114年新竹市國中教甄聯招-數學詳解

 新竹市114學年度國民中學第二次教師聯合甄選

一、單選題:100分(共50題，每題2分)

解答:$$a_n={n\over a_{n-1}} \Rightarrow \cases{a_2=\displaystyle {2\over a_1} \\a_4=\displaystyle {4\over a_3} \\a_6=\displaystyle {6\over a_5} \\a_8=\displaystyle {8\over a_7} \\a_{10}=\displaystyle {10\over a_9}} \Rightarrow \cases{a_1a_2=2\\ a_3a_4=4\\ a_5a_6= 6\\ a_7a_8=8\\ a_9a_{10}=10} \Rightarrow a_1a_2\cdots a_{10}=2\cdot 4\cdot 6\cdots10=3840，故選\bbox[red, 2pt]{(C)}$$

解答:$$(7\sqrt 2)^2=98 \lt 100\\ (6\sqrt 3)^2=108 \gt 100\\ (5\sqrt 5)^2=125 \gt 100\\ (4\sqrt 7)^2=112 \gt 100\\ (3\sqrt{11})^2=99 \lt 100\\ (3\sqrt 5+\sqrt{11})^2= 56+6\sqrt{55 } 又(6\sqrt{55})^2=1980\gt 1936=44^2 \Rightarrow 3\sqrt 5+\sqrt{11}\gt 10\\ 因此共有四個數(6\sqrt 3,5\sqrt 5,4\sqrt 7, 3\sqrt 5+\sqrt{11})大於10，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{z_1=\displaystyle {-1+i\sqrt 3\over 2} =\cos{2\pi\over 3}+i\sin {2\pi\over 3} =e^{2 \pi i/3} \\z_2=\displaystyle {-1-i\sqrt 3\over 2} =\cos{4\pi\over 3}+i\sin {4 \pi\over 3} =e^{4 \pi i/3} } \Rightarrow \cases{z_1^3=1\\ z_2^3=1} \\(A)\bigcirc: \cases{z_1^7 =z_1\\ z_2^7=z_2} \Rightarrow z_1^7+z_2^7 =z_1+z_2=-1 \\(B)\times: \cases{z_1^9 = 1\\ z_2^9=1} \Rightarrow z_1^9+z_2^9 = 2\ne -1 \\(C) \bigcirc: \cases{z_1^{11} =z_2\\ z_2^{11}=z_1} \Rightarrow z_1^{11}+z_2^{11} =z_1+z_2= -1\\(D) \bigcirc: \cases{z_1^{13} =z_1\\ z_2^{13}=z_2} \Rightarrow z_1^{13}+z_2^{13} =z_1+z_2=-1\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{\log_8 a+\log_4b^2 =5 \\\log_8b+ \log_4a^2 =7} \Rightarrow \cases{\displaystyle {\log a\over 3\log 2}+  {2\log b\over 2\log 2} =5\\ \displaystyle {\log b\over 3\log 2}+ {2\log a\over 2\log 2} =7} \Rightarrow \cases{2\log a+6\log b=30\log 2\\ 2\log b+6\log a=42\log 2} \\ \Rightarrow \cases{\log a+3\log b=15\log 2\\ 3\log a+\log b=21\log 2} \Rightarrow \cases{\log a=6\log 2\\ \log b=3\log 2} \Rightarrow \log a+\log b=\log ab=9\log 2 \\ \Rightarrow ab=2^9=512，故選\bbox[red, 2pt]{(D)}$$

解答:$$a_1+a_2+\cdots +a_{100} =(2a_1+99)50=250 \Rightarrow a_1=-47 \\ \Rightarrow a_2+a_4+ \cdots+a_{100}=(a_2+a_{100})\cdot 25 =(2a_1+100)\cdot 25 =150，故選\bbox[red, 2pt]{(D)}$$

