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2021年2月16日 星期二

107年中正預校教甄-數學詳解

 中正國防幹部預備學校107年教師甄選測驗試題

第壹部分:單選題


解答
$$x^{10} +x^8+x^6+x^4 +x^2+1=0 \Rightarrow (x^2-1)(x^{10} +x^8+x^6+x^4 +x^2+1)=0 \\ \Rightarrow x^{12}=1 \Rightarrow 其解為單位圓上正12邊形的頂點,但需扣除x^2-1=0,即x=\pm 1兩點\\ 因此所求凸多邊形面積=8個三角形及2個正三角形=8\times {1\over 2}\cdot 1\cdot 1\cdot\sin 30^\circ+ 2\times {1\over 2}\cdot 1\cdot 1 \cdot\sin 60^\circ \\ =2+{\sqrt 3\over 2},故選\bbox[red,2pt]{(2)}$$
解答:$$3^{4x-1} =2^{4x}+16^{x-3/4} \Rightarrow {1\over 3}\cdot 3^{4x} =2^{4x}+2^{4x-3} =2^{4x}+{1\over 8} \cdot 2^{4x} ={9\over 8}\cdot 2^{4x} \\ \Rightarrow {8\over 27} =({2\over 3})^{4x} \Rightarrow 3=4x \Rightarrow {3\over 4}={b\over a} \Rightarrow a+b= 7,故選\bbox[red,2pt]{(4)}$$
解答:$$z=x+yi,其中x,y\in R \Rightarrow |z-2i|+|z+4i| = \sqrt{x^2+(y-2)^2}+ \sqrt{x^2+(y+4)^2} \le 10\\ 由於\sqrt{x^2+(y-2)^2}+ \sqrt{x^2+(y+4)^2} = 10 為一橢圓,其中\cases{2a=10\\ 2c=2-(-4)=6} \Rightarrow \cases{a=5\\ c=3\\ b=4}\\ \Rightarrow 橢圓面積= ab\pi = 20\pi,故選\bbox[red,2pt]{(5)}$$
解答
$$20個頂點可決定C^{20}_3=1140個三角形,但需扣除以凸邊形的邊所形成的三角形\\(1)以連續兩個凸邊形的邊所形成的三角形:有20個,如上圖的紅色\triangle OPQ\\(2)只有一邊是凸邊形的邊所形成的三角形:對任意凸邊形的邊,可形成16個三角形;\\以上圖的\overline{IJ}為例,三角形另一個頂點不可以是I,J及相鄰的K,H,只能是剩下16個頂點;\\此類三角形有20\times 16=320個;\\因此共有1140-20-320=800個,故選\bbox[red,2pt]{(4)}$$
解答:$$\cases{ab+a+b= 524 \\ bc+b+c= 146\\ cd + c+d=104} \Rightarrow \cases{ab+a+b+1= 525 \\ bc+b+c +1= 147\\ cd + c+d +1=105} \Rightarrow \cases{(a+1)(b+1)= 525 \cdots(1)\\ (b+1)(c+1)= 147 \cdots(2)\\ (c+1)(d+1)=105  \cdots(3)} \\ \Rightarrow \cases{(1) \& (2) \Rightarrow b+1是525與147的公因數\\ (2) \& (3) \Rightarrow c+1是147與105的公因數} \qquad \Rightarrow \cases{b+1 \mid  \gcd(525,147)\\ c+1 \mid \gcd(147,105)} \\ \Rightarrow \cases{b+1\mid 21\\ c+1 \mid 21} \Rightarrow \cases{b+1=3,7,21\\ c+1=3,7,21\\ (b+1)(c+1)=147} \Rightarrow (b+1,c+1)=(7,21)或(21,7) \\ \Rightarrow \cases{(a+1,b+1, c+1,d+1) =(75,7,21,5)\\ (a+1,b+1, c+1,d+1) =(25,21,7,15) } \Rightarrow \cases{(a,b,c,d)=(74,6,20,4)\\ (a,b,c,d)=(24,20, 6,14)}\\ \Rightarrow \cases{a-d=70\\ a-d=10},故選\bbox[red,2pt]{(2)}$$
解答:$$(x+(y+z))^{2018} +(x-(y+z))^{2018} \\=\sum_{k=0}^{2018}C^{2018}_k\left( x^k(y+z)^{2018-k} +x^k(-1)^{2018-k}(y+z)^{2018-k}\right)\\ =2\sum_{k=0}^{1009}C^{2018}_{2k}x^{2k}(y+z)^{2018-2k}\\由於(y+z)^{2018-2k}有2018-2k+1個不同的項,因此共有\sum_{k=0}^{1009}(2018-2k+1) \\=\sum_{k=0}^{1009}(2019-2k)= 2019\times 1010-2\times {1010\times 1009\over 2}=1010^2=2020100 項,故選\bbox[red,2pt]{(1)}$$
解答
    第2個圓與第1個圓最多有2個交點;
    第3個圓與其它2個圓最多有4個交點;
    第4個圓與其它3個圓最多有6個交點;
    第5個圓與其它4個圓最多有8個交點;
    因此五個不同的圓最多有2+4+6+8=20個交點,故選\(\bbox[red,2pt]{(5)}\)

