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2023年5月9日 星期二

112年警專42期數學科(乙組)詳解

臺灣警察專科學校 112 學年度專科警員班第 42 期正期學生組
新生入學考試乙組數學科試題

壹、單選題:(一)30 題,題號自第 1 題至第 30 題,每題 2 分,計 60 分。
(二)未作答者不給分,答錯者不倒扣。
(三)請將正確答案以 2B 鉛筆劃記於答案卡內。

解答:$$與x-y+2=0平行之直線方程式可寫作x-y+k=0,且過(3,1)\Rightarrow 3-1+k=0 \Rightarrow k=-2\\,故選\bbox[red, 2pt]{(B)}$$
解答:$$\tan \theta={5\over 12} \Rightarrow 斜邊長=\sqrt{5^2+12^2}=13 \Rightarrow \sin\theta= {5\over 13},故選\bbox[red, 2pt]{(A)}$$


解答:$${A牌手機數量\over 師生總人數}= {1\times 80\%+7\times 40\% \over 1+7} ={3.6\over 8}={45\over 100},故選\bbox[red, 2pt]{(D)}$$
解答:$$x \to x-{\pi\over 3} 代表向右平移{\pi\over 3},故選\bbox[red, 2pt]{(D)}$$
解答:$$f(-2)=0 \Rightarrow (4+2a-2)(4a+2+2)= 8(a+1)^2 =0 \Rightarrow a=-1,故選\bbox[red, 2pt]{(A)}$$
解答:$$(x+3)(x+2)(x-1)(x-2) \le 0 \Rightarrow \cases{1\le x\le 2 \Rightarrow x=1,2\\ -3\le x\le -2 \Rightarrow x=-3,-2} \Rightarrow x=1,2,-2,-3\\,共四個整數解,故選\bbox[red, 2pt]{(C)}$$
解答:$$任一張被抽中的機率都是{1\over 20},抽一張的期望值={1\over 20}(100\cdot 15+ 500\cdot 4+ 1000\cdot 1)\\=225 \Rightarrow 抽二張的期望值=450,故選\bbox[red, 2pt]{(C)}$$
解答:$$A=\begin{bmatrix} 1& 2\\ 0 & 1\end{bmatrix} \Rightarrow A^2= \begin{bmatrix} 1& 2\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1& 2\\ 0 & 1\end{bmatrix} =\begin{bmatrix} 1& 4 \\ 0 & 1\end{bmatrix} \\\Rightarrow A^4= \begin{bmatrix} 1& 4 \\ 0 & 1\end{bmatrix}\begin{bmatrix} 1& 4 \\ 0 & 1\end{bmatrix} =\begin{bmatrix}1& 8 \\ 0 & 1 \end{bmatrix} \Rightarrow k=8,故選\bbox[red, 2pt]{(C)}$$
解答

