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2023年5月8日 星期一

112年警專42期數學科(甲組)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
新生入學考試甲組數學科試題

壹、單選題:30 題,題號自第 1 題至第 30 題,每題 2 分,計 60 分。

解答:$$令f(x,y)=11x-7y-1 \Rightarrow f(7,11)=77-77-1\lt 0\\(A) \bigcirc:f(6,10)=66-70-1 \lt 0 \Rightarrow f(6,10)\cdot f(7,11)\gt 0 \Rightarrow 同側\\ (B)\times: f(8,12)=88-84-1\gt 0 \Rightarrow f(8,12)\cdot f(7,11) \lt 0 \Rightarrow 異側\\ (C)\times:f(9,9) \gt 0\\ (D)\times: f(11,7)= \gt 0\\ 故選\bbox[red, 2pt]{(A)}$$
解答:$$Y=-{1\over 10}X \Rightarrow \sigma(Y)= |{1\over 10}|\sigma(X) ={a\over 10},故選\bbox[red, 2pt]{(D)}$$
解答:$$15x=2\pi \Rightarrow x={2\over 15}\pi ,故選\bbox[red, 2pt]{(B)}$$
解答:$${(3+2+2)!\over 3!2!2!} ={7!\over 6\cdot 2\cdot 2}={5040\over 24}=210,故選\bbox[red, 2pt]{(A)}$$
解答:$$P={2B+3A\over 5} ={1.6\times 2+(-5.9)\times 3 \over 5}=-{14.5\over 5}=-2.9,故選\bbox[red, 2pt]{(B)}$$
解答:$$10000=100^2 \Rightarrow \sqrt{100^2+n}=a為整數 \Rightarrow 最小的a=101 \Rightarrow 100^2 +a = 101^2 \\ \Rightarrow n=101^2-100^2 = 201,故選\bbox[red, 2pt]{(D)}$$
解答:$$\log_3 0.3={\log 0.3\over \log 3} ={\log {3\over 10} \over \log 3} ={\log 3-1\over \log 3}=1-{1\over \log 3},故選\bbox[red, 2pt]{(D)}$$
解答:$$1-{\sin 79^\circ \times \cos  79^\circ\over \tan  79^\circ} =1-{\sin 79^\circ \times \cos  79^\circ\over \sin  79^\circ/\cos 79^\circ} = 1-\cos^2 79^\circ= \sin^2 79^\circ  \\ =\sin^2(90^\circ -11^\circ) =\cos^2 11^\circ,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{a_3=a_1r^2=3 \\ a_6= a_1r^5=-24} ,兩式相除\Rightarrow {1\over r^3}=-{1\over 8} \Rightarrow r=-2 \Rightarrow a_3=a_1(-2)^2=3 \Rightarrow a_1={3\over 4} \\ \Rightarrow a_1+ a_2+\cdots +a_{10} =a_1(1+r+r^2 +\cdots +r^9) ={3\over 4}\cdot {1-(-2)^{10}\over 1-(-2)} \\ = {3\over 4}\cdot {1-1024\over 3}=-{1023\over 4},故選\bbox[red, 2pt]{(A)}$$
解答:$$兩直線\cases{3x-4y=5\\ 4x+ky=5}平行\Rightarrow {3\over 4}={-4\over k} \Rightarrow k=-{16\over 3} \Rightarrow 兩直線\cases{3x-4y=5\\ 4x-{16\over 3}y=5} \\ \Rightarrow \cases{3x-4y=5\\ 3x-4y={15\over 4}} 距離={5-15/4\over \sqrt{3^2+4^2}} ={5/4\over 5} ={1\over 4},故選\bbox[red, 2pt]{(A)}$$