解答:$$假設下方格子的代號是「下」,上方格子的代號是「上」,即:\begin{array}{|c|c|c|c|c|c|c|}\hline 上&上& 上&上&上 &上&上\\ \hline 下 & 下& 下& 下& 下& 下& 下\\\hline\end{array};\\ 對任意符合要求的結果,如\begin{array}{|c|c|c|c|c|}\hline 1& 2&5&6 &7 & 11 & 12\\ \hline 3 & 4& 8& 9& 10 & 13 & 14\\\hline\end{array}。\\按照1-10的順序寫下代號,此例的代號序列就是:上上下下上上上下下下上上下下。\\反是符合要求的代號序列一定是7個下7個上,且第1個一定是上,最後1個是下;\\而且由左向右觀察,上的數量大於等於下的數量,這也就是n=7 \text{的卡特蘭數(Catalan number)}\\ 即C(n=7)={1\over n+1}C^{2n}_n ={1\over 8}C^{14}_7=429,故選\bbox[red,2pt]{(3)}$$

解答
$$假設小圓半徑為a,且\overline{BC}與小圓的切點為D,則\overline{OD}\parallel \overline{AB} \Rightarrow {\overline{OD} \over \overline{AB} } ={\overline{CO}\over \overline{CA}} \\ \Rightarrow {a\over 12} ={3a\over 6a} \Rightarrow a=6 \Rightarrow 大圓半徑3a=18,故選\bbox[red,2pt]{(2)}$$

解答:$$A=\{1,2,3,\dots,60\}= A_0\cup A_1\cup \cdots \cup A_6,其中A_i=\{7k+i\mid 0\le 7k+i\le 60,k\in N\}\\ \Rightarrow\cases{ n(A_0) =n(A_5)=n(A_6)=8\\ n(A_1)=n(A_2) =n(A_3)=n(A_4)=9 } \Rightarrow A_1-A_6個別集合符合要求\\,但A_1\cup A_6,A_2\cup A_5,A_3\cup A_4就不符合要求。為求n(S)之最大數,可取S= A_1\cup A_2\cup A_3\\,則n(S)=3\times 9=27,再加上A_0任一元素,仍符合要求,因此最大的n(S)=28,故選\bbox[red,2pt]{(5)}$$
解答:$$三角函數公式: \cos(3\theta) = 4\cos^3\theta -3\cos \theta\\令f(x)=x^3-3x+1 = (x-2\cos \alpha)(x-2\cos \beta)(x-2\cos \gamma) \\ \Rightarrow \cases{f(2\cos \alpha) =8\cos^3\alpha-6\cos\alpha+ 1=0 \\f(2\cos \beta) =8\cos^3\beta-6\cos\beta+ 1=0 \\f(2\cos \gamma) =8\cos^3\gamma-6\cos\gamma+ 1=0 } \Rightarrow \cases{2\cos(3\alpha)=-1 \\2\cos(3\beta)=-1 \\2\cos(3\gamma)=-1 }\Rightarrow \cases{\cos(3\alpha)=-1/2 \\\cos(3\beta)=-1/2 \\\cos(3\gamma)=-1/2 } \\ \Rightarrow \cases{3\alpha =120^\circ,240^\circ, 480^\circ \\3\beta =120^\circ,240^\circ, 480^\circ \\3\gamma =120^\circ,240^\circ, 480^\circ }\Rightarrow \cases{\alpha =40^\circ,80^\circ, 160^\circ \\ \beta =40^\circ,80^\circ, 160^\circ \\ \gamma = 40^\circ,80^\circ, 160^\circ }\\ 由於 \alpha \lt \beta \lt \gamma \Rightarrow \cases{\alpha=40^\circ \\ \beta=80^\circ \\ \gamma=160^\circ} \Rightarrow \sin(\gamma-\alpha)= \sin(160^\circ-40^\circ)=\sin(120^\circ) = \bbox[red,2pt]{\sqrt 3\over 2}$$
解答


