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2020年6月20日 星期六

98年警專28期乙組數學科詳解


臺灣警察專科學校專科警員班二十八期(正期學生組)
新生入學考試乙組數學科試題
壹、單選題


:$$本題\bbox[red,2pt]{(送分)}$$


:$$z=x+yi,其中\;x,y\in R \Rightarrow |z+2-3i|=|(x+2)+(y-3)i| = \sqrt{(x+2)^2+(y-3)^2}=1 \\ \Rightarrow (x+2)^2+(y-3)^2=1\;為一圓,故選\bbox[red,2pt]{(D)}$$


:$$令f(x)=x^6-18x^5-16x^4-50x^3-132x^2+4 = (x-19)(x^5+x^4+3x^3+7x^2+x+19)+365\\ \Rightarrow f(19)=365,故選\bbox[red,2pt]{(C)}$$



$$x=a\times 10^n\;其中1\le a\le 10 \Rightarrow \log x=n+\log a,其中\cases{n是首數\\ \log a是尾數}\\\cases{\log x的首數與\log325.78的首數相同 \Rightarrow n=2\\ \log x的尾數與\log5.082的尾數相同\Rightarrow a=5.082} \Rightarrow x= 5.082\times 10^2 =508.2,故選\bbox[red,2pt]{(B)}$$



:$$\cos \angle C={a^2+b^2 -c^2 \over 2ab} ={9+16-13 \over 24} =\frac{1}{2} \Rightarrow \sin \angle C={\sqrt 3\over 2} \Rightarrow {c\over \sin \angle C}=2R \Rightarrow {\sqrt{13}\over \sqrt 3/2}=2R \\\Rightarrow R=\sqrt{13\over 3},故選\bbox[red,2pt]{(C)}$$


:$${\sin(270^\circ +\theta)\tan(180^\circ+\theta) \cos(90^\circ-\theta) \over \cos(270^\circ +\theta) \tan(540^\circ +\theta)} = {\sin(270^\circ +\theta) \cos(90^\circ-\theta) \over \cos(270^\circ +\theta) } ={-\cos\theta\sin \theta \over \sin\theta} \\= -\cos \theta,故選\bbox[red,2pt]{(D)}$$




$$\cases{\cos\theta =4/5 \\ 7\pi/2 < \theta < 4\pi} \Rightarrow \cases{\sin \theta = -3/5\\ 7\pi/4 < \theta/2 < 2\pi} \Rightarrow \sin (\theta/2) <0; \\又\cos \theta = 2\cos^2{\theta\over 2} -1 \Rightarrow  {4\over 5} =2\cos^2{\theta\over 2} -1 \Rightarrow \cos{\theta\over 2} = {3\over \sqrt{10}} \Rightarrow \sin{\theta\over 2}=-{1\over \sqrt{10}},故選\bbox[red,2pt]{(D)}$$



:$$令\overrightarrow{AM}=t\overrightarrow{AP},由 \overrightarrow{AM}= x\overrightarrow{AB} +y\overrightarrow{AC} \\\Rightarrow t\overrightarrow{AP}= x\overrightarrow{AB} +y\overrightarrow{AC} \Rightarrow \overrightarrow{AP}={x\over t}\overrightarrow{AB} +{y\over t} \overrightarrow{AC}={4\over 7}\overrightarrow{AB} +{6\over 7}\overrightarrow{AC} \\ \Rightarrow \cases{{x\over t}=4/7\\ {y \over t}= 6/7}\overrightarrow{AP} \Rightarrow \cases{x={4\over 7}t \\y={6\over 7}t };\\由B,M,C在一直線上,可得x+y=1 \Rightarrow {4\over 7}t+{6\over 7}t =1 \Rightarrow t={7\over 10} \\ \Rightarrow x={4\over 7}t ={4\over 7}\times {7\over 10} ={2\over 5},故選\bbox[red,2pt]{(C)}。 $$