解答:$$\overline{AB}與\overline{CD}的距離即為立方體邊長=1,故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{A(0,1,-1) \\B(1,-3,-8) \\ C(0,-1,-5)} \Rightarrow \cases{\overrightarrow{AB} =(1,-4,-7)\\ \overrightarrow{AC}= (0,-2,-4)} \Rightarrow \vec n= \overrightarrow{AB} \times \overrightarrow{AC}=(2,4,-2) \\ \Rightarrow \triangle ABC面積={1\over 2}|\vec n|={1\over 2}\sqrt{4+16+4} =\sqrt 6,故選\bbox[red, 2pt]{(A)}$$
解答:$$y=x(3-2x^2) =0 \Rightarrow x=0,\pm\sqrt{3\over 2},又x^3係數為負值 \Rightarrow 圖形為左上右下,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{(x+2)^2 (x-2)^2 =((x+2)(x-2))^2 =(x^2-4)^2= x^4-8x^2+16\\ (x+1)(x-1)=x^2-1}\\ 再利用長除法可得(x+2)^2 (x-2)^2 =(x+1)(x-1)(x^2-7)+9 \Rightarrow 餘式為9,故選\bbox[red, 2pt]{(B)}$$
解答

$$x^2+y^2+8x+12y+25=0 \Rightarrow (x^2+8x+16) +(y^2+12y+36)+25-16-36=0\\ \Rightarrow (x+4)^2+ (y+6)^2 = 27=(3\sqrt 3)^2 \Rightarrow \cases{圓心O(-4,-6) \\半徑r=3\sqrt 3}\\ 過P(-3,-5)作弦\overline{AB},使其\overline{AB} \bot \overline{OP},則\overline{AB}就是最短的弦長\\ 在直角\triangle APO中,\overline{OP}=\sqrt{1^2+1^2}=\sqrt 2 \Rightarrow \overline{AP}= \sqrt{r^2-\overline{OP}^2} =\sqrt{27-2}=5\\ \Rightarrow \overline{AB}=2\overline{AP}= 10,故選\bbox[red, 2pt]{(B)}$$
解答:$$y=f(x-1)=f(x)+3 \Rightarrow a(x-1)+b = ax+b+3 \Rightarrow ax-a+b=ax+b+3\\ \Rightarrow -a+b=b+3 \Rightarrow a=-3,故選\bbox[red, 2pt]{(B)}$$
解答


$$令\cases{L_1:y=0\\ L_2:x+2y=4\\ L_3:x-y=1} ,則所圍區域各頂點坐標\cases{A=L_1\cap L_3=(1,0)\\B=L_2\cap L_3=(2,1) \\ C=L_1\cap L_2 = (4,0)} \\ \Rightarrow \triangle ABC面積={1\over 2} \cdot 3\cdot 1=1.5,故選\bbox[red, 2pt]{(A)}$$
解答:$$甲被選中後剩下九人,九人中選中乙或丙的機率都是{1\over 9},因此機率為{1\over 9}+ {1\over 9}={2\over 9},故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{a_1=1\\ a_n= 3a_{n-1}+1 } \Rightarrow a_2=4 \Rightarrow a_3=13 \Rightarrow a_4=40 \Rightarrow a_5=121 \Rightarrow a_6=364,故選\bbox[red, 2pt]{(C)}$$