$$假設正六邊形邊長為1,A(0,0)及B(1,0),圖形如上\\(A) \overrightarrow{AB} \cdot \overrightarrow{AB} =\overline{AB}^2=1\\(B)\overrightarrow{AB} \cdot \overrightarrow{AC} =\sqrt 3\cos 30^\circ ={3\over 2} \\(C)\overrightarrow{AB} \cdot \overrightarrow{AD} = 2\cos 60^\circ=1\\ (D)\cos 120^\circ =-{\sqrt 3\over 2}\\,故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{A(a,a,0)\\ B(0,-a,-a)} \Rightarrow \overline{AB} =\sqrt{a^2+4a^2+a^2} =\sqrt 6a=3\sqrt 2 \Rightarrow a={3\sqrt 2\over \sqrt 6} =\sqrt 3,故選\bbox[red, 2pt]{(B)}$$
解答:$$(\vec u-\vec v)\cdot (\vec u-\vec v)=|\vec u|^2-2\vec u\cdot \vec v+|\vec v|^2=9^2+(-8)^2 \Rightarrow 144-2\cdot 0+ |\vec v|^2=145\\ \Rightarrow |\vec v|^2=1 \Rightarrow |\vec v|=1,故選\bbox[red, 2pt]{(A)}$$
解答:$$y=f(x)= (m^2+1)x^2 +(m+1)x+1 \Rightarrow \cases{m^2+1 \gt 0\\ f(0)=1} \Rightarrow 圓形凹向上且通過(0,1),故選\bbox[red, 2pt]{(A)}$$
解答:$$圓C:x^2+y^2-6x-8y=0 \Rightarrow (x-3)^2+(y-4)^2 =5^2 \Rightarrow \cases{圓心O(3,4) \\ 圓半徑r=5}\\ 又\overline{OA}=\sqrt{5^2+(-12)^2} =13 \gt r \Rightarrow A在圓外 \Rightarrow \overline{AP}最短距離=\overline{OA}-r=13-5=8,故選\bbox[red, 2pt]{(B)}$$
解答:$$\sqrt{19+ 6\sqrt{10}} =\sqrt{19+2\sqrt{90}} =\sqrt{9}+\sqrt{10}=3+\sqrt{10},整數部分為3+3=6,故選\bbox[red, 2pt]{(D)}$$
解答:$$令公比為r ,則\cases{a_1+a_2+ a_3 =a_1+a_1r+a_1r^2=5 =a_1(1+r+r^2) =5\\ a_4+a_5+a_6 =a_1r^3+ a_1r^4+a_1r^5= a_1r^3(1+r+r^2) =-40} \\ 兩式相除\Rightarrow {1\over r^3}=-{1\over 8} \Rightarrow r=-2 \Rightarrow a_1(1-2+(-2)^2) =5 \Rightarrow a_1={5\over 3} \\因此a_1+\cdots+ a_9 = a_1(1+r+\cdots +r^8)=a_1\cdot {1-r^9\over 1-r}={5\over 3} \cdot {1+512\over 3} =5\cdot 57=285,故選\bbox[red, 2pt]{(C)}$$
解答:$$在九節沒有充電站的車廂間有十個間隔,可在十個間隔任選三個設置充電站的車廂\\ 因此有C^{10}_3種設置方法,故選\bbox[red, 2pt]{(B)}$$
解答:$$假設\cases{黃球有a個\\ 藍球有b個} \Rightarrow \cases{C^a_2/C^9_2=1/6\\ C^b_2/C^9_2=1/12} \Rightarrow \cases{a=4\\ b=3} \Rightarrow 紅球有9-4-3=2個,故選\bbox[red, 2pt]{(B)}$$
解答:$$\log x=-12.34 \Rightarrow x=10^{-12.34} =10^{-13}\cdot 10^{0.66} =4.57\times 10^a\\ 由於0\lt 10^{0.66} \lt 10,因此a=-13,故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{\log 2=a\\ \log 3=b} \Rightarrow \cases{10^a=2 \\10^b=3} \Rightarrow 10^{3b-a+1} =(10^b)^3\cdot (10^a)^{-1} \cdot 10=3^3\cdot {1\over 2}\cdot 10=135,故選\bbox[red, 2pt]{(D)}$$
