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2020年6月8日 星期一

99年警專29期乙組數學科詳解


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
壹、單選題


:$$x+k為f(x)的因式,必須滿足k是5的因數,故選\bbox[red,2pt]{(A)}$$


:$$\cases{\log_{11}a=11 \\ \log_{11}b=5} \Rightarrow \cases{a=11^{11} \\ b=11^5} \Rightarrow \log_{11}(a+b)= \log_{11}(11^5+11^{11}) = \log_{11}(11^5(1+11^{6})) \\ =5+\log_{11}(1+11^6) \approx 5+\log_{11}(11^6) =5+6=11,故選\bbox[red,2pt]{(C)}$$


:$$此題相當於求y=\sin x\;與y=\log_2 x\;兩圖形的交點數量,故選\bbox[red,2pt]{(B)}$$



$$f(x)=x^5-8x^4+9x^3-12x^2-13x-4 \\\Rightarrow f(7)= 7^5-8\times 7^4+9\times 7^3-12\times 7^2-13\times 7-4 \\ =7^4(7-8)+9\times 7^3-12\times 7^2-13\times 7-4 \\ =-7^4+9\times 7^3-12\times 7^2-13\times 7-4 =7^3(-7+9)-12\times 7^2-13\times 7-4 \\ =2\times 7^3-12\times 7^2-13\times 7-4 =7^2(14-12)-13\times 7-4\\ =2\times 7^2-13\times 7-4 = 7(14-13)-4 =7-4=3,故選\bbox[red,2pt]{(C)}$$


:$$g(x)=x^3-4x^2+5x-5=(x-2)f(x)+2x+k \Rightarrow g(2)=8-16+10-5=-3=4+k \\\Rightarrow k=-7,故選\bbox[red,2pt]{(A)}$$


:$$\begin{vmatrix}a & 3 & 7 \\2 & -1 & 1\\ 3 & 1 & 4 \end{vmatrix} =0 \Rightarrow -4a+14+9+21-24-a=0 \Rightarrow 5a=20 \Rightarrow a=4,故選\bbox[red,2pt]{(D)}$$



:$$(A)\bigcirc: (A+I)(A-I)=A^2-A+A-I^2 =A^2-I\\(B) \times: (A-B)^2 =A^2-AB-BA+B^2 (AB不一定等於BA)\\(C)\times: (AB)^t=B^tA^t \\(D)\times: A=0 \Rightarrow AB=AC,但B不一定要等於C,故選\bbox[red,2pt]{(A)}$$



:$${C^n_2 \over C^{n+5}_2} ={7\over 22} \Rightarrow {n(n-1) \over (n+5)(n+4)}={7\over 22} \Rightarrow {n^2-n \over n^2+9n+20} ={7\over 22} \Rightarrow 22n^2-22n=7n^2+63n+140\\ \Rightarrow 15n^2-85n-140=0 \Rightarrow 3n^2-17n-28=0 \Rightarrow (3n+4)(n-7)=0 \Rightarrow n=7,故選\bbox[red,2pt]{(B)}。 $$



:$$每個獎品有5種選擇,因此共有5\times 5\times 5=5^3種分法,故選\bbox[red,2pt]{(D)}。$$





$$P(甲中乙沒中)+P(乙中甲沒中) ={2\over 3}\times (1-{3\over 4})+ {3\over 4}\times (1-{2\over 3})={1\over 6} +{1\over 4}={5\over 12},故選\bbox[red, 2pt]{(C)}$$



:$$x+y為偶數 \Rightarrow \cases{x,y都是偶數\\ x,y都是奇數} \Rightarrow \cases{P(x,y都是偶數)={1\over 2}\times{1\over 3}={1\over 6} \\ P(x,y都是奇數)={1\over 2}\times{2\over 3}={1\over 3}}\\ \Rightarrow P(x+y為偶數)= {1\over 6}+{1\over 3} ={1\over 2},故選\bbox[red, 2pt]{(A)}$$