解答:$$a\le x\le 15 \Rightarrow f(x)=x-a+15-x+a+15-x =30-x \Rightarrow 最小值為f(15)=15，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{\angle HLB=6\cdot 180^\circ/8=135^\circ \\\angle HLK=3\cdot 180^\circ/5=108^\circ\\ \angle KLA=60^\circ} \Rightarrow \angle ALB=360^\circ-135^\circ-108^\circ-60^\circ=57^\circ\\ \Rightarrow \angle ABL ={1\over 2}(180^\circ-57^\circ)={123^\circ\over 2}，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cot x+\cot y={1\over \tan x}+{1\over \tan y}={\tan x+\tan y\over \tan x \tan y} ={25\over \tan x\tan y}=30 \Rightarrow \tan x\tan y={5\over 6 } \\ \Rightarrow \tan(x+y)= {\tan x+ \tan y\over 1-\tan x\tan y}={25\over 1-5/6}=150，故選\bbox[red, 2pt]{(C)}$$

解答:$$41^k 除以100的餘數=\cases{41,k=1\\ 81,k=2\\21,k=3\\ 61,k=4\\ 1,k=5\\ 41,k=6\\ \cdots} \Rightarrow 循環數為5 \Rightarrow 41^{48} \equiv 41^3 \equiv 21 \mod 41，故選\bbox[red, 2pt]{(B)}$$

解答:$$a,b,c,d均為正整數\Rightarrow \cases{(2^4)^5=(2^5)^4 \Rightarrow \cases{a=16\\ b=32} \Rightarrow c=16+19=35 \Rightarrow d無正整數解\\ (3^4)^5=(3^5)^4 \Rightarrow \cases{a=81\\ b=243} \Rightarrow c=81+19=100 \Rightarrow d=1000} \\ \Rightarrow d-b=1000-243=757，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x)={9x^2 \sin^2 x+4\over x\sin x} \Rightarrow f'(x)={ -4\sin x-4x \cos x+9x^2\sin^3 x+9x^3 \cos x \sin^2x\over x^2 \sin^2 x} \\f'(x)=0 \Rightarrow -4(\sin x+x\cos x)+9x^2 \sin^2x(\sin x+x\cos x)=0\\ \Rightarrow (9x^2\sin^2 x-4)(\sin x+x\cos x)=0 \Rightarrow 9x^2\sin^2 x=4 \Rightarrow x\sin x={2\over 3} \Rightarrow f(x)={4+4\over 2/3}=12\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x)=x^{2025}(x^2+ax+b) =(x-2)^2 p(x)+2^{2025}(x-2) \\ \Rightarrow f'(x)=2025x^{2024}(x^2+ax+b)+x^{2025}(2x+a)= 2(x-2)p(x)+(x-2)^2p'(x)+2^{2025} \\ \Rightarrow \cases{f(2)=2^{2025}(4+2a+b)= 0 \cdots(1)\\ f'(2)=2025\cdot 2^{2024}(4+2a+b)+2^{2025}(4+a)=2^{2025} \cdots(2)} \\(1)\Rightarrow 4+2a+b= 0代入(2) \Rightarrow 4+a=1 \Rightarrow a=-3 \Rightarrow 4-6+b=0 \Rightarrow b=2，故選\bbox[red, 2pt]{(C)}$$

解答:$$任20個連續的數之和都是75，又gcd(269,20)=4 \Rightarrow 循環數為4 \Rightarrow \sum_{k=i+1}^{i+20} a_k =5\sum_{k=i+1}^{i+5} a_k =75 \\ \Rightarrow \sum_{k=i+1}^{i+5} a_k =15 \Rightarrow \cases{a_{17}=a_1=3\\ a_{83} =a_3=4\\ a_{144}=a_0=9\\ a_{210}=a_2} \Rightarrow a_1+a_2+a_3+a_0=15\\ \Rightarrow a_{210}=a_2=15-(3+4+9)=-1，故選\bbox[red, 2pt]{(A)}\\註:此題為2015TRML 數學競試團體賽試題$$

解答:$$A=\begin{bmatrix} 3& 3& 5\\ -1& 2& 0\\ 5& -1& 5\end{bmatrix} \Rightarrow rref(A)= \begin{bmatrix} 1& 0&10/9\\ 0& 1& 5/9\\ 0& 0& 0\end{bmatrix} \Rightarrow rank(A)=2，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{n^2f(n)=f(1)+f(2)+\cdots +f(n) \\ (n-1)^2 f(n-1)=f(1)+f(2) +\cdots +f(n-1)} \Rightarrow f(n)=n^2f(n)-(n-1)^2f(n-1) \\ \Rightarrow (n^2-1)f(n) =(n-1)^2f(n-1) \Rightarrow f(n)={n-1\over n+1}f(n-1) \\ \Rightarrow f(100)={99\over 101}f(99)= {99\over 101}\cdot {98\over 100}f(98)= {99\over 101}\cdot {98\over 100} \cdot {97\over 99}f(97)= \cdots={99\cdot 98\cdots 1\over 101\cdot 100\cdots3}f(1) \\={2\over 101\cdot 100}\cdot 101={1\over 50}，故選\bbox[red, 2pt]{(B)}$$

