107年全國科學班資格考-數學詳解

107 年度全國科學班聯合學科資格考試 數學科

一、 單選題：

  解答：

$$假設A(0,0)，則\cases{\overline{AB} =2\sqrt 2\\ \angle EAB= 45^\circ } \Rightarrow \cases{B(-2 , 2) \\ \angle ABC= 75^\circ} \Rightarrow \cases{C(2\sqrt 3-2 ,4)\\ \angle BCD=120^\circ} \Rightarrow D(2 \sqrt 3-2,10) \\ 因此\cases{\overrightarrow{AB} =(-2,2)\\ \overrightarrow{AD}=(2\sqrt 3-2,10)} \Rightarrow \overrightarrow{AB}\cdot \overrightarrow{AD} =-4\sqrt 3+4+20 =24-4\sqrt 3，故選 \bbox[red, 2pt]{(D)}$$

  解答：$$由於M值未定，所以(A),(B),(C)無法確定；\\又\cases{週期為\pi\\ x=2\pi/3 為對稱軸} \Rightarrow 對稱中心位於x={2\pi\over 3} \pm {n\pi\over 4},n\in \mathbb{N}，取n=1 \Rightarrow x={2\pi\over 3} - { \pi\over 4}={5\pi\over 12}\\ \Rightarrow ({5\pi\over 12},0) 為對稱中心， 故選\bbox[red,2pt]{(D)}$$

二、多選題:

  解答：$$已知\cases{\bar x= 60 = {1\over 30}\sum_{k=1}^{30} x_k\\ \sigma =10 = \sqrt{{1\over 30}\sum_{k=1}^{30}(x_k-\bar x)^2}}\\(A) \bigcirc:f(60)= f(\bar x)=\sum_{k=1}^{30}(\bar x-x_k)^2 =30\cdot \sigma^2 =3000 \\(B)\times: \sigma^2 ={1\over 30}\sum_{k=1}^{30} x_k^2 -\bar x^2 \Rightarrow {1\over 30}\sum_{k=1}^{30} x_k^2= \sigma^2+ \bar x^2 =100+60^2=3700\\ \qquad \Rightarrow \sum_{k=1}^{30} x_k^2=3700\times 30= 111000 \ne 12100 \\(C) \bigcirc: f(50) =\sum_{k=1}^{30}(50-x_k)^2 =\sum_{k=1}^{30}((60-x_k)-10)^2 =\sum_{k=1}^{30}((60-x_k)^2 -20(60-x_k)+100)\\ \qquad =\sum_{k=1}^{30}(60-x_k)^2- 20\sum_{k=1}^{30}(\bar x-x_k) + 30\times 100 =\sum_{k=1}^{30}(60-x_k)^2 +3000 =30\sigma^2+3000\\ \qquad =6000 \\(D) \bigcirc: f(61)= \sum_{k=1}^{30}((60-x_k)^2+30 =f(60)+30 \Rightarrow f(61)\gt f(60)\\ (E) \times: f(x)=\sum_{k=1}^{30} (x-x_k)^2 \Rightarrow f'(x)=2 \sum_{k=1}^{30} (x-x_k) \Rightarrow f(59) =2 \sum_{k=1}^{30} (60-x_k-1) \\ \qquad = 2 \sum_{k=1}^{30}(60-x_k)-2\times 30 =0-2\times 30 \lt 0 \\ 故選\bbox[red,2pt]{(ACD)}$$

  解答：$$(A)\times: y=g(x)為面積積分，顯然g(2)\gt (1)，g(1)不是最大值\\(B)\bigcirc: g(2)至g(4)的面積是負值，因此g(4)為極小值\\(C)\bigcirc: g(2)至g(3)面積為負值，且|g(x)|越來越大，面積越積越小\\(D)\times: g'(x)=f(x) \Rightarrow g''(x)=f'(x)，而f'(x)\lt 0, x\in (2,3)，因此為凹向下\\(E) \times: g''(x)=f'(x)=0 \Rightarrow 有三個相異解x=1,3,5，且f'(x)在此三解左右正負號均改變\\\qquad，因此皆為反曲點\\ 故選\bbox[red, 2pt]{(BCE)}$$

