112年大學學測-數學A詳解

112學年度學科能力測驗試題-數學A考科

第壹部分、選擇（填）題（占85分 ）
一、單選題（占 30 分 ）

解答: $$\sqrt {\sqrt {{\sqrt N}}} = 2 \Rightarrow \sqrt{\sqrt N} =2^2 \Rightarrow \sqrt N=(2^2)^2=2^4 \Rightarrow N=(2^4)^2= 2^8，故選\bbox[red,2pt]{(4)}$$

解答:

$$\cases{\angle COE+ \theta=90^\circ\\ \angle COE+\angle COD=90^\circ} \Rightarrow \angle COD=\theta \Rightarrow \tan \theta =\tan \angle COD = {\overline{CD}\over \overline{CO}} ={\overline{CD}\over 1} =\overline{CD}，故選\bbox[red,2pt]{(5)}$$

解答: $$(s_k,\log t_k)軌跡近乎一直線，且該直線經過(0,0)及(1,1/2)因此直線為y=x/2\\ 也就是\log t =s/2\ \Rightarrow t=10^{s/2} \Rightarrow t^2=10^s，故選\bbox[red,2pt]{(4)}$$

解答: $$9個數字ABCDEFGHI需滿足A \lt B\lt C\lt D\lt E且E\gt F\gt G\gt H\gt I，因此E=9;\\接著在剩下8個數字中，任取4個數字(有C^8_4種取法)，依序由小至大作為A,B,C,D;\\ 最後剩下的4個數字再依序由大至小作為F,G,H,I;\\因此總共有C^8_4={8!\over 4!4!}個九位數，故選\bbox[red,2pt]{(1)}$$

解答: $$依題意(\overrightarrow{PQ} \times \overrightarrow{PR}) \parallel (2,-3,5) \Rightarrow \overrightarrow{PQ} \times \overrightarrow{PR}= k(2,-3,5)= (2k,-3k,5k),k不為0的常數\\ (1) \begin{Vmatrix} -1& 1 & 1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{Vmatrix} =|(2k,-3k,5k)\cdot (-1,1,1)|=0\\ (2) \begin{Vmatrix} 1& -1 & 1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{Vmatrix} =|(2k,-3k,5k)\cdot (1,-1,1)|= |10k|\\ (3) \begin{Vmatrix} 1& 1 & -1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{Vmatrix} =|(2k,-3k,5k)\cdot (1,1,-1)|= |-6k|\\ (4) \begin{Vmatrix} -1& -1 & 1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{Vmatrix} =|(2k,-3k,5k)\cdot (-1,-1,1)|=|6k|\\ (5) \begin{Vmatrix} -1& -1 & -1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{Vmatrix} =|(2k,-3k,5k)\cdot (-1,-1,-1)|=|-4k|\\ ，故選\bbox[red,2pt]{(2)}$$

解答:

$$將\overrightarrow{OA}縮寫成\vec a，即\overrightarrow{OA}\cdot \overrightarrow{OB} 縮寫成\vec a\cdot \vec b\\ \cases{ \vec a\cdot \{\vec c,\vec f,\vec g\}=0\\ \vec a\cdot \{ \vec b,\vec d,\vec e\}=1},\cases{\vec b \cdot  \vec g =0 \\ \vec b\cdot \{\vec c,\vec d, \vec f\} =1\\\vec b\cdot  e=2} ,\cases{\vec c \cdot \{\vec d,\vec g\}=0\\ \vec c\cdot \{ \vec e,\vec f\}=1},\cases{\vec d \cdot \vec e=2\\ \vec d\cdot \{\vec f,\vec g\}=1 } ,\cases{\vec e\cdot \vec f=2\\ \vec e\cdot \vec g=1},\vec f \cdot \vec g=1\\ 共有\cases{6個0\\ 12個1\\ 3個2} \Rightarrow 期望值={0+12+3\cdot 2\over 21}={18\over 21} ={6\over 7}，故選\bbox[red,2pt]{(3)}$$