解答:$$|x-10\pi |\le \pi \Rightarrow -\pi \le x-10\pi \le \pi  \Rightarrow 9\pi \le x\le 11\pi \\ \Rightarrow 28.3 \le x\le 34.5 \Rightarrow x=29,30,31,32,33,34 \Rightarrow 共六個解,故選\bbox[red, 2pt]{(C)}$$
解答:$$f(x)= (x-1)(x+1)(x+3)+5 =(x^2-1)(x+3)+5 =x^3+3x^2-x+2 \\ \Rightarrow f'(x)=3x^2+6x-1 \Rightarrow f''(x)=6x+6\\ 若f''(x)=0 \Rightarrow x=-1 \Rightarrow 對稱中心(-1,f(-1))=P(-1,5) \\ 又f'(-1)=-4 \Rightarrow 通過P且斜率為-4的直線:y=-4(x+1)+5=-4x+1,故選\bbox[red, 2pt]{(D)}$$
解答:$$ \begin{bmatrix} a & b\\ c& d\end{bmatrix}  \begin{bmatrix} 12 & 24\\ 36& 48\end{bmatrix} =  \begin{bmatrix} 0 & 2\\ 3& 0 \end{bmatrix}  \Rightarrow 24a+48b=2 \Rightarrow 12a+24b=1   \\  因此 \begin{bmatrix} a & b\\ c& d\end{bmatrix}  \begin{bmatrix} 12 & 24\\ 24& 72\end{bmatrix} = \begin{bmatrix} 12a+24b=1 & ? \\ ? & ?\end{bmatrix} ,故選\bbox[red, 2pt]{(A)}$$
解答:$$\triangle ABC= \triangle ABD+ \triangle ACD \Rightarrow {1\over 2}\cdot 3\cdot 4\sin 120^\circ ={1\over 2}\cdot 3\cdot \overline{AD}\sin 60^\circ +{1\over 2}\cdot 4\cdot \overline{AD} \sin 60^\circ\\ \Rightarrow 3\sqrt 3= {3\over 4}\sqrt 3\overline{AD}+ \sqrt 3 \overline{AD} \Rightarrow \overline{AD}={12\over 7},故選\bbox[red, 2pt]{(B)}$$
解答:$$\overrightarrow{OP} =\overrightarrow{OB}+t\vec v=(2,-1)+(-t,3t) =(2-t,-1+3t) \Rightarrow P(2-t,-1+3t) \\ \Rightarrow \overrightarrow{AP} =(-t-13,-6+3t);\\由於\overrightarrow{OP} \bot \overrightarrow{AP} \Rightarrow \overrightarrow{OP} \cdot \overrightarrow{AP} = (2-t,-1+3t) \cdot (-t-13,-6+3t) =0\\ \Rightarrow 10t^2-10t-20=0 \Rightarrow 10(t-2)(t+1)=0 \Rightarrow \cases{t=2\\t=-1} \Rightarrow \cases{P=(0,5)\\ P=(3,-4)} \\ \Rightarrow \cases{\overline{OP}=5 \\ \overline{AP}= 15}\Rightarrow \triangle OAP ={1\over 2}\cdot \overline{OP}\cdot \overline{AP}={1\over 2}\cdot 5\cdot 15={75\over 2},故選\bbox[red, 2pt]{(C)}$$
解答:$$A(0,0,1)\in L:{x\over 2}={y\over 2}={z-k\over -1} \Rightarrow k=1,因此P\in L \Rightarrow P(2t,2t,-t+1),t\in \mathbb R \\ 將P代入E\Rightarrow 2t+ 4t-2(1-t)=14\Rightarrow t=2 \Rightarrow P(4,4,-1) \Rightarrow \overline{AP}= \sqrt{4^2+4^2 +2^2} =6\\,故選\bbox[red, 2pt]{(B)}$$
解答:$$x^2+y^2-4x-6y-12=0 \Rightarrow (x-2)^2 +(y-3)^2 =5^2 \\ 令\cases{x=5\cos \theta+2 \\ y=5\sin \theta+3} \Rightarrow 3x-4y=15\cos \theta-20\sin \theta-6 =25({3\over 5}\cos \theta-{4\over 5}\sin \theta)-6 \\ =25\sin(\theta+\alpha)-6 \Rightarrow |3x-4y|最大值=|-25-6|=31,故選\bbox[red, 2pt]{(C)}$$
解答:$$500\cdot 2^{n/2}\gt 10^5 \Rightarrow 2^{n/2 }\gt 200 \Rightarrow {n\over 2}\ge 8 \Rightarrow n\ge 16,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{E法向量\vec n=(2,-4,5)\\ L方向向量\vec u=(-2,1,0)},假設 \vec n與\vec u夾角\phi \Rightarrow \cos \phi={\vec n\cdot \vec u\over |\vec n||\vec u|} ={-8\over 15}\\ \phi 與\theta互餘\Rightarrow \sin \theta =|\cos \phi|={8\over 15},故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{\triangle OBC/\triangle OAB=4\\ \triangle OAC/\triangle OAB=2} \Rightarrow {\triangle ABC\over \triangle OAB} =4+2-1=5,故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{P\in L_1 \Rightarrow P(-3+t,1+2t,1+2t) \\ Q\in L_2 \Rightarrow Q(s,1+2s,-5+2s)} \Rightarrow \overline{PQ}^2 =(s-t+3)^2 +(2s-2t)^2+(2s-2t-6)^2\\ 取x=s-t,則f(x)=\overline{PQ}^2 =(x+3)^2 +(2x)^2 +(2x-6)^2 = 9x^2-18x+45\\ \Rightarrow 當x=1時,\overline{PQ}^2有最小值9-18+45=36 \Rightarrow \overline{PQ}=6,故選\bbox[red, 2pt]{(B)}$$