  解答：$$(A)\bigcirc: \alpha^2 -2\alpha \beta + 4\beta^2 =0 \Rightarrow ({\alpha \over \beta})^2-2{\alpha \over \beta} +4=0 \Rightarrow {\alpha \over \beta} =1\pm \sqrt 3i \\(B)\times: {\alpha \over \beta} =1\pm \sqrt 3i =2(\cos {\pi \over 3} \pm i \sin {\pi \over 3}) \Rightarrow \overrightarrow{OP}是\overrightarrow{OQ}繞原點旋轉\pm 60^\circ，且|\overrightarrow{OP}| =2|\overrightarrow{OQ}|\\(C)\bigcirc: |\overrightarrow{OP}| =2|\overrightarrow{OQ}| \Rightarrow \overline{OP} = 2\overline{OQ} \\(D)\bigcirc: \cases{\angle POQ=60^\circ \\ \overline{OP} =2\overline{OQ}} \Rightarrow \triangle POQ = 30^\circ-60^\circ-90^\circ\\(E)\times: \triangle POQ為直角\triangle \\ 故選\bbox[red,2pt]{(ACD)}$$

三、填充題:

  解答：$$甲比乙大的情形:(2,1),(3,1-2),(4,1-3),(5,1-4),(6,1-5)，共有15種情形；\\ 其中點數和為7的情形:(4,3),(5,2),(6,1)，有3種情形，因此機率為{3\over 15} = \bbox[red, 2pt]{1\over 5}$$

  解答：$$f(x)={a\over 3}x^3+{b\over 3}x 通過(1,1) \Rightarrow {a\over 3} +{b\over 3}=1 \Rightarrow a+b=3 \Rightarrow a=3-b;\\ G= 2\int_0^1 (x-f(x))\,dx ={1\over 3} \Rightarrow \int_0^1 (x-f(x))\,dx =\int_0^1 -{a\over 3}x^3 +{3-b\over 3}x\,dx\\  = \int_0^1 -{a\over 3}x^3 +{a\over 3}x\,dx={1\over 6}  \Rightarrow -{a\over 12}+{a\over 6}={1\over 6} \Rightarrow a=2 \Rightarrow b=3-a=1 \Rightarrow (a,b)= \bbox[red,2pt]{(2,1)}$$

  解答：

$$假設雙曲線中心點在原點，則雙曲線方程式為{x^2\over 12^2}-{y^2\over b^2}=1，且通過\cases{A(15,k)\\ C(20,k-50)},k\gt 0；\\ 將\cases{x=15\\ x=20}代入方程式\Rightarrow \cases{{k\over b} ={9\over 12} ={3\over 4}\\ {k-50\over b} =-{16\over 12}= -{4\over 3}}  \Rightarrow  {3\over 4}b-(-{4\over 3}b)=50 \Rightarrow b=24 \\ \Rightarrow 正交弦長={2b^2 \over a} ={ 2\times 24^2  \over 12} = \bbox[red, 2pt]{96}$$

  解答：

$$假設\triangle OBC的外心為P，則\overline{OP}為\overline{BC}的中垂線，並令\angle BAC=\angle POC= \theta \\ 餘弦定理:\cases{ \triangle OPC \Rightarrow \cos \theta ={R_1^2 +R_2^2-R_2^2\over 2R_1R_2} ={R_1\over 2R_2} ={R_1\over 4R_1} ={1\over 4} \\[1ex] \triangle ABC \Rightarrow \cos \theta ={14^2+11^2 -\overline{BC}^2 \over 2\cdot 14\cdot 11} ={317-\overline{BC}^2\over 308}} \Rightarrow {317-\overline{BC}^2\over 308}={1\over 4}\\ \Rightarrow \overline{BC}^2 =240 \Rightarrow \overline{BC}= 4\sqrt{15}\\ 再由正弦定理:\triangle ABC \Rightarrow {\overline{BC} \over \sin \theta } ={4\sqrt{15} \over \sqrt{15}/4}=2R_1 \Rightarrow R_1= \bbox[red, 2pt]{8}$$

  解答：$$\begin{array}{} n & \lfloor \log n\rfloor & 數量 & 小計\\\hline 1-9 & 0 & 9 &  9\times 0=0\\ 10-99 & 1 & 90 & 90\times 1 =90\\ 100-999 & 2 & 900 & 900\times 2=1800\\ 1000-1018 & 3 & 1019 & 1019\times 3= 3057\\\hline &   \sum &2018 & 4947\end{array} \Rightarrow \sum_{n=1}^{2018} a_n =\bbox[red, 2pt]{4947}$$

