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2022年3月11日 星期五

109年全國科學班資格考-數學詳解

109年度全國科學班聯合學科資格考-數學科

第壹部分:單選題、多選題及填充題
一、單選題:(共二題,每題5分,共10分)

解答:$$\cases{2+\log_2 a= k\\ 3+\log_3 b=k\\ \log_6(a+b)=k} \Rightarrow \cases{a=2^{k-2} \\ b=3^{k-3}\\ a+b= 6^k} \Rightarrow {1 \over a}+ {1\over b} ={a+b\over ab} ={6^k \over 2^{k-2}\cdot 3^{k-3}}  ={6^k \over 2\cdot 6^{k-3}} ={6^3\over 2} ={216\over 2}\\ =108,故選\bbox[red,2pt]{(4)}$$
解答:$$假設\cases{p=1/6\\ q=2/6 \\ r=3/6}\\ p_1= p +(1-p)(1-q)(1-r)p +((1-p)(1-q)(1-r))^2p +\cdots  = {p\over 1-(1-p)(1-q)(1-r)}   \\ p_2= (1-p)q +(1-p)(1-q)(1-r)(1-p)q + ((1-p)(1-q)(1-r))^2 (1-p)q+\cdots \\\quad = {(1-p)q\over 1-(1-p)(1-q)(1-r)} \\ p_3=(1-p)(1-q)r + (1-p)(1-q)(1-r)(1-p)(1-q)r \\\quad +(1-p)(1-q)(1-r))^2 (1-p)(1-q)r+ \cdots ={(1-p)(1-q)r\over 1-(1-p)(1-q)(1-r)}\\ 由於三者分母相同,僅需考慮分子,即\cases{p=1/6 =3/18\\ (1-p)q=5/18\\ (1-p)(1-q)r = 5/18},因此 p_2=p_3 \gt p_1,故選\bbox[red,2pt]{(4)}$$

二、多選題: (共三題,每題 5 分,共 15 分)

解答:$$(1)\bigcirc: 每個禮物有3種分法,共有3^{12}種分法\\(2)\bigcirc: 12選4給第1人,剩下8個選4個給第2人,剩下4個全給第3人,故C^{12}_4 C^8_4C^4_4種分法\\ (3)\times: 方法同(2),但需乘上3,3,6的排列數,即C^{12}_3C^9_3C^6_6 \times 3\\ (4)\bigcirc: x+ y+z=12的自然數解個數,也就是x+y+z=9的非負整數解個數,即H^3_9= C^{11}_2\\ (5)\bigcirc: x+y+z+w= 12的非負整數解再扣除x,y,z其中之一\ge 8(不可能同時有兩個超過7)\\ \qquad,因此共有 H^4_{12}-C^3_1H^4_{12-8} =C^{15}_3-C^3_1C^7_3\\ 故選\bbox[red,2pt]{(1245)}$$
解答:$$(1)\bigcirc: \sin(A+B) =\sin A\cos B+ \sin B\cos A= 2\sin A\cos B \Rightarrow \sin A\cos B- \sin B\cos A=0\\ \qquad \Rightarrow \sin(A-B)=0 \Rightarrow A=B \Rightarrow 等腰 \\(2) \times:\cos(A+B) = \cos A\cos B-\sin A\sin B= 2\cos A\cos B \Rightarrow \cos A\cos B+\sin A\sin B=0\\ \qquad \Rightarrow \cos (A-B)=0 \Rightarrow A-B=\pm 90^\circ ,不一定是等腰\\(3) \bigcirc: \sin(A+B) = \sin B \Rightarrow 180^\circ-(A+B)=B \Rightarrow A+2B=180^\circ = A+B+C \Rightarrow B=C \\(4)\times: \cases{A=30^\circ\\ B=60^\circ \\ C=90^\circ }符合\sin (2A)=\sin (2B),但不是等腰\\ (5)\bigcirc: \cos(2A)= \cos(2B) \Rightarrow A=B\\,故選\bbox[red,2pt]{(135)}$$
解答:$$f(x)=ax^4+bx^3 +cx^2 +dx+e \Rightarrow f'(x)= 4ax^3 +3bx^2 +2cx +d \\(1) \bigcirc: f'(x)=0的三根為\alpha,\beta, \gamma, 又\cases{y=f(x)圖形凹向上\Rightarrow a\gt 0\\ |\beta|\lt |\alpha|\lt |\gamma| \Rightarrow \alpha+\beta+ \gamma =-3b/4a \gt 0} \Rightarrow b\lt 0\\(2)\bigcirc: \alpha\beta +\beta\gamma+ \gamma\alpha =2c/4a \lt 0 \Rightarrow c\lt 0 \\(3)\times: \alpha \beta \gamma =-d/4a \lt 0 \Rightarrow d\gt 0 \\(4)\bigcirc: f''(x)=12ax^2 +6bx+2c =0 有相異實根\Rightarrow 36b^2-96ac \gt 0 \Rightarrow 3b^2-8ac\gt 0 \\(5)\bigcirc: \alpha,\beta,\gamma 為f'(x)=0之三相異實根\\ 故選\bbox[red,2pt]{(1245)}$$

