臺灣警察專科學校專科警員班二十九期（正期學生組）
新生入學考試甲組數學科試題
新生入學考試甲組數學科試題
壹、單選題
解：$$a_2+a_3+a_{10}+a_{11} =(a_1+d)+(a_1+2d)+ (a_1+9d) +(a_1+10d)= 4a_1+ 22d \\= 2(a_1+11d) = 2(a_1+5d+a_1+6d) = 2(a_6+a_7) =48 \Rightarrow a_6+a_7 = 24，故選\bbox[red,2pt]{(D)}$$
解：$$f(x)=-x^2+2x-5 = -(x^2-2x+1)-4 =-(x-2)^2-4 \Rightarrow 最大值為f(2)=-4，故選\bbox[red,2pt]{(C)}$$
解：$$g(x)=f(f(x)) = (x^3-2x^2-x+5)^3-2(x^3-2x^2-x+5)^2-(x^3-2x^2-x+5)+5 \\ \Rightarrow g(1)=3^3-2\times 3^2-3+5=27-18-3+5=11，故選\bbox[red,2pt]{(D)}$$
解：
$$由圖形可知:m_2> m_1 > m_3 > m_4，故選\bbox[red,2pt]{(B)}$$
解：$$\log_3 54 + \log_3 6-2\log_3 2 =\log_3(54\times 6\div 2^2)= \log_3 81=4，故選\bbox[red,2pt]{(A)}$$
解：$$\cases{a=3-1=2 \\ a^0=b-1 \Rightarrow 1=b-1 \Rightarrow b=2 \\a^c=9-1=8 \Rightarrow 2^c=8 \Rightarrow c=3} \Rightarrow a+b+c = 2+2+3=7，故選\bbox[red,2pt]{(A)}$$
解：$$\cases{\log x的首數=\log 2468的首數=3 \Rightarrow x是四位數\\ \log x的尾數= \log 0.1357的尾數} \Rightarrow x=0.1357\times 10^4= 1357，故選\bbox[red,2pt]{(D)}$$
解：$$\cos (-1680^\circ) =\cos 1680^\circ = \cos (360^\circ\times 4+240^\circ) = \cos 240^\circ =-\cos 60^\circ =-{1\over 2}，故選\bbox[red,2pt]{(B)}。 $$
解：$$利用正弦定理\Rightarrow \cases{\triangle ADC中: {\overline{DC} \over \sin 30^\circ}=2R \\ \triangle ABC中: {\overline{AB} \over \sin 45^\circ}=2R} \Rightarrow {4\over \sin 30^\circ } =2R= {\overline{AB} \over \sin 45^\circ} \\\Rightarrow \overline{AB} =4\times {\sin 45^\circ \over \sin 30^\circ} =4\sqrt 2，故選\bbox[red,2pt]{(C)}。$$
解：
$$此二點的x坐標對稱於x={3\over 2}\pi，因此和為2\times {3\over 2}\pi =3\pi，故選\bbox[red, 2pt]{(D)}$$
解：$$\cases{\cos \theta =3/5>0 \\ \pi < \theta < 2\pi} \Rightarrow {3\over 2}\pi < \theta < 2\pi \Rightarrow {3\over 4}\pi < {\theta \over 2}< \pi \Rightarrow \cos {\theta \over 2} < 0\\ \Rightarrow \cos \theta= \cos({\theta \over 2}+{\theta \over 2})=2\cos^2{\theta \over 2}-1= {3\over 5} \Rightarrow \cos{\theta \over 2} =- {2\over \sqrt 5}，故選\bbox[red, 2pt]{(B)}$$
解：$${z_2\over z_1} ={4\over 2}(\cos \angle AOB+i\sin \angle AOB) = 2(\cos 60^\circ +i\sin 60^\circ)，故選\bbox[red, 2pt]{(C)}$$
