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:
{∠COE+θ=90∘∠COE+∠COD=90∘⇒∠COD=θ⇒tanθ=tan∠COD=¯CD¯CO=¯CD1=¯CD,故選(5)
解答:(sk,logtk)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logt=s/2 ⇒t=10s/2⇒t2=10s,故選(4)
解答:9個數字ABCDEFGHI需滿足A<B<C<D<E且E>F>G>H>I,因此E=9;接著在剩下8個數字中,任取4個數字(有C84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C84=8!4!4!個九位數,故選(1)
解答:依題意(→PQ×→PR)∥(2,−3,5)⇒→PQ×→PR=k(2,−3,5)=(2k,−3k,5k),k不為0的常數(1)‖−111a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(−1,1,1)|=0(2)‖1−11a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(1,−1,1)|=|10k|(3)‖11−1a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(1,1,−1)|=|−6k|(4)‖−1−11a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(−1,−1,1)|=|6k|(5)‖−1−1−1a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(−1,−1,−1)|=|−4k|,故選(2)
解答:
解答:(sk,logtk)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logt=s/2 ⇒t=10s/2⇒t2=10s,故選(4)
解答:9個數字ABCDEFGHI需滿足A<B<C<D<E且E>F>G>H>I,因此E=9;接著在剩下8個數字中,任取4個數字(有C84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C84=8!4!4!個九位數,故選(1)
解答:依題意(→PQ×→PR)∥(2,−3,5)⇒→PQ×→PR=k(2,−3,5)=(2k,−3k,5k),k不為0的常數(1)‖−111a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(−1,1,1)|=0(2)‖1−11a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(1,−1,1)|=|10k|(3)‖11−1a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(1,1,−1)|=|−6k|(4)‖−1−11a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(−1,−1,1)|=|6k|(5)‖−1−1−1a1b1c1a2b2c2‖=|(2k,−3k,5k)⋅(−1,−1,−1)|=|−4k|,故選(2)
解答:
將→OA縮寫成→a,即→OA⋅→OB縮寫成→a⋅→b{→a⋅{→c,→f,→g}=0→a⋅{→b,→d,→e}=1,{→b⋅→g=0→b⋅{→c,→d,→f}=1→b⋅e=2,{→c⋅{→d,→g}=0→c⋅{→e,→f}=1,{→d⋅→e=2→d⋅{→f,→g}=1,{→e⋅→f=2→e⋅→g=1,→f⋅→g=1共有{6個012個13個2⇒期望值=0+12+3⋅221=1821=67,故選(3)
二、多選題( 占30 分)
解答:假設{甲在第i月的薪資為ai−1元乙在第i月的薪資為bi−1元,且a0=b0⇒{ak=a0+200m,m=⌊k/3⌋bk=a0+1000m,m=⌊k/12⌋,k∈N(1)×:⌊8/3⌋=2⇒a8=a0+200⋅2=a0+400⇒增加400元,不是600元(2)×:{a12=a0+200⋅4=a0+800b12=a0+1000⋅1=a0+1000⇒a12<b12⇒甲比乙低(3)◯:{a18=a0+200⋅6=a0+1200b18=a0+1000⋅1=a0+1000⇒a18<b18⇒甲比乙高(4)×:{∑17i=0ai=3(a0+(a0+200)+(a0+400)+⋯+(a0+1000))=18a0+9003∑17i=0bi=12a0+(a0+1000)⋅6=18a0+6000⇒甲比乙多(5)◯:第三年:{甲薪資:1−3月:a0+1600,4−6月:a0+1800,7−9:a0+2000,10−12月:a0+2200乙薪資:a0+2000⇒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(35)解答:pn=1−(1−0.1)n=1−0.9n(1)◯:{pn=1−0.9npn+1=1−0.9n+1⇒pn+1>pn(2)×:p3=1−0.93=1−0.729=0.271≠0.3(3)×:{p2−p1=(1−0.92)−(1−0.9)=0.9−0.92p3−p2=(1−0.93)−(1−0.92)=0.92−0.93⇒p2−p1≠p3−p2⇒非等差(4)◯:{第1次未中獎且第2次中獎機率=0.9⋅0.1=0.09p2−p1=1−0.92−(1−0.9)=0.9−0.92=0.09(5)×:至少中2次=全部-全沒中-只中1次=1−0.9n−Cn1⋅0.1⋅0.9n−1≠2pn,故選(14)
解答:令S(n)=log3a1−log3a2+log3a3−⋯+(−1)n+1log3an(1)×:n=23⇒S(23)=log3a1⋅a3⋯a23a2⋅a4⋯a22=log3312(3√3)2+4+⋯+22311(3√3)1+3+⋯+21=log33(3√3)11=log3335/2=352≯
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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20題BQ的斜率有問題
回覆刪除謝謝提醒,已修訂完畢
刪除選填題13題的z,分子的行列式第(1,3)有問題
回覆刪除少了一個零,已修訂,謝謝!
刪除