110年全國科學班資格考-數學詳解

110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：

$$假設\cases{\angle EFC=\angle DCF =\theta \\ \overline{FB}=a} \Rightarrow \angle BFC=\angle DCF =\theta (\because \overline{AB} \parallel \overline{CD} ) \Rightarrow \angle AFE=\pi-2\theta\\ 又\cases{\overline{AE}:\overline{ED} =2:1\\ \overline{AD}=2} \Rightarrow \cases{\overline{AE}=4/3\\ \overline{ED}= 2/3} \Rightarrow \cases{\tan \angle AFE = \overline{AE}/\overline{AF} \\ \tan \angle BFC=\overline{BC} /\overline{BF}} \Rightarrow \cases{\tan (\pi-2\theta)= 4/3(3-a)\\ \tan \theta =2/a}\\ \Rightarrow \tan (\pi-2\theta)= -\tan(2\theta)=-{2\tan \theta\over 1-\tan^2\theta } =- {4/a \over 1-4/a^2} = -{4a\over a^2-4}={4\over 3(3-a)}\\ \Rightarrow 2a^2-9a+4=0 \Rightarrow (2a-1)(a-4)=0 \Rightarrow a=1/2(a=4違反a=\overline{FB}\lt \overline{AB}=3)\\ \Rightarrow \tan \angle AFE={4/3\over 3-1/2} = {8\over 15}，故選\bbox[red,2pt]{(3)}$$

解答：$$(A)\times: G= (A+B+C)\div 3=({1+2+0\over 3},{2+0+1\over 3},{0+1+2\over 3}) =(1,1,1) \ne (1,1,-1)\\(B)\bigcirc: \cases{D(9,5,1)\\ G(1,1,1)} \Rightarrow \overleftrightarrow{DG}:{x-1\over 8}={y-1\over 4},z=1 \Rightarrow (8s+1,4s+1,1) =(2t+1,t+1,1)\\(C)\times: H=(B+C+D)/3 =(11/3,2,4/3) \Rightarrow \overleftrightarrow{AH}:(2r+1,2,r),r\in \mathbb{R}\\ \qquad當r=t=1時 (3,2,1)在\overleftrightarrow{AH}上，也在\overleftrightarrow{DG}上，兩直線有交點 \\(D)\times: \triangle ABC構成的平面E，其法向量\vec n作為直線L的方向向量，且L通過重心G\\\qquad，則L上的點均符合要求(不只一個點)\\，故選\bbox[red,2pt]{(2)}$$

二、多選題：(共三題，每題 5 分，共 15 分)

解答：$$(1)\bigcirc: \sum_{n=1}^\infty k^{2n} \lt \infty \Rightarrow |k^2|\lt 1 \Rightarrow |k|\lt 1 \Rightarrow \sum_{n=1}^\infty k^{n} \lt \infty \\(2)\bigcirc: \lim_{x\to a} {f(x)\over (x-a)^3} = L\lt \infty \Rightarrow  \lim_{x\to a}{f(x) \over x^3-a^3} =  \lim_{x\to a}{f(x)\over (x-a)^3} \cdot  \lim_{x\to a}{(x-a)^2\over x^2+ax+ a^2} =L\cdot 0=0\\ (3)\times: 若唯一極值點為P(a,f(a))\Rightarrow f'(a)=0 且a為f'(x)=0之重根\Rightarrow f''(a)=0\\\qquad \Rightarrow P為反曲點非極值點\\ (4)\bigcirc: f為三次式\Rightarrow f''為一次式 \Rightarrow f''(x)=0必有一根，即為反曲點；\\\qquad 又若f(x)=x^3，則f有反曲點(0,0)，但(0,0)非極值點\\ (5)\bigcirc: f為四次式，其圖形為凹向上或凹向下，有極小值或極大值；\\\qquad 又若f(x)=x^4，則(0,0)為極小值點，但無反曲點；\\ 故選\bbox[red,2pt]{(1245)}$$

解答：$$A為轉移矩陣\Rightarrow A=\begin{bmatrix} \alpha & \beta \\ \gamma & \delta\end{bmatrix} =\begin{bmatrix} \alpha & \beta \\ 1-\alpha & 1-\beta \end{bmatrix}；\\又A為不可逆\Rightarrow \det(A)=0 \Rightarrow \alpha-\alpha\beta-\beta +\alpha\beta =0 \Rightarrow \alpha=\beta \Rightarrow A=\begin{bmatrix} \alpha & \alpha \\ 1-\alpha & 1-\alpha \end{bmatrix} \\ \Rightarrow A^2= \begin{bmatrix} \alpha & \alpha \\ 1-\alpha & 1-\alpha \end{bmatrix}\begin{bmatrix} \alpha & \alpha \\ 1-\alpha & 1-\alpha \end{bmatrix} =\begin{bmatrix} \alpha & \alpha \\ 1-\alpha & 1-\alpha \end{bmatrix} =A \Rightarrow A^n= A, n\in \mathbb{N} \\ \Rightarrow a_n+b_n= a_1+b_1 =a_5+ b_5={2\over 3}+{4\over 3}=2,n\in \mathbb{N}，故選\bbox[red,2pt]{(45)}$$