解：
$$\left| { -8-9+2\over \sqrt{4^2+(-3)^2}} \right| ={15 \over 5}=3，故選\bbox[red,2pt]{(B)}。$$
解：
$$3\overrightarrow{AD} =2\overrightarrow{AB} +\overrightarrow{AC} \Rightarrow \overrightarrow{AD} ={2\over 3} \overrightarrow{AB} +{1\over 3}\overrightarrow{AC} \Rightarrow \overline{BD}:\overline{DC}= 1:2 \Rightarrow {\triangle ABD \over \triangle ABC} = {\overline{BD} \over \overline{BC}} ={1\over 3}\\，故選\bbox[red,2pt]{(A)}。$$
解：$$\cases{A(1,2,1) \\B(0,-1,1) \\C(-1,0,0)} \Rightarrow \cases{\overrightarrow{AB} =(-1,-3,0) \\ \overrightarrow{AC} =(-2,-2,-1)} \Rightarrow \vec n= \overrightarrow{AB} \times \overrightarrow{AC} =(3,-1,-4)，故選\bbox[red,2pt]{(B)}$$
解：$$x^2+y^2 +2(m+2)x-2(m+3)y +3m^2+2=0 \\\Rightarrow (x+m+2)^2+(y-(m+3))^2= (m+2)^2+(m+3)^2-3m^2-2 \\ \Rightarrow (x+m+2)^2+(y-(m+3))^2=-(m-5)^2+36最大值為36 \\\Rightarrow 半徑最大值為\sqrt{36}=6，故選\bbox[red,2pt]{(C)}$$
解：$$x^2+y^2-6x +2ay+b=0 \Rightarrow (x-3)^2+(y+a)^2=9+a^2-b \Rightarrow \cases{圓心O(3,-a) \\ 半徑r=\sqrt{a^2-b+9}}\\ 又\cases{切點P(4,1)\\ 直線L:y={x\over 2}-1} \Rightarrow \overline{OP}=\text{dist}(O,L) \Rightarrow \sqrt{1+(a+1)^2} = \left| { 2a+1\over \sqrt 5}\right| \\ \Rightarrow (a+1)^2+1 ={(2a+1)^2\over 5} \Rightarrow a^2+6a+9=0 \Rightarrow (a+3)^2=0 \rightarrow a=-3;\\另，P在圓上\Rightarrow 16+1-24+2a+b=0 \Rightarrow b=13 \Rightarrow a+b=13-3=10，故選\bbox[red, 2pt]{(A)}$$
解：$$圓心(2,3,0) \Rightarrow 球心(2,3,a) \Rightarrow 球方程式:(x-2)^2 +(y-3)^2+(z-a)^2=r^2;\\ (6,6,4)在球上 \Rightarrow 4^2+3^2+(4-a)^2=r^2 \Rightarrow r^2=a^2-8a+41 \cdots(1); \\又球心、圓心與球圓相交處呈直角三角形，即r^2=a^2+(\sqrt{17})^2\cdots(2);\\ 由(1)與(2)可得a^2-8a+41=a^2+17 \Rightarrow a=3 \Rightarrow r^2=3^2+17=26 \Rightarrow r=\sqrt{26}，故選\bbox[red,2pt]{(B)}$$
解：$$C^n_2=55 \Rightarrow {n(n-1)\over 2} =55 \Rightarrow n^2-n-110=0 \Rightarrow (n-11)(n+10)=0 \Rightarrow n=11，故選\bbox[red,2pt]{(D)}$$
解：
$$\cases{取到2個紅球:C^3_2=3 \\ 取到2個黃球:C^5_2=10 \\ 取到2個白球:C^2_2=1\\ 取到2個球:C^{10}_2=45} \Rightarrow 取到2個同色球的機率={3+10+1 \over 45}= {14\over 45}，故選\bbox[red,2pt]{(C)}$$
