111年香山高中教甄-數學詳解

新竹市立香山高級中學111學年度教師甄選

一、單選題(每題4分，共計60分)

解答：$$1+2+\cdots +9=45為3的倍數 \Rightarrow 任取2數a與b，只要a+b是3的倍數\\，剩下的7個數字和也是3的倍數；因此(a,b)= (1,2),(1,5),(1,8),(2,4), (2,7), (3,6),(3, 9), (4,5)\\, (4,8), (5,7), (6,9),(7,8)，共12組，故選\bbox[red,2pt]{(D)}$$

解答：$$正弦定理:{27\over \sin A}={48\over \sin 3A} \Rightarrow {9\over \sin A}={16\over 3\sin A-4\sin^3 A} \Rightarrow 27\sin A-36\sin^3 A= 16\sin A\\ \Rightarrow 36\sin^3 A-11\sin A=0 \Rightarrow \sin A(36\sin^2 A-11)=0 \Rightarrow \sin A={\sqrt{11}\over 6} \Rightarrow \cos A={5\over 6}\\ \Rightarrow {5\over 6}={48^2+\overline{AC}^2 -27^2\over 96\overline{AC}} \Rightarrow \overline{AC}^2-80\overline{AC}+1575=0 \Rightarrow (\overline{AC}-35)(\overline{AC}-45)=0\\ \Rightarrow \overline{AC}=35或45，故選\bbox[red,2pt]{(B)}$$

解答：

$$\cases{\sin^2 x +\cos^2 y= y^2 \cdots(1)\\ \sin^2y + \cos^2 x= x^2\cdots(2)}，\cases{(1)+(2)\\ (2)-(1)} \Rightarrow \cases{x^2 +y^2=2 :一圓\\x^2-y^2=\cos(2x)-\cos(2y):兩直線}\\ \Rightarrow 兩圖形交於四點\cases{A(1,1)\\ B(-1,1)\\ C(-1,-1)\\ D(1,-1)}，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{z_1= \cfrac{-1+ i\sqrt 3}{2} =e^{2\pi i/3}\\ z_2=\cfrac{-1-i\sqrt 3}{2} =e^{4\pi i/3}} \Rightarrow \cases{z_1^{101} =e^{202\pi i/3} =e^{66\pi +4\pi i/3} =e^{4\pi i/3} =z_2\\[1ex] z_2^{101} = e^{404\pi i/3} =e^{134\pi +2\pi i/3} =e^{2\pi i/3} =z_1} \\ \Rightarrow z_1^{101}+ z_2^{101}= z_2+z_1= -1，故選\bbox[red,2pt]{(A)}$$

解答：$$\cos x=1-{1\over 2}x^2 +{1\over 4!}x^4-\cdots \Rightarrow \cos x^2 = 1-{1\over 2}x^4+{1\over 4!}x^8-\cdots \\ \Rightarrow \cases{1-\cos x= {1\over 2}x^2-{1\over 4!}x^4+ \cdots\\ 1-\cos x^2 = {1\over 2}x^4-{1\over 4!}x^8 + \cdots } \Rightarrow \lim_{x\to 0}{\sqrt{1-\cos  x^2} \over 1-\cos x} =\cfrac{\sqrt{(1/2)}}{1/2} = \sqrt 2，故選\bbox[red,2pt]{(D)}$$

解答：$$x^2+18x+30 =2 \sqrt{(x+9)^2-36} \Rightarrow (x+9)^2 -36-15 =2 \sqrt{(x+9)^2-36}\\ \Rightarrow a-15=2\sqrt a,\text{ where }a= (x+9)^2 -36   \Rightarrow a^2-34a+225=0 \Rightarrow (a-25)(a-9)=0\\ \Rightarrow \cases{(x+9)^2 -36 =a =25\\ (x+9)^2 -36= a= 9，不合(\because a-15 = 2\sqrt a \lt 0)} \Rightarrow  (x+9)^2 = 61 \Rightarrow x=-9\pm \sqrt{61}\\ \Rightarrow 兩根之積=81-61=20，故選\bbox[red,2pt]{(A)}$$