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:$$f(x)=(2x-3)^{12} =\sum_{k=0}^{12}C^{12}_k(2x)^k(-3)^{12-k} \\(A)\bigcirc:常數項=(-3)^{12}=3^{12} \\(B)\times: 有x^{12},x^{11},\dots,x,常數項,共13類\\ (C)\bigcirc: x^9係數=C^{12}_9\cdot 2^9 \cdot(-3)^3 \\(D)\bigcirc: f(1)=(-1)^{12}=1 \\(E)\bigcirc: g(x)=f(2x)= (4x-3)^{12} \Rightarrow g(1)=1\\,故選\bbox[red, 2pt]{(ACDE)}$$

解答:$$\begin{array}{} & X & Y &X^2 & Y^2 & XY\\\hline & 1& 1& 1& 1& 1\\ & 3 & 4& 9 & 16 & 12 \\ & 3 & 3& 9 & 9 & 9\\ & 4 & 2& 16 & 4& 8\\ & 5 & 3& 25& 9 & 15\\ & 5 & 2& 25& 4& 10\\ & 7 & 1& 49 &1 & 7 \\\hline \sum &28 & 16 & 134 & 44 & 62\end{array} \\(A)\times: \cases{\mu_x =28/7=4\\ \mu_y=16/7   } \Rightarrow \mu_X\gt \mu_Y \\(B) \times:\cases{E(X^2)=134/7 \Rightarrow \sigma_X=\sqrt{22\over 7} \\E(Y^2)=44/7 \Rightarrow \sigma_Y=2\sqrt{13}/7}\Rightarrow \sigma_X \gt \sigma_Y \\(C)\bigcirc: X越大Y越小\Rightarrow r\lt 0 \\(D) \times: 斜率變小,相關係數變得更小\\(E) \bigcirc: 斜率變正值(6\times 55-21\times 15 \gt 0)\\,故選\bbox[red, 2pt]{(CE)}$$
解答:$$\begin{bmatrix} 0 & 1\\ 1& 0\end{bmatrix} \begin{bmatrix} a\\ b\end{bmatrix} = \begin{bmatrix} 2\\ 1\end{bmatrix} \Rightarrow \cases{a=1\\ b=2} \\(A)\bigcirc: A(2,1),B(1,2)對稱於x=y \\ (B) \times:\overrightarrow{OA} \cdot \overrightarrow{OB} =(2,1)\cdot (1,2)=4\ne 0 \\(C) \times: B(1,2)在第一象限\\ (D) \bigcirc: \overline{OA}=\sqrt 5 =\overline{OB} \\(E) \times: \overleftrightarrow{AB}斜率={2-1\over 1-2}=-1 \ne -2\\,故選\bbox[red, 2pt]{(AD)}$$
解答:$$(A) \bigcirc: \cases{f(\pi/6)= \cos (\pi/6)=\sqrt 3/2 \\ g(\pi/6) =\cos(-\pi/6) =\sqrt 3/2} \\(B)\bigcirc: g(x)為f(x)的平移,兩者週期相同,f+g週期仍是2\pi \\ (C)\bigcirc: f+g=\cos x+\cos (x-\pi/3) =2 \cos(x-\pi/6)\cos (\pi/6) =\sqrt 3\cos(x-\pi/6)最大值=\sqrt 3 \\(D) \times: f-g週期仍為2\pi \\ (E)\times: f-g=\cos x-\cos (x-\pi/3) =-2\sin(x-\pi/6)\sin \pi/6=-\sin(x-\pi/6)最大值1\\,故選\bbox[red, 2pt]{(ABC)}$$
解答:$$(A) \bigcirc: a_{10}=S_{10}-S_9=10^2-9^2 =19 \\(B) \bigcirc: a_n= S_n-S_{n-1} =n^2 -(n-1)^2=2n-1為正奇數 \\ (C) \bigcirc: \Rightarrow a_n=2n-1 \Rightarrow 公差為2的等差數列 \\(D)\times: a_{10}+a_{11}+\cdots+ a_{20} =S_{20}-S_9=20^2-9^2=319\ne 300 \\ (E)\times: a_2+a_4 +\cdots +a_{20}= \sum_{k=1}^{10}a_{2k}= \sum_{k=1}^{10} 4k-1=4\cdot 55-10=210\\,故選\bbox[red, 2pt]{(ABC)}$$
解答:$$\begin{vmatrix}12 & 13\\ 22& 23 \end{vmatrix}+ 2\begin{vmatrix}11 & 13\\ 21& 23 \end{vmatrix} +3\begin{vmatrix}11 & 12\\ 21& 22 \end{vmatrix} =-10+2\cdot (-20)+3\cdot (-10)=-80\\(A)\times: 2應該為(-2) \\(B)\bigcirc: 正負號正確\\(C)\bigcirc: 剛好也是-10+2\cdot (-20)+3\cdot (-10)\\(D)\times: 其值為2+2\cdot 1+3\cdot 1,顯然錯誤\\ (E)\times: 正負號剛好相反\\,故選\bbox[red, 2pt]{(BC)}$$
解答:$$(A)\times: f(\alpha)=6 \Rightarrow 2^{-\alpha}=6 \Rightarrow -\alpha=\log_2 6 \Rightarrow \alpha\lt 0 \\(B)\times: f(2\alpha)= 2^{-2\alpha} =(2^{-\alpha})^2= 6^2=36 \ne 12 \\(C)\times: \cases{2^{-\alpha}=6 \Rightarrow 2^\alpha =1/6\\ 2^{-\beta}=12} \Rightarrow 2^{\alpha-\beta} =2 \ne{1\over 2} \\(D) \bigcirc:f({\alpha+\beta\over 2}) =2^{-(\alpha+ \beta)/2} =(2^{-(\alpha+\beta)})^{1/2} =\sqrt{6\cdot 12} \lt \sqrt{81}=9\\ (E)\bigcirc: \overline{AB}斜率={12-6\over \beta-\alpha} ={6\over 1}=6\\,故選\bbox[red, 2pt]{(DE)}$$
解答:$$(A)\bigcirc: 第2次只能是5,機率為{1\over 6} \\(B)\times: 期望值={1\over 36}(2\cdot 1+ 3\cdot 2+ 4\cdot 3+ 5\cdot 4+ 6\cdot 5+ 7\cdot 6+ 8\cdot 5+ 9\cdot 4+ 10\cdot 3+ 11\cdot 2+ 12\cdot 1) \\={253\over 36} \ne 6 \\(C) \bigcirc:{6\over 36}={1\over 6} \\(D)\bigcirc: 都是{3\over 36} \\ (E)\times: 乘積為奇數必須兩數皆為奇數,而乘積為偶數只兩數之一為偶數即可,因此偶數奇率較大\\,故選\bbox[red, 2pt]{(ACD)}$$