解答：$$(1)\times: P在\triangle ABC內部 \Rightarrow \alpha+\beta \lt 1,其中\alpha,\beta \gt 0\\ (2)\bigcirc: \cases{ \overrightarrow{AP} \cdot \overrightarrow{BC}=0 \\ \overrightarrow{BP} \cdot \overrightarrow{AC}=0} \Rightarrow P為垂心 \Rightarrow \overrightarrow{CP} \cdot \overrightarrow{AB}=0 \\(3)\bigcirc: 3\overrightarrow{AP} = \overrightarrow{AB} +\overrightarrow{AC} \Rightarrow \overrightarrow{AP} = {1\over 3}\overrightarrow{AB} +{1\over 3}\overrightarrow{AC} \Rightarrow P為重心 \Rightarrow \overrightarrow{BP} = {1\over 3}\overrightarrow{BC} +{1\over 3}\overrightarrow{BA} \\ \qquad \Rightarrow 3\overrightarrow{BP} =  \overrightarrow{BC} + \overrightarrow{BA} \\(4)\bigcirc: |\overrightarrow{PA} | =|\overrightarrow{PB} | =|\overrightarrow{PC} | \Rightarrow P為外心，又P在\triangle ABC內部 \Rightarrow \triangle ABC為銳角\triangle \\(5) \times: \alpha\left( {\overrightarrow{AB} \over |\overrightarrow{AB}| }+{\overrightarrow{AC} \over |\overrightarrow{AC}|} \right) 平分\angle A，且\overrightarrow{PM}-\overrightarrow{PA} = \overrightarrow{AM}，因此\overline{AM}為中線且為角平分線 \\\qquad \Rightarrow \overline{AB}=\overline{AC}  \Rightarrow \triangle ABC為等腰，不一定是全等。 \\故選\bbox[red,2pt]{(234)}$$

三、填充題： (共五題，每題 5 分，共 25 分)

解答：$$此題可看成BCDEF排列(頭尾再加上A)，滿足BD不相鄰且CF不相鄰的排列數\\\begin{array}{} BD不相鄰& CF不相鄰 & 數量\\\hline B   \bigcirc D \bigcirc\bigcirc & BCD\bigcirc \bigcirc & 2\\ & BFD\bigcirc\bigcirc & 2\\\hdashline B\bigcirc\bigcirc D \bigcirc& B\bigcirc\bigcirc DC & 2\\& B\bigcirc\bigcirc DF & 2 \\\hdashline  B\bigcirc \bigcirc \bigcirc D & BCEFD & 1\\ & BFECD & 1\\\hdashline \bigcirc B\bigcirc D\bigcirc & \bigcirc B\bigcirc D\bigcirc & 6\\\hdashline \bigcirc B\bigcirc \bigcirc D & CB\bigcirc \bigcirc D & 2\\ & FB\bigcirc \bigcirc D & 2 \\\hdashline \bigcirc\bigcirc B\bigcirc D & \bigcirc\bigcirc BC D & 2 \\ &  \bigcirc\bigcirc BF D & 2  \\\hline\end{array} \Rightarrow 共24種，BD互換再乘2，共24\times 2=\bbox[red,2pt]{48}種$$

解答：$$由x\gt 0，f(x)\gt 0 可知圖形為凹向上且y截距大於0；\\也就是同時滿足\cases{5-m \gt 0\\ m-5\ge 0} \Rightarrow m的解集合為\bbox[red,2pt]{\varnothing}$$