三、 填充題: (共五題,每題 5 分,共 25 分)

解答:$$令\cases{\vec u=\overrightarrow{AB} \\ \vec v= \overrightarrow{AC}},則\cases{|\vec u|=|\vec v|=2\\ \vec u\cdot \vec v= |\vec u||\vec v|\cos 60^\circ = 2\cdot 2\cdot {1\over 2} =2} \Rightarrow \cases{\overrightarrow{AP} =20\vec u+\vec v\\ \overrightarrow{AQ}= \vec u+20\vec v} \\ \Rightarrow \cases{ |\overrightarrow{AP}|^2 =400|\vec u|^2 +|\vec v|^2 +40\vec u\cdot \vec v=1600 +4+80 = 1684 \\ |\overrightarrow{AQ}|^2 = |\overrightarrow{AP}|^2 =1684\\ (\overrightarrow{AP} \cdot \overrightarrow{AQ})^2 =20|\vec u|^2 +20|\vec v|^2 +401\vec u\cdot \vec v =80+80+802 = 962} \\ \Rightarrow \triangle APQ 面積= {1\over 2}\sqrt{ |\overrightarrow{AP}|^2 |\overrightarrow{AQ}|^2 -(\overrightarrow{AP} \cdot \overrightarrow{AQ})^2} ={1\over 2}\sqrt {1684^2-962^2} ={1\over 2}\cdot 798 \sqrt 3 =\bbox[red,2pt]{399\sqrt 3}$$
B. 丟擲一個公正骰子(即點數 1,2,3,4,5,6每個點數出現的機率皆為\({1\over 6}\))三次,設依序出現的點數為 \(a,b,c\),定義隨機變數\(X\)如下:若\(a\)是3的倍數,則\(X=b+c\),若\(a\)不是3的倍數,則\(X=|b-c|\),試求\(P(X=4)\)的值為_____。