解答:$${7^{\alpha+ \beta} +7^{\beta+ \gamma} +7^{\gamma+ \alpha} \over 49^\alpha +49^\beta+ 49^\gamma} ={7^{\alpha+ \beta} +7^{\beta+ \gamma} +7^{\gamma+ \alpha} \over 7^{2\alpha} +7^{2\beta}+ 7^{2\gamma}} = 1 \Rightarrow \alpha=\beta=\gamma \Rightarrow 3\alpha-5\beta+2\gamma=0，故選\bbox[red, 2pt]{(A)}$$


解答:$$x^2+y^2=20 \Rightarrow \cases{x= 2\sqrt 5 \cos \theta\\ y=2\sqrt 5 \sin \theta} \Rightarrow xy=20\cos \theta\sin \theta=10\sin 2\theta \\ \Rightarrow 4xy-7\sqrt{xy+1}=40\sin 2\theta -7\sqrt{10\sin 2\theta+1}=11 \Rightarrow (40\sin 2\theta-11)^2= (7\sqrt{10\sin 2\theta+1})^2 \\ \Rightarrow 800\sin^2 2\theta-685\sin 2\theta+36=0 \Rightarrow (160\sin 2\theta-9)(5\sin 2\theta-4)=0 \\ \textbf{Case I }\sin 2\theta={4\over 5} \Rightarrow \cases{\cos 2\theta=3/5 \Rightarrow \cos \theta=\pm 2/\sqrt 5 \Rightarrow (\cos \theta,\sin \theta) =(\pm 2/\sqrt 5, \pm 1/\sqrt 5)\\ \cos 2\theta=-3/5 \Rightarrow \cos \theta=\pm 1/\sqrt 5 \Rightarrow (\cos \theta, \sin \theta)=(\pm 1/\sqrt 5, \pm 2/\sqrt 5)} \\\qquad \Rightarrow (x,y) =(4,2),(-4,-2),(2,4),(-2,-4) \\\textbf{Case II }\sin 2\theta={9\over 160} \Rightarrow xy={9\over 16} \Rightarrow 4xy-7\sqrt{xy+1}={9\over 4}-7\cdot {5\over 4} \ne 11 不合\\ 因此共有4組解，故選\bbox[red, 2pt]{(C)}$$

解答:$$3^2 \times 4^{16} \times 5^{25}=3^2 \times 2^{32}\times 5^{25}=3^2 \times 2^{7}\times 10^{25}= 9\times 128 \times 10^{25}=1152\times 10^{25}為29位數\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{4位數的A:1111,1112,1121,1211,2111,1122,2211,2112\Rightarrow a_4=8 \\4位數的B:  1000,  1100,1001 \Rightarrow b_4=3} \\ \Rightarrow a_4+b_4=11，故選\bbox[red, 2pt]{(B)}$$

解答:$$正整數:\cases{1\triangle \triangle \triangle \triangle :有3^4=81個\\ 01\triangle \triangle \triangle   :有3^3=27個\\ 001\triangle   \triangle :有3^2=9個\\ 0001\triangle :有3^1=3個\\ 00001:有1個} \Rightarrow 合計:81+27+9+3+1= 121，故選\bbox[red, 2pt]{(A)}$$

解答:

$$ABCD為正方形\Rightarrow PQRS為正方形，作\overline{FT} \parallel \overline{DE} \Rightarrow \angle TFC=\angle HCB=\theta \\ \Rightarrow \tan \theta={\overline{BH} \over \overline{BC}}={n-1\over n } \Rightarrow \cos \theta={n\over \sqrt{2n^2-2n+1}} \Rightarrow \overline{FT}=\overline{FC}  \cos \theta={1\over \sqrt{2n^2-2n+1}} \\ \Rightarrow PQRS面積=\overline{FT}^2 ={1\over  {2n^2-2n+1} }={}{1\over 421} \Rightarrow n^2-n-210 =0 \Rightarrow (n-15)(n+14)=0\\ \Rightarrow n=15，故選\bbox[red, 2pt]{(B)}$$