二、多選題（ 占30 分）

解答: $$假設\cases{甲在第i月的薪資為a_{i-1}元\\ 乙在第i月的薪資為b_{i-1}元},且a_0=b_0 \Rightarrow \cases{a_k =a_0+ 200m,m= \lfloor k/3 \rfloor \\ b_k = a_0+1000m,m=  \lfloor k/12 \rfloor}, k\in \mathbb{N}\\ (1)\times:  \lfloor 8/3 \rfloor =2 \Rightarrow a_8= a_0+200\cdot 2= a_0+400\Rightarrow 增加400元，不是600元 \\(2) \times: \cases{a_{12}= a_0+ 200\cdot 4=a_0+ 800\\ b_{12} = a_0+1000\cdot 1=a_0+1000} \Rightarrow a_{12} \lt b_{12} \Rightarrow 甲比乙低 \\(3)\bigcirc: \cases{a_{18}= a_0+ 200\cdot 6=a_0+ 1200\\ b_{18} = a_0+1000\cdot 1=a_0+1000} \Rightarrow a_{18} \lt b_{18} \Rightarrow 甲比乙高 \\(4) \times:\cases{\sum_{i=0}^{17} a_i =3(a_0+(a_0+200)+ (a_0+400)+ \cdots+ (a_0+1000)) =  18a_0+ 9003\\ \sum_{i=0}^{17}b_i=12a_0+ (a_0+1000)\cdot 6 =18a_0+ 6000}\\ \qquad \Rightarrow 甲比乙多 \\(5)\bigcirc:第三年：\cases{甲薪資：1-3月：a_0+1600，4-6月:a_0+1800，7-9:a_0+2000,\\\qquad  10-12月:a_0+2200\\ 乙薪資:a_0+2000}\\ \qquad \Rightarrow 甲在第三年的10,11,12月薪資比乙高,其它1-9月都小於等於乙\\ 故選\bbox[red, 2pt]{(35)}$$

解答: $$p_n=1-(1-0.1)^n= 1-0.9^n\\ (1) \bigcirc: \cases{p_n= 1-0.9^n\\ p_{n+1}=1-0.9^{n+1}} \Rightarrow p_{n+1}\gt p_n \\(2) \times: p_3=1-0.9^3 =1-0.729= 0.271 \ne 0.3 \\(3)\times:\cases{p_2-p_1=(1-0.9^2)-(1-0.9)=0.9-0.9^2\\ p_3-p_2 = (1-0.9^3)-(1-0.9^2) =0.9^2-0.9^3} \Rightarrow p_2-p_1\ne p_3-p_2\Rightarrow 非等差 \\(4) \bigcirc: \cases{第１次未中獎且第２次中獎機率＝0.9\cdot 0.1=0.09\\ p_2-p_1 = 1-0.9^2-(1-0.9)=0.9-0.9^2=0.09} \\(5)\times:至少中２次＝全部－全沒中－只中１次=1-0.9^n-C^n_1 \cdot 0.1\cdot 0.9^{n-1}\ne 2p_n\\，故選\bbox[red,2pt]{(14)}$$

解答: $$令S(n)＝\log_3 a_1-\log_3a_2 +\log_3 a_3-\cdots +(-1)^{n+1}\log_3 a_n\\(1) \times: n=23 \Rightarrow S(23)=\log_3 \cfrac{a_1\cdot a_3\cdots a_{23}}{a_2\cdot a_4\cdots a_{22}} \\\qquad =\log_3 \cfrac{3^{12}(3\sqrt 3)^{2+4+\cdots +22}}{3^{11}(3\sqrt 3)^{1+3+\cdots +21}} =\log_3 3(3\sqrt 3)^{11}=\log_3 3^{35/2}={35\over 2} \not \gt 18 \\ (2) \times: S(24)=S(23)-\log_3 a_{24}={35\over 2}-\log_3 3(3\sqrt 3)^{23} \lt S(23) \lt 18\\ (3)\bigcirc: S(25)=S(23)-\log_3 a_{24}+ \log_3 a_{25}={35\over 2}+\log_3{a_{25}\over a_{24}} ={35\over 2}+{3\over 2}=19\gt 18 \\(4)\times: S(26)=S(25)-\log_3 a_{26}=19-\log_3 3^{1+{3\over 2}\cdot 25} =19-{77\over 2}\lt 18\\ (5)\bigcirc: S(27)=S(25)-\log_3 a_{26}+\log_3a_{27} = 19+{3\over 2}\gt 18\\，故選\bbox[red,2pt]{(35)}$$