解：$$a_1=1 \Rightarrow a_2={3-1 \over 4-1}={2\over 3} \Rightarrow a_3={2-1 \over 5/3} ={3\over 5} \Rightarrow a_4={4/5 \over 7/5}={4\over 7} \\ \Rightarrow a_n={n\over 2n-1} \Rightarrow a_8={8\over 15}，故選\bbox[red,2pt]{(C)}$$
解：$${1\over x}< x \Rightarrow {1\over x}-x <0 \Rightarrow {1-x^2 \over x}< 0 \Rightarrow x(1-x^2)< 0 ，故選\bbox[red,2pt]{(C)}$$
解：$$和為偶數:(1,3),(1,5),(1,7), (1,9),(2,4),(2,6),(2,8), (3,5),(3,7),(3,9),\\(4,6),(4,8),(5,7),(5,9),(6,8),(7,9)，共有16個，其中兩數皆為偶數的有6個\\，因此機率為6/16=3/8，故選\bbox[red,2pt]{(A)}。$$
解：$$\begin{vmatrix} 1 & 2 & 33 \\ -2 & 0 & 8\\ 14 & 28 & 7\end{vmatrix} =\begin{vmatrix} 1 & 0 & 33 \\ -2 & 4 & 8\\ 14 & 0 & 7\end{vmatrix} = 28-4\times 14\times 33 = 28 (1-66) = -28\times 65= -1820\\，故選\bbox[red,2pt]{(B)}$$
解：$$A=\begin{bmatrix} 3 & -1 \\ 2 & 1\end{bmatrix} \Rightarrow det(A)=3+2=5 \Rightarrow c=-2/det(A)=-2/5，故選\bbox[red,2pt]{(D)}$$
解：$$\lim_{x\to a^+}f(x) \ne \lim_{x\to a^-}f(x)，故選\bbox[red, 2pt]{(D)}$$
解：$$f(x)=x^3-kx^2+x-3 \Rightarrow f'(x)= 3x^2-2kx+1 \Rightarrow f''(x)=6x-2k \Rightarrow f''(-2)=0\\ \Rightarrow -12-2k=0 \Rightarrow k=-6，故選\bbox[red,2pt]{(D)}$$
解：$$f(x)=x^3-3x^2-x+2 \Rightarrow f'(x)= 3x^2-6x-1 \Rightarrow f'(1)=3-6-1=-4，故選\bbox[red,2pt]{(C)}$$
解：$$\int_{-1}^3(3x^2-4x)\;dx = \left. \left[ x^3-2x^2\right] \right|_{-1}^3 =(27-18)-(-1-2) =9+3=12，故選\bbox[red,2pt]{(C)}$$
解：$$\int_1^2 (f(x))^2\pi \;dx = \pi\int_1^2(2x+1)\;dx = \pi \left .\left[ x^2+x\right] \right|_1^2 =\pi(6-2)=4\pi，故選\bbox[red,2pt]{(D)}$$
貳、多重選擇題
解：$$f(x)=x^3+ax^2 +bx+c \Rightarrow \cases{f(-1)=-1 \\ f(2)=2 \\ f(4)=4} \Rightarrow \cases{-1+a-b+c=-1 \\ 8+4a+2b+c=2 \\ 64+16a+4b+c=4} \\ \Rightarrow \cases{a-b+c=0 \\ 4a+2b+c=-6 \\ 16a+4b+c=-60} \Rightarrow \cases{a=-5 \\ b=3 \\c=8} \Rightarrow f(x)=x^3-5x^2+3x+8 \Rightarrow \cases{f(0)=8>0 \\f(1)=7 > 0 \\f(3)=-1< 0\\ f(5)=23 > 0}\\ \Rightarrow \cases{(A)f(-\infty)f(0) < 0有實根 \\ (B)f(0)f(1) >0 \\(C) f(2)f(3)<0 有實根\\ (D) (2,3)有實根\Rightarrow (2,4)有實根 \\ (E)f(5)f(\infty)>0}$\\，故選\bbox[red,2pt]{(ACD)}$$