$$延長\overline{AB},並在其上找一點D,使得\overline{BC}= \overline{BD}=a,則\triangle BCD為等腰,如上圖;\\ 令\angle D=\theta \Rightarrow \angle ABC=2\theta \Rightarrow \angle ACB = 180^\circ-42^\circ-2\theta=138^\circ-2\theta;\\由於b^2-c^2=ac \Rightarrow b^2=c(a+c) \Rightarrow {b\over c} ={a+c\over b} \Rightarrow {\overline{AC} \over \overline{AB}} ={\overline{AD} \over \overline{AC}} \Rightarrow \triangle ABC \sim \triangle ACD \\ \Rightarrow \angle ACB=\angle D \Rightarrow 138^\circ-2\theta = \theta \Rightarrow \theta=138^\circ\div 3= \bbox[red,2pt]{46}^\circ$$
解答:$${N^2+7 \over N+4} =N-4 + {23 \over N+4}不是最簡分數 \Rightarrow N+4 是23的倍數;\\ 由於1 \le N \le 2018 \Rightarrow 5 \le N+4 \le 2022 \Rightarrow N+4=23,46, 69,...,23\times 87=2001\\ \Rightarrow 共有\bbox[red,2pt]{87}個$$
解答:$$\begin{vmatrix} a+b & b+c & c+a \\ x+y & y+z & z+x \\ 3 & 4 & 3\end{vmatrix} \underrightarrow{c_2+c_1,c_3+c_1} \begin{vmatrix} 2(a+b+c) & b+c & c+a \\ 2(x+y+z) & y+z & z+x \\ 10 & 4 & 3\end{vmatrix} =2 \begin{vmatrix} a+b+c & b+c & c+a \\ x+y+z & y+z & z+x \\ 5 & 4 & 3\end{vmatrix} \\\underrightarrow{-c_1+c_2 ,-c_1+c_3}\quad 2 \begin{vmatrix} a+b+c & -a & -b \\ x+y+z & -x & -y \\ 5 & -1 & -2\end{vmatrix} \underrightarrow{c_2+c_1,c_3+c_1}\quad 2 \begin{vmatrix} c  & -a & -b \\  z & -x & -y \\ 2 & -1 & -2\end{vmatrix}=2 \begin{vmatrix} c  & a & b \\  z & x & y \\ 2 & 1 & 2\end{vmatrix};\\ 由於\begin{vmatrix} c  & a & b \\  z & x & y \\ 2 & 1 & 2\end{vmatrix}的最大值為\sqrt{c^2+a^2+b^2} \times \sqrt{z^2+x^2+y^2} \times \sqrt{2^2+1^2+2^2} = \sqrt{1} \times \sqrt {4}\times \sqrt {9}\\=6 \Rightarrow \begin{vmatrix} a+b & b+c & c+a \\ x+y & y+z & z+x \\ 3 & 4 & 3\end{vmatrix}的最大值為2\times 6=\bbox[red,2pt]{12}$$:$$\cases{\vec u=(u_1,u_2,u_2)\\ \vec v=(v_1,v_2,v_3)\\ \vec w=(w_1,w_2,w_3)} \Rightarrow 平行六面體體積V= \begin{vmatrix} u_1 & u_2 & u_3\\ v_1 & v_2 & v_3\\ w_1& w_2 & w_3\end{vmatrix}\\ 什麼情況下V值最大?就是在三向量\vec u,\vec v,\vec w,互相垂直,\\此時V=|\vec u||\vec v||\vec w| = \sqrt{u_1^2 +u_2^2 +u_3^2} \times \sqrt{v_1^2+v_2^2+v_3^2} \times \sqrt{w_1^2 +w_2^2 +w_3^2}$$
