115年警專45期數學科(乙組)詳解

臺灣警察專科學校 115學年度專科警員班第45期
正期學生組新生入學考試乙組數學科

※注意：（一）本科目為單選題，共 40 題，每題 2.5 分，計 100 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:$$斜率={2-3\over 5-1}=-{1\over 4}，故選\bbox[red, 2pt]{(A)}$$

解答:$$f(x)=2x^3+3x^2-1 \Rightarrow f(-1/2)=-{1\over 4}+{3\over 4}-1=-{1\over 2}，故選\bbox[red, 2pt]{(A)}$$

解答:$$|x+2| \lt 4 \Rightarrow -4\lt x+2\lt 4 \Rightarrow -6\lt x\lt 2 \Rightarrow x\in (-6,2)，故選\bbox[red, 2pt]{(D)}$$

解答:$$x^2+y^2-4x+6y-12=0 \Rightarrow (x^2-4x+4)+(y^2+6y+9)-4-9-12=0\\ \Rightarrow (x-2)^2+ (y+3)^2 =25 \Rightarrow 圓心(2,-3)，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cos 67^\circ= \sin(90^\circ-67^\circ)= \sin 23^\circ =x，故選\bbox[red, 2pt]{(A)}$$

解答:$$A+B=2I \Rightarrow B=2I-A= \begin{bmatrix}2& 0\\0&2 \end{bmatrix} - \begin{bmatrix}1&5\\2& 3 \end{bmatrix} = \begin{bmatrix}1&-5\\ -2& -1 \end{bmatrix}，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{\log a=3.1\\ \log b=1.1} \Rightarrow \log a-\log b=\log {a\over b}=2 \Rightarrow {a\over b}=10^2=100 \Rightarrow a=100b，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{x=4^{0.3 } =2^{0.6} \\y=2^{0.4}} \Rightarrow \cases{x^2=2^{1.2} \\xy= 2^1} \Rightarrow x^2最大，故選\bbox[red, 2pt]{(D)}$$

解答:$$3a_{n+1}+a_n=0 \Rightarrow a_{n+1} =-{1\over 3}a_n \Rightarrow a_5=-{1\over 3}a_4 ={1\over 9}a_3 =-{1\over 27}a_2=-{1\over 27}\cdot \left( -{4\over 3} \right)={4\over 81}\\，故選\bbox[red, 2pt]{(A)}$$

解答:$$取出的2個數差絕對值為2的情形:(1,3) ,(2,4)，有2種情況，機率為{2\over C^4_2} ={2\over 6}={1\over 3}，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{a_1=35\\d=5\\ a_n=100} \Rightarrow 100=35+(n-1)5 \Rightarrow n=14 \Rightarrow a_1+a_2+ \cdots +a_{14} =7(2\cdot 35+13\cdot 5) =945\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$x^2+4x+5 =(x+2)^2+1 \ge 1 \Rightarrow 對任意實數x,x^2+4x+5 \ge 0，故選\bbox[red, 2pt]{(A)}$$

解答:$${1\over 3} \left( 3^{3/2} +3^{5/2} +3^{7/2}  \right) = 3^{1/2} +3^{3/2} +3^{5/2} =\sqrt 3+3\sqrt 3+9\sqrt 3=13\sqrt 3，故選\bbox[red, 2pt]{(C)}$$

解答:$$30\times {65\over 100}=19.5 \Rightarrow 第65百分位數就是第20個數字，即40，故選\bbox[red, 2pt]{(D)}$$

解答:$$甲向北移動了90^\circ-20^\circ =70^\circ \Rightarrow 乙向北移動了{70^\circ\over 2}=35^\circ \Rightarrow 90^\circ-35^\circ =55^\circ\\ 乙從北緯55^\circ 向北移動了 35^\circ，故選\bbox[red, 2pt]{(D)}$$