解答: $$(1)\bigcirc: k=4 \Rightarrow L:5y+4x=40通過A(10,0) \\(2)\times: C(0,6)代入L\Rightarrow 30-10k=0 \Rightarrow k=3 \Rightarrow 5y+2x=30 \Rightarrow y=-{2\over 5}x+6 \\\qquad \Rightarrow 斜率＝-{2\over 5}\ne -{5\over 2} \\(3)\bigcirc: \cases{經過C點\Rightarrow k=3\\ 經過O點\Rightarrow k=0} \Rightarrow 0\le k\le 3\\ (4)\times: k={1\over 2} \Rightarrow L:5y-3x=5與x=10交於(10,7)\not \in \overline{AB}\\ (5)\bigcirc: L斜率＝{4-2k\over 5}={3\over 10} \Rightarrow k={5\over 4} \Rightarrow L:5y-{3\over 2}x={25\over 2}與x=10交於(10,{11\over 2})\in \overline{AB}\\，故選\bbox[red,2pt]{(135)}$$

解答: $$A=\begin{bmatrix} \cos (-90^\circ) & -\sin (-90^\circ)\\ \sin (-90^\circ) & \cos (-90^\circ)\end{bmatrix} =\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}\\ B＝\begin{bmatrix} \cos 90^\circ & -\sin 90^\circ\\ \sin 90^\circ & \cos 90^\circ\end{bmatrix} =\begin{bmatrix} 0 & -1\\ 1 & 0\end{bmatrix} \\ x=y \Rightarrow 斜率＝1＝\tan 45^\circ \Rightarrow C＝\begin{bmatrix} \cos 90^\circ & \sin 90^\circ\\ \sin 90^\circ & -\cos 90^\circ\end{bmatrix} =\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix} \\ x=-y \Rightarrow 斜率＝-1＝\tan (-45^\circ) \Rightarrow D= \begin{bmatrix} \cos -90^\circ & \sin -90^\circ\\ \sin -90^\circ & -\cos (-90^\circ)\end{bmatrix} =\begin{bmatrix} 0 & -1\\ -1 & 0\end{bmatrix}\\ (1) \times: \cases{\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}\begin{bmatrix}  1\\  0\end{bmatrix} =\begin{bmatrix} 0  \\ -1  \end{bmatrix} \\\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix} \begin{bmatrix}  1\\   0\end{bmatrix} =\begin{bmatrix} 0  \\ 1  \end{bmatrix}} \Rightarrow \begin{bmatrix} 0  \\ -1  \end{bmatrix} \ne\begin{bmatrix} 0  \\ 1  \end{bmatrix} \\(2)\bigcirc: -B=-\begin{bmatrix} 0 & -1\\ 1 & 0\end{bmatrix} =\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix} =A \Rightarrow A=-B\\ (3)\times: D^{-1}=\begin{bmatrix} 0 & -1\\ -1 & 0\end{bmatrix}^{-1}=\begin{bmatrix} 0 & -1\\ -1 & 0\end{bmatrix}=D \ne C \\(4) \times: \cases{AB= \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix} \\CD= \begin{bmatrix} -1 & 0\\ 0 & -1\end{bmatrix} } \Rightarrow AB\ne CD \\(5) \bigcirc:AC=\begin{bmatrix} 1 & 0\\ 0 & -1\end{bmatrix} =BD\\，故選\bbox[red,2pt]{(25)}$$

