114年警專44期數學科(乙組)詳解

 臺灣警察專科學校 114學年度專科警員班第44期正期學生組新生入學考試乙組數學科試題

本科目為單選題，共 40 題，每題 2.5 分，計 100 分。

解答: $${6^{20}\over 12^{10}} ={(2\cdot 3)^{20}\over (3\cdot 2^2)^{10}} = {2^{20} \cdot 3^{20} \over 3^{10} \cdot 2^{20}} =3^{10}，故選\bbox[red, 2pt]{(D)}$$

解答: $$3x+4y+5=0 \Rightarrow y=-{3\over 4}x-{5\over 4} \Rightarrow 斜率=-{3\over 4}，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{a_1=2\\ r=\sqrt 3} \Rightarrow a_1=2, a_2=2\sqrt 3, a_3=6,a_4=6\sqrt 3, a_5=18, a_6=18\sqrt 3,  a_7=54  \\ \Rightarrow a_1,a_3,a_5,a_7為有理數，共4個，故選\bbox[red, 2pt]{(B)}$$

解答: $$\sigma(X)=\sigma(X+15) \Rightarrow 標準差不變，故選\bbox[red, 2pt]{(D)}$$

解答: $$\sin 30^\circ+\sqrt 2\cos 45^\circ-\cos 120^\circ ={1\over 2}+\sqrt 2\cdot {1\over \sqrt 2}-(-{1\over 2}) =2，故選\bbox[red, 2pt]{(D)}$$

解答: $$f(-2)=0 \Rightarrow x+2為因式，故選\bbox[red, 2pt]{(C)}$$

解答: $$可能的走法:A[BCD]A ,其中BCD排列數=3!=6，故選\bbox[red, 2pt]{(A)}$$

解答: $$A\begin{bmatrix}1 \\0\end{bmatrix} =\begin{bmatrix}2 \\4\\7\end{bmatrix} =\begin{bmatrix}a \\b \\c \end{bmatrix} \Rightarrow a+b+c =2+4+7=13，故選\bbox[red, 2pt]{(C)}$$

解答: $$10^2= (-6)^2+8^2 \Rightarrow (2,-7)+(-6,8) =(-4,1)，故選\bbox[red, 2pt]{(A)}$$

解答: $$阿明命中且阿添不命中+阿添命中且阿明不命中=0.75\times (1-0.8) +(1-0.75)\times 0.8 =0.35\\，故選\bbox[red, 2pt]{(B)}$$

解答: $$|x-\sqrt{15}|\lt 4 \Rightarrow -4\lt x-\sqrt{15}\lt 4 \Rightarrow -4+\sqrt {15} \lt x\lt 4+\sqrt{15} \\ \Rightarrow -0.XX\lt x\lt 7.XX \Rightarrow x=0,1,2,\dots,7，共8個整數，故選\bbox[red, 2pt]{(B)}$$

解答: $$第2,3次為(正,反)或(反正)，機率為{1\over 2}\times {1\over 2}+{1\over 2}\times {1\over 2}={1\over 2} ，故選\bbox[red, 2pt]{(C)}$$

解答: $$圓心=(A+B)/2 ={1\over 2}((1,4,7)+(-5,6,-3)) =(-2,5,2)，故選\bbox[red, 2pt]{(D)}$$

解答: $$f(x)=-2(x+1)^2+3 \Rightarrow \cases{f(10) =-2\cdot 11^2+3\\ f(7)=-2\cdot 8^2+3\\ f(-3)=-2\cdot 2^2+3\\ f(-5)=-2\cdot 4^2+3} \Rightarrow f(-3)最大，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{a_1=1\\a_2= 2\\ a_{n+2} =a_{n+1}-2a_n} \Rightarrow a_3=a_2-2a_1=0 \Rightarrow a_4=a_3-2a_2=-4 \Rightarrow a_5=a_4-2a_3=-4\\故選\bbox[red, 2pt]{(A)}$$