解答:$$\cases{O(0,0,0)\\ A(a,0,0)\\ B(0,b,0)\\ C(0,0,c)} \Rightarrow \triangle ABC面積={1\over 2} \sqrt{|\overrightarrow{AB}|^2 |\overrightarrow{AC}|^2 -(\overrightarrow{AB} \cdot \overrightarrow{AC})^2} \\={1\over 2}\sqrt{|(-a,b,0)|^2|(-a,0,c)|^2 -((-a,b,0) \cdot (-a,0,c))^2} ={1\over 2}\sqrt{(a^2+b^2)(a^2+c^2)-a^4} \\ ={1\over 2}\sqrt{a^2c^2 +a^2b^2 +b^2c^2} =4 \Rightarrow a^2c^2 +a^2b^2 +b^2c^2=64;\\ 令k=|\overrightarrow{OA} \times \overrightarrow{OB}| +2|\overrightarrow{OB} \times \overrightarrow{OC}| +2|\overrightarrow{OC} \times \overrightarrow{OA}| = |(0,0,ab)| +2|(bc,0,0)| + 2|(0,ac,0)| \\ =ab+2bc+ 2ca\\ 柯西不等式:((ab)^2+ (bc)^2 +(ca)^2)(1^2 +2^2 +2^2) \ge (ab+2bc+2ca)^2 =k^2\\ \Rightarrow 64\times 9\ge k^2 \Rightarrow 24 \ge k \Rightarrow 最大值=\bbox[red,2pt]{24}$$
解答$$\cases{a_1=1= \tan(45^\circ)\\ a_2={\sqrt 3\over 3} =\tan(30^\circ)\\ } ,令\cases{\theta_1=45^\circ\\ \theta_2= 30^\circ \\},由a_{n+2} ={a_n+a_{n+1} \over 1-a_na_{n+1}} \\\Rightarrow  a_{n+2} =\tan \theta_{n+2} =\tan(\theta_{n+1}+\theta_n) ={\tan \theta_n +\tan \theta_{n+1} \over 1-\tan \theta_n\tan \theta_{n+1}};\quad \tan \theta的周期為180^\circ,由此可得下表\\ \begin{array}{}\hline n & 1 & 2 & 3 & 4& 5 & 6& 7 & 8& 9 & 10& 11 & 12& 13 & 14& 15 \\  \theta_n & \color{blue}{45} & \color{blue}{30} & 75 &105 & 0 & 105 & 105 & 30 & 135 & 165 & 120 & 105 & 45 & 150 & 15  \\\hdashline n &16& 17 & 18& 19 & 20& 21 & 22 & 23 & 24 & 25 & 26\\ \theta_n & 165 & 0 & 165 & 165 & 150 &135 & 105 & 60 & 165 & \color{blue}{45} & \color{blue}{30}\\\hline\end{array}\\ 由上表可知此數列的周期為24,因此2018=24\times 84+2 \Rightarrow a_{2018}= a_2= \bbox[red,2pt]{\sqrt 3\over 3}$$
解答:$$|z+3-3i|=|z-(-3+3i)|=2 \Rightarrow 以O_1(-3,3)為圓心,半徑r_1=2的圓;\\|iw-1|=|i(w-{1\over i})|= |i||w+i|=1 \Rightarrow |w+i|=1 \Rightarrow 以O_2(0,-1)為圓心,半徑r_2=1的圓\\|z-w| 即為兩圓上點的距離,最大值為\overline{O_1O_2}+r_1+r_2 = \sqrt{9+16}+2+1 =\bbox[red,2pt]{8}$$
解答