解：$$(A)\times: \tan \theta= 4/3 \\(B) \bigcirc:第3象限，\sin \theta <0 \\(C) \bigcirc:\cos (\theta+180^\circ) = -\cos \theta=3/5 \\(D) \bigcirc: \sin (90^\circ+\theta)=\cos \theta =-3/5 \\(E)\times: \sin(360^\circ+\theta) = \sin \theta= -4/5\\，故選\bbox[red,2pt]{(BCD)}$$
解：$$(A) \bigcirc: \vec a-2\vec b=(2,-3)-(8,16) =(-6,-19) \\(B) \times:\vec a\cdot \vec c = (2,-3)\cdot (2,-1)=4+3=6 \ne 1 \\(C) \bigcirc: \vec b\cdot \vec c=(4,8) \cdot (2,-1)=0 \Rightarrow \vec b\bot \vec c \\(D) \bigcirc: 依定義\\ (E) \bigcirc:\cos \theta ={\vec a\cdot \vec b\over |\vec a||\vec b|} ={8-24 \over \sqrt{13}\times \sqrt{80}}<0 \Rightarrow \theta > 90^\circ\\，故選\bbox[red,2pt]{(ACDE)}$$
解：$$\sqrt{(x-1)^2+(y-2)^2} +\sqrt{(x+1)^2+(y+2)^2} =6 \Rightarrow \cases{焦點F_1(1,2)\\ 焦點F_2(-1,-2)\\ 2a=6}\\ (A) \bigcirc: 中心位於\overline{F_1F_2}的中點\Rightarrow 中點坐標({1-1\over 2},{2-2\over 2})=(0,0) \\(B)\bigcirc: (1,2)及(-1,-2)為其兩焦點\\ (C)\bigcirc: 2a=6為長軸長\\ (D) \times: \overline{F_1F_2}不垂直也不在直線x=y上 \\(E)\bigcirc: 橢圓對稱兩焦點的連線\\，故選\bbox[red,2pt]{(ABCE)}$$
解：
$$假設D為坐標原點(0,0,0)，各頂點坐標如上圖；\overline{OD}=2 \Rightarrow \sqrt{1+1+a^2}=2 \Rightarrow a=\sqrt 2\\(A)\times: \cases{ \overrightarrow{AC} =(2,-2,0) \\\overrightarrow{BD} =(-2,-2,0)} \Rightarrow \overrightarrow{AC} \cdot\overrightarrow{BD} =-4+4=0 \ne 8 \\(B) \bigcirc: \cases{\overrightarrow{OA}-\overrightarrow{OC} =(-1, 1, -\sqrt 2)- (1,-1, -\sqrt 2) = (-2,2,0) \\\overrightarrow{CB}+\overrightarrow{CD} =(0,2,0)+(-2,0,0)=(-2,2,0)}\Rightarrow 兩者相同 \\(C) \bigcirc: \cases{\overrightarrow{OA}\cdot \overrightarrow{OD} =(-1, 1, -\sqrt 2)\cdot (-1,-1, -\sqrt 2) = 2 \\\overrightarrow{OB} \cdot \overrightarrow{OC} =(1,1,-\sqrt 2) \cdot (1,-1,-\sqrt 2)= 2}\Rightarrow 兩者相同\\(D) \times:\cases{\vec u=\overrightarrow{OA}\times \overrightarrow{OB} =(-1, 1, -\sqrt 2)\times (1,1, -\sqrt 2) = (0,-2\sqrt 2,-2) \\ \vec v=\overrightarrow{OB} \times \overrightarrow{OC} =(1,1,-\sqrt 2 \times (1,-1,-\sqrt 2)= (-2\sqrt 2,0,-2)} \\ \qquad \Rightarrow \cos \theta ={\vec u\cdot \vec v \over |\vec u||\vec v|} ={4\over 12} ={1\over 3} \Rightarrow \theta 為銳角 \\(E)\bigcirc: a=\sqrt 2\\，故選\bbox[red,2pt]{(BCE)}$$