解答: $$\angle C=180^\circ-87^\circ-48^\circ=45^\circ \Rightarrow {\overline{AB} \over \sin \angle C}=2R \Rightarrow {4\sqrt 2\over 1/\sqrt 2}=8=2R \Rightarrow R=4，故選\bbox[red, 2pt]{(C)}$$


解答: $$\cases{a_3+a_5=6\\ a_4+a_6=9} \Rightarrow \cases{a_1r^2+ a_1r^4=6\\ a_1r^3+a_1r^5=9} \Rightarrow 兩式相除{1 \over r }={2\over 3} \Rightarrow r={3\over 2}，故選\bbox[red, 2pt]{(B)}$$

解答: $$圓C: (x-2)^2+y^2=4-k \Rightarrow \cases{圓心O(2,0) \\ 半徑r=\sqrt{4-k}} \\ 圓與直線相切\Rightarrow d(O,L)=r \Rightarrow {6\over 5}=\sqrt{4-k} \Rightarrow k={64\over 25}，故選\bbox[red, 2pt]{(A)}$$


解答: $$假設\cases{首項a=a_1\\ 公差d} \Rightarrow {(2a+6d)7\over 2}=14 \Rightarrow a+3d=2 \\ \Rightarrow a_3+a_5=a+2d+a+4d=2a+6d=2(a+3d)=4，故選\bbox[red, 2pt]{(B)}$$


解答: $$選到男生的機率為p \Rightarrow 選到女生的機率為{5\over 8}p \Rightarrow p+{5\over 8}p=1 \Rightarrow p={8\over 13} \\ \Rightarrow {5\over 8}p={5\over 13}，故選\bbox[red, 2pt]{(A)}$$

解答: $$(A)\times: 算術平均數=(1+2+\cdots +9)/9=45/9=5 \ne 4.5\\ (B)\times: 中位數=排序第5的數=5\\ (C)\times: 全距=9-1=8 \ne 9\\ (D)\bigcirc: 9/4=2.25 \Rightarrow Q_1= 排序第3的數=3\\，故選\bbox[red, 2pt]{(D)}$$

解答: $$\vec u\cdot \vec v=3\times 5+2\times(-2)=15-4=11，故選\bbox[red, 2pt]{(A)}$$

解答: $$f(x)=3(x-2)^3-5x+7 \Rightarrow f'(x)=9(x-2)^2-5 \Rightarrow f''(x)=18(x-2)\\ f''(x)=0 \Rightarrow x=2 \Rightarrow 對稱中心(2,f(2)) =(2,-3)，故選\bbox[red, 2pt]{(C)}$$

解答: $$2000\times {3\over 20}+1000\times {5\over 20}={11000\over 20}=550，故選\bbox[red, 2pt]{(A)}$$


解答: $$\cases{A(-2,2) \\B(1,3) \\C(3,-5)} \Rightarrow \cases{\overrightarrow{AB} =(3,1) \\ \overrightarrow{BC} =(2,-8)} \Rightarrow 2\overrightarrow{AB}-3\overrightarrow{BC}=(6,2)-(6,-24)=(0,26)，故選\bbox[red, 2pt]{(A)}$$

解答: $$\rho(X',Y')={Cov(X',Y') \over \sigma(X')\sigma(Y')} ={Cov(X+5,10Y/9) \over \sigma(X+5)\sigma(10Y/9)} = {{10\over 9}Cov(X,Y) \over {10\over 9}\sigma (X) \sigma(Y)} =\rho(X,Y)=0.63，故選\bbox[red, 2pt]{(B)}$$

解答: $$\sin x:\cases{遞增, x\in [0,\pi/2]\\ 遞減,x\in[\pi/2,3\pi/2]} \Rightarrow \sin({\pi\over 3})到\sin ({11\pi\over 9}):先遞增再遞減，故選\bbox[red, 2pt]{(A)}$$


解答: $$\cases{男女各1人:C^3_1C^4_1=12\\ 男女各2人:C^3_2C^4_2=18\\ 男女各3人:C^3_3C^4_3= 4 } \Rightarrow 合計34種可能，故選\bbox[red, 2pt]{(D)}$$