:$$P(A\cup B)= P(A)+P(B)- P(A\cap B) = P(A)+P(B)- P(A)P(B) \\ \Rightarrow {7\over 12}={1\over 3}+P(B)- {1\over 3}\times P(B) \Rightarrow {1\over 4}={2\over 3}\times P(B) \Rightarrow P(B)={3\over 8},故選\bbox[red, 2pt]{(A)}$$




$$本題\bbox[red,2pt]{(送分)}。$$




$$平均值\mu=(65+66+70+75+80+82)\div 6 =73 \Rightarrow 當k=\mu時,有最小的標準差,故選\bbox[red,2pt]{(D)}。$$



:$$依題意迴歸直線通過(\bar x,\bar y)=(5,3)及(0,2),將此兩點代入y=mx+b\\ 可得\cases{3=5m+b\\ 2=b} \Rightarrow m=1/5,故選\bbox[red,2pt]{(B)}$$

16. 有一公司錄取新進員工20人,在上班前作了職前訓練。公司對這20名新進員工進行職前訓練前的測驗與職前訓練後的測驗,若訓練前的成績X與訓練後的成績Y經統計的結果後知:\(\bar x=35,\bar y=50\),\(\sum_{i=1}^{20}(x_i-\bar x)(y_i-\bar y)=72\),\(\sum_{i=1}^{20} (x_i-\bar x)^2=81\),\(\sum_{i=1}^{20}(y_i-\bar y)^2=100\)。則X與Y的相關係數最接近哪個選項?
(A) 0.1   (B) 0.3  (C) 0.5  (D) 0.8

:$$r={\sum_{i=1}^{20}(x_i-\bar x)(y_i-\bar y) \over \sqrt{\sum_{i=1}^{20}(x_i-\bar x)^2}\times \sqrt{\sum_{i=1}^{20}(y_i-\bar y)^2}} ={72 \over \sqrt{81}\times \sqrt{100}} = {72 \over 9\times 10}=0.8,故選\bbox[red,2pt]{(D)}$$




$$直線L:y=mx-5必過C(0,-5),L與線段\overline{AB}相交,代表L的斜率介於兩直線\overline{CA}與\overline{CB}的斜率之間;\\ \cases{\overline{AC}: y=-{3\over 2}x-5 \\ \overline{BC}: y={5\over 3}x-5} \Rightarrow -{3\over 2}\le m \le {5\over 3} \Rightarrow m=-1,0,1,共三個整數值,故選\bbox[red, 2pt]{(B)}$$



:$$f(x)=x^2+2x-3 = (x+1)^2-4 \Rightarrow f(-1)=-4為最小值,但-1不在區間[0,1]內;\\因此最小值出現在x最靠近-1的地方,即x=0 \Rightarrow 最小值為f(0)=-3,故選\bbox[red,2pt]{(C)}$$


:$$\sum_{k=1}^{11}k^2= {11\times 12\times 23 \over 6}=506\Rightarrow 506-x^2=47\times 10 \Rightarrow x^2=36,故選\bbox[red,2pt]{(C)}$$



$${1\over a+bi}+{1\over 3-i}= { 3-i+a+bi\over (a+bi)(3-i)} ={ (a+3)+(b-1)i\over (3a+b)+(3b-a)i} ={3\over 5} \Rightarrow \cases {5a+15=9a+3b \\ 5b-5=9b-3a} \\\Rightarrow \cases{4a+3b=15 \\ 3a-4b=5} \Rightarrow \cases{a=3\\ b=1} \Rightarrow a+b=3+1=4,故選\bbox[red,2pt]{(A)}$$




$$此題相當於問:\overline{OP}與哪一條線的長度最接近?\cases{\overline{QA}?\\  \overline{QB}?\\ \overline{QC}?\\ \overline{QD}?\\ },故選\bbox[red,2pt]{(D)}$$





$$6-8-10為直角\triangle,同樣以兩邊邊長為6與8的三角形,以直角三角形面積最大,故選\bbox[red,2pt]{(B)}$$




$$令\overline{BD}=a,見上圖;則\cases{\triangle ABC: \cos \angle B={\overline{AB}^2 +\overline{BC}^2-\overline{AC}^2 \over
 2\times \overline{AB}\times \overline{BC}} ={36+25-16 \over 2\times 6\times 5}\\\triangle ABD: \cos \angle B={\overline{AB}^2 +\overline{BD}^2-\overline{AD}^2 \over  2\times \overline{AB}\times \overline{BD}} ={36+a^2-16 \over 2\times a\times 6}} \\ \Rightarrow {45\over 60} = {a^2+20\over 12a} \Rightarrow (a-5)(a-4)=0 \Rightarrow a=4(5不合,違反D\ne C),故選\bbox[red,2pt]{(D)}。$$