解答:$$晚上11時至隔日早上七時，經過了八小時，時針每一小時旋轉了{\pi\over 6}，共旋轉了8\times {\pi\over 6}={4\pi\over 3}\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$假設\cases{A(0,0) \\B(1,0)\\ C(3,0) \\D(6,0)}  \Rightarrow \cases{(A)\times: \cases{\overrightarrow{AB} =(1,0) \\ \overrightarrow{CD}=(3,0)} \Rightarrow \overrightarrow{AB} \ne 3\overrightarrow{CD} \\(B) \times: \cases{ \overrightarrow{BC} =(2,0)\\ \overrightarrow{BA}=(-1,0)} \Rightarrow \overrightarrow{BC} \ne 2\overrightarrow{BA} \\(C) \bigcirc: \cases{ \overrightarrow{AC}= (3,0)\\ \overrightarrow{DC}=(-3,0)} \Rightarrow \overrightarrow{AC} +\overrightarrow{DC} = \vec 0 \\(D)\times: \overrightarrow{BC}+ \overrightarrow{CA}+ \overrightarrow{CD} =\overrightarrow{BA}+ \overrightarrow{CD} =(-1,0)+(3,0)\ne \vec 0}\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$假設\cases{A(0,0) \\B(2,0) \\C(2,2) \\D(0,2)} \Rightarrow E={C+D\over 2} =(1,2) \Rightarrow \overrightarrow{AE} \cdot \overrightarrow{DB} =(1,2) \cdot (2,-2)=2-4=-2，故選\bbox[red, 2pt]{(A)}$$

解答:$$三個主題排列數:3!=6, 又\cases{防火排列數2!=2\\ 防震排列數2!=2\\ 防空排列數3!=6} ,合計6\times 2\times 2\times 6=144，故選\bbox[red, 2pt]{(B)}$$

解答:$$3正面機率={1\over 2^3}={1\over 8}、2正1反排列數為{3!\over 2!}=3\Rightarrow 機率為{3\over 8} \\ 期望伂=3000\times {1\over 8}+1200\times {3\over 8}+ 600\times \left( 1-{1\over 8}-{3\over 8} \right) =375+450+300=1125，故選\bbox[red, 2pt]{(B)}$$

解答:$$假設\cases{E(0,0,0)\\ H(a,0,0) \\ F(0,b,0)\\ A(0,0,c) \\ G(a,b,0)\\B(0,b,c) \\C(a,b,c) \\D(a,0,c)} \Rightarrow \cases{(A) \times:\overrightarrow{AB} \cdot \overrightarrow{AC} =(0,b,0) \cdot (a,b,c) =b^2\ne 0 \\(B) \times: \overrightarrow{AB} \cdot \overrightarrow{DH} =(0,b,0) \cdot (0,0,-c) =0,但兩直線不相交 \\(C)\times: \overrightarrow{AG} \cdot \overrightarrow{DF} =(a,b,-c) \cdot (-a,b,-c)=-a^2+b^2+c^2 \ne 0 \\(D) \bigcirc: \overrightarrow{AE} \cdot \overrightarrow{EG} =(0,0,-c) \cdot (a,b,0) =0,且兩直線交於E點}\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{a^2=3^2+4^2 \Rightarrow a=5\\b=|3|=3} \Rightarrow a+b=8，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{x-y+1\ge 0\\ x+y-1\ge 0\\ x\le 3} \Rightarrow 所圍區域頂點為\cases{A(0,1) \\B(3,4) \\C(3,-2)} \Rightarrow \triangle ABC面積= {1\over 2} \begin{Vmatrix} 0&1& 1\\3&4& 1\\3&-2& 1 \end{Vmatrix}= 9\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$\begin{bmatrix}6& 5\\4& 3 \end{bmatrix} \begin{bmatrix}p&q\\ r& s \end{bmatrix} = \begin{bmatrix}1&0\\0& 1 \end{bmatrix} \Rightarrow  \begin{bmatrix}p&q\\ r& s \end{bmatrix} = \begin{bmatrix}6& 5\\4& 3 \end{bmatrix}^{-1 }\begin{bmatrix}1&0\\0& 1 \end{bmatrix}  = \begin{bmatrix}\frac{-3}{2} & \frac{5}{2} \\2 & -3 \end{bmatrix} \\ \Rightarrow p=-{3\over 2}，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cos A={3^2+4^2-6^2\over 2\cdot 3\cdot4}=-{11\over 24} \Rightarrow \cos(B+C)=\cos(\pi-A)=-\cos A={11\over 24}，故選\bbox[red, 2pt]{(D)}$$

