113年警專43期數學科(甲組)詳解

臺灣警察專科學校113學年度專科警員班第43期
正期學生組新生入學考試-甲組數學科

解答: $$x\to -\infty \Rightarrow y=-2，故選\bbox[red, 2pt]{(D)}$$

解答: $$\log 2\times 10^{34.5}=34.5+\log 2=34.5+0.301=34.801 \Rightarrow 35位數，故選\bbox[red, 2pt]{(B)}$$

解答: $$\sqrt{10}^{\log 20} =10^{{1\over 2}\log 20} =10^{\log \sqrt{20}} =\sqrt{20}=2\sqrt 5，故選\bbox[red, 2pt]{(B)}$$

解答: $$\log a_4=\log a_1r^3=\log a_1+3\log r=\sqrt{10}+3\log \sqrt{10} =\sqrt{10}+{3\over 2}，故選\bbox[red, 2pt]{(D)}$$

解答: $$A(5,0)對稱於y=x的對稱點A'(0,5) \Rightarrow \overline{PA}+ \overline{PB}的最小值=\overline{A'B}= \sqrt{5^2+12^2}=13，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{10^x=300\\ 10^y=30} \Rightarrow \cases{x=\log 300\\ y=\log 30} \Rightarrow x-2y=\log 300-2\log 30=\log {300 \over 30^2} \\=\log {1\over 3} =-\log 3，故選\bbox[red, 2pt]{(A)}$$

解答: $$圓心P(2,3)至直線x=7的距離=半徑 \Rightarrow r=5\\ 同理P至3x-4y=k的距離={|6-12-k|\over 5}=5 \Rightarrow -6-k=-25 \Rightarrow k=19，故選\bbox[red, 2pt]{(B)}$$

解答: $$圓x^2+y^2-8x+4y=0 \Rightarrow (x-4)^2+(y+2)^2=20 \Rightarrow \cases{圓心O(4,-2)\\ 半徑r=2\sqrt 5} \\ \Rightarrow \cases{\overline{OQ}=r= 2\sqrt 5\\ \overline{OP}= \sqrt{6^2+ 2^2} =2\sqrt {10}} \Rightarrow \overline{PQ}^2= \overline{OP}^2-\overline{OQ}^2= 40-20=20 \\ \Rightarrow \overline{PQ}=2\sqrt 5，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{|\vec a|=4\\ |\vec b|=6 \\ \vec a\cdot \vec b=0} \Rightarrow 取\cases{(x_1,y_1)=(4,0)\\ (x_2,y_2)=(0,6)} \Rightarrow \begin{Vmatrix}x_1& y_1\\ x_2& y_2 \end{Vmatrix} =|x_1y_2-x_2y_1|=|24-0|=24，故選\bbox[red, 2pt]{(D)}$$

解答: $$\cos \theta ={10^2+12^2-10^2 \over 2\cdot 10\cdot 12} ={144\over 240} ={3\over 5}，故選\bbox[red, 2pt]{(A)}$$

解答: $$此題相當於10顆相同的藍球和6顆相同的排球和4顆相同的足球排成一列,\\ 有{20!\over 10!6!4!}排法，故選\bbox[red, 2pt]{(D)}$$

解答: $$\triangle CBB' \sim \triangle CAA' \Rightarrow {\triangle AA'C\over \triangle BB'C} ={d(A,L)^2\over d(B,L)^2} ={64\over 4}=16，故選\bbox[red, 2pt]{(D)}$$

解答: $$A\begin{bmatrix}1 & 0& -3 \\0 & 1& 4 \end{bmatrix} =\begin{bmatrix}2 & 1& x \\3 & 2& y \end{bmatrix} \Rightarrow A=\begin{bmatrix}2 & 1 \\3 & 2 \end{bmatrix} \Rightarrow \begin{bmatrix}x \\y \end{bmatrix} =A\begin{bmatrix}-3 \\4 \end{bmatrix} =\begin{bmatrix}-2 \\-1 \end{bmatrix}，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{{x-2\over -3} ={y+3\over 2} \\ {y+3\over 2}={z-1\over -18}} \Rightarrow \cases{x={-3y-5\over 2} \\ z=-9y-26} 代入x-3y+2z={-3y-5\over 2}-3y-18y-52=4\\ \Rightarrow y=-{13\over 5} \Rightarrow \cases{x=7/5\\ y=-13/5} \Rightarrow (x,y,z)=({7\over 5},-{13\over 5},-{13\over 5})，故選\bbox[red, 2pt]{(B)}$$

解答: $$a_n=a_{n-1}-2=a_{n-2}-4 =\cdots = a_2-2(n-2)=a_1-2(n-1)=9-2(n-1)\\ \Rightarrow a_{27}-a_{37}=9-2\cdot 26-(9-2\cdot 36)=20，故選\bbox[red, 2pt]{(C)}$$

解答: $$\sin(90^\circ-\theta)+ \cos(90^\circ+\theta)+ \cos(180^\circ-\theta)= \cos\theta-\sin \theta-\cos \theta=-\sin \theta=-{3 \over 5}，故選\bbox[red, 2pt]{(A)}$$