解答: $$(1)\bigcirc: f(x)=2(\cos{\pi\over 3}\sin x+\sin{\pi\over 3}\cos x) = 2\sin(x+{\pi\over 3}) \Rightarrow 其中之一的對稱軸x+{\pi\over 3} ={\pi\over 2}\Rightarrow x={\pi\over 6} \\(2) \times:\cases{x+\pi/3=\pi/2\\ x+\pi/3=0}皆可為對稱軸，此時\cases{a=\pi/6\\ b=-\pi/3} \Rightarrow f(a)\ne f(b) \\(3)\times: f(x)的週期為2\pi,因此有兩個x滿足f(x)=\sqrt 3 \\(4)\times: f(x)={1\over 2} \Rightarrow \sin(x+{\pi\over 3})={1\over 4},假設 \sin(x_1+\pi/3)= \sin(x_2+\pi/3) = 1/4,其中0\lt x_1\lt x_2\lt 2\pi\\\qquad \Rightarrow \cases{\sin(\pi/2+\pi/3)={1\over 2}\gt {1\over 4}且f(x)遞減，x\in[\pi/2,2\pi/3] \Rightarrow x_1\gt \pi/2\\ \sin(3\pi/2+\pi/2)=0 \lt {1\over 4}且f(x)遞增，x\in[3\pi/2,2\pi ] \Rightarrow x_2\gt 3\pi/2} \\ \qquad \Rightarrow x_1+x_2 \gt 2\pi \\(5)\bigcirc: 4\sin^2{x\over 2}=2(1-\cos x)=2(1-\sin({\pi\over 2}-x))=2+ 2\sin(x-{\pi\over 2}) \xrightarrow{向下平移2}2\sin(x-{\pi\over 2}) \\ \qquad \xrightarrow{向左平移5\pi/6} 2\sin(x-{\pi\over 2}+{5\pi\over 6})= 2\sin(x+{\pi\over 3})=f(x)\\，故選\bbox[red,2pt]{(15)}$$

三、選填題（ 占25 分）

解答: $$假設\cases{果汁每杯x元 \\奶茶每杯y元 \\咖啡每杯z元 },則\cases{60x+80y+ 50z=12900\\ 30x+40y+30z= 6850\\ 50x+ 70y+40z = 10800} \Rightarrow \cases{6x+8y+5z=1290\\ 3x+4y+3z = 685\\ 5x+7y+4z= 1080} \\ \Rightarrow z=\cfrac{ \begin{vmatrix}6 & 8 & 1290\\ 3 & 4 & 685\\ 5 & 7 & 1080\end{vmatrix}}{\begin{vmatrix}6 & 8 & 5\\ 3 & 4 & 3\\ 5 & 7 & 4\end{vmatrix}} = \cfrac{-80}{-1}=\bbox[red,2pt]{80}$$

解答: $$ax^2+(2a+b)x-12 = a\left( x^2+(2-a)x-2a\right)+6 \Rightarrow \cases{2a+b=a(2-a)\\ -12=-2a^2+6} \\ \Rightarrow \cases{a^2+b=0 \cdots(1)\\ a^2=9 \cdots(2)} ,由(2)得a=3(a\gt 0,a\ne -3)代入(1) \Rightarrow b=-9\Rightarrow (a,b)=\bbox[red, 2pt]{(3,-9)}$$

解答: $${3\over 5}+{2\over 5}=1 \Rightarrow \cases{C\in \overline{AB} \\ \overline{AC}:\overline{CB}=2:3} \Rightarrow 令\cases{\overline{AC}=2k\\ \overline{BC}=3k}\\ 假設\overline{BD}=a,再由3\overline{AD} =8\overline{BD} \Rightarrow 3(2k+3k+a)=8a \Rightarrow a=3k \Rightarrow B為\overline{CD}中點\\ \Rightarrow \overline{OB}為\triangle OCD的中線定理 \Rightarrow \cases{\overline{OC}^2 +\overline{OD}^2=2(\overline{BC}^2+\overline{OB}^2) \\ \overline{OC}^2 +\overline{OD}^2= \overline{CD}^2} \Rightarrow 2(9k^2+\overline{OB}^2)=(6k)^2 \\ \Rightarrow \overline{OB}=3k;\\再加上\overline{OA}^2 +\overline{OB}^2 = \overline{AB}^2  \Rightarrow \overline{OA}^2 +9k^2 =(2k+3k)^2 \Rightarrow \overline{OA}=4k\\ 因此{\overline{OB}\over \overline{OA}}={3k\over 4k}=\bbox[red,2pt]{3\over 4}$$

解答: $$假設P在E的投影點為Q\Rightarrow Q=(a,b,2-a) \Rightarrow \overline{QA}=\overline{QB}=\overline{QC} \\ \Rightarrow (a-2)^2+(b+1)^2 +(2-a)^2 = a^2+(b-1)^2+(-a)^2 = (a+2)^2+(b-1)^2 +(-a-2)^2\\ \Rightarrow \cases{a=-1\\ b=-4} \Rightarrow Q(-1,-4,3) \Rightarrow 過Q且方向向量為(1,0,1)的直線L:(-1,-4,3)+t(1,0,1)\\=(-1+t,-4,3+t) \Rightarrow L與平面z=1的交點P=(-3,-4,1)\Rightarrow P與E的距離＝\left|{-2-2\over \sqrt{1+1}} \right| = \bbox[red,2pt]{2\sqrt 2}$$