解答：$$假設A(0,1)，即\cases{橢圓\Gamma_1的b=1\\ 拋物線\Gamma_2的c=-1} \Rightarrow \Gamma_2:x^2 =-4(y-1) \Rightarrow y=1-x^2/4 \Rightarrow \cases{F_1(-2,0)\\ F_2(2,0)} \\ \Rightarrow 橢圓\Gamma_1的c=2 \Rightarrow 橢圓\Gamma_1的a=\sqrt 5\Rightarrow \Gamma_1: {x^2\over 5}+y^2=1 \\求兩圖形的交點:\cases{\Gamma_1: x^2/5+y^2 =1 \cdots(1) \\ \Gamma_2: x^2=4-4y \cdots(2)}，將(2)代入(1) \Rightarrow 5y^2-4y-1=0 \\\Rightarrow (5y+1)(y-1)=0 \Rightarrow \cases{y=1 \Rightarrow x=0\\ y=-1/5 \Rightarrow x=\pm {2\sqrt 6\over \sqrt 5}} \Rightarrow \cases{A(0,1)\\ P(-{2\sqrt 6\over \sqrt 5},-{1\over 5})\\ Q({2\sqrt 6\over \sqrt 5},-{1\over 5})}\\ \Rightarrow {\overline{OP} \over \overline{OA}} ={11/5\over 1} =\bbox[red,2pt]{11\over 5}$$

解答：

$$令\cases{A(0,0)\\ \angle A=\theta} \Rightarrow \cases{B(7,0)\\ C(3\cos\theta, 3\sin \theta)\\ P(3,0)\\ Q(2\cos\theta, 2\sin \theta)} \Rightarrow M=(B+C)/2 = \left({7+3\cos\theta \over 2},{3\over 2}\sin \theta \right) \\ \Rightarrow \cases{\overrightarrow{MP} =({1-3\cos\theta \over 2},-{3\over 2}\sin\theta)\\ \overrightarrow{MQ}=({\cos\theta -7\over 2},{1\over 2}\sin \theta)}；又\angle PMQ=90^\circ \Rightarrow \overrightarrow{MP} \cdot \overrightarrow{MQ}=0 \\ \Rightarrow {1\over 4}(3\cos^2\theta-22\cos\theta+7+ 3\sin^2\theta)=0 \Rightarrow 22\cos\theta = 10 \Rightarrow \cos A =\cos\theta= \bbox[red,2pt]{5\over 11}$$

解答：$$\begin{array}{} 總和& 三張牌& 組合數 & 累計\\\hline 5 &(1,1,3) & C^3_2C^3_1=9 & 9\\ &(1,2,2) & C^3_1C^3_2=9 &18 \\10 &(1,4,5) & C^3_1C^3_1C^3_1= 27 & 45 \\ &(2,3,5) &C^3_1C^3_1C^3_1= 27 & 72\\&(2,4,4) &C^3_1C^3_2= 9 &81 \\ &(3,3,4) &C^3_2C^3_1= 9 & 90\\ 15& (5,5,5)&1 & 91\\\hline \end{array} \\ \Rightarrow 機率為{91\over C^{15}_3} ={91\over 455} = \bbox[red,2pt]{1\over 5}$$

四、數學寫作能力：(共二題，共 12 分)

解答：$$X\sim G(p) \Rightarrow P(X=x)= \bbox[red,2pt]{p(x)=  p(1-p)^{x-1}} \Rightarrow E(X)= \sum_{x=1}^\infty xp(1-p)^{x-1} \\ =p\sum_{x=1}^\infty x(1-p)^{x-1} =-p\sum_{x=0}^\infty {d\over dp}(1-p)^{x}  =-p {d\over dp}\sum_{x=0}^\infty (1-p)^{x} =-p {d\over dp}{1\over p} =-p\cdot (-{1\over p^2}) ={1\over p}\\ \Rightarrow \bbox[red,2pt]{E(X)={1\over p}}$$

解答：

(1)$$假設對所有接近但不一定等於a的x，滿足h(x)\le f(x)\le g(x)，\\若\lim_{x\to a}h(x) = \lim_{x\to a}g(x) = L，則\lim_{x\to a}f(x) = L$$(2)$$取\cases{h(x)= x^2({1\over x}-1)=x-x^2\\[1ex] f(x)=x^2 \lfloor {1\over x}\rfloor\\[1ex] g(x)= x^2\cdot {1\over x}=x} \Rightarrow \lim_{x\to 0}h(x) =\lim_{x\to 0}g(x)= 0 \Rightarrow \lim_{x\to 0}f(x)=0$$

五、計算證明題：(共五題，共 38 分)