解答: $$假設a=k+b,k為整數且k\ge 0 \Rightarrow a^2+10b^2=(k+b)^2+10b^2=k^2+2kb+11b^2=20 \\ \Rightarrow \cases{k=0 \Rightarrow 11b^2=20\Rightarrow b=\sqrt{20\over 11}\gt 1不合\\k=1 \Rightarrow 11b^2+2b-19=0 \Rightarrow b={-1+\sqrt{210} \over 11} \gt 1不合\\ k=2 \Rightarrow 11b^2+4b-16=0 \Rightarrow b={-2+ 6\sqrt 5\over 11} \gt 1不合\\ k=3 \Rightarrow  11b^2+6b-11=0 \Rightarrow b={-3+\sqrt{130}\over 11} \lt 1 \\ k=4 \Rightarrow 11b^2+8b-4=0 \Rightarrow b={-4+ 2\sqrt{15} \over 11} \lt 1 \\k=5 \Rightarrow 11b^2+10b+5=0 \Rightarrow b無實數解} \\ \Rightarrow k=a-b =3或4有兩種可能，故選\bbox[red, 2pt]{(B)}$$

解答: $$f(x)=(3x-6)P(x)+12=2xQ(x)+4=x(x-2)+ax+b\\ \Rightarrow \cases{f(2)=12=2a+b\\ f(0)=4=b} \Rightarrow a=b=4 \Rightarrow 餘式:4x+4，故選\bbox[red, 2pt]{(D)}$$

解答: $$y=3x^3-x-1 \Rightarrow 平移後 y=f(x)=3(x-h)^3-(x-h)-1 \\ 對稱y軸 \Rightarrow f(x)=f(-x) \Rightarrow 3(x-h)^3-(x-h)-1=3(-x-h)^3-(-x-h)-1 \\ \Rightarrow -6hx-x=6xh+x \Rightarrow 2x(6h+1)=0 \Rightarrow h=-{1\over 6}，故選\bbox[red, 2pt]{(B)}$$

解答: $$|x-a|=b \ge 0 \Rightarrow \cases{x=a+b\\ x=a-b} \Rightarrow a-b=1.2，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{\overrightarrow{OA}=(2,5,1)\\ \overrightarrow{OB}= (3,11,3)} \Rightarrow \vec n=\overrightarrow{OA} \times \overrightarrow{OB} =(4,-3,7) \Rightarrow 平面E:4x-3y+7z=0 \\ C在E上 \Rightarrow 8+6+7k=0 \Rightarrow k=-2，故選\bbox[red, 2pt]{(A)}$$

解答: $$x^2+y^2+7x-5y+4=0 \Rightarrow 圓心O(-7/2,5/2) \Rightarrow O=(A+B)\div 2 \\ \Rightarrow B=2O-A=(-7,5)-(0,4)=(-7,1)，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{A(\sqrt 3)\\ B({8\over \sqrt 3+\sqrt 5})=(4(\sqrt 5-\sqrt 3))\\ C(\sqrt 5)} \Rightarrow \overline{AB}: \overline{BC}=4\sqrt 5-5\sqrt 3:-3\sqrt 5+4\sqrt 3 \\=\sqrt{15}:3=\sqrt 5:\sqrt 3，故選\bbox[red, 2pt]{(B)}$$

解答:

$$所圍區域頂點坐標\cases{A(6,0)\\ B(3,1)\\ C(4,0)} \Rightarrow 所圍面積=\triangle ABC={1\over 2}\times 2\times 1=1，故選\bbox[red, 2pt]{(A)}$$

解答: $$f(x)=(x-1)^3+(x-1)^2+2(x-1)+1 \Rightarrow f(0.99)=-0.01^3+0.01^2-0.02+1 \approx 0.98\\，故選\bbox[red, 2pt]{(B)}$$

解答: $$(x^2-3)(x^2-33)\le 0 \Rightarrow 3\le x^2\le 33 \Rightarrow x=\pm 2,\pm 3 ,\pm 4,\pm 5,共8個整數解，故選\bbox[red, 2pt]{(B)}$$

解答: $$f(x)=ax+b \Rightarrow \cases{{f(1113)-f(113)\over 1000}=a={1\over 2}\\ f(-1)=-a+b=1} \Rightarrow \cases{a=1/2\\ b=3/2} \\ \Rightarrow f(1)=a+b=2，故選\bbox[red, 2pt]{(D)}$$

解答: $$a+ar+ar^2+\cdots+ar^7={a(1-r^8)\over 1-r} ={-3(1-256)\over 3 } =255，故選\bbox[red, 2pt]{(A)}$$

