115年宜蘭高中教甄-數學詳解

 國立宜蘭高級中等學校 115 學年度教師甄選

第一部份： 填充題（共 16 題， 每題 5 分，占 80 分）

解答:$$P(n)的公式: P(n)=n^{d(n)/2}, 其中 d(n)為n的正因數個數\\ P(n)=2^{36} \times 3^{24} \times 7^{12} \Rightarrow n= 2^a\times 3^b\times 7^c \Rightarrow d(n)=(a+1) (b+1)(c+1) \\ \Rightarrow P(n)= \left( 2^a\times 3^b\times 7^c \right)^{d(n)/2} \Rightarrow \cases{36= a\cdot d(n)/2 \\24=b\cdot d(n)/2\\ 12=c\cdot d(n)/2} \Rightarrow a:b:c=36:24:12 =3:2:1 \\\Rightarrow \cases{a=3k\\ b=2k\\ c=k} \Rightarrow d(n)=(3k+1)(2k+1)(k+1) \Rightarrow 12=k\cdot {(3k+1)(2k+1)(k+1) \over 2} \\ \Rightarrow k(k+1)(2k+1)(3k+1)=24 \Rightarrow k=1 \Rightarrow \cases{a=3\\b=2\\c=1} \Rightarrow n=2^3\times 3^2\times 7^1= \bbox[red, 2pt]{504}$$

解答:$$y=3^x \Rightarrow x=\log_3 y \Rightarrow f(3^x)= 4x\log_2 3+233 \Rightarrow f(y)=4(\log_3 y)(\log_2 3)+233 =4\log_2 y+233 \\ \Rightarrow f(2^k)=4\log_2 2^k+233=4k+233 \Rightarrow \sum_{k=1}^{10}f(2^k) =4\sum_{k=1}^{10}k+ \sum_{k=1}^{10}233 =4\cdot 55+2330= \bbox[red,2pt]{2550}$$

解答:$$M= \begin{bmatrix}1&5\\0&3a \end{bmatrix} \Rightarrow \cases{A'=MA  = (2,3a) \\B'=MB =(20,6a) \\ C'=MC=(18,9a)} \Rightarrow \cases{\overrightarrow{B'A'} =(-18,-3a) \\ \overrightarrow{B'C'}=(-2,3a)} \\ \Rightarrow \overrightarrow{B'A'} \cdot \overrightarrow{B'C'}=0 \Rightarrow 36-9a^2=0\Rightarrow a= \bbox[red, 2pt]2 \\也可以假設A為直角,或C為直角，結果a都不是整數$$

解答:$$P \in \Gamma:{x^2\over 16}+{y^2\over 4}=1 \Rightarrow P(4\cos \theta, 2\sin \theta) \Rightarrow P'= \begin{bmatrix}1&3\\0& 1 \end{bmatrix} \begin{bmatrix}4\cos \theta\\ 2\sin \theta \end{bmatrix} = \begin{bmatrix}4\cos \theta+6\sin \theta\\ 2\sin \theta \end{bmatrix} \\ \Rightarrow P'與y軸的距離=|4\cos\theta+ 6\sin \theta|=|2\sqrt{13}\sin(\alpha+\theta)| \Rightarrow 最大值為 \bbox[red, 2pt]{2\sqrt{13}}$$

解答:$$\cases{\vec a=(4,-3,1) \\ \vec b=(2,2,k)} \Rightarrow \vec a\times \vec b=(-3k-2,2-4k,14) \Rightarrow (1,2,2) \parallel \vec a\times \vec b \\ \Rightarrow {-3k-2\over 1} ={2-4k\over 2}={14\over 2}  \Rightarrow k=-3 \Rightarrow \vec a\times \vec b=(7,14,14)\\ \Rightarrow \cases{|\vec a\times \vec b|=\sqrt{441} =21\\ |(1,2,2)|=3} \Rightarrow 體積=3\times 21=\bbox[red, 2pt]{63}$$

解答:$$\cases{a,c\in \{1,2,3,4,5,6\} \\ b\in \{2,4,6\}} \Rightarrow 總樣本有6\times 3\times 6=108種可能 \\ ax^2-bx+c=0有實根\Rightarrow \Delta=b^2-4ac\ge 0 \Rightarrow b^2\ge 4ac \\\textbf{Case I }b=2: 4\ge 4ac \Rightarrow ac=1 \Rightarrow (a,c)=(1,1) \Rightarrow 只有1種\\ \textbf{Case II }b=4: 16\ge 4ac \Rightarrow ac\le 4 \Rightarrow (a,c)=(1,1-4),(2,1-2)(3,1),(4,1) \Rightarrow 8種 \\ \textbf{Case III }b=6: 36\ge4ac \Rightarrow ac\le9 \Rightarrow (a,c)=(1,1-6),(2,1-4),(3,1-3),(4,1-2),(5-6,1)\\\qquad  \Rightarrow 共17種 \\ 合計:1+8+17=26種\Rightarrow 機率={26\over 108} = \bbox[red, 2pt]{13\over 54}$$