解答: $$\cases{T\in L_1\Rightarrow T=(1+t,1-t,2+t)\\ S\in L_2\Rightarrow S=(2+2s,5+s,6-s)} \Rightarrow \vec n=\overrightarrow{TS}=(2s-t+1,s+t+4,-s-t+4)\\ \Rightarrow \cases{\vec n\cdot (1,-1,1)=0\\ \vec n\cdot (2,1,-1)=0} \Rightarrow \cases{s=-1/3\\ t=1/3} \Rightarrow \cases{T=(4/3,2/3,7/3)\\ S=(4/3,14/3,19/3)} \\ 又\cases{\overline{PT}=3\\ \overline{QS}=3} \Rightarrow \cases{P=T＋\sqrt 3(1,-1,1)\\ Q=S+{\sqrt 6\over 2}(2,1,-1)} \Rightarrow \overline{PQ}= \sqrt{50}=\bbox[red,2pt]{5\sqrt 2}$$

解答: $$餘弦定理：\cos\theta ={\overline{OA}^2+\overline{OP}^2- \overline{AP}^2 \over 2 \cdot\overline{OA} \cdot \overline{OP}} ={1+\overline{OP}^2-1\over 2\overline{OP}} ={\overline{OP}\over 2} \Rightarrow \overline{OP}=2 \cos \theta，故選\bbox[red,2pt]{(4)}$$

解答: $$\cos \angle QOB = \cos(90^\circ-\theta)= \sin\theta ={3\over 5} \Rightarrow {3\over 5}={\overline{OB}^2+ \overline{OQ}^2 -\overline{BQ}^2 \over 2\cdot \overline{OQ}\cdot \overline{OB}} ={\overline{OQ}\over 4} \Rightarrow \overline{OQ}={12\over 5}\\ \Rightarrow Q(-\overline{OQ}\cos \angle QOB,\overline{OQ}\sin \angle QOB) = (-{12\over 5}\cdot {3\over 5},{12\over 5}\cdot {4\over 5})= \bbox[red,2pt]{（-{36\over 25},{48\over 25})}\\ 又\cases{\angle A=180^\circ-2\theta\\ \angle B=180^\circ-2(90^\circ-\theta)=2\theta} \Rightarrow \angle A+\angle B=180^\circ \Rightarrow \overline{AP}\parallel \overline{BQ},\\再加上\cases{\overline{BQ}= \overline{OB}=2 \\ \overline{AP}=\overline{OA}=1},即\overline{BQ}= 2\overline{AQ};因此由\cases{\overline{AP}\parallel \overline{BQ}\\ \overline{BQ}= 2\overline{AQ}} 可得\bbox[red, 2pt] {\overrightarrow{BQ}= 2\overrightarrow{AP}}$$

解答: $$\overleftrightarrow{BQ}斜率＝{48\over 25}\div (2-{36\over 25})={24\over 7} \Rightarrow \overleftrightarrow{BQ}:y={24\over 7}(x+2) \Rightarrow 24x-7y+48=0 \\\Rightarrow d(A,\overleftrightarrow{BQ})= {24+48\over \sqrt {24^2+ 7^2}} =\bbox[red,2pt]{{72\over 25}}\\ PABQ面積＝\triangle OAP+\triangle OPQ+\triangle OBQ = {1\over 2}\left( \overline{OA}\cdot \overline{OP}\sin \theta + \overline{OP}\cdot \overline{OQ} + \overline{OB}\cdot \overline{BQ}\sin (2\theta) \right) \\={1\over 2}({8\over 5} \cdot {3\over 5}+{8\over 5}\cdot {12\over 5}+2\cdot 2\cdot (2\cdot {3\over 5}\cdot {4\over 5}))= \bbox[red,2pt]{108\over 25}$$

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解題僅供參考，其他升大學歷屆試題及詳解