解答：$${5n-23\over n-7}= 5+{12\over n-7} \Rightarrow n-7=\pm 1,\pm 2,\pm 3,\pm 4,\pm 6,\pm 12 \Rightarrow 共12個解，故選\bbox[red,2pt]{(C)}$$

解答：$$取\cases{a=3\\ b=4\\ c=5}滿足a,b,c成等差；因此\cases{\tan A=3/4 \Rightarrow \tan(A/2)=1/3\\ \tan (C/2) = \tan 45^\circ =1} \Rightarrow \tan{A\over 2}\cdot \tan{C\over 2}={1\over 3}\\，故選\bbox[red,2pt]{(B)}$$

解答：$$\cases{3a^2+ 2022a+8=0 \\ 8b^2+2022b + 3=0} \Rightarrow \cases{a=\cfrac{-2022\pm \sqrt{k}}{6} \\b=\cfrac{-2022\pm \sqrt{k}}{16}},\text{where }k=2022^2-4\cdot 3\cdot 8 \\ \Rightarrow (a,b)=(\cfrac{-2022+ \sqrt{k}}{6},\cfrac{-2022+ \sqrt{k}}{16})  或(\cfrac{-2022- \sqrt{k}}{6},\cfrac{-2022- \sqrt{k}}{16}) \Rightarrow {a\over b} ={16\over 6}={8\over 3} \\ 另兩組解將造成ab=1，不符要求，故選\bbox[red,2pt]{(C)}$$

解答：$$\lim_{x\to 0} \cfrac{x^3 \sin{1\over x}}{\sin x} =\lim_{x\to 0} \cfrac{(x^3 \sin{1\over x})'}{(\sin x)'} =\lim_{x\to 0} \cfrac{3x^2 \sin{1\over x} -x\cos{1\over x}}{\cos x} ={0\over 1}=0，故選\bbox[red,2pt]{(B)}$$

解答：$$\lim_{n\to \infty} {1\over \sqrt n}\left(1 +{1\over \sqrt 2}+{1\over \sqrt 3}+ \cdots+{1\over \sqrt n} \right)=\lim_{n\to \infty} \sum_{k=1}^n{1\over \sqrt n}\cdot {1\over \sqrt k} =\lim_{n\to \infty} \sum_{k=1}^n{1\over   n}\cdot {1\over \sqrt {(k/n)}}\\ =\int_0^1 {1\over \sqrt x}dx = \left. \left[ 2\sqrt x\right] \right|_0^1 =2，故選\bbox[red,2pt]{(C)}$$

解答：$$令\overline{AB}=a，則 \cos C = {\sqrt 3\over 2}={(2+2\sqrt 3)^2+16-a^2\over 8(2+2\sqrt 3)} \Rightarrow a^2=8 \Rightarrow \overline{AB} =2\sqrt 2\\ \cos B={4^2+(2\sqrt 2)^2-(2+2\sqrt 3)^2\over 16\sqrt 2} = {\sqrt 2-\sqrt 6\over 4} \lt 0 \Rightarrow B=180^\circ-75^\circ = 105^\circ，故選\bbox[red,2pt]{(D)}$$

解答：$$y=8^nx^2-2^n(2^n+1)x+1 = (2^{2n}x-1)(2^nx-1) \Rightarrow L_n= \overline{A_nB_n} ={1\over 2^n}-{1\over 2^{2n}} \\ \Rightarrow \sum_{n=1}^\infty L_n= \sum_{n=1}^\infty \left( {1\over 2^n}-{1\over 2^{2n}} \right) ={1/2\over 1-1/2}-{1/4\over 1-1/4} =1-{1\over 3}={2\over 3}，故選\bbox[red,2pt]{(D)}$$