解答:$$(A) \log_4 9 ={\log 9\over \log 4} ={2\log 3\over 2\log 2}={\log 3\over \log 2}=\log_2 3\\ (B) \log_{20}30 ={\log 30\over \log 20} ={\log 10+\log 3\over \log 10+\log 2}={1+\log 3\over 1+\log 2} \ne {\log 3\over \log 2} \\(C) \log_{\sqrt 2}\sqrt 3={\log \sqrt 3\over \log \sqrt 2} ={(1/2) \log 3\over (1/2) \log 2} ={\log 3\over \log 2}=\log_2 3\\ (D){\log 3\over 1-\log 5} ={\log 3\over 1-(1-\log 2) } = ={\log 3\over \log 2}=\log_2 3\\,故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{a_1+ a_2+ a_3=51\\ a_6=7} \Rightarrow \cases{3a_1+ 3d=51\\ a_1+5d=7} \Rightarrow \cases{a_1=39/2\\ d=-5/2} \\ \Rightarrow a_n= {39\over 2}-(n-1) \cdot {5\over 2} \lt 0 \Rightarrow 5n\gt 44 \Rightarrow n\ge 9,故選\bbox[red, 2pt]{(B)}$$
解答:$${3次出現1正面\over 5次出現3正面} ={C^3_1\over C^5_3} ={3\over 10},故選\bbox[red, 2pt]{(C)}$$
解答:$${戴眼鏡的男生\over 所有戴眼鏡} ={60\%\times 60\% \over 60\%\times 60\%+ 40\%\times 40\%} ={36\over 52} ={9\over 13},故選\bbox[red, 2pt]{(D)}$$
解答:$$r\theta ={1\over 3}\pi r \Rightarrow \theta={\pi\over 3}=60^\circ \Rightarrow 甲在北緯(90^\circ-60^\circ) =30^\circ,故選\bbox[red, 2pt]{(B)}$$
解答:$$1={180^\circ\over \pi} \approx 57^\circ \Rightarrow \sin 60^\circ \gt \sin 1\gt \cos 1 又\cases{(\tan \pi)/2= \sqrt 3/2\\ \cos(-\pi/6) =\sqrt 3/2} \\ 因此{1\over 2}\tan {\pi\over 3}=\cos {-\pi\over 6} \gt \sin 1\gt \cos 1 \Rightarrow \cos 1最小,故選\bbox[red, 2pt]{(B)}$$
解答:$$假設A=\begin{bmatrix}s & t\\ u& v\end{bmatrix} \Rightarrow \cases{A\begin{bmatrix}1\\2 \end{bmatrix} =\begin{bmatrix}5\\ 6 \end{bmatrix} \\[1ex] A \begin{bmatrix} 3\\ 4\end{bmatrix} = \begin{bmatrix}7\\ 8\end{bmatrix}} \Rightarrow \cases{s+2t=5\\ u+2v=6\\ 3s+4t= 7 \\3u+ 4v= 8} \Rightarrow \cases{s=-3\\ t=4\\ u=-4\\ v=5 } \\ \Rightarrow A=\begin{bmatrix} -3& 4\\ -4& 5\end{bmatrix} \Rightarrow A\begin{bmatrix} 2& 6\\ 4& 8 \end{bmatrix} =\begin{bmatrix} 10& 14\\ 12 & 16\end{bmatrix} \Rightarrow a+b+c+d= 52,故選\bbox[red, 2pt]{(D)}$$
解答:$$P(\sin (\theta+45^\circ), \sin 2\theta)在第二象限\Rightarrow \cases{\sin(\theta+45^\circ)= {\sqrt 2\over 2}(\sin \theta+\cos\theta)\lt 0\\ \sin 2\theta =2\sin \theta\cos \theta\gt 0}\\ \Rightarrow \cases{\cos\theta \lt 0\\ \sin \theta\lt 0},故選\bbox[red, 2pt]{(C)}$$
解答:$$餘弦定理:\cos \angle A= {\overline{AB}^2 +\overline{AC}^2 -\overline{BC}^2 \over 2\cdot \overline{AB}\cdot \overline{AC}} \Rightarrow \cos 60^\circ={1\over 2}={25+64-\overline{BC}^2 \over 2\cdot 5\cdot 8} \\ \Rightarrow \overline{BC}^2=89-40=49 \Rightarrow \overline{BC}=7,故選\bbox[red, 2pt]{(C)}$$
解答:$$需刪除中間的數字2或3,刪除3後變異較大,故選\bbox[red, 2pt]{(C)}$$
解答:$$【兩半】一定是封閉曲線,不可能是拋物線,故選\bbox[red, 2pt]{(D)}$$