解答:$$\begin{array}{} a & (b,c) & 數量\\\hline 3,6 & (1,3),(2,2),(3,1) & 2\times 3=6\\\hdashline 1,2 &(1,5),(2,6) & 4\times 4=16\\4, 5& (5,1), (6, 2)\\\hline\end{array} \Rightarrow P(X=4)={6+16\over 6^3} =\bbox[red,2pt]{11\over 108}$$
C. 在 \(xy\) 平面上有一個圓 \(C:x^2+y^2=16\),\(O\)為圓心,且在\(y\)軸上有一點 \(A(0,8)\),若有一光線自\(A\)點射向第一象限中圓\(C\)上某一點 \(P\),經反射之後平行 \(x\)軸射出經過\(A'\)點,且\(A、A'\)兩點對稱於\(\overleftrightarrow{OP}\),試問 \(P\)點的 \(y\)座標為___。
解答
$$P在圓C:x^2+y^2=4^2上\Rightarrow P(4\cos\theta, 4\sin\theta) \Rightarrow \overleftrightarrow{OP}:y=\tan \theta \cdot x\\ 又A'與P有相同的y軸標,取A'(a,4\sin\theta);因此MA=A',其中M為鏡射矩陣\\ 即\begin{bmatrix}\cos 2\theta & \sin 2\theta\\ \sin 2\theta & -\cos 2\theta \end{bmatrix}\begin{bmatrix} 0\\ 8 \end{bmatrix} =\begin{bmatrix} a\\ 4\sin\theta\end{bmatrix} \Rightarrow  -8 \cos 2\theta = 4\sin \theta \Rightarrow -8(1-2\sin^2\theta)=4\sin \theta \\ \Rightarrow 4\sin^2\theta- \sin \theta -2=0 \Rightarrow \sin\theta ={1+\sqrt{33}\over 8}(P在第一象限,負值不合)\\ \Rightarrow 4\sin\theta =\bbox[red,2pt]{1+\sqrt{33}\over 2}$$
解答
$$z^{12}-z^9+z^6 -z^3+1=0 \Rightarrow (z^3+1)(z^{12}-z^9+z^6 -z^3+1)=0 \Rightarrow z^{15}=-1\\ 又\cases{z^{15}=-1的15根(z_k= \cos(\pi+{2k\pi\over 15}) + i\sin(\pi+{2k\pi\over 15}),k=0-14)代表正15邊形的15個頂點\\ z^3=-1的3根(z_k= \cos(\pi+{2k\pi\over 3}) + i\sin(\pi+{2k\pi\over 3}),k=0-2)代表正三角形的三個頂點} \\\Rightarrow 正15邊形的15個頂點扣除正三角形的三個頂點,即為z^{12}-z^9+z^6 -z^3+1=0的12個根\\也就是上圖15個頂點扣除頂點A,F,K,因此周長為9\overline{IJ}+ 3\overline{JL}\\ 由於\cases{\angle IOJ=24^\circ \Rightarrow \overline{IJ} =2\sin 12^\circ = 2\sqrt{(1-a)/2}\\ \angle JOL=48^\circ \Rightarrow \overline{JL} =2\sin 24^\circ =2 \sqrt{1-a^2}} \Rightarrow 9\overline{IJ}+ 3\overline{JL}= \bbox[red,2pt]{18\sqrt{1- a\over 2} +6\sqrt{1-a^2}}$$

解答:$$橢圓\cases{中心(0,0)\\ 2a=4\\ 2b=3} \Rightarrow 橢圓方程式:{x^2\over 4}+{y^2\over 9/4}=1 \Rightarrow y={3\over 4}\sqrt{4-x^2} (只考慮第一象限)\\ \Rightarrow 陰影面積= \int_1^\sqrt 2 {3\over 4}\sqrt{4-x^2}\,dx ={3\over 4}\int_{\pi/3}^{\pi/4} \sqrt {4-4\cos^2\theta}(-2)\sin\theta \,d\theta(\because x=2\cos \theta \Rightarrow dx=-2\sin \theta d\theta)\\ =-3 \int_{\pi/3}^{\pi/4} \sin^2\theta \,d\theta =-{3\over 2}\int_{\pi/3}^{\pi/4}  1-\cos 2\theta\,d\theta =-{3\over 2} \left.\left[ \theta-{1\over 2}\sin 2\theta\right]\right|_{\pi/3}^{\pi/4} \\=-{3\over 2}\left( ({\pi\over 4}-{1\over 2}) -({\pi\over 3} -{\sqrt 3\over 4})\right)  = \bbox[red,2pt]{{1\over 8}(6+\pi -3\sqrt 3)}$$

第貳部分:非選擇題 (數學寫作能力、計算證明題共 50 分)

四、 數學寫作能力: (共二題,共計 12 分)