解答:$$(A) \times:{1\over 3}+{1\over 2}\lt 1 \\(B)\bigcirc: {2\over 3}+{1\over 2} \gt 1\\ (C)\times: -{1\over 2}+{2\over 3} \lt 1\\ (D)\bigcirc:\cos\theta+\sin \theta=\sqrt 2 \sin(\theta+45^\circ)可能\gt 1\\ (E) \bigcirc: \log 3+\log 5= \log 15 \gt 1\\,故選\bbox[red, 2pt]{(BDE)}$$
解答:$$(A) \bigcirc: \cases{\theta_3=60^\circ\\ \theta_6=120^\circ} \Rightarrow \sin\theta_3=\sin \theta_6={\sqrt 3\over 2} \\(B)\times: \theta_8 = 135^\circ \Rightarrow \cases{\cos 135^\circ=-1/\sqrt 2\\ \sin 135^\circ =1/\sqrt 2} \Rightarrow \cos \theta_8 \ne \sin \theta_8 \\(C) \times:\tan 135^\circ =-1\lt \cos 135^\circ=-1/\sqrt 2\\ (D)\bigcirc: 90^\circ\lt \theta_7 \lt \theta_8 \lt 180^\circ \Rightarrow \cos \theta_8\lt \cos \theta_7 \\(E)\times: 90^\circ\lt \theta_7 \lt \theta_8 \lt 180^\circ \Rightarrow \tan \theta_8 \gt \tan \theta_7\\,故選\bbox[red, 2pt]{(AD)}$$

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