+\theta)+ i\sin(\alpha+ \theta))為小於0的實數 \Rightarrow \alpha+\theta =\pi \Rightarrow \theta 為第二象限角(\alpha 為第一象限角)\\,故選\bbox[red, 2pt]{(2)}$$
解答:$$\overrightarrow{AP}={1\over 5}\overrightarrow{AB}+ {2\over 5}\overrightarrow{AC} \Rightarrow {1\over 5}+{2\over 5} \lt 1 \Rightarrow P在\triangle ABC內部;\\假設D=\overleftrightarrow{AP} \cap\overline{BC} 及\overrightarrow{AD}= t \overrightarrow{AP} ={1\over 5}t\overrightarrow{A B}+ {2\over 5}t\overrightarrow{AC} ,B、D、C三點在一直線上\\ \Rightarrow {1\over 5}t+{2\over 5}t=1 \Rightarrow t={5\over 3}  \Rightarrow \cases{\overrightarrow{AD}={1\over 3}\overrightarrow{A B}+ {2\over 3}\overrightarrow{AC} \Rightarrow \overline{BD}:\overline{DC}=2:1 \Rightarrow {\triangle ABD \over \triangle ABC}={2\over 3}\\\overrightarrow{AD}= {5\over 3} \overrightarrow{AP} \Rightarrow \overline{AP}:\overline{AD} =3:5 \Rightarrow {\triangle ABP\over \triangle ABD} ={3\over 5}} \\\Rightarrow {\triangle ABP \over \triangle ABC}={2\over 3}\times {3\over 5}={2\over 5},故選\bbox[red, 2pt]{(3)}$$
解答:$$100(1-2^{-3k})=70 \Rightarrow 2^{-3k}={3\over 10} \Rightarrow 2^{-k}= ({3\over 10})^{1/3}\\ 因此 100(1-2^{-Tk})=99 \Rightarrow 2^{-Tk}={1\over 100} \Rightarrow 2^{-k}=({1\over 100})^{1/T} =({3\over 10})^{1/3} \Rightarrow {1\over T}\log {1\over 100}={1\over 3}\log {3\over 10}\\ \Rightarrow {-2\over T}={1\over 3}(\log 3-1) \Rightarrow T=-{6\over \log 3-1} ={6\over 1-0.4771} \approx 11.475,故選\bbox[red, 2pt]{(4)}$$