貳、多重選擇題:

(一)共 10 題,題號自第 31 題至第 40 題,每題 4 分,計 40 分。
(二)每題 5 個選項各自獨立其中至少有 1 個選項是正確的,每題皆不倒扣,5 個選項全部
答對得該題全部分數,只錯 1 個選項可得一半分數,錯 2 個或 2 個以上選項不給分。
(三)請將正確答案以 2B 鉛筆劃記於答案卡內。

解答:$$(B)\times: \cases{a=3\\b=2\\ c=-2},符合a\gt b,但ac=-6\lt -4=bc\\ (C)\times: 若ab\lt 0,則\sqrt{ab}非實數 \\(E)\times: \cases{a=\sqrt 2\\ b=-\sqrt 2} \Rightarrow \cases{a+b=0為有理數\\ ab=-2也是有理數},但a,b均不是有理數\\,故選\bbox[red, 2pt]{(AD)}$$
解答:$$(A)\times: y={3\over 4}x+3 \Rightarrow 斜率為{3\over 4} \ne -{3\over 4}\\(B)\times:y=0時,x=-4 \ne 4 \\(C)\bigcirc: x=0時,y=12/4=3 \\(D) \bigcirc: {1\over 2}\cdot 4\cdot 3=6\\(E)\times: (0,0)與(-3,4)的中點(-3/2,2)不在L上\\,故選\bbox[red, 2pt]{(CD)}$$
解答:$$(A) \bigcirc:\cases{AB為3\times 3\\ BA為2\times 2} \Rightarrow AB\ne BA \\(B) \times: A=\begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix} \Rightarrow AB=BA=\begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix},但A^{-1}不存在 \\(C)\times: 若AB\ne BA,則(A+B)(A-B)\ne A^2-B^2 \\(D)\bigcirc: (A+I)(A+I)(A+I)= (A^2+2A+ I)(A+I)= A^3+3A^2+3A+ I\\ (E)\bigcirc: AB=I \Rightarrow A,B互為反矩陣,即\cases{A^{-1}=B\\ B^{-1}=A} \Rightarrow BA=A^{-1}A=I\\,故選\bbox[red, 2pt]{(ADE)}$$
解答:$$(D)\times: (a-b)^8中a^3b^5係數=-C^8_3\\,故選\bbox[red, 2pt]{(ABCE)}$$
解答:$$10^x=8 \Rightarrow x=\log 8\\ (A)\bigcirc:100^x=(10^2)^x =(10^x)^2 =8^2=64 \\(B)\times: (0.1)^x= ({1\over 10})^x= {1\over 10^x}={1\over 8} \ne -8\\ (C) \times:10^{\sqrt x} =2\sqrt 2 \Rightarrow \sqrt x=\log(2\sqrt 2) ={3\over 2}\log 2 \Rightarrow x=({3\over 2}\log 2 )^2 \ne \log 8 \\(D) \times:(\log 2)^3 \ne \log 8 \\(E) \bigcirc: x=\log 8 =3\log 2=3\cdot 0.301 = 0.903\Rightarrow 0.8\lt x\lt 1\\,故選\bbox[red, 2pt]{(AE)}$$
解答:$$(A)\bigcirc: 相關係數介於\pm 1之間\\ (B)\times:垂直線的斜率不在\pm 1之間\\ (C)\times: 垂直線的相關係數不是\pm 1\\ (D)\bigcirc: 仍維持線性關性\\ (E)\bigcirc:標準化亦為線性運算\\,故選\bbox[red, 2pt]{(ADE)}$$
解答:$$(A)\times: \sin\theta-\cos\theta={1\over 3} \Rightarrow \sqrt 2({1\over \sqrt 2}\sin \theta-{1\over \sqrt 2}\cos \theta)={1\over 3} \Rightarrow \sqrt 2\sin(\theta-45^\circ)={1\over 3}\\ \qquad \Rightarrow \sin(\theta-45^\circ)={1\over 3\sqrt 2} \gt \sin 0^\circ \Rightarrow \theta \gt 45^\circ \\(B) \bigcirc:(\sin\theta-\cos\theta)^2 =1-2\sin\theta\cos \theta={1\over 9} \Rightarrow \sin\theta \cos\theta={4\over 9} \\(C)\times: (\sin\theta +\cos \theta)^2= 1+2\sin\theta \cos\theta=1+{8\over 9}={17\over 9} \Rightarrow \sin\theta+\cos \theta={\sqrt{17}\over 3} \ne {5\over 3} \\(D) \bigcirc: \sin^3\theta- \cos^3\theta =(\sin \theta-\cos\theta)(\sin^2\theta +\sin\theta\cos \theta+ \cos^2\theta) ={1\over 3}\cdot (1+{4\over 9}) ={13\over 27} \\(E)\times: \cases{\sin \theta-\cos\theta=1/3\\ \sin \theta +\cos \theta=\sqrt{17}/3} \Rightarrow \cases{\sin \theta =(1+\sqrt{17})/6\\ \cos \theta=(\sqrt{17}-1)/6} \Rightarrow \tan \theta={\sqrt{17}+1\over \sqrt{17}-1}\ne {9\over 8}\\ ,故選\bbox[red, 2pt]{(BD)}$$
解答:$$(A)\times:f(x)=(x-1)^4= \sum_{k=0}^4 C^4_k x^k(-1)^{4-k} \Rightarrow x^2係數=C^4_2(-1)^2=6\ne -6 \\(B)\bigcirc: f(1)=0^4=0\\ (C)\bigcirc: f(-1)=(-2)^4=16 \\ (D)\bigcirc: (x-1)^4=(x^2-2x+1)^2 = ((x+1)^2-4x)^2 = (x+1)^4-8x(x+1)^2 +16x^2\\ =((x+1)^2-4x)^2 = (x+1)^4-8x(x+1)^2 +16(x+1)^2-32x-16 \\(E)\times: (x-1)^4=(x^2-2x+1)^2 = ((x^2-1)-2x+2)^2 \\\quad=(x^2-1)^2+2(-2x+2)(x^2-1)+ 4x^2-8x+4 \\\quad =(x^2-1)^2+2(-2x+2)(x^2-1)+ 4(x^2-1)-8x+8 \Rightarrow 餘式為-8x+8\\,故選\bbox[red, 2pt]{(BCD)}$$
解答:$$(A)\bigcirc: a_n=a_1+(n-1)d \Rightarrow \cases{a_n+3=(a_1+3)+(n-1)d \\ 3a_n= 3a_1+(n-1)3d} \\ \quad \Rightarrow \cases{\langle a_n+3 \rangle為首項a_1+3,公差為d的等差數列\\ \langle 3a_n\rangle 為首項3a_1,公差為3d的等差數列} \\(B)\times: b_n= b_1r^{n-1} \Rightarrow {b_{n+1}+3\over b_n+3} ={b_1r^n+3\over b_1r^{n-1}+3} 非定值\Rightarrow \langle b_n+3\rangle 非等比數列\\ (C)\times:a_nb_n -a_{n-1}b_{n-1}= (a_1+(n-1)d)b_1r^{n-1} -(a_1+(n-2)d)b_1r^{n-2}非定值 \\(D)\times: {a_{n+1}b_{n+1}\over a_nb_n} ={(a_1+nd)b_1r^n \over (a_1+(n-1)d)b_1r^{n-1}}非定值\\ (E)\bigcirc: \langle a_n^2\rangle 為等差 \Rightarrow a_n^2-a_{n-1}^2 = (a_n+a_{n-1})(a_n-a_{n-1}) =(2a_1+(2n-3)d)d 為一常數\\\quad \Rightarrow d=0\\,故選\bbox[red, 2pt]{(AE)}$$
解答:$$(A)\bigcirc: \cases{A(0,-1)\\ B(7,0)} \Rightarrow \overrightarrow{AB}=(7-0,0-(-1)) =(7,1) \\(B)\times: \cases{\overrightarrow{AB}=(7,1) \\\overrightarrow{BC}=(-5,5)} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{BC}= -35+5=-30\ne 30 \\(C)\bigcirc \cases{\overline{AB}= \overline{BC}= 5\sqrt{2} \\ \overline{CA}=2\sqrt{10}} \Rightarrow \cos \angle A={50+40-50\over 2\cdot 5\sqrt 2\cdot 2\sqrt{10} }={1\over \sqrt 5}\gt 0 \Rightarrow 最大的角\angle A為銳角 \\(D)\times:\overline{BH}=\overline{AB} \sin \angle A=5\sqrt 2\cdot {2\over \sqrt 5}=\sqrt{40} \ne \sqrt{30} \\(E)\bigcirc: {\overrightarrow{AB} \cdot \overrightarrow{AC}\over \overline{AC}^2}\times \overrightarrow{AC}  ={20\over 40}\times (2,6) =(1,3)\\,故選\bbox[red, 2pt]{(ACE)}$$


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