:$$3+4i=5(\cos \alpha+i\sin \alpha),其中\sin \alpha=4/5 \Rightarrow 45^\circ <\alpha < 60^\circ (\because \sin 60^\circ = {\sqrt 3\over 2}\approx 0.86,\sin 45^\circ \approx 0.71)\\ (3+4i)(\cos \theta+i\sin \theta )<0 \Rightarrow \alpha+\theta = \pi (3\pi,5\pi,...) \Rightarrow 180^\circ-60^\circ <\theta <180^\circ-45^\circ \\\Rightarrow 120^\circ < \theta < 135^\circ \Rightarrow \theta 在第二象限\\,故選\bbox[red,2pt]{(B)}$$




$$各向量長度相等,因此兩向量所夾角度越大則內積越小,故選\bbox[red,2pt]{(C)}$$


:$$|4\vec a-2\vec b|^2=(4\vec a-2\vec b)\cdot (4\vec a-2\vec b) = 16|\vec a|^2-16\vec a\cdot \vec b+4|\vec b|^2= 16|\vec a|^2+4|\vec b|^2 (\because \vec a\bot \vec b\Rightarrow \vec a\cdot \vec b=0)\\ =16\times 4+4\times 9= 100 \Rightarrow |4\vec a-2\vec b|=\sqrt{100} =10,故選\bbox[red, 2pt]{(D)}$$



:$$\cases{A(2,0,-1) \\B(6,-1,4) \\C(1,-5,3)} \Rightarrow \cases{\overrightarrow{CA} =(1,5,-4) \\ \overrightarrow{CB} =(5,4,1)} \Rightarrow \cos \angle C={\overrightarrow{CA} \cdot \overrightarrow{CB}\over |\overrightarrow{CA}|| \overrightarrow{CB}|} = {5+20-4 \over \sqrt{1+25+16}\times \sqrt{25+16+1}}\\ ={21 \over 42}={1\over 2} \Rightarrow \theta = {\pi \over 3},故選\bbox[red,2pt]{(B)}$$



:$$P在E上\Rightarrow 3a-4b+12c=8,由柯西不等式: \\\Rightarrow ((a-1)^2+(b+2)^2+(c-3)^2)(3^2+(-4)^2+12^2) \ge (3(a-1)-4(b+2)+12(c-3))^2\\ \Rightarrow ((a-1)^2+(b+2)^2+(c-3)^2)\times 169 \ge (3a-4b+12c-47)^2 =(8-47)^2=39^2 \\ \Rightarrow (a-1)^2+(b+2)^2+(c-3)^2\ge {39^2 \over 13^2} \Rightarrow \sqrt{(a-1)^2+(b+2)^2+(c-3)^2} \ge {39\over 13}=3\\ \Rightarrow 最小值為3,故選\bbox[red,2pt]{(C)}$$



:$$離球心O(0,0,0)的距離越近,則所交圓面積越大;球心在平面x-y+z=0上,距離為0\\,故選\bbox[red,2pt]{(A)}$$


:$$球:(x+1)^2+(y-1)^2+(z-1)^2=27 \Rightarrow \cases{球心O(-1,1,1)\\ 球半徑R=3\sqrt 3}\\ \Rightarrow \text{dist}(O,E)= \left|{-2-2+1-6\over \sqrt{4+4+1}}\right| =3  \Rightarrow R^2= r^2+3^2,r為圓半徑 \Rightarrow 27=r^2+9 \Rightarrow r^2=18 \\\Rightarrow 圓面積=r^2\pi = 18\pi,故選\bbox[red,2pt]{(B)}$$