$$此題相當於求兩圖形\cases{y=f(x)=\sin x-3\cos x\\ y=k},在0\le x \le \pi 限制下,有兩個交點的條件;\\ f(x)=\sin x-3\cos x = \sqrt{10}\sin (x-\theta) \Rightarrow f(x)的最大值為\sqrt{10},最小值為-\sqrt{10};\\ 又f'(x)=\cos x+3\sin x \Rightarrow f'(x)=0 \Rightarrow \tan x = -{1\over 3} \Rightarrow \tan {5\over 6}\pi <\tan x < \tan \pi \\ 因此若要兩個交點,則 f(\pi) \le k< f的最大值 \Rightarrow \bbox[red, 2pt]{3 \le k < \sqrt{10}}$$


解答:$$\cases{z_1= \cos{\pi\over 3}+i\sin{\pi\over 3} = e^{{\pi\over 3}i} \\z_1= \cos{\pi\over 4}+i\sin{\pi\over 4} = e^{{\pi\over 4}i} } \Rightarrow \cases{z_3 = z_1\cdot z_2 =e^{({1\over 3}+{1\over 4})\pi i} =e^{{1\over 12}\pi i} =\cos {1\over 12}\pi +i\sin {1\over 12}\pi\\z_4 ={z_1\over z_2} =e^{({1\over 3}- {1\over 4})\pi i} =e^{{7\over 12}\pi i} =\cos {7\over 12}\pi +i\sin {7\over 12}\pi} \\ 令z_3'= \cos {1\over 12}\pi -i\sin {1\over 12}\pi,則|a-z_3|+ |a-z_4|的最小值= |z_3'-z_4| \\=\sqrt{\left(\cos{7\pi\over 12} -\cos{\pi\over 12}\right)^2 + \left(\sin{7\pi\over 12} +\sin{\pi\over 12}\right)^2} =\sqrt{2-2\left(\cos{7\pi\over 12}\cos{\pi\over 12} -\sin{7\pi\over 12}\sin{\pi\over 12}\right)} \\ =\sqrt{2-2\cos ({7\pi\over 12} +{\pi\over 12})} =\sqrt{2-2\cos {2\pi\over 3}} =\sqrt{2+1} =\bbox[red,2pt]{\sqrt 3}$$
解答:$$由於(\sqrt 3-\sqrt 2)^6非常小,因此 (\sqrt 3+\sqrt 2)^6 \approx (\sqrt 3+\sqrt 2)^6 +(\sqrt 3-\sqrt 2)^6  \\而(\sqrt 3+\sqrt 2)^6 +(\sqrt 3-\sqrt 2)^6 =  ((\sqrt 3+\sqrt 2)^2)^3 + ((\sqrt 3-\sqrt 2)^2)^3 =(5+2\sqrt 6)^3 +(5-2\sqrt 6)^3\\ = 10(49+20\sqrt 6-1+49-20\sqrt 6)=970,因此大於(\sqrt 3+\sqrt 2)^6的最小整數為\bbox[red,2pt]{970}$$
$$\cases{x+y=5 \cdots(1)\\ x^2+z^2 +xz =16\cdots(2) \\ y^2+z^2-yz= 9\cdots(3)},由(1) \Rightarrow x=5-y 代入(2) \Rightarrow (5-y)^2+z^2 +(5-y)z=16 \\ \Rightarrow y^2+z^2-yz +9-10y+5z=0 \cdots(4); 將(3)代入(4) \Rightarrow 9+9-10y+5z=0 \\ \Rightarrow y={18+5z\over 10} 代入(3) \Rightarrow \left({18+5z\over 10}\right)^2 +z^2-\left({18+5z\over 10}\right)z=9 \Rightarrow {3\over 4}z^2 ={144\over 25} \Rightarrow z=\pm {8\sqrt 3\over 5} \\ \Rightarrow xz+yz= z(x+y)=\pm {8\sqrt 3\over 5} \times 5 = \bbox[red,2pt]{\pm 8\sqrt 3}$$: 學校公布的答案是\(\bbox[red,2pt]{ 8\sqrt 3}\)
解答:$$x_n= x_{n-1}-x_{n-2}+x_{n-3}-x_{n-4} \Rightarrow x_5=x_4-x_3+x_2-x_1 \Rightarrow x_6=x_5-x_4+x_3-x_2 \\ =(x_4-x_3+x_2-x_1) -x_4+x_3-x_2 = -x_1;\\同理可得x_7=-x_2,x_8=-x_3,\dots,x_{11}=-x_6=x_1,x_{12}=x_2,\dots,x_{15}=x_5,x_{16}=-x_1\\因此循環數為10 \Rightarrow x_{31}+ x_{53} +x_{1975}= x_1+x_3+x_5 =21+42+ (23-42+37-21)= \bbox[red,2pt]{60}$$