解答：$$令\cases{f(x,y,z)=\cos 2x+\cos 2y+\cos 2z\\ g(x,y,z)=\cos x+\cos y+\cos z} \Rightarrow -3\le f,g,\le 3，則原式af+2(1-a)g=9a-6\\ \Rightarrow a(f-2g)+2g=9a-6 \Rightarrow a={6+2g\over 9-f+2g} 為整數\Rightarrow a=1,0\\ \cases{a=0 \Rightarrow 2g=-6 \Rightarrow g=\cos x+\cos y+\cos z=-3 \Rightarrow (x,y,z)=(\pi,\pi,\pi)\\ a=1 \Rightarrow f= \cos 2x+\cos 2y+\cos 2z=3 \Rightarrow (x,y,z)=(0,0,0),(\pi,\pi,\pi),(2\pi,2\pi,2\pi)}\\ 因此共有三組解，故選\bbox[red,2pt]{(C)}\\ 但a=1時，(x,y,z)=(0,0,2\pi),(0,2\pi,0),...也成立，應該有2^3+1(\pi,\pi,\pi)=\bbox[blue,2pt]{9}組解$$

解答：$$假設m\gt n+1 \Rightarrow 4m\gt 4n+4 \Rightarrow m^2-4m \lt m^2-4n-4 \Rightarrow (m-2)^2 \lt m^2-4n \\ 又m^2-4n\lt m^2，因此(m-2)^2 \lt m^2-4n\lt m^2 \Rightarrow m^2-4n= (m-1)^2 = m^2-2m+1\\ \Rightarrow 2m= 4n+1 \Rightarrow 偶數=奇數，矛盾，因此m\not \gt n+1;只剩下m=n+1,m=n兩種情況;\\ 當m=n+1時，則\cases{m^2-4n= (n+1)^2-4n= (n-1)^2 為一完全平方數\\ n^2-4m= n^2-4n-4 = (n-2)^2-8 \Rightarrow n=5 \Rightarrow m=6}\\當m=n時，則n^2-4m= m^2-4n=n^2-4n \Rightarrow m=n=4\\ 同理，n\gt m+1亦矛盾，僅剩n=m+1與n=m兩種情形；\\因此(m,n)= (4,4),(5,6),(6,5)三組正整數解，故選\bbox[red,2pt]{(C)}$$

二、多選題(每題8分，共計40分；每題有5個選項，至少有一個是正確的選項)

解答：

$$(A)\bigcirc: \angle EAF= \angle A-\angle BAE-\angle DAF = 90^\circ-30^\circ-15^\circ=45^\circ \\(B)\times: \cases{\overline{DF}= 2\sqrt 3\cdot \tan 15^\circ =  4\sqrt 3-6 \Rightarrow \overline{CF}= 2\sqrt 3-(4\sqrt 3-6)=6-2\sqrt 3 \\\overline{BE} =2\sqrt 3\cdot \tan 30^\circ = 2 \Rightarrow \overline{CE}= 2\sqrt 3-2} \\ \qquad \Rightarrow \overline{CF}/\overline{CE}= \sqrt 3 \Rightarrow \angle FEC=60^\circ \Rightarrow \angle AEF= 60^\circ \ne 75^\circ\\ (C)\bigcirc: \angle CFE = 180^\circ -\angle AEF-\angle C=30^\circ \\(D)\bigcirc: 由(B)可知:\overline{CE}= 2\sqrt 3-2 \\(E) \bigcirc: \overline{EF}= 2\overline{CE}= 4\sqrt 3-4\\，故選\bbox[red,2pt]{(ACDE)}$$