貳、多重選擇題


:$$(A)\bigcirc: g(3)=2 \Rightarrow \log_a 3=2 \Rightarrow a^2=3 \Rightarrow f(2)=3 \\(B)\times: f(3)=2 \Rightarrow a^3=2 \Rightarrow a^6 = 4 \Rightarrow \log_a 4=6 \Rightarrow g(4)=6 \ne 9 \\(C)\times: f與g互為反函數,對稱於x=y \\(D)\bigcirc: {f(2010)\over f(1993)} ={a^{2010} \over a^{993}} =a^{17} ={a^{99} \over a^{82}} = {f(99)\over f(82)} \\(E) \times: \cases{g(2010)-g(1993) = \log_a 2010-\log_a 1993 =\log_a{2010\over 1993} \\ g(99)-g(82) = \log_a 99-\log_a 82 =\log_a{99\over 82}} \Rightarrow g(2010)-g(1993) \ne g(99)-g(82)\\,故選\bbox[red,2pt]{(AD)}$$



:$$(A)\bigcirc: 第n年\left[\matrix{甲\\乙\\ 丙} \right] \Rightarrow 第n+1年\left[\matrix{甲'\\乙'\\ 丙'} \right] \Rightarrow  \cases{0.7甲+0.1乙+0.1丙=甲' \\0.2甲+0.8乙+0.1丙=乙' \\0.1甲+0.1乙+0.8丙=丙' \\} \\\qquad \Rightarrow \left[\matrix{0.7 & 0.1 & 0.1\\ 0.2 & 0.8 & 0.1\\ 0.1 & 0.1 & 0.8} \right] 為轉移矩陣 \\(B)\times: \left[\matrix{0.7 & 0.1 & 0.1\\ 0.2 & 0.8 & 0.1\\ 0.1 & 0.1 & 0.8} \right]\left[\matrix{a\\b\\ c} \right]= \left[\matrix{a\\b\\ c} \right] \Rightarrow \cases{0.7a+0.1b+0.1c=a \\ 0.2a+0.8b+0.1c=b \\ 0.1a+0.1b+0.8c=c} \Rightarrow \cases{a=1/4\\b=5/12 \\ c=1/3} \\(C)\bigcirc: 理由同上\\(D) \times: c=1/3\ne 1/4 \\(E) \bigcirc: 三家報社占有率為100\%\\,故選\bbox[red,2pt]{(ACE)}$$


:$$(A)\times: 點數和至少為2,不可能為1 \Rightarrow P(1)=0\ne {1\over 36}\\ (B)\bigcirc: 5=(1,4),(2,3),(3,2),(4,1),有4種情形,機率為{4\over 36}\\(C) \bigcirc: 7=(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)有6種情形,機率為{6\over 36} \\(D)\bigcirc: 12=(6,6),只有1種情形,機率為{1\over 36} \\(E)\times: 3=(1,2),(2,1)有2種情形;10=(4,6),(5,5),(6,4)有3種情形,兩者機率不同\\,故選\bbox[red,2pt]{(BCD)}$$



:$$a=H^3_{10} = C^{12}_{10}=C^{12}_2;\\正整數解相當於x+y+z=7的非負整數解,即b=H^3_7\\,故選\bbox[red,2pt]{(BCE)}$$




$$算術平均數與中位數都增加5,其餘不變,故選\bbox[red,2pt]{(CDE)}$$




:$$(A)\bigcirc: [0.544,0.736]=[p-2\sigma,p+2\sigma] \Rightarrow p=(0.544+0.736)\div 2=0.64\\ (B)\bigcirc: (0.736-0.544)\div 2= 2\sigma = 2\sqrt{p(1-p)\div n} =2\sqrt{0.64(1-0.64)\div n} \Rightarrow n=100\\ (C)\bigcirc: p=0.64=m/100 \Rightarrow m=64 \\(D)\times: 信賴區間並非此意 \\(E)\times: 並非「所有選民」
\\故選\bbox[red,2pt]{(ABC)}$$