解答:$$假設A(x_1,y_1, z_1)在平面E上,且\overleftrightarrow{PA}\bot E,則\overline{PA}即為所求;\\ E的法向量\vec n= (a,b,c) 與\overrightarrow{PA} =(x_1-x_0,y_1-y_0, z_1-z_0)平行 \\,即{ x_1-x_0 \over a} ={ y_1-y_0 \over b} ={z_1-z_0 \over c}=t \Rightarrow \cases{x_1=at+ x_0\\ y_1=bt+y_0 \\ z_1=ct+z_0};\\又A在E上\Rightarrow a(at+x_0 )+ b(bt+y_0) +c(ct+z_0)+ d=0 \Rightarrow t=-{ax_0+ by_0 +cz_0+ d\over a^2+b^2+c^2} \\\Rightarrow \left| \overrightarrow{PA}\right| =\left| t \vec n\right| =\left|-{ax_0+ by_0 +cz_0 +d\over a^2+b^2+c^2}\cdot \sqrt{a^2+b^2 +c^2} \right| =\left|{ax_0+ by_0 +cz_0 +d\over\sqrt{a^2+b^2 +c^2}}   \right| \\\bbox[red,2pt]{故得證}$$
解答:$$(1)(\cos \theta+i\sin\theta)^n = \cos n\theta + i\sin n\theta\\ (2)利用歸納法:\\ n=1時,顯然成立\\ 假設n=k成立,即z^k+{1\over z^k}=2\cos k\theta;\\ 當n=k+1時,z^{k+1}+{1\over z^{k+1}} =(z^k+{1\over z^k}) (z+{1\over z}) - (z^{k-1}+ {1\over z^{k-1}})  =2\cos k\theta \cdot 2\cos \theta-2\cos(k-1)\theta \\ =4\cos k\theta\cos \theta-2\cos (k\theta-\theta) =4\cos k\theta\cos \theta-2(\cos k\theta\cos \theta +\sin k\theta \sin \theta)\\ =2(\cos k\theta \cos\theta -\sin k\theta\sin \theta)=2 \cos(k+1)\theta  \Rightarrow z^{k+1}+{1\over z^{k+1}} =2 \cos(k+1)\theta \\ n=k+1時亦成立,\bbox[red,2pt]{故得證}$$

五、計算證明題: (共 5 題,共計 38 分)

解答:$$圓:x^2+(y-b)^2=a^2 \Rightarrow \cases{圓心P(0,b) \\ 圓半徑r=a} \Rightarrow \cases{圓面積A=a^2\pi \\ P繞x軸一圈的圓周長=2b\pi} \\ \Rightarrow 圓繞x軸旋轉體積=(a^2\pi)\times (2b\pi)= 2a^2b\pi^2 ,\bbox[red,2pt]{故得證}$$

解答:$$\cos \angle A= {\overline{AB}^2 +\overline{AC}^2 -\overline{BC}^2 \over 2\cdot \overline{AB}\cdot \overline{AC}} ={4^2+ 6^2 - (2\sqrt{19})^2 \over 2\cdot 4\cdot 6} =-{1\over 2} \Rightarrow \angle A=120^\circ \\ 令\angle BAP=\theta,則\angle CAP=120^\circ-\theta \Rightarrow \cases{\triangle ABP ={1\over 2}\cdot \overline{AB}\cdot \overline{AP}\sin \theta=4\sin \theta \\ \triangle ACP ={1\over 2}\cdot \overline{AC}\cdot \overline{AP}\sin (120^\circ-\theta) = 6 \sin (120^\circ-\theta)} \\ \Rightarrow \triangle ABP+\triangle ACP =4\sin\theta +6 \sin (120^\circ-\theta) =4\sin\theta +3\sqrt 3\cos\theta +3\sin\theta\\ =7\sin\theta +3\sqrt 3\cos \theta \Rightarrow 最大值為\sqrt{7^2 +(3\sqrt 3)^2} =\sqrt{76} =\bbox[red,2pt]{2\sqrt{19}}$$
解答
(1)$$由定義可知:\cases{g(2n)=g(n) \\ g(2n-1)=2n-1},n\in \mathbb{N},因此S_n = g(1)+ g(2)+\cdots +g(2^n) \\=\sum_{k=1}^{2^{n-1}} \left(g(2k-1)+g(2k) \right) = \sum_{k=1}^{2^{n-1}} \left(g(2k-1)+g(k) \right) =S_{n-1}+ \sum_{k=1}^{2^{n-1}}(2k-1) =S_{n-1} + {2^n\times 2^{n-1}\over 2}\\ =S_{n-1}+ 4^{n-1}\\ 當n=1時,S_1= g(1)+ g(2^1)= 1+1=2\Rightarrow \bbox[red,2pt]{ S_n= \begin{cases} 2,&n=1\\ S_{n-1}+ 4^{n-1}, &n\ge 2\end{cases} };$$(2)$$S_{2019}= S_{2018} +4^{2018} = S_{2017} +4^{2017} +4^{2018} = S_1 +4+ 4^2 +\cdots 4^{2018} \\=1+ (1+ 4+ 4^2 +\cdots 4^{2018})=1+{1-4^{2019}\over 1-4} =1+{1\over 3}(4^{2019}-1)\\ \Rightarrow S_{2019}= \bbox[red,2pt]{{1\over 3}(2+4^{2019})}$$
4.某個班級有36個學生,某次考試成績的算術平均數為50分,標準差為10分,試問最少有多少學生的分數介於35分到65分之間(不包含35分及65分)?