解答:$$\triangle ABC={1\over 2}\cdot 3 \cdot 4\sin(268^\circ-118^\circ)=6\sin150^\circ =6\cdot {1\over 2}=3，故選\bbox[red, 2pt]{(A)}$$

解答:$$2\sin(3x-{\pi\over 3})+1的週期與\sin(3x)一樣，都是{2\pi\over 3}，故選\bbox[red, 2pt]{(B)}$$

解答:$$L與直線12x+5y=0垂直\Rightarrow L:5x-12y=k \\ (1,k)在直線2x-y=3上\Rightarrow 2-k=3\Rightarrow k=-1 \Rightarrow (1,-1)在L上 \Rightarrow 5+12=k \Rightarrow k=17 \\ \Rightarrow L: 5x-12y=17，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{取球順序為紅-藍的機率:(2/5)\times (3/5)=6/25\\ 取球順序為藍-紅的機率:(3/5)\times (2/5)=6/25 } \Rightarrow {6\over 25}+{6\over 25}= 0.48，故選\bbox[red, 2pt]{(C)}$$

解答:$$點數和小於7的情形:(1,1), (1,2), (1,3),(1,4), (1,5),(2,1),(2, 2), (2,3) ,(2,4),\\(3,1), (3,2), (3,3),(4,1), (4,2),(5,1)，共 15種;其中出現3的有5種，機率為{5\over 15}={1\over 3}，故選\bbox[red, 2pt]{(D)}$$

解答:$${甲命中乙沒命中 \over 甲命中乙沒命中+乙命中甲沒命中} ={0.6\times(1-0.5) \over 0.6\times(1-0.5)+(1-0.6)\times 0.5}= 0.6，故選\bbox[red, 2pt]{(D)}$$

解答:$$假設15秒內\cases{長鳴a次\\ 短鳴b次} \Rightarrow 間隔a+b-1次\Rightarrow 需時2a+b+(a+b-1)=15 \Rightarrow 3a+2b=16 \\ \Rightarrow (a,b)=(0,8), (2,5),(4,2) \Rightarrow \cases{a=0,b=8:排列數1\\ a=2,b=5:排列數{7!\over 2!5!} =21\\ a=4,b=2:排列數{6!\over 4!2!}=15} \Rightarrow 合計:37，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x=1)=3 將x=1代入各選項，只有(D)符合，故選\bbox[red, 2pt]{(D)}$$

解答:$$取g(x)=f(x)+11 \Rightarrow g(-3)=g(5)=0 \Rightarrow g(x)=f(x)+11=a(x+3)(x-5) \\ \Rightarrow y= f(x)=a(x+3)(x-5)-11   =ax^2-2ax-15a-11=a(x-1)^2-16a-11 \\ \Rightarrow 頂點坐標(1,-16a-11)在直線y=2x+3 上5\Rightarrow 5=-16a-11\Rightarrow a=-1 \\ \Rightarrow f(x)=-(x+3)(x-5)-11 \Rightarrow f(0)=15-11=4，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{\vec u=(5,10) \\ \vec v=(-4,2)  } \Rightarrow \cases{|\vec u|=5\sqrt 5\\ |\vec v|=2\sqrt 5\\ \vec u\cdot \vec v=0}\\ \cases{(A) |3\vec u|=3|\vec u|=15\sqrt 5\\ (B)|6\vec v|=6|\vec v|=12\sqrt 5\\ (C) |2\vec u-5\vec v|^2=4|\vec u|^2+25|\vec v|^2=4\cdot 125+25\cdot 20=1000 \Rightarrow |2\vec u-5\vec v|=10\sqrt{10} \\(D)|\vec u+2 \vec v|^2=|\vec u|^2+4|\vec v|^2 =125+80=205 \Rightarrow |\vec u+2\vec v|=\sqrt{205}} \\ \Rightarrow 15\sqrt 5最大\\，故選\bbox[red, 2pt]{(A)}$$