解答: $$平均值E(X)={4(60+80)\over 8} =70 \\ E(X^2)={1\over 8}\cdot 4(60^2+ 80^2)=5000 \Rightarrow Var(X)=5000-70^2=100 \Rightarrow \sigma=\sqrt{100}=10，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{A(2,-3,5)\\ B(-3,p,2)\\ C(5,2,q)} \Rightarrow \cases{\overrightarrow{AB}= (-5,p+3,-3)\\ \overrightarrow{AC}=(3,5,q-5)} \Rightarrow {-5\over 3}={p+3\over 5} ={-3\over q-5} \Rightarrow \cases{p=-34/3\\ q=34/5} \\ \Rightarrow (p,q)=(-{34\over 3},{34\over 5})，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{X'=2X-0.1\\ Y'=-Y+0.2} \Rightarrow \cases{\sigma(X')=2\sigma(X)\\ \sigma(Y')=\sigma(Y)\\ Cov(X',Y')=-2Cov(X,Y)} \\ \Rightarrow 0.3={Cov(X,Y)\over \sigma(X) \sigma(Y)} \Rightarrow {Cov(X',Y')\over \sigma(X') \sigma(Y')}=-{Cov(X,Y)\over \sigma(X) \sigma(Y)} =-0.3，故選\bbox[red, 2pt]{(C)}$$

解答: $$\cases{3男3女: C^7_3C^4_3 =140\\ 2男4女:C^7_2C^4_4=21} \Rightarrow 140+21=161，故選\bbox[red, 2pt]{(A)}$$

解答: $$9顆球期望值6 \Rightarrow 編號總和=9\times 6=54\\ \Rightarrow 4\times 2+6\times 3+4k=54 \Rightarrow k=7，故選\bbox[red, 2pt]{(C)}$$

解答: $$假設\angle AOB=\theta \Rightarrow B為A逆時針旋轉\theta \Rightarrow B=\begin{bmatrix}\cos \theta & -\sin \theta \\\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix}3 \\4 \end{bmatrix} =\begin{bmatrix}3/5 & -4/5 \\4/5 & 3/5 \end{bmatrix} \begin{bmatrix}3 \\4 \end{bmatrix} \\=\begin{bmatrix}-7/5 \\24/5 \end{bmatrix} \Rightarrow B=(-{7\over 5},{24\over 5})，故選\bbox[red, 2pt]{(A)}$$

解答: $$\angle ADB+\angle BDC=180^\circ \Rightarrow \sin \angle ADB=\sin \angle BDC \Rightarrow \cases{{7\over \sin \angle ADB}=2r_1\\ {7\over \sin \angle BDC}=2r_2} \\ \Rightarrow r_1=r_2\Rightarrow r_1:r_2=1:1，故選\bbox[red, 2pt]{(A)}$$

解答: $$\angle B=(2\pi-2\cdot {\pi\over 4})\div 2={3\over 4}\pi \Rightarrow \cos B=-{1\over \sqrt 2}={8+9-\overline{AC}^2 \over 12\sqrt 2} \Rightarrow \overline{AC}^2 =29\\ \Rightarrow \overline{AC}=\sqrt {29}，故選\bbox[red, 2pt]{(D)}$$

解答: $$\sin(x+{\pi\over 3})+ \cos(x+{\pi\over 6})={1\over 2}\sin x+{\sqrt 3\over 2}\cos x+ {\sqrt 3\over 2}\cos x-{1\over 2}\sin x =\sqrt 3\cos x\le \sqrt 3\\，故選\bbox[red, 2pt]{(C)}$$

解答:

$$假設正六邊形邊長為1,且E(0,0),如上圖,則\cases{A(-{1\over 2},{3\over 2}\sqrt 3)\\ B(0,\sqrt 3)\\ C(0,2\sqrt 3)\\ D(1,2\sqrt 3)} \Rightarrow \cases{\overrightarrow{DE}=(-1,-2\sqrt 3)\\ \overrightarrow{AB}=({1\over 2},-{1\over 2}\sqrt 3)\\ \overrightarrow{CD}=(1,0)} \\ \Rightarrow (-1,-2\sqrt 3)=x({1\over 2},-{1\over 2}\sqrt 3)+ y(1,0) \Rightarrow (x,y)=(4,-3)，故選\bbox[red, 2pt]{(D)}$$

解答: $$假設O=\overline{BD}與\overline{CE}的交點,並令\cases{O(0,0,0)\\ B(-3/2,-3/2,0)\\ D(3/2,3/2,0)} \Rightarrow A(0,0,\sqrt{46}/2) \\\Rightarrow \cases{\overrightarrow {AB}=(-3/2,-3/2,-\sqrt{46}/2)\\ \overrightarrow{AD}=(3/2,3/2,-\sqrt{46}/2)} \Rightarrow \overrightarrow{AB}\cdot \overrightarrow{AD}=-{9\over 2}+{46\over 4} =7，故選\bbox[red, 2pt]{(C)}$$

解答: $$\begin{vmatrix}2a+3b & b-a \\2c+3d & d-c \end{vmatrix} =\begin{vmatrix}5a & b-a \\5c & d-c \end{vmatrix} =\begin{vmatrix}5a & b \\5c & d \end{vmatrix} = 5 \begin{vmatrix}a & b \\c & d \end{vmatrix} =5\cdot 10=50，故選\bbox[red, 2pt]{(C)}$$

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解題僅供參考,警專歷年試題及詳解