解：$$(A)\times: 含糖量平均約為40，最大值不到70，最小值為30，與平均值的差距都不到40，因此標準差小於40\\ (B)\bigcirc: 低於50的有6個樣本，剛好50的有5個樣本，因此中位數落在50\\ (C)\bigcirc: 由圖形知:熱量越高，則含糖量越高，兩者為正向關\\ (D)\times: 正比必須剛好一直線\\ (E)\bigcirc: 正相關\Rightarrow 斜率大於0\\故選\bbox[red,2pt]{(BCE)}$$
解：$$(A)\bigcirc: 符合算幾不等式{a+b\over 2} > \sqrt{ab}(此時a\ne b) \\(B)\bigcirc: \cases{\sqrt{10}+\sqrt{20} \approx 3.X+4.X =7.X\\ \sqrt{30}=5.X} \Rightarrow \sqrt{10}+\sqrt{20} > \sqrt{30} \\(C) \bigcirc: \log 10+\log 20= \log 200 > \log 30 \\(D) \bigcirc: 理由同(A) \\(E) \bigcirc: \cases{(10^2+20^2)/2=500/2=250 \\ ({10+20 \over 2})^2 =15^2=225} \Rightarrow {10^2+20^2\over 2} >({10+20 \over 2})^2\\，故選\bbox[red,2pt]{(ABCDE)}$$
解：$$(B)\times: det(5M)=5^3det(M)\\其餘皆正確，故選\bbox[red,2pt]{(ACDE)}$$
解：$$g(x)=x^3+ax^2+bx+c \Rightarrow g'(x)=3x^2+2ax+b \Rightarrow g''(x)=6x+2a\\ g(-2)及g(4)有極值 且g(-2)=29\Rightarrow \cases{g'(-2)=0 \\ g'(4)=0 \\ g(-2)=29} \Rightarrow \cases{12-4a+b=0 \\ 48+8a+b=0 \\ -8+4a-2b+c=29} \\ \Rightarrow \cases{a=-3\\b=-24 \\c=1} \Rightarrow g(x)=x^3-3x^2-24x+1\\(A)\times: g(-2)=29 \ne 0 \Rightarrow -2不是g(x)=0的根\\ (B)\bigcirc: \cases{g(-2)為極大值\\ g(4)為極小值\\ g(x)為三次式} \Rightarrow g(x)在區間(-2,4)遞減\\ (C)\times: g''(x)=0 \Rightarrow 6x-6=0 \Rightarrow x=1 \Rightarrow g(x)在x>1凹向上 \\(D)\bigcirc: 由上述聯立方程組可知:c=1 \\(E) \times: g(\infty)=\infty \not \le -79\\，故選\bbox[red,2pt]{(BD)}$$註: 公布的答案是ABD
解：$$(A) \times: f(x)=x^3-ax^2=x^2(x-a) \Rightarrow f(x)=0至少有重根0 \\(B)\bigcirc: f'(x)=3x^2-2ax \Rightarrow f''(x)=6x-2a \Rightarrow f''(1)=0 \Rightarrow 6-2a=0 \Rightarrow a=3\\ (C) \bigcirc: \cases{f'(0)=0\\ f''(0)=-2a=-6} \Rightarrow f(0)為極大值 \\(D)\times: f'(a)=f'(3)=27-18 \ne 0 \Rightarrow f(a)非極值\\ (E)\bigcirc: f(0)=0為極大值\Rightarrow 在x\in [0,3]，f(x)為遞減且f(x)<0，因此所圍面積= -\int_0^a f(x)\;dx\\故選\bbox[red, 2pt]{(BCE)}$$
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