解答:

$$\overline{AB}=\overline{BC} \Rightarrow \overline{OB}為\triangle ABC中線\Rightarrow \overline{OA}^2+\overline{OC}^2=2(\overline{OB}^2+\overline{AB}^2) \\ \Rightarrow 8^2+18^2=2(8^2+\overline{AB}^2) \Rightarrow \overline{AB}=\sqrt{130}，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{\overline{AB}=10a+b\\ \overline{CD}=10b+a}, 1\le a,b\le 9 \Rightarrow \cases{\overline{OC}=\overline{AB}/2=5a+b/2\\ \overline{CE} =\overline{CD}/2=5b+a/2} \\\Rightarrow \overline{OE}^2=\overline{OC}^2-\overline{CE}^2 ={99\over 4}(a^2-b^2) \Rightarrow \overline{OE}={3\over 2} \sqrt{11(a^2-b^2)} 為有理數\\ \Rightarrow a^2-b^2=11 \Rightarrow \cases{a=6\\ b=5} \Rightarrow \overline{AB}=65，故選\bbox[red, 2pt]{(C)}$$

解答:$$\triangle ADE \sim \triangle EFB (AAA) \Rightarrow {\overline{AD}\over \overline{DE}=12}={ \overline{EF}=12\over \overline{BF}} \Rightarrow \overline{AD}\cdot \overline{BF}=144 \Rightarrow \cases{\overline{AD}=9\\ \overline{BF}=16} \\ \Rightarrow \overline{AC}+\overline{BC}=9+12+12+16= 49，故選\bbox[red, 2pt]{(D)}$$

解答:$$\lim_{x\to \infty}{f(x)\over x} =6 \Rightarrow \lim_{x\to \infty}f'(x)=6 \Rightarrow \lim_{x\to  \infty} {f(x)+ 4x\over xf(x)-5x^2+2} = \lim_{x\to  \infty} {f(x)+ 4x\over x(f(x)-5x)+2} \\=  \lim_{x\to  \infty} {f(x)+ 4x\over 2x+2} = {1\over 2}\lim_{x\to  \infty}{f(x)+4x\over x+1}= {1\over 2}\lim_{x\to  \infty}{f'(x)+4\over 1}={1\over 2}\cdot 10=5，故選\bbox[red, 2pt]{(D)}$$

解答:$$假設a=-2,b=-5,c=-1 \Rightarrow \cases{\displaystyle {b\over a}=5/2\\ \displaystyle {b-c\over a-c}={-4\over -1}=4\\ \displaystyle {b+c\over a+c}={-6\over -3}=2} \Rightarrow \cases{(B)\times: \displaystyle {b\over a}\ne {b-c\over a-c} \\(C)\times: \displaystyle {b\over a} \not \lt {b+c\over a+c} \\ (D)\times:\displaystyle {b\over a}\ne {b+c\over a+c}}，故選\bbox[red, 2pt]{(A)}$$

解答:$$絕對值越來越大，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{英文的z分數=\displaystyle {84-70\over 7}=2\\數學的z分數=\displaystyle {75-65\over 5}=2} \Rightarrow 兩科一樣好，故選\bbox[red, 2pt]{(B)}$$

解答:$$第一次翻滾，以C為旋轉中心，順時針旋轉120^\circ;第二次翻滾，以B為旋轉中心，順時針旋轉120^\circ;\\ \Rightarrow 路線長=9\cdot {2\pi\over 3} +9\cdot {2\pi\over 3} =12\pi，故選\bbox[red, 2pt]{(D)}$$

解答:$$甲出租23輛，其中3輛歸還至B，乙出租18輛, 歸還至乙的有18+3=21\\因此從甲出租23比從乙出租18，多了5輛，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{8男8女\Rightarrow C^{10}_8C^8_8 =45\\ 9男7女\Rightarrow C^{10}_9C^8_7=80} \Rightarrow 共有45+80=125種可能，故選\bbox[red, 2pt]{(B)}$$

解答:$$母群平均數不偏估計值=樣本平均值,\\母群體標準差不偏估計值=\sqrt{n\over n-1}\cdot \sigma=\sqrt{30\over 29}\cdot 10 \approx 10.17，故選\bbox[red, 2pt]{(D)}$$