:$$S(n)=4n^2-3\\(A) \bigcirc: a_1=S(1)=4-3=1 \\(B) \times: a_2=S(2)-S(1)=13-1=12\ne 13 \\(C)\bigcirc: a_3=S(3)-S(2)=33-13=20 \\(D) \times: a_n=S(n)-S(n-1) = 4n^2-3-(4(n-1)^2-3)= 8n-4, n\ge 2\\\qquad \Rightarrow 從n=2開始為公差為8的等數列 \\(E)\bigcirc: 理由同(D) \\,故選\bbox[red,2pt]{(ACE)}$$


:$$(A)\bigcirc: a=\cos{2\pi \over 5} +i\sin {2\pi \over 5} \Rightarrow a^5=\cos 2\pi +i\sin 2\pi=1 \\(B) \times: a^2=\cos{4\pi \over 5} +i\sin {4\pi \over 5} \ne 1+\cos{2\pi \over 5} +i\sin {2\pi \over 5} \\(C) \bigcirc: a^5=1 \Rightarrow a^5-1=0 \Rightarrow (a-1)(a^4+a^3+a^2+a+1)=0 \\\qquad\Rightarrow a^4+a^3+a^2+a+1=0(a\ne 1)\Rightarrow f(a)=0 \\(D)\bigcirc: 由\cases{a^5=1\\ a^4+a^3+a^2+a+1=0} \Rightarrow f(a^2)=a^8+a^6+a^4+a^2+1 \\\qquad =a^3+a+a^4+a^2+1=0; 同理,f(a^3)=f(a^4)=0,因此a,a^2,a^3,a^4 均為f(x)=0的四根\\\qquad \Rightarrow f(x)=(x-a)(x-a^2)(x-a^3)(x-a^4) \\(E)\times: f(x)=1+x+x^2+x^3+x^4=(x-a)(x-a^2)(x-a^3)(x-a^4)\\\qquad \Rightarrow f(1)=5=(1-a)(1-a^2)(1-a^3)(1-a^4) \ne 0\\,故選\bbox[red,2pt]{(ACD)}$$




$$(A)\times: |\overrightarrow{OA}+\overrightarrow{OB}|^2 =(\overrightarrow{OA}+\overrightarrow{OB}) \cdot (\overrightarrow{OA} +\overrightarrow{OB}) =(10,11)\cdot (10,11)=100+121=221\\\qquad |\overrightarrow{OA}+\overrightarrow{OB}|=\sqrt{221} \ne 21 \\(B)\bigcirc: \overrightarrow{OA} \cdot \overrightarrow{OB} =(6,8) \cdot (4,3)=24+24=48 \\(C)\bigcirc: C=(6\div 2,8\div 2)=(3,4) \Rightarrow \overrightarrow{BC} =(3-4,4-3)=(-1,1) \\(D) \times:\cases{\vec u=\overrightarrow{AB}=(-2,-5) \\\vec v=\overrightarrow {AC}=(-3,-4)} \Rightarrow \triangle ABC面積={1\over 2} \sqrt{|\vec u|^2|\vec v|^2-(\vec u\cdot \vec v)^2} ={1\over 2}\sqrt{29\times 25-26^2} ={7\over 2}\ne 7 \\(E)\times: 直線\overline{OA}方程式: 4x-3y=0,B(4,3)至此直線距離為{16-9\over 5}={7\over 5}\ne 2\\,故選\bbox[red,2pt]{(BC)}$$


:$$若圓心(1,1)至直線的距離等於半徑(r=1),則該直線與圓相切\\(A)\bigcirc: {3+4-2\over 5}=1 \\(B)\bigcirc: {4+3-2\over 5}=1 \\(C) \bigcirc: {|4-3-6|\over 5} ={5\over 5} =1 \\(D) \times: {|3-4-6|\over 5} ={7\over 5} \ne 1 \\(E)\times: {|4-3-5|\over 5} ={4\over 5} \ne 1\\故選\bbox[red, 2pt]{(ABC)}$$


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