一、數學寫作能力：

  解答：$$體積公式為 \left| \vec a\cdot (\vec b\times \vec c)\right|\\先證明由\vec b及\vec c所張開的平行四邊形面積為|\vec b\times \vec c|\\假設\cases{\vec b=(b_1,b_2,b_3)\\ \vec c=(c_1,c_2,c_3)} \Rightarrow \vec b及\vec c所張開的平行四邊形面積= \sqrt{|\vec b|^2 |\vec c|^2 -(\vec b\cdot \vec c)^2} \\ =\sqrt{(b_1^2+b_2^2 +b_3^2)(c_1^2+c_2^2 +c_3^2)-(b_1c_1+ b_2c_2+b_3c_3)^2} =\sqrt{\begin{vmatrix} a_1 & a_2\\ b_1& b_2 \end{vmatrix}^2 +\begin{vmatrix} a_2 & a_3\\ b_2& b_3 \end{vmatrix}^2 +\begin{vmatrix} a_3 & a_1\\ b_3& b_1 \end{vmatrix}^2 }\\ =|\vec b\times \vec c|\\ 體積=|\vec b\times \vec c| \times 高，而高=\vec a在\vec b\times \vec c上的正射影長，即|\vec a| |\cos \theta| = {|\vec a\cdot (\vec b\times \vec c)|\over |\vec b\times \vec c|} \\ \Rightarrow 體積= |\vec b\times \vec c| \times  {|\vec a\cdot (\vec b\times \vec c)|\over |\vec b\times \vec c|}=\left| \vec a\cdot (\vec b\times \vec c)\right|，\bbox[red,2pt]{故得證}$$

  解答：

(1)$$當a=0或0\lt r\le 1時，數列收斂，收斂值為\begin{cases}0, & a=0或0\lt r\lt 1\\ a,& r=1 \end{cases}$$(2)$$等比級數前n項和S(n)={a(1-r^n)\over 1-r} \Rightarrow \lim_{n\to \infty}S(n) =\begin{cases} a & r=0\\ {a\over 1-r} & 0\lt |r|\lt 1 \\ 發散& |r|\ge 1\end{cases}$$

二、計算證明題：

  解答：$$ (2^x+ 2^{-x})^2 = 2^{2x}+ 2^{-2x} +2 \Rightarrow 2^{1+2x}+ 2^{1-2x} = 2\cdot 2^{2x} +2\cdot 2^{-2x} =2(2^{2x}+ 2^{-2x})
\\=2(2^x+2^{-x})^2-4 \Rightarrow 2^{1+2x}+2^{1-2x}-7(2^x+2^{-x})+ 9 = 2(2^x+2^{-x})^2-7(2^x+2^{-x})+5\\ =(2(2^x+2^{-x})-5) ((2^x+2^{-x})-1) \lt 0 \Rightarrow 1\lt 2^x+2^{-x}\lt 5/2 \Rightarrow 2\le 2^x+2^{-x}\lt 5/2\\ 令f(x)= 2^x +2^{-x}，則2\le f(x)\lt 5/2 \Rightarrow -1\lt x\lt 1\\ 令g(x)=10^x +10^{-x}，則g(1)=g(-1)={101\over 10} \Rightarrow \bbox[red, 2pt]{2\le 10^x+10^{-x} \lt {101\over 10}}$$

  解答：$$\cases{P(X=0) = (1-p)(1-q)\\ P(X=1)=p(1-q) +(1-p)q\\ P(X=2)=pq} \Rightarrow \cases{E(X)= 0+p(1-q) +(1-p)q+ 2pq ={p+q}\\ E(X^2)= 0+p(1-q) +(1-p)q+  4pq =p+q+2pq} \\ \Rightarrow \sigma(X)=\sqrt{ E(X^2)-(EX)^2 }= \sqrt{p+q+2pq -(p+q)^2} =\sqrt{p(1-p)+ q(1-q)} \\ \Rightarrow \cases{期望值=\bbox[red,2pt]{p+q} \\ 標準差=\bbox[red,2pt]{p(1-p)+ q(1-q)}}$$