:$$|\vec a+2\vec b|^2 =(\vec a+2\vec b)\cdot (\vec a+2\vec b) = |\vec a|^2+4\vec a\cdot \vec b+4|\vec b|^2 \Rightarrow (2\sqrt{13})^2 = 2^2+ 4\vec a\cdot \vec b+4\times 3^2 \\ \Rightarrow 52=40+ 4\vec a\cdot \vec b \Rightarrow \vec a\cdot \vec b=3 \Rightarrow \cos \theta={\vec a\cdot \vec b \over |\vec a||\vec b|} ={3 \over 2\times 3}={1\over 2} \Rightarrow \theta =60^\circ,故選\bbox[red,2pt]{(C)}。$$




$$(A) \times: |-3|=3\\(B)\bigcirc: \sqrt{2^2+(-3)^2} =\sqrt{13} \\(C)\times: \left|{ -1-4-3+3\over \sqrt{1^2+(-2)^2+1^2}} \right|={5\over \sqrt 6} \\(D)\times: \sqrt{(-1)^2+1^2+(-1)^2} =\sqrt 3\\,故選\bbox[red, 2pt]{(B)}$$



:$$\cases{E_1:2x-y+2z-1=0 \\E_2:2x-y+2z+3/2=0} \Rightarrow \text{dist}(E_1,E_2)= \left| { -1-3/2\over \sqrt{2^2+(-1)^2+2^2}}\right| ={5/2 \over 3} =5/6,故選\bbox[red, 2pt]{(B)}$$


:$$\cases{E_1:x-2y+3z+1=0 \\E_2:2x-y+z+2=0} \Rightarrow \cases{E_1的法向量\vec n_1=(1,-2,3) \\ E_2的法向量\vec n_2=(2,-1,1)} \Rightarrow  \vec n_1\times \vec n_2=(1,5,3) =(1,a,b)\\ \Rightarrow a+b=5+3=8,故選\bbox[red, 2pt]{(A)}$$



$$x^2+y^2-4x+2y+1=0 \Rightarrow (x-2)^2 +(y+1)^2=4 \Rightarrow 圓心O(2,-1) \\ \Rightarrow 圓方程式(x-2)^2+(y+1)^2=r^2,經過(4,5) \Rightarrow 2^2+6^2=40=r^2 \Rightarrow r=2\sqrt{10}\\(A)\times: 圓心(2,-1) \ne (-2,1) \\(B)\bigcirc: r=2\sqrt{10} \\(C)\times: (2,-1)=(a,b) \Rightarrow a+b=1\ne 2 \\(D) \times: 40\pi-4\pi=36\pi \approx 113 \not \gt 150,故選\bbox[red,2pt]{(B)}。$$




$$\overline{F_1F_2}= |-4-2|=6 =\overline{PF_1}+\overline{PF_2} \Rightarrow P在線段\overline{F_1F_2}上,P軌跡為一線段,故選\bbox[red,2pt]{(C)}。$$

15. 已知雙曲線\({x^2\over 16}- {y^2\over 9}=1\)上一點P到其中一焦點\(\;F_1\;\)的距離為6,那麼P到另一焦點\(\;F_2\;\)的焦點距離是多少?
(A) 10  (B) 12   (C) 14   (D) 16

:$${x^2\over 16}-{y^2 \over 9}=1 \Rightarrow a=4 \Rightarrow |\overline{PF_1}-\overline{PF_2}|=2a=8 \Rightarrow |6-\overline{PF_2}| =8 \Rightarrow \overline{PF_2}=14,故選\bbox[red,2pt]{(C)}$$


:$$a_3=2a_2+a_1 = 6-1=5 \Rightarrow a_4=2a_3+a_2 =10+3=13 \Rightarrow a_5= 2a_4+a_3 =26+5=31 \\ \Rightarrow a_6=2a_5+a_4 = 62+13=75,故選\bbox[red,2pt]{(A)}$$