解答:$$6x^3+11x^2-3x-3=x+1 \Rightarrow 6x^3+11x^2-4x-4=0 \Rightarrow (x+2)(3x-2)(2x+1)=0 \\ \Rightarrow x=\bbox[red, 2pt]{{3\over 2}或-{1\over 2}} \; (x=-2 \Rightarrow x+1\lt 0不合)$$

解答:$$\cases{(5\sin A+ 6\cos B)^2= 49 \\ (6\sin B+ 5\cos A)^2=16} \Rightarrow 兩式相加\\\Rightarrow 25(\sin^2A+ \cos^2 A)+36(\sin^2B+ \cos^2B)+60(\sin A\cos B+ \sin B\cos A)=65 \\ \Rightarrow 25+36+ 60\sin(A+B)=65 \Rightarrow \sin (A+B) = \sin(\pi-C)= {1\over 15} \Rightarrow \sin C= \bbox[red, 2pt]{1\over 15}$$

解答:$$S_n=n^3+2n^2-503n \Rightarrow a_n=S_n-S_{n-1} =n^3+2n^2-503n- \left( (n-1)^3+2(n-1)^2-503(n-1) \right) \\=3n^2+n-504 \Rightarrow \cases{a_{12}= -60\lt 0\\a_{13}=16\gt 0} \Rightarrow \sum_{n=1}^{30} |a_n|= \sum_{n=1}^{12} (-a_n)+ \sum_{n=13}^{30} a_n =-S_{12}+(S_{30}-S_{12}) \\=S_{30}-2S_{12} =13710-2 \times(-4020) = \bbox[red, 2pt]{21750}$$

解答:$$L_1: \cases{x+y+z=1\\ x+y=1} \Rightarrow \cases{x+y=1\\ z=0} \Rightarrow L_1:(t,1-t,0) \Rightarrow L_1方向向量 \vec v_1=(1,-1,0) \\ \Rightarrow E_1:x+y+z=1的法向量\vec n_1=(1,1,1) \Rightarrow \vec n_1\times \vec v_1= (1,1,-2) \\ \Rightarrow 通過(1,0,0)且法向量為(1,1,-2)的投影面P_1:(x-1)+(y-0)-2(z-0)=0  \\ \Rightarrow P_1: x+y-2z=1\\ 同理, E_2:z=0 \Rightarrow \vec n_2=(0,0,1), 又L_2:\cases{z=0\\ x+2y+z=3} \Rightarrow \cases{z=0\\ x+2y=3} \Rightarrow P_2:x+2y=3 \\ \Rightarrow L=P_1\cap P_2 :\cases{x+y-2z=1\\ x+2y=3} , 取z=t \Rightarrow \cases{x+y=1+2t\\ x+2y=3} \Rightarrow \cases{x=4t-1\\ y=2-2t} \\ \Rightarrow L: \bbox[red, 2pt]{\cases{x=4t-1\\ y=-2t+2\\z=t} , t\in \mathbb R}$$

解答:$$\cases{\log_5 x-2y=1 \Rightarrow x=5^{2y+1} =5\cdot (5^y)^2\\ x-5^y=6} \Rightarrow \cases{x=5u^2\\ x-u=6}\; (取u=5^y)  \\\Rightarrow 5u^2=u+65u^2-u-6=0 \Rightarrow (5u-6)(u+1)=0 \\ \Rightarrow u={6\over 5} \Rightarrow x =5\cdot (6/5)^2= \bbox[red, 2pt]{36\over 5} \quad(u=-1 \Rightarrow 5^y=-1 不合)$$

解答:

$$取\cases{P(x,y) \\A(0,0) \\B(8,0) \\C(10,4) \\D(4,6)} \Rightarrow \cases{\sqrt{x^2+y^2} =\sqrt{(x-0)^2+ (y-0)^2} =\overline{PA} \\ \sqrt{x^2+y^2-16x+64} =\sqrt{(x-8)^2+(y-0^2)} =\overline{PB} \\ \sqrt{x^2+y^2-20x-8y+116} =\sqrt{(x- 10)^2 +(y-4)^2} =\overline{PC} \\ \sqrt{x^2+y^2-8x-12y+52} =\sqrt{(x-4)^2+(y-6)^2} =\overline{PD}}\\ \Rightarrow \cases{P \in \overline{AC} \Rightarrow \overline{PA}+ \overline{PC}最小\\ P\in \overline{BD} \Rightarrow \overline{PB}+\overline{PD}最小} \Rightarrow P=\overline{AC}\cap \overline{BD} \Rightarrow \overline{PA}+\overline{PB} +\overline{PC} +\overline{PD}= \overline{AC} + \overline{BD} \\= \bbox[red, 2pt]{2\sqrt{29}+2\sqrt{13}} 為最小值$$

解答:$$字數很少，要求很多，直接窮舉:\\\textbf{全家: }全家校讚一、一全家校讚、一讚全家校、讚一全家校、校讚一全家 \\ \textbf{家全: }家全一讚校、校家全一讚、校家全讚一、讚校家全一、一讚校家全 \\ 總共\bbox[red, 2pt]{10}種$$