  解答：$$y=f(x)= x^3+ Px^2+1 \Rightarrow f'(x)= 3x^2+2Px \\假設切點為(k,f(k)) =(k,k^3+Pk^2+1) \Rightarrow 切線斜率=f'(k)=3k^2+2Pk \\ \Rightarrow 切線L方程式:y=(3k^2+2Pk)(x-k)+k^3+Pk^2+1 \\ 又L通過(0,0)\Rightarrow 0=(-k)(3k^2+2Pk)+k^3+Pk^2+1 =-2k^3-Pk^2+1\\ 令g(k)=2k^3+Pk^2-1 \Rightarrow g'(k)=6k^2+2Pk=0 \Rightarrow k=0,-P/3\\ g(k)=0 有三相異實根\Rightarrow g(0)g(-P/3) \lt 0 \Rightarrow -{2\over 27}P^3 +{1\over 9}P^3-1 \gt 0 \Rightarrow P^3\gt 27 \Rightarrow \bbox[red,2pt]{ P\gt 3}$$

  解答：

(1)$$\cases{P\in L_1 \Rightarrow P(2s+1,-ks+2,ks+1),s\in \mathbb{R}\\ Q\in L_2 \Rightarrow Q(3t+2,kt+2,-5t+4/5),t\in \mathbb{R}}，若P=Q，則\cases{3t+2=2s+1 \cdots(1)\\ kt+2=-ks+2 \cdots(2)\\ -5t+4/5=ks+1\cdots(3)}；\\由(2)可得k(s+t)=0 \Rightarrow s=-t代入(1) \Rightarrow \cases{s=1/5\\ t=-1/5}再代入(3) \Rightarrow 1+{4\over 5} = {k\over 5}+1 \Rightarrow k= \bbox[red, 2pt]{4}$$(2)$$k=4 \Rightarrow P(2s+1,-4s+2,4s+1) \Rightarrow \cases{\overline{PA}^2 = (2s-2)^2 +(-4s+1)^2 +(4s+1)^2\\ \overline{PB}^2 = (2s)^2 +(-4s+2)^2 +(4s+2)^2}\\ \Rightarrow f(s)=\overline{PA}^2+\overline{PB}^2 =72s^2 -8s+14 \Rightarrow f'(s)=144s-8=0 \Rightarrow s={1\over 18} \\ \Rightarrow \bbox[red, 2pt]{P\left({10\over 9},{16\over 9},{11\over 9}\right)}$$

  解答：

(1)$$\cases{甲有a公升\\ 乙有b公升}，即\begin{bmatrix} a\\ b\end{bmatrix} \xrightarrow{甲一半給乙}\begin{bmatrix} a/2\\ b+a/2\end{bmatrix} \xrightarrow{乙一半給乙} \begin{bmatrix} 3a/4 +b/2\\ a/4+ b/2\end{bmatrix} \Rightarrow M= \bbox[red, 2pt]{\begin{bmatrix} 3/4 & 1/2\\ 1/4 & 1/2\end{bmatrix}}$$(2)$$\begin{bmatrix} a_n\\ 1-a_n\end{bmatrix} =M \begin{bmatrix} a_{n-1}\\ 1-a_{n-1}\end{bmatrix} \Rightarrow a_n={3\over 4}a_{n-1}+{1\over 2}(1-a_{n-1}) ={1\over 4}a_{n-1}+{1\over 2} \\ \Rightarrow \bbox[red,2pt]{遞迴式a_n=\begin{cases}a_{n-1}/4+1/2,& n\ge 1\\ 0.6& n=0\end{cases}}\\ a_n={1 \over 4}a_{n-1} +{1\over 2} ={1\over 4^2}a_{n-2} + {1\over 4}\cdot {1\over 2}+{1\over 2} ={1\over 4^3} a_{n-3} +{1\over 4^2}\cdot {1\over 2}  + {1\over 4}\cdot {1\over 2}+{1\over 2}  \\ ={1\over 4^n} a_0 +{1\over 2}(1+{1\over 4}+{1\over 4^2}+\cdots +{1\over 4^{n-1}}) ={1\over 4^n}\times 0.6+{1\over 2}\times {4 \over 3}\cdot (1-{1\over 4^{n}}) = {2\over 3}-{1\over 15}\cdot {1\over 4^n}\\ \Rightarrow \bbox[red,2pt]{一般式a_n= \begin{cases}2/3-1/(15\cdot 4^n), & n\ge 1\\ 0.6 & n=0 \end{cases} }$$

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