解答：$$\sin A與\sin B為(m+5) x^2-(2m-5)x+12=0的二實根 \Rightarrow \cases{\sin A+\sin B=(2m-5)/(m+5)\\ \sin A\sin B=12/(m+5)} \\因此 (\sin A+\sin B)^2 =\sin^2 A+\sin ^2B + 2\sin A\sin B =1+2\sin A\sin B\\ \Rightarrow \left({2m-5\over m+5}\right)^2 =1+2 \cdot {12\over m+5} ={m+29\over m+5} \Rightarrow (2m-5)^2 = (m+29)(m+5) \Rightarrow m^2-18m-40=0\\ \Rightarrow (m-20)(m+2)=0 \Rightarrow m=20(\because m\gt 0, m=-2不合) \Rightarrow 25x^2-35x+12=0\\ \Rightarrow (5x-3)(5x-4)=0 \Rightarrow \cases{\sin A=3/5\\ \sin B=4/5} \Rightarrow {a\over \sin A}={b\over \sin B}=2R= 4 \Rightarrow \cases{a=12/5\\ b=16/5} \\ \Rightarrow \cases{a^2+b^2=c^2 = {400\over 25} \Rightarrow c={20\over 5}=4 \\a+b=28/5\\ ab= 192/25}，故選\bbox[red,2pt]{(BCD)}$$

解答：$$由題意可知:\cases{a^2+b^2=c^2 \\ a+b+c=126 \\ ab=630\times 2= 1260} \Rightarrow \cases{(a+b)^2 =(126-c)^2 =c^2-252c+ 126^2\\ (a+b)^2 = a^2+b^2 +2ab=c^2+2ab =c^2+2520} \\ \Rightarrow c^2-252c+ 126^2 =c^2+2520 \Rightarrow 252c = 13356 \Rightarrow \color{blue}{c=53} \Rightarrow   a+b=126-53=73 \\ \Rightarrow ab= a(73-a)=1260 \Rightarrow a^2-73a+1260=0 \Rightarrow (a-45)(a-28)=0 \\\Rightarrow \cases{(a,b)=(28,45)\\ (a,b)=(45,28)不合，違反b\gt a} \Rightarrow \cases{ b-a=45-28=17\ne 15\\ c-b=53-45=8  \\ c-a= 53-28=  25\\ a+b+c=周長=126\ne 120}\\，故選\bbox[red,2pt]{(BCD)}$$

解答：$$(D)\times:令\cases{g(x)=2/|x|\\ f(x)=1/|x|}，滿足 g(x)\gt f(x) \;\forall x\ne 0，但\cases{\lim_{x\to 0} f(x)= \infty\\ \lim_{x\to 0} g(x)= \infty}\\\qquad，因此\lim_{x\to 0} f(x)\not \lt \lim_{x\to 0} g(x) \\(E)\times: 令\cases{g(x)=2/x\\ f(x)=x} \Rightarrow \lim_{x\to 0}(f(x)\times g(x)) = \lim_{x\to 0}2=2，但\lim_{x\to 0} g(x)不存在\\其餘皆正確，故選\bbox[red,2pt]{(ABC)}$$

解答：$$a=\sqrt 2+b 代入2ab+ 2\sqrt 2c^2+1=0 \Rightarrow 2(\sqrt 2+b)b+ 2\sqrt 2c^2+1=0 \\ \Rightarrow 2b^2 +2\sqrt 2b+ 2\sqrt 2c^2+1=0 有實數解\Rightarrow (2\sqrt 2)^2-4\cdot 2\cdot (2c^2+1)\ge 0 \Rightarrow 16c^2\le 0 \Rightarrow c=0\\ \Rightarrow  2b^2 +2\sqrt 2b+1=0 \Rightarrow b=-{\sqrt 2\over 2} \Rightarrow a= \sqrt 2-{\sqrt 2\over 2}={\sqrt 2\over 2} \Rightarrow \cases{a=\sqrt 2/2\\ b=-\sqrt 2/2\\ c=0\\ a+b+c=0}，故選\bbox[red,2pt]{(BCE)}$$

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解題僅供參考，其他教甄歷年試題及詳解