解答:$$第一階段:黑白猜\\ 每個人有2種選擇，共有2^4=16種, 成功條件:\cases{一人手心、三人手背:有4種情形\\一人手背、三人手心:有4種情形} , 共8種\\ \Rightarrow 單次成功機率P_1={8\over 16}={1\over 2} \Rightarrow 期望值E_1=1/P_1=2 \\ 第二階段:剪刀、石頭、布\\剩下三個人，每個人有3種選擇，共有3^3=27種, 成功條件:\cases{輸家一人:有3種可能\\ 輸家有三種選擇(剪刀、石頭、布)} \\ \qquad \Rightarrow 有3\times 3=9種情形，單次成功機率P_2= {9\over 27}={1\over 3} \Rightarrow E_2=3 \\ 期望值=E_1+E_2= \bbox[red, 2pt]5$$

解答:$$取\cases{P(0,0,0) \\A(1,0,1) \\B(1,1,0) \\C(0,1,1)} \Rightarrow \cases{平面ABC:x+y+z=2\\ A_0=\triangle PBC重心} \Rightarrow \cases{P'=(4/3,4/3,4/3) \\A_0=(1/3,2/3,1/3)} \\ \Rightarrow \overline{P'A_0} =\sqrt{1^2+(2/3)^2+1^2}= \bbox[red, 2pt]{\sqrt{22} \over 3}$$

解答:$$\Gamma: y=x^3+x \Rightarrow \Gamma_1:y=x^3+x+k \Rightarrow 假設切點\cases{P(a,a^3+a)\in \Gamma \\Q(b,b^3+b+k)\in \Gamma_1} \\ \Gamma與\Gamma_1有相同的y'=3x^2+1 \Rightarrow 3a^2+1=3b^2+1 \Rightarrow b=\pm a \\ 若b=a , P與Q同在一垂直線\Rightarrow k=0 \Rightarrow \Gamma=\Gamma_1 \\ 若b=-a \Rightarrow \cases{過P的切線: y=(3a^2+1)(x-a)+a^3+a\\ 過Q(-a,-a^3-a+k)的切線:y= (3a^2+1)(x+a)-a^3-a+k} \\ 兩切線相同\Rightarrow k=-4a^3 \Rightarrow 公切線斜率=3a^2+1 \gt {7\over 4} \Rightarrow a^2\gt {1\over 4} \\ \Rightarrow \cases{a\gt 1/2 \Rightarrow k=-4a^3 \lt -4\cdot (1/2)^3=-1/2 \\ a\lt -1/2 \Rightarrow k=-4a^3 \gt -4(-1/2)^3=1/2} \Rightarrow \bbox[red, 2pt]{k\lt -{1\over 2}或k\gt{1\over 2}}$$

第二部份： 計算證明題（共 2 題，每題 10 分，占 20 分）

解答:$$柯西不等式: \left( {b^2\over a-2}+{a^2\over b-2} \right) \left( (a-2)+(b-2) \right) \ge (b+a)^2 \\ \Rightarrow {b^2\over a-2}+{a^2\over b-2} \ge {(a+b)^2 \over a+b-4} =(a+b-4)+{16\over a+b-4}+8 \ge 8+ 2\cdot \sqrt{16}=16 \\ \Rightarrow \textbf{(1) }最小值為\bbox[red, 2pt]{16} \\ a+b-4={16\over a+b-4} \Rightarrow (a+b-4)^2=16 \Rightarrow a+b-4=4\Rightarrow a+b=8\\ 又{b\over a-2}={a\over b-2} \Rightarrow (b-a)(a+b-2)=0 \Rightarrow (b-a)(8-2)=0 \Rightarrow a=b\\\Rightarrow \textbf{(2) }(a,b)= \bbox[red, 2pt]{(4,4)}$$

解答:$$f(n)= \sum_{k=1}^{2n} {1\over (2k-1)(2k)} =\sum_{k=1}^{2n} \left( {1\over 2k-1}-{1\over 2k} \right) = 1-{1\over 2}+{1\over 3}-{1\over 4}+\cdots+{1\over 4n-1}-{1\over 4n} \\=\left( 1+{1\over 2}+{1\over 3} +\cdots+{1\over 4n} \right)- \left( 1+ {1\over 2}+ \cdots+{1\over 2n} \right) = {1\over 2n+1}+{1\over 2n+2}+ \cdots+{1\over 4n} \\= \sum_{k=1}^{2n}{1\over 2n+k} = \sum_{k=1}^{2n}{1\over 2n}\cdot {1\over 1+(k/2n)}  \Rightarrow \lim_{n\to \infty} f(n)= \lim_{N\to \infty} {1\over N}\cdot {1\over 1+k/N} =\int_0^1 {1\over 1+x}\,dx \\= \bbox[red, 2pt]{\ln 2}$$

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解題僅供參考，其他教甄試題及詳解