:$$只有(D)正確,故選\bbox[red, 2pt]{(D)}$$



:$$(B)\times: 測量高度無需抽樣\\(C)\times: 樣本多的結果不一定比樣本少的結果更接近母體\\ (D)\times:各學期抽樣是獨立的,機率相同\\,故選\bbox[red,2pt]{(A)}$$

19. 若某校1000位學生的國文成績平均分數是70分,標準差是5分。若已知成績呈常態分配,試問全校約有多少同學的國文成績低於60分?
(A) 約320人  (B) 約160人  (C) 約50人  (D) 約25人

:$$P(X\le 60) = P(X\le \mu-2\sigma) = (1-P(\mu-2\sigma \le X \le \mu+2\sigma))\div 2={1\over 2}(1-95\%) \\ = 2.5\% \Rightarrow 人數為1000\times 2.5\%=25,故選\bbox[red,2pt]{(D)}$$

20. 同時投擲3個硬幣,若出現3正面可得18元、2正面可得12元、1正面可得6元,為了公平起見(期望值為0),出現3反面時應賠多少元?
(A) 18元  (B) 36元   (C) 54 元   (D) 72元


$$假設3反面應賠a元,則\cases{3正面機率為1/8,期望值為18\times {1\over 8}=18/8 \\2正面機率為3/8,期望值為12\times {3\over 8}=36/8 \\1正面機率為3/8,期望值為6\times {3\over 8}=18/8 \\ 3反面機率為1/8,期望值為-a/8\\} \\ \Rightarrow 總期望值為0 \Rightarrow {18\over 8} + {36\over 8} + {18\over 8}-{a\over 8}=0 \Rightarrow a=72,故選\bbox[red,2pt]{(D)}$$

21. 某種診斷方法可有效的檢驗出初期的癌症,依過去的經驗知道,該方法對於癌症患者的檢出率高達0.95,同時對於健康的人誤判為罹癌者的比例亦低至0.05。假設一群人中有5%的人罹患癌症,現從中任選一人加以檢驗,若此人被檢驗出患有癌症,求此人確實罹癌的機率。
(A) 0.05  (B) 0.5  (C) 0.75  (D) 0.95

:$${ 檢出有癌症且確實罹癌\over 檢出有癌症} ={ 5\%\times 0.95\over 5\%\times 0.95 + 95\%\times 0.05} ={1 \over 2},故選\bbox[red,2pt]{(B)}$$



:$$身長越長則體重越重,兩者正相關,斜率為正值,故選\bbox[red,2pt]{(A)}$$


:$$(D)\begin{vmatrix} 2 & 12 \\1 & 6\end{vmatrix} =12-12=0,故選\bbox[red,2pt]{(D)}。$$



:$$3A-2X=4B \Rightarrow 3\left[\matrix{-1 & 5 \\3 & 2} \right] -2\left[\matrix{a & b \\c & d} \right] =4\left[\matrix{1 & -3 \\2 & 1/2} \right] \\\Rightarrow \left[\matrix{-3-2a & 15-2b \\9-2c & 6-2d} \right] =\left[\matrix{4 & -12 \\8 & 2} \right] \Rightarrow \cases{-3-2a=4\\ 15-2b=-12\\ 9-2c=8 \\ 6-2d=2} \Rightarrow \cases{a=-7/2\\ b=27/2\\ c=1/2 \\d=2},故選\bbox[red,2pt]{(A)}$$



$$(A)\times:A=\left[\matrix{1 & 0 \\0 & -1} \right] \Rightarrow A^2=I,但A\ne I且A\ne -I\\ (B)\times:A=\left[\matrix{0 & 1 \\0 & 0} \right] \Rightarrow A^2=0,但A\ne 0 \\(C)\times:\left[\matrix{0 & 1 \\0 & 0} \right] \left[\matrix{0 & 0 \\0 & 2} \right] =\left[\matrix{0 & 1 \\0 & 0} \right] \left[\matrix{0 & 1 \\0 & 2} \right] =\left[\matrix{0 & 2 \\0 & 0} \right] ,但\left[\matrix{0 & 1 \\0 & 0} \right] \ne \left[\matrix{0 & 0 \\0 & 2} \right]\\,故選\bbox[red,2pt]{(D)}$$