解答：

(1)$$(x+y)^n = \bbox[red,2pt]{\sum_{k=0}^n C^n_kx^ky^{n-k}}$$(2)$$\cases{(1+1)^n = C^n_0+ C^n_1+C^n_2+ C^n_3+\cdots +C^n_n \cdots(1)\\ (1+\omega)^n =C^n_0+ C^n_1 \omega +C^n_2 \omega^2+ C^n_3\omega^3+\cdots +C^n_n\omega^n \cdots(2) \\(1+\omega^2)^n =C^n_0+ C^n_1 \omega^2 +C^n_2 \omega^4+ C^n_3\omega^6+\cdots +C^n_n\omega^{2n} \cdots(3) } \\ (1)+(2)+(3) \Rightarrow 2^n+ (1+\omega)^n+(1+\omega^2)^n =3(C^n_0+ C^n_3+C^n_6+\cdots +C^n_n)\\ \Rightarrow C^n_0+ C^n_3+C^n_6+\cdots +C^n_n={1\over 3}(2^n+ (1+\omega)^n+(1+\omega^2)^n) \cdots(4)\\ 若n=6k,k\in \mathbb{N}，則\cases{(1+\omega)^n = (-\omega^2)^{6k} = (\omega^3)^{4k} =1\\ (1+\omega^2)^n = (-\omega)^{6k} = (\omega^3)^{2k}=1} \\\qquad \Rightarrow  C^n_0+ C^n_3+C^n_6+\cdots +C^n_n={1\over 3}(2^n+2);\\ 若n=6k-3,k\in \mathbb{N}，則\cases{(1+\omega)^n = (-\omega^2)^{6k-3} = -(\omega^3)^{4k-2} =-1\\ (1+\omega^2)^n = (-\omega)^{6k-3} = -(\omega^3)^{2k-1}= -1}\\\qquad \Rightarrow  C^n_0+ C^n_3+C^n_6+\cdots +C^n_n={1\over 3}(2^n-2)\\，\bbox[red,2pt]{故得證}$$

解答：$$S_n= 2\int_0^1 x^n-x^{n+2}\,dx \Rightarrow \sum_{n=1}^\infty S_n =2\int_0^1 x-x^3+x^2-x^4+x^3-x^5+\cdots\,dx \\ =2\int_0^1 x +x^2 \,dx =2\left. \left[ {1\over 2}x^2+{1\over 3}x^3 \right] \right|_0^1 =2\cdot {5\over 6} =\bbox[red, 2pt]{5\over 3}$$

解答：$$$$

解答：

(1)$$\omega= \cos{2\pi \over n} +i\sin {2\pi \over n} \Rightarrow \left|1-\omega^k \right| =\left|1-\cos{2k\pi \over n} -i\sin {2k\pi \over n} \right| = \sqrt{\left( 1-\cos{2k\pi \over n}\right)^2 + \sin^2{2k\pi \over n}}\\ = \sqrt{\left( 1-2\cos{2k\pi \over n} +\cos^2{2k\pi \over n}\right)  + \sin^2{2k\pi \over n}} =\sqrt{2-2\cos{2k\pi \over n}}  =\sqrt{2-2 (1-2\sin^2{k\pi \over n})} \\= \sqrt{4\sin^2{k\pi \over n}} =2\sin{k\pi \over n}，\bbox[red,2pt]{故得證}$$(2)$$z^n-1=(z-1)(z^{n-1}+z^{n-2} +\cdots +z+1)\\，因此z^{n-1}+z^{n-2} +\cdots +z+1=0的根為\omega^k,k=1-(n-1) \\ \Rightarrow f(z)= z^{n-1}+z^{n-2} +\cdots +z+1= (z-\omega)(z-\omega^2)\cdots (z-\omega^{n-1})\\ \Rightarrow f(1)= \color{blue}{n= (1-\omega)(1-\omega^2)\cdots (1-\omega^{n-1})}\\ 由題(1)可知:\sin{\pi\over n}\cdot \sin{2\pi\over n}\cdots\sin{(n-1)\pi\over n}  ={1\over 2}|1-\omega| \cdot{1\over 2}|1-\omega^2| \cdots {1\over 2}|1-\omega^{n-1}| \\={1\over 2^{n-1}}|\color{blue}{(1-\omega)(1-\omega^2) \cdots (1-\omega^{n-1})}| ={n\over 2^{n-1}}，\bbox[red,2pt]{故得證}$$

解答：$$(1)\cases{L_1: {x+1\over 2}= {y-2\over -2} ={z\over -1} \\L_2: {x-3\over 1} ={y-1\over -4} ={z-1\over 1}} \Rightarrow \cases{L_1方向向量\vec u=(2,-2,-1)\\ L_2方向向量\vec v=(1,-4,1)} \Rightarrow \vec n=\vec u\times \vec v=(-6,-3,-6)\\ \qquad\Rightarrow 包含L_1且與L_2平行之平面E:-6(x+1)-3(y-2)-6z=0 \Rightarrow \bbox[red,2pt]{2x+y+2z=0}\\(2)d(L_1,L_2) = d(L_2,E)= d((3,1,1),E) ={6+1+2\over \sqrt{4+1+4}} =\bbox[red,2pt]{3}$$
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解題僅供參考，主辦單未公布「四、數學寫作能力」及「五、計算證明題」的答案!!