解答:$$36位考生成績依序為:x_1\le x_2\le x_3 \le \cdots \le x_{36},且\cases{\bar x= (x_1+x_2+\cdots +x_{36})/36= 50\\[1ex] \sigma= \sqrt{{1\over 36}\sum_{i=1}^{36}(x_i-\bar x)^2} =10} \\ 令A=\{x_i\mid x_i\ge 65或x_i\le 35,i=1-36\},則我們有(x-\bar x)^2=(x-50)^2  \le 15^2, \forall x\in A\\若A有n個元素,則 \sum_{x_i\in A}(x_i-50)^2 \le 225n \le \sum_{i=1}^{36}(x_i-50)^2 =10^2 \times 36=3600\\ \Rightarrow 225n \le 3600 \Rightarrow n\le 16 \Rightarrow 至少有36-16=\bbox[red,2pt]{20}位成績介於36與64之間$$
解答
(1)$$\cases{P(x_0,y_0)\\ F_1(c,0) \\[1ex]{x^2\over a^2} +{y^2\over b^2}=1} \Rightarrow \overline{PF_1}^2 =  (x_0-c)^2 +y_0^2 = (x_0-c)^2 + b^2\left(1-{x_0^2\over a^2} \right)\\ = (x_0-c)^2 + (a^2-c^2)\left(1-{x_0^2\over a^2} \right) = x_0^2-2cx_0+ c^2 +a^2-x_0^2-c^2+{c^2\over a^2}x_0^2 ={c^2\over a^2}x_0^2 -2cx_0+a^2\\ =({c\over a}x_0-a)^2  \Rightarrow \overline{PF_1} =\left|{c\over a}x_0-a \right|\\ 又{x_0^2\over a^2} +{y_0^2\over b^2}=1 \Rightarrow {x_0^2\over a^2}\lt 1 \Rightarrow -1\lt {x_0\over a} \lt 1 \Rightarrow   -c\lt c\cdot {x_0\over a} \lt c \lt a \Rightarrow \left|{c\over a}x_0-a \right| =a-{c\over a}x_0\\ \Rightarrow \overline{PF_1} =a-{c\over a}x_0,\bbox[red,2pt]{故得證}$$(2)$${x^2\over 25} +{y^2\over 16}=1 \Rightarrow \cases{a=5\\ b=4} \Rightarrow c=3 \Rightarrow \cases{F_1(3,0)\\ F_2(-3,0)} \\ 令P(s,t),由\cases{\overline{PF_1} +\overline{PF_2} =2a=10\\ \overline{PF_1} :\overline{PF_2}= 2:3} \Rightarrow \cases{\overline{PF_1}=4 \\ \overline{PF_2}=6} \Rightarrow \cases{(s-3)^2+ t^2 =16\\ (s+3)^2+t^2=36} \\ \Rightarrow (s+3)^2 -(s-3)^2 =20 \Rightarrow 12s =20 \Rightarrow s=5/3 \Rightarrow {16\over 9}+t^2 =16 \Rightarrow t^2={128\over 9} \\ \Rightarrow t=\pm {8\sqrt 2\over 3} \Rightarrow \bbox[red,2pt]{P\left({5\over 3},\pm {8\sqrt 2\over 3}\right)}$$

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