:$$柯西不等式: (x^2+(2y)^2)(2^2+({3\over 2})^2) \ge (2x+3y)^2 \Rightarrow (x^2+4y^2)\times {25\over 4} \ge 25 \\ \Rightarrow x^2+4y^2\ge 4,故選\bbox[red, 2pt]{(B)}$$



:$$x^2-6x+9 = (x-3)^2 \ge 0,故選\bbox[red,2pt]{(B)}$$


:$$-2\le x\le {1\over 2} \Rightarrow (x-{1\over 2})(x+2) \le 0 \Rightarrow x^2+{3\over 2}x-1 \le 0 \Rightarrow -2x^2-3x+2 \ge 0 \\ \Rightarrow \cases{a=-2 \\ b=-3} \Rightarrow a+b=-5,故選\bbox[red,2pt]{(B)}$$





$$所圍區域包含原點\Rightarrow \cases{x-y\ge -2\\ 2x+3y\le 6\\ x-2y \le 5} \Rightarrow \cases{a-b\ge -2\\ 2a+3b\le 6\\ a-2b \le 5},故選\bbox[red,2pt]{(C)}$$


:$${x+x+x+y \over 4} \ge \sqrt[4]{x^3y} \Rightarrow 2\ge \sqrt[4]{x^3y} \Rightarrow 16\ge x^3y,故選\bbox[red,2pt]{(C)}$$

貳、多重選擇題


:$$\cases{甲\to  癸=1\to 10\\ 子\to 亥= 1\to 12 } \Rightarrow 農歷年=(1,1)(2,2)..(10,10)(1,11)(2,12)(3,1)...\\
(辛己)=(8,6),往前推60年\Rightarrow \\ \begin{array} {}農曆年 & 數量\\\hline (8,6)\cdots(3,1)(2,12) \color{blue}{ (1,11) (10,10)} (9,9) & 10 \\ (8,8)\dots (1,1)(10,12)(9,11) & 10 \\ (8,10)\dots (1,3)(10,2) (9,1) & 10\\ (8,12) \cdots (1,5) (10, 4)(9,3)& 10 \\ (8,2) (7,1)(6,12)\dots (1,7)(10,6)(9,5) & 10 \\ (8,4)\color{blue}{(7,3)} (6,2)(5,1)(4,12)\dots(1,9) (10,8) (9,7) & 10\\\hline & 60\end{array}\\\cases{(A)(甲戌)=(1,11) \\(B)己午=(6,7) \\(C)庚寅=(7,3) \\(D)壬巳 =(9,6) \\(E) 癸酉=(10,10)},故選\bbox[red,2pt]{(BD)}$$


$$(A)\times: < a_n> 不一定是等差數列 \\(B) \times: 未定義S_0,無法計算a_1=S_1-S_0 \\(C)\times: < a_n>=1,1,1,...,則< a_n>是等差(d=0),也是等比(r=1)數列 \\(D) \bigcirc: < a_n>為等差\Rightarrow \cases{a_4=a_1+3d\\ a_7=a_1+6d \\ a_{10} =a_1+9d} \Rightarrow a_1,a_4, a_7, a_{10}為等差(公差=3d) \\(E) \bigcirc: < a_n>為等比\Rightarrow \cases{a_1+ a_2=a_1+a_1r\\ a_3+a_4 =a_1r^2+a_1r^3 =(a_1+a_2)r^2\\ a_5+a_6 =a_1r^4+a_1r^5 =(a_3+a_4)r^2\\a_7+a_8 =a_1r^6+a_1r^7 =(a_5+a_6)r^2} \Rightarrow  為等比數列,公比為r^2\\,故選\bbox[red,2pt]{(DE)}$$