解答
$$\cases{|z|=1\\ z^{28}-z^8-1=0} \Rightarrow z在單位圓上,且z^{28}在z^8右邊,水平距離為1的地方;也就是z^8=e^{{2\over 3}\pi i},e^{{4\over 3}\pi i},如上圖;\\ z=e^{\theta i} \Rightarrow \cases{z^8=e^{8\theta i} \\z^{28}=e^{28\theta i}},其中\cases{(8\theta, 28\theta)= ( 120^\circ +360^\circ \times k,60^\circ+360^\circ \times t), k,t=0,1,2,\dots\\ (8\theta, 28\theta)=(240^\circ+360^\circ \times k,300^\circ+360^\circ \times t) ,k,t=0,1,2,\dots}\\ \begin{array}{} \hline 8\theta & 120^\circ & 480^\circ & 840^\circ & 1200^\circ & 1560^\circ & 1920^\circ& 2280^\circ & 2640\\\hdashline  \theta & 15^\circ & 60^\circ & 105^\circ & 150^\circ & 195^\circ & 240^\circ &285^\circ & 330^\circ \\\hdashline 28\theta & 420^\circ &1680^\circ &2940^\circ &4200^\circ &5460^\circ &6720^\circ &7980^\circ &9240^\circ \\ & (60^\circ) &(240^\circ) &(60^\circ) &(240^\circ) &(60^\circ) &(240^\circ) & (60^\circ) &(240^\circ) \\\hline\end{array} \\ 以上符合條件者: \theta = 15^\circ,105^\circ,195^\circ,285^\circ;\\\quad \\\begin{array}{} \hline 8\theta & 240^\circ & 600^\circ & 960^\circ & 1320^\circ & 1680^\circ & 2040^\circ& 2400^\circ & 2760\\\hdashline  \theta & 30^\circ & 75^\circ & 120^\circ & 165^\circ & 210^\circ & 255^\circ &300^\circ & 345^\circ \\\hdashline 28\theta & 840^\circ &2100^\circ &3360^\circ &4620^\circ &5880^\circ &7140^\circ &8400^\circ &9660^\circ \\ & (120^\circ) &(300^\circ) &(120^\circ) &(300^\circ) &(120^\circ) &(300^\circ) & (120^\circ) &(300^\circ) \\\hline\end{array} \\ 以上符合條件者: \theta = 75^\circ,165^\circ,255^\circ,345^\circ;\\所有符合條件,依大小順序為15^\circ,75^\circ,105^\circ, 165^\circ, 195^\circ, 255^\circ,285^\circ, 345^\circ\\依題意只取偶數位置,所求之值為75^\circ+ 165^\circ+ 255^\circ+ 345^\circ = \bbox[red,2pt]{840}^\circ$$
解答:$$假設一銳角三角形,其三邊長為x,y,z,相對應的高分別為{1\over 4},{1\over 5}及{1\over 6};\\因此\triangle 面積={1\over 2} \cdot x\cdot {1\over 4} ={1\over 2} \cdot y\cdot {1\over 5} ={1\over 2} \cdot z\cdot {1\over 6} \Rightarrow x:y:z = 4:5:6 \\ 因此假設\cases{x=4k\\ y=5k\\ z=6k} \Rightarrow s= (x+y+z)\div 2= {15\over 2}k \\ \Rightarrow \triangle 面積=\sqrt{s(s-x)(s-y)(s-z)} ={1\over 2} \cdot x\cdot {1\over 4} \Rightarrow \sqrt{{15\over 2}k \cdot {7\over 2}k \cdot {5\over 2}k \cdot {3\over 2}k} =  {1\over 2}k \\ \Rightarrow k={4\over 30\sqrt 7} \Rightarrow x+y+z = 15k = 15\cdot {4\over 30\sqrt 7} ={2\over \sqrt 7} = {m\over \sqrt n} \Rightarrow m+n= \bbox[red,2pt]{9}$$