解答:$$|2\vec a+\vec b|^2=(2\vec a+\vec b) \cdot (2\vec a+\vec b)=4|\vec a|^2+4\vec a\cdot \vec b+|\vec b|^2=4\cdot 12^2+0+|\vec b|^2=20^2+15^2=625\\ \Rightarrow |\vec b|^2=49 \Rightarrow |\vec b|=7，故選\bbox[red, 2pt]{(B)}$$

解答:$$滿足(A+B)^2= A^2+2AB +B^2 \Rightarrow AB=BA \Rightarrow \begin{bmatrix}1&-2\\3&-1 \end{bmatrix} \begin{bmatrix}2& a\\b&4 \end{bmatrix}= \begin{bmatrix}2& a\\b&4 \end{bmatrix} \begin{bmatrix}1&-2\\3&-1 \end{bmatrix}  \\ \Rightarrow \begin{bmatrix}2-2b& a-8\\ 6-b& 3a-4 \end{bmatrix} = \begin{bmatrix}2+3a& -4-a\\ b+12& -2b-4 \end{bmatrix} \Rightarrow \cases{a=2\\ b=-3} \Rightarrow a+b=-1，故選\bbox[red, 2pt]{(A)}$$

解答:$$假設公差d \Rightarrow \log 1250=\log 162+4d \Rightarrow 4d=\log{1250\over 162} =\log {625\over 81} =4\log {5\over 3} \Rightarrow d=\log{5\over 3} \\ \Rightarrow \log a=\log 162+d=\log 162+\log {5\over 3} =\log \left( 162\cdot {5\over 3} \right)=\log 270 \Rightarrow a=270，故選\bbox[red, 2pt]{(A)}$$

解答:$$A(2,4)至直線x+y-10=0的距離= {4\over \sqrt 2} \Rightarrow 半徑= \sqrt{ \left( {4\over \sqrt 2} \right)^2+4^2}= \sqrt{24}=2\sqrt 6，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{情況一:甲乙都抽到白球，機率為{4\over 7}\times {3\over 6}={2\over 7} \\情況二:甲乙都抽到黑球，機率為{3\over 7}\times {2\over 6}={1\over 7} } \Rightarrow 甲乙抽到同色球的機率={3\over 7} \\ 甲乙抽到同色球且m丁抽到白球的情形:\\ (甲, 乙,丙,丁)=\cases{(白,白,白,白):機率={4\cdot 3\cdot 2\cdot 1\over 7\cdot 6\cdot 5\cdot 4} \\(白,白,黑,白) :機率={4\cdot 3\cdot 3\cdot 2\over 7\cdot 6\cdot 5\cdot 4} \\ (黑,黑,白,白):機率={3\cdot 2\cdot 4\cdot 3\over 7\cdot 6\cdot 5\cdot 4} \\(黑,黑,黑,白):機率={3\cdot 2\cdot 1\cdot 4\over 7\cdot 6\cdot 5\cdot 4}} \Rightarrow 合計:{8\over 35} \\ \Rightarrow 條件機率={8/35\over 3/7}={8\over 15}，故選\bbox[red, 2pt]{(D)}$$

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