:$$(A)\bigcirc: \lim_{n\to \infty} 6=6 \\(B)\times: \lim_{n\to \infty} 3\times(-1)^n =\pm 3\\(C) \bigcirc: \lim_{n\to\infty}{7n^2+17^{100} \over 3n^2-1} ={7\over 3} \\(D)\times: \lim_{n\to \infty} {3^{2n} \over 5^n} =\lim_{n\to \infty} {9^{n} \over 5^n} = \lim_{n\to \infty} (9/5)^n = \infty \\(E) \bigcirc: \lim_{n\to \infty} {\log n\over 2^n} =0\\,故選\bbox[red,2pt]{(ACE)}$$



:$$f(x)={2x^2-ax+b \over 3x^2+x+3} \Rightarrow \cases{f(0)=b/3\\ f(1)={2-a+b \over 7}\\ f(-1)={2+a+b \over 5}} \Rightarrow \cases{f(0)=f(1)\\ f(0)=f(-1)} \Rightarrow \cases{3a+4b=6 \\ 3a-2b=-6}\\ \Rightarrow \cases{a=-2/3\\b=2}  \Rightarrow k=b/3=2/3 \Rightarrow \cases{3a+b=-2+2=0 \\ ab=-4/3 <0},故選\bbox[red,2pt]{(CD)}$$



$$(A)\bigcirc: \log_2 3\times \log_3 5= {\log 3\over \log 2}\times {\log 5\over \log 3}={\log 5\over \log 2}=\log_2 5 \\(B) \bigcirc: \log_3 2+\log_3 4 = \log_3 (2\times 4) =\log_3 8\\ (C)\times: \log_5 7 -\log_5 9 = \log_5 {7\over 9} \ne {\log_5 7 \over \log_5 9} \\(D) \times: \log_2 35 = \log_2 (5\times 7) = \log_2 5+\log_2 7 \ne \log_2 5 \times \log_2 7 \\(E) \bigcirc: \log_4 27 = {\log_2 27 \over \log_2 4} ={\log_2 3^3 \over \log_2 2^2} ={3\log_2 3\over 2\log_2 2} ={3\over 2}\log_2 3\\,故選\bbox[red,2pt]{(ABE)}$$


:$$(B)\times: a+b=1+2=3 \not \gt c=3 \\(E)\times: \cos \angle B={a^2+c^2-b^2 \over 2ac} \Rightarrow {1\over 2}={a^2+16-9 \over 8a} \Rightarrow a^2-4a+7=0 \Rightarrow 判別式16-28<0 無實解,\\其餘皆正確,故選\bbox[red,2pt]{(ACD)}$$


:$$\cases{(A)[3,{\pi \over 3}] =(3\cos {\pi \over 3},3\sin {\pi \over 3}) = ({3\over 2}, {3\sqrt 3\over 2}) \\(B)[3,-{\pi \over 3}] =(3\cos (-{\pi \over 3}),3\sin (-{\pi \over 3})) = ({3\over 2}, -{3\sqrt 3\over 2}) \\ (C)[3,{2\pi \over 3}] =(3\cos {2\pi \over 3},3\sin {2\pi \over 3}) = (-{3\over 2}, {3\sqrt 3\over 2}) \\(D) [3,-{5\pi \over 3}] =[3,{\pi \over 3}] = ({3\over 2}, {3\sqrt 3\over 2}) \\ (E)[3,{7\pi \over 3}]=[3,{\pi \over 3}] = ({3\over 2}, {3\sqrt 3\over 2})  }\\,故選\bbox[red,2pt]{(ADE)}$$