解答: $$5\theta=10 \Rightarrow \theta =2 弳={360^\circ\over \pi} \approx 114.6^\circ，故選\bbox[red, 2pt]{(B)}$$

解答: $$P(A\mid B) ={P(A\cap B) \over P(B)} \Rightarrow {3\over 4} ={P(A\cap B) \over 3/8} \Rightarrow P(A\cap B)={9 \over 32}，故選\bbox[red, 2pt]{(C)}$$


解答: $$y=f(x)=3^x \Rightarrow 向左平移2單位:f(x+2) =3^{x+2}=9\times 3^x，故選\bbox[red, 2pt]{(C)}$$

解答: $$5個節目有6個間隔讓2個新節目插入\Rightarrow \cases{兩個新節目不相鄰有C^6_2\times 2!=30\\ 兩個新節目相鄰有C^6_1\times 2!=12} \\ \Rightarrow 共有30+12=42種安排，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cases{2\log 2=a\\ \log 3+\log 2=b} \Rightarrow \log 3=b-{1\over 2}a \Rightarrow \log 9=2\log 3=2b-a，故選\bbox[red, 2pt]{(B)}$$


解答: $$\begin{bmatrix}a & 3 \\5 &1 \end{bmatrix}+b \begin{bmatrix}4 & c \\-2 &-1 \end{bmatrix} =\begin{bmatrix}a +4b& 3 +bc\\5-2b &1-b \end{bmatrix} =\begin{bmatrix}-1 & 1 \\7 &2 \end{bmatrix} \Rightarrow \cases{a=3\\ b=-1\\c=2} \Rightarrow a+b+c =4\\，故選\bbox[red, 2pt]{(D)}$$


解答: $$f(x,y)=2x-y-2 \Rightarrow f(A)=f(5,3)=5\gt 0\\ (A) \times:f(0,-2)=0 \\(B)\times: f(6,0)=10\gt 0 \\(C)\bigcirc: f(-1,-1) =-3 \lt 0 \\(D)\times f(3,-4) =8\gt 0\\，故選\bbox[red, 2pt]{(C)}$$



解答: $$(x-2)(x^2-3x-4) =(x-2)(x-4)(x+1)\gt 0 \Rightarrow x\gt 4或 -1\lt x\lt 2，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{抽中甲袋白球的機率={1\over 3}\times {6\over 10}={1\over 5} \\ 抽中乙袋白球的機率={2\over 3}\times {2\over 5} ={4\over 15}} \\ \Rightarrow {抽中甲袋白球的機率\over 抽中甲袋白球的機率+抽中乙袋白球的機率} ={1/5\over 1/5+4/15} ={3\over 7}，故選\bbox[red, 2pt]{(B)}$$

解答: $$地球半徑r \Rightarrow 2\pi r=120 \Rightarrow r={60\over \pi} \Rightarrow 北緯60^\circ 的半徑={60\over \pi }\cos 60^\circ={30\over \pi} \\ \Rightarrow A到B弧長={30\over \pi}\cdot {\pi \over 6} =5，故選\bbox[red, 2pt]{(B)}$$

解答: $$A^{-1}=A \Rightarrow AA=I \Rightarrow \begin{bmatrix}1 & 2 \\a &b \end{bmatrix} \begin{bmatrix}1 & 2 \\a &b \end{bmatrix} =\begin{bmatrix}1+2a & 2+2b \\a+ab &2a+b^2 \end{bmatrix} =\begin{bmatrix}1 & 0 \\0 &1 \end{bmatrix} \Rightarrow \cases{a=0\\ b=-1} \\ \Rightarrow a+b=-1，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{無人機P(0,0,300) \\ A(300\tan 45^\circ,0,0)=(300 ,0,0) \\ B(0,300\tan 60^\circ,0) =(0,300\sqrt 3,0)} \Rightarrow \overline{AB} =600，故選\bbox[red, 2pt]{(D)}$$

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解題僅供參考，其他歷年試題及詳解