解答:$$(a+b):(b+c):(c+a)=4:5:6 \Rightarrow \cases{a+b=4k\\ b+c=5k \\ c+a=6k} \Rightarrow a+b+c={15\over 2}k \Rightarrow \cases{a=5k/2\\ b=3k/2\\ c=7k/2} \\ \Rightarrow \cos C={a^2+b^2-c^2 \over 2ab}=-{1\over 2} \Rightarrow \angle C=120^\circ，故選\bbox[red, 2pt]{(B)}$$

解答:$$z={82-68.2\over 7.5}=1.84，故選\bbox[red, 2pt]{(B)}$$

解答:$$1\triangle \triangle \triangle :排列數3!=6\Rightarrow 個位數字總和= 十位數字總和= 百位數字總和 =2\times (0+2+3)=10 \\\qquad \Rightarrow 四位數總和=1000\cdot 6+10(100+10+1)=7110 \\2\triangle \triangle \triangle :排列數3!=6\Rightarrow 個位數字總和= 十位數字總和= 百位數字總和 =2\times (0+1+3)=8 \\\qquad \Rightarrow 四位數總和=2000\cdot 6+8(100+10+1)=12888 \\3\triangle \triangle \triangle :排列數3!=6\Rightarrow 個位數字總和= 十位數字總和= 百位數字總和 =2\times (0+1+2)=6 \\\qquad \Rightarrow 四位數總和=3000\cdot 6+6(100+10+1)=18666 \\ 所有四位數總和=7110+12888+18666= 38664，故選\bbox[red, 2pt]{(A)}$$

解答:$$ n=0, 面積為A_0 \Rightarrow n=1,面積為A_1=A_0+{3\over 4}A_0\cdot {4\over 9} \Rightarrow n=k, A_k= A_0+ \sum_{i=1}^k{3\over 4}\cdot ({4\over 9})^i  A_0\\ \Rightarrow  A_n={\sqrt 3\over 20}\left( 8-3({4\over 9})^n\right) \Rightarrow \lim_{n\to \infty}A_n ={\sqrt 3\over 20}\cdot 8={2\sqrt 3\over 5}，故選\bbox[red, 2pt]{(C)}$$

解答:$$n=0 ,邊長為1 \Rightarrow n=1, 邊長為1+{1\over 3}={4\over 3} \Rightarrow n=2, 邊長為({4\over 3})^2 \Rightarrow \cdots \Rightarrow n=k,邊長=({4\over 3})^k\\公比大於1 \Rightarrow 邊長無限大，故選\bbox[red, 2pt]{(D)}$$

解答:$$已知年齡依序為18,20,24,28,28,36,40,44,50 \Rightarrow \cases{x\le 28 \Rightarrow 中位數=28 =a\\ x\ge 36 \Rightarrow 中位數=(28+36)/2=32 =b}  \\ \Rightarrow b-a=4，故選\bbox[red, 2pt]{(B)}$$

解答:$$c_1=0 \Rightarrow f(x)=f(c)+c_2(x-c)^2+c_3(x-c)^3 +\cdots+ c_n(x-c)^n \\\Rightarrow f'(x)=2c_2(x-c)+ 3c_3(x-c)^2+ \cdots+nc_n(x-c)^{n-1} \\ \Rightarrow f''(x)=2c_2+6c_3(x-c)+ \cdots+n(n-1)c_n(x-c)^{n-2} \\ \Rightarrow \cases{f'(c)=0\\ f''(c)=2c_2}  \Rightarrow 若c_2=0, 則f(c)非極值，故選\bbox[red, 2pt]{(D)}$$

解答:$${13\cos x+1\over 2-4\cos x}=3 \Rightarrow 13\cos x+1=6-12\cos x \Rightarrow \cos x={1\over 5} \\\Rightarrow \sin x=-{2\sqrt 6\over 5} \;(\tan x\lt 0 \Rightarrow \sin x\lt 0)，故選\bbox[red, 2pt]{(C)}$$