貳 、 多 重 選 擇 題

解答:$$(1)\times: L_1斜率為正\Rightarrow -1/a\gt 0 \Rightarrow a\lt 0\\(2) \times: L_1的x截距為正\Rightarrow -b\gt 0 \Rightarrow b\lt 0\\ (3) \times: 理由同(1)\\ (4)\bigcirc: L_2的x截距為負\Rightarrow -d\lt 0\Rightarrow d\gt 0\\ (5)\bigcirc: L_1斜率\gt L_2斜率\Rightarrow -{1\over a} \gt -{1\over c} \Rightarrow a\gt c\\,故選\bbox[red, 2pt]{(45)}$$
解答:$$假設六面體三邊長分別為\alpha,\beta,\gamma,令\cases{D(0,0,0)\\ C(\alpha,0,0)\\ A(0,\beta,0)\\ B(\alpha,\beta,0)\\ E(0,\beta,\gamma)\\ H(0,0,\gamma)\\ G(\alpha,0,\gamma)\\ F(\alpha,\beta,\gamma)} \Rightarrow J=(B+C+F+G)\div 4=(\alpha,\beta/2,\gamma/2)\\ \Rightarrow \cases{\overrightarrow{AJ} =(\alpha,-\beta/2,\gamma/2) \\ \overrightarrow{AB}= (\alpha,0,0)\\ \overrightarrow{AD}= (0,-\beta,0)\\ \overrightarrow{AE}= (0,0,\gamma)} \Rightarrow \overrightarrow{AJ}=a\overrightarrow{AB}+b\overrightarrow{AD} +c\overrightarrow{AE} \Rightarrow (\alpha,-\beta/2,\gamma/2) =(a\alpha,-b\beta,c\gamma)\\ \Rightarrow \cases{a=1\\ b=1/2\\ c=1/2 } \Rightarrow \cases{(1)\bigcirc: 1/3\lt 1/2\lt 2/3 \\(2) \bigcirc: a+b+c =1+1/2+1/2=2\\ (3)\bigcirc: a=1\\ (4) \bigcirc: \cases{a=1\\ c=1/2} \Rightarrow a=2c\\ (5)\times: \cases{a=1\\b=1/2} \Rightarrow a\ne b}\\,故選\bbox[red, 2pt]{(1234)}$$
解答:$$(1)\bigcirc: \sqrt 2\gt \sqrt[3]2 \Rightarrow \sqrt 2-\sqrt[3]2\gt 0\\ (2)\bigcirc: \log_2 3\gt \log_2 2=1 \Rightarrow \log_2 3-1\gt 0\\ (3) \times: \log_3 2\lt \log_3 3=1\Rightarrow \log_3 2-1\lt 0\\(4) \times:\log_{1/2}3 ={\log_2 3\over \log_2 1/2} =-\log_2 3\lt 0 \\(5)\bigcirc: \log_{1/3}{1\over 2} ={\log 1/2\over \log 1/3} ={\log 2\over \log 3} \gt 0\\,故選\bbox[red, 2pt]{(125)}$$
解答:$$(1)\times: \sin x+\cos x= \sqrt 2\sin(x+\alpha) \Rightarrow 週期為2\pi\\(2)\times:\sin x-\cos x= \sqrt 2\sin(x-\alpha) \Rightarrow 週期為2\pi \\(3)\bigcirc: |\sqrt 2\sin(x+\alpha)|週期為\pi\\ (4)\bigcirc: |\sqrt 2\sin(x-\alpha)|週期為\pi\\ (5)\times: |\sin x|與|\cos x| 週期 為\pi且對稱於{\pi/4}+{k\pi/2} \Rightarrow 週期為\pi/2\\,故選\bbox[red, 2pt]{(34)}$$


解答:$$(1)\times: x\le 0 無法推論y值\\ (2)\bigcirc:等價敘述\\ (3)\times: 若p就q不代表若q則p \\(4) \bigcirc: x \gt 1 符合x\gt 0 \\(5)\bigcirc: y\lt 0符合y\le 0\\,故選\bbox[red, 2pt]{(245)}$$