現在來解釋一下,為什麼可以這樣假設。以上圖為例:$$x=\overline{BC} =\overline{BD} +\overline{DC} = \sqrt{y^2-{1\over 16}} + \sqrt{z^2-{1\over 16}} \\ \Rightarrow x=\sqrt{y^2-{1\over 16}} + \sqrt{z^2-{1\over 16}},即為題意聯立方程式之一;\\另二式也可用相同方法求得。$$


解答


$$令\cases{f(x)=\log_{107}x \\g(x)=107^x} \Rightarrow f(x)=g^{-1}(x) \Rightarrow y=f(x)與y=g(x)對稱於直線y=x;\\ 兩直線\cases{y=x\\ y=-x+3}交於P(3/2,3/2) \Rightarrow {f(\alpha)+g(\beta)\over 2} ={3\over 2} \Rightarrow f(\alpha)+g(\beta)= \bbox[red,2pt]{3}$$
解答:$$a=\sqrt[3]{3-\sqrt 5\over 2} +\sqrt[3]{3+\sqrt 5\over 2} \Rightarrow a^3=3+3a \Rightarrow a^6=(3+3a)^2 = 9a^2+18a+9 \\ \Rightarrow a^6-9a^2-18a-4 = 9-4=\bbox[red,2pt]{5}$$
解答:$$令兩等差數列\cases{\langle a_n\rangle =a,a+d,a+2d,\dots\\ \langle b_n \rangle= b,b+e,b+2e,\dots},由題意知:\cases{ab=1440\\ (a+d)(b+e)= 1716\\ (a+2d)(b+2e) =1848} \\ \Rightarrow \cases{ab=1440 \cdots(1)\\ ab+ae+bd+ de=1716 \cdots(2) \\ ab+2(ae+bd)+ 4de=1848 \cdots(3)},將(1)代入(2)及(3) \Rightarrow \cases{ae+bd+de= 276\cdots(4)\\ ae+bd+ 2de= 204 \cdots(5)} \\ (5)-(4) \Rightarrow de=-72 代入(4) \Rightarrow ae+bd= 348 \Rightarrow a_8\times b_b =(a+7d)(b+7e) \\ =ab+7(ae+bd) + 49de = 1440+7\times 348+49\times (-72)= \bbox[red,2pt]{348}$$
解答:$$1000\sum_{n=3}^{10000}{1\over n^2-4} =1000\sum_{n=3}^{10000}{1\over 4}\left({1\over n-2}-{1\over n+2}\right) = 250\sum_{n=3}^{10000} \left({1\over n-2}-{1\over n+2}\right) \\ =250\left[\left({1\over 1}+{1\over 2}+\cdots+{1\over 9998} \right) -\left( {1\over 5} +{1\over 6}+\cdots +{1\over 10002}\right)\right] \\ =250 \left({1\over 1}+{1\over 2}+ {1\over 3} +{1\over 4} -{1\over 9999}-{1\over 10000}-{1\over 10001}-{1\over 10002} \right) \approx 250\left({25\over 12}-{1\over 2500} \right)\\ \approx 520.8-0.1= 520.7 \approx 521 \Rightarrow a+b+c=5+2+1=\bbox[red,2pt]{8}$$
解答
$$大塊面積=扇形OAB-扇形OCD = {20\over 360}(80^2-60^2)\pi\\小塊面積=扇形ODE-扇形OFG = {20\over 360}(60^2-40^2)\pi\\大塊面積-小塊面積 ={20\over 360}(80^2-2\times 60^2+40^2)\pi =\bbox[red,2pt]{400 \over 9}\pi$$


解答


$$\overline{AE}=\overline{AD} \Rightarrow \angle AEC =\angle ADE=\theta \Rightarrow \angle DAE =180^\circ-2\theta\\ \Rightarrow \angle C= (180^\circ-(30^\circ+180^\circ-2\theta))\div 2=\theta -15^\circ\\ \angle AED=\angle C+\angle EDC \Rightarrow \theta = x+ \theta-15^\circ \Rightarrow x=\bbox[red,2pt]{15}^\circ$$











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  1. 您好:請問第13題答案是不是怪怪的,因為我代答案幾乎都是最簡分數喔,謝謝

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