解答:$$3^{x+2}\gt 5(2^{2x-1}) \Rightarrow (x+2)\log 3 \gt \log 5+(2x-1)\log 2 \\ \Rightarrow x\log 3+ \log 9 \gt \log 5+x\log 4-\log 2 \Rightarrow x(\log 4-\log 3)\lt \log 9+\log 2-\log 5=\log {18\over 5 }\\ \Rightarrow x\lt{ \log (18/5) \over \log (4/3)} \Rightarrow x\lt \log_{4/3}{18\over 5}，故選\bbox[red, 2pt]{(A)}$$

解答:$$假設點數為k,機率為kp \Rightarrow \sum_{k=1}^6 kp=1 \Rightarrow p={1\over 21} \Rightarrow 點數出現3的機率={ 3\over 21}={1\over 7} \\ 擲骰子4次點數3出現2次，共有C^4_2=6種情形，每次的機率為({1\over 7})^2({6\over 7})^2 \\ \Rightarrow P(X=2)=6\cdot ({1\over 7})^2({6\over 7})^2={6^3\over 7^4} ={216\over 2401}，故選\bbox[red, 2pt]{(A)}$$

解答:$$擲幤2次只得1元的情形:正反, 反正，機率分別為p(1-p),(1-p)p，\\機率合計為2p(1-p)=2(p-p^2)，故選\bbox[red, 2pt]{(B)}$$

解答:$${(x+1)^2\over 4}+{(y-2)^2\over 16}=1 \Rightarrow \cases{a=2\cos \theta-1\\ b=4\sin \theta+2} \Rightarrow 3a+2b-2=6\cos \theta+8\sin \theta-1 \\=10\sin(\theta+\alpha)-1 \Rightarrow 最小值=-10-1=-11，故選\bbox[red, 2pt]{(D)}$$

解答:$$假設一週可完成窗簾a件及桌布b件,則滿足條件\cases{50a+60b \le 16\times 60\\ 75a+45b\le 15\times 60\\ a\le b}, 即 \cases{5a+6b\le 96\\ 15a+9b\le 180\\ a\le b}, \\ 將可獲利潤f(a,b)=140a+120b \\ 所圍區域頂點\cases{A(0,16)\\ B({24\over 5},12)\\C({15\over 2},{15\over 2}) \\O(0,0)} , 由於a,b為整數, 因此取\cases{a=6\\ b=10} \Rightarrow f(6,10)=2040，故選\bbox[red, 2pt]{(A)}$$

解答:$${陽性且患癌人數\over 陽性人數} ={1\% \times 80\%\over 1\% \times 80\%+99\%\times 9.6\%} ={25\over 322}=0.0776，故選\bbox[red, 2pt]{(C)}$$

解答:$$\overrightarrow{AB} \cdot \overrightarrow{AC}=24-24=0 \Rightarrow \angle A=90^\circ \Rightarrow \triangle ABC面積={1\over 2}\cdot \overline{AB}\cdot \overline{AC}={1\over 2}\cdot 5\cdot 10=25 \\又{\triangle ABD\over \triangle ABC}={\overline{BD}\over \overline{BC}} \Rightarrow \triangle ABD={2\over 3}\cdot 25 ={50\over 3}，故選\bbox[red, 2pt]{(B)}$$

解答:

$$假設\cases{A(-a,0)\\ B(a,0)} \Rightarrow \cases{C(a,12-a^2\\ D(-a,12-a^2)} \Rightarrow 矩形ABCD面積=f(a)=2a\times (12-a^2) \\ \Rightarrow f(a)=24a-2a^3 \Rightarrow f'(a)=24-6a^2 =0 \Rightarrow a= 2 \Rightarrow  f(2)=4\times 8=32，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{L_1\bot E\\ L_2\bot E} \Rightarrow \cases{L_1方向向量與E的法向量平行 \\L_2方向向量與E的法向量平行} \Rightarrow L_1\parallel L_2，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{L_1方向向量\vec u=(2,1,-2) \\ L_2方向向量\vec v=(1,-2,1)} \Rightarrow \vec n=\vec u\times \vec v=(-3,-4,-5) \\ \Rightarrow 包含L_1且法向量為\vec n的平面E:-3(x-7)-4(y-2)-5(z-10)=0 \Rightarrow 3x+4y+5z=79 \\ P(3,9,2) \in L_2 \Rightarrow d(P,E)={|9+36+10-79| \over \sqrt{9+16+25}} ={24\over 5\sqrt 2} ={12\sqrt 2\over 5}，故選\bbox[red, 2pt]{(B)}$$

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