$$(A)\bigcirc: 2x-3y+1=0 \Rightarrow y={2\over 3}x+{1\over 3}\Rightarrow 斜率為2/3 \\(B) \bigcirc: 法向量為(2,-3) \Rightarrow 2\times (2,-3) =(4,-6)為 L之法向量 \\(C)\bigcirc: L:2x-3y+1=0 \Rightarrow L上的點可表示成({3t-1\over 2},t),t\in R \Rightarrow ({3\over 2},1)為方向向量; \\(D) \times:\cases{x=1+2t\\ y=1-3t} \Rightarrow 2x-3y+1= 2+4t-3+9t+1=13t \ne 0 \\(E)\times: M:3x-2y+1=0 \Rightarrow M之斜率為{3\over 2} \Rightarrow M斜率\times L斜率 ={3\over 2} \times {2\over 3}\ne -1\\,故選\bbox[red,2pt]{(ABC)}$$


:$$(A) \times: P(A\cup B)= P(A)+P(B)-P(A\cap B)= P(A)+P(B)-P(A)P(B)\\\qquad \Rightarrow {3\over 4}= {1\over 3} +P(B)- {1\over 3} P(B) \Rightarrow P(B)={5\over 8} \ne {5\over 12}\\ (B)\bigcirc: P(A\cap B)=P(A)P(B) ={1\over 3}\times {5\over 8} ={5\over 24} \\(C)\bigcirc: P(A'\cup B')=1-P(A\cap B)=1-{5\over 24} ={19\over 24} \\ (D)\times: P(A\mid B) = {P(A\cap B) \over P(B)} = {P(A)P(B)\over P(B)} =P(A)={1\over 3} \not \gt {1\over 3} \\(E) \bigcirc: P(B)={5\over 8} > {1\over 2}\\,故選\bbox[red,2pt]{(BCE)}$$

40. 某座海島上所有的食物皆由A,B,C 三間海產店供應,根據在地人提供的調查記錄,原本選擇在A店消費的顧客,次月仍選擇在A消費的比例有80%,有10%會改至B店消費,剩餘10%則選擇至C店消費;原本選擇在B店消費的顧客,次月仍選擇在B店消費的比例有50%,有30%會改至A店消費,剩餘20%則選擇至C店消費;原本選擇在C店消費的顧客,次月仍選擇在C店的比例有60%,有20%會改至A店消費,剩餘20%則選擇至B店消費。請問長期而言,下列關於三間海產店市佔率\(P_A,P_B,P_C\)的估計何者正確?

(A) \(P_A> P_B > P_C\)  (B)\(P_A > P_B+P_C\)  (C) \(P_A+P_B+P_C=1\) (D) \(P_A>0.6\)  (E)\( |P_B-P_C| < 0.1\)

:$$\cases{A\rightarrow \cases{0.8A\\ 0.1B\\ 0.1C} \\B\rightarrow \cases{0.8A\\ 0.1B\\ 0.1C} \\ C\rightarrow \cases{0.8A\\ 0.1B\\ 0.1C}} \Rightarrow 轉移矩陣P=\begin{bmatrix}0.8 & 0.3 & 0.2 \\ 0.1 & 0.5 & 0.2 \\ 0.1 & 0.2 & 0.6 \end{bmatrix} \Rightarrow 穩定狀態: PX=X \\ \Rightarrow \begin{bmatrix}0.8 & 0.3 & 0.2 \\ 0.1 & 0.5 & 0.2 \\ 0.1 & 0.2 & 0.6 \end{bmatrix}\begin{bmatrix}A \\ B \\ C \end{bmatrix} = \begin{bmatrix}A \\ B \\ C \end{bmatrix} \Rightarrow \cases{-0.2A+ 0.3B+0.2C=0 \\0.1A-0.5B +0.2C=0 \\ 0.1A+ 0.2B-0.4C=0} \Rightarrow \cases{A=8B/3\\ C=7B/6}\\ 又A+B+C= 1 \Rightarrow {8\over 3}B+B+{7\over 6}B=1 \Rightarrow B={6\over 29} \Rightarrow \cases{A=16/29 \\ B=6/29 \\ C=7/29}\\故選\bbox[red, 2pt]{(BCE)}$$


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