解答:$$(1)\times: 若\pi_a \parallel E_1 \Rightarrow {1\over 1}={-4\over -2}={a\over 1},不存在a滿足左式\\ (2)\bigcirc: 若\pi_a \bot E_1 \Rightarrow (1,-4,a) \cdot (1,-2,1)=0 \Rightarrow 1+8+a=0 \Rightarrow a=-9存在\\ (3)\bigcirc: \left| \begin{matrix} 1 & -4 & a \\ 1 & -2 & 1 \\ 2 & -5 & 4 \end{matrix} \right| =5-a, 只要a\ne 5就有一解, 例a=2 \Rightarrow \cases{x=0\\ y=-23/3\\ z=-31/3} \\(4)\times: 三平面交於一直線代表有無限多解,此時a=5, 但a=5時,聯立方程式無解,因此a不存在\\ (5)\bigcirc: 如(4), a=5聯立方程式無解,即三平面無交點,故選\bbox[red, 2pt]{(235)}$$
解答:$$假設\cases{-1有m個\\ 0有n個\\ 1有50-m-n個},因此\cases{\sum a_i=9 \Rightarrow m\times(-1)+n \times 0+ (50-m-n)=9\\[1ex] \sum(a_i+1)^2=107 \Rightarrow m\cdot 0^2+n\cdot 1^2+(50-m-n)2^2=107} \\ \Rightarrow \cases{2m+n=41\\ 4m+3n=93} \Rightarrow \cases{m= 15\\ n=\bbox[red,2pt]{11}}$$
解答:$$3389的排列數={4!\over 2!}=12 \Rightarrow 試一次成功機率=\bbox[red,2pt]{1\over 12}$$
解答:$$P\in \Gamma \Rightarrow P(a^2,2a); 又\triangle ABP為一正\triangle 且\overline{AB}在x軸上\\\Rightarrow \cases{\overline{PA}=\overline{AB} \Rightarrow (a^2-1)^2+ 4a^2= (b-1)^2 \cdots(1)\\ P的x坐標為A、B兩點x坐標的平均值\Rightarrow a^2=(b+1)/2 \cdots(2)}\\ 將(2)代入(1) \Rightarrow \left({b+1\over 2}-1\right)^2 +4\cdot {b+1\over 2}=(b-1)^2 \Rightarrow 3b^2-14b-5=0 \Rightarrow (b-5)(3b+1)=0\\ \Rightarrow b=\bbox[red, 2pt]{5}(-1不合,違反b\gt 1)$$
解答:$${x^2\over 9}-{y^2\over 16}=1 \Rightarrow \cases{a=3\\ b=4} \Rightarrow c=5 \Rightarrow \cases{F_1(5,0)\\ F_2(-5,0)};又\overline{PF_1}:\overline{PF_2}= 1:3 \Rightarrow \cases{\overline{PF_1}=k\\ \overline{PF_2}=3k}, k\in \mathbb{R}\\ \Rightarrow 3k-k=2a \Rightarrow 2k=6 \Rightarrow k=3 \Rightarrow \triangle F_1PF_2= 4k+\overline{F_1F_2}= 12+2c= \bbox[red, 2pt]{22}$$
解答:$$該平面通過球心,因此相交於一個單位圓\Rightarrow \cos \angle NOP = \cfrac{\overrightarrow{ON} \cdot \overrightarrow{OP}}{|\overrightarrow{ON}||\overrightarrow{OP}|} ={-1/2\over 1\cdot 1} =-{1\over 2}\\ \Rightarrow \angle NOP={2\pi\over 3} \Rightarrow \stackrel{\Large{\frown}}{NP} =2\pi \times {1\over 3}= \bbox[red,2pt]{2\over 3}\pi$$
解答:$$kx^2+7x+1=0\cases{有兩相異實根\Rightarrow 49-4k\gt 0 \Rightarrow k\lt 49/4\\ 兩根之積介於5/71與6/71之間\Rightarrow 5/71\lt 1/k\lt 6/71 \Rightarrow 71/6\lt k\lt 71/5}\\ \Rightarrow \cases{k\le 12\\ 12\le k\le 14} \Rightarrow k= \bbox[red, 2pt]{12}$$
解答

$$假設拱門底部中心為原點 \Rightarrow \cases{拱門右底部A(3,0)、左底部B(-3,0)\\ 左上C(-5/2,3/2)、右上D(5/2,3/2)} \\ \Rightarrow 拋物線方程式y=ax^2+b \Rightarrow \cases{0=9a+b\\ 3/2=25a/4+b} \Rightarrow \cases{a=-6/11 \\ b=54/11} \\ \Rightarrow y=-{6\over 11}x^2+{54\over 11} \Rightarrow 拱門高度=\bbox[red,2pt]{54\over 11}$$
解答:$$16題答對可得16\times 4=64分;\\6題在3個選項猜一題\Rightarrow\cases{答對機率{1\over 3} \Rightarrow 6\times 4\times {1\over 3}=8\\ 答錯機率{2\over 3} \Rightarrow 6\times (-1)\times {2\over 3}=-4};\\ 3題在5個選項猜一題\cases{答對機率{1\over 5} \Rightarrow 3\times 4\times {1\over 5}=2.4\\ 答錯機率{4\over 5} \Rightarrow 3\times (-1)\times {4\over 5}=-2.4}\\ 期望值=64+8-4+2.4-2.4=\bbox[red,2pt]{68}$$
解答:$$\cases{E(X)=16\\ \sigma(X)=3.5} \Rightarrow \cases{E({9\over 5}X+32)= {9\over 5}E(X)+ 32=\bbox[red, 2pt]{60.8} \\ \sigma({9\over 5}X+32)= {9\over 5}\sigma(X)=\bbox[red, 2pt]{6.3}}$$

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