108年臺中二中教甄-數學詳解

臺中市立臺中第二高級中等學校108學年度

第一次教師甄選數學科題目卷

一、填充題：10 題共 50 分

解：$$\cfrac{1}{(\log_n 2n)-1} =\cfrac{1}{\log_n 2+ \log_n n-1} =\cfrac{1}{\log_n 2 } = \log_2 n \\ \Rightarrow \sum_{n=2}^{2056}\lfloor \log_2 n\rfloor =(1+1)+\overbrace{2+2+2+2}^{2^2=4個}+\overbrace{3+\cdots+3}^{2^3=8個}+ \cdots+ \overbrace{10+\cdots+10}^{2^{10}=1024個} +\overbrace{11+\cdots+11}^{9個} \\ =\left(\sum_{k=1}^{10}k\cdot 2^k\right)+ 9\times 11= \bbox[red,2pt]{18533}$$

2. 設\(k\in Z\)，若 \(x^2+y^2=2k^2\) 與 \(xy=4k\) 的圖形不相交，則\(k\)=_____

解：$$\cases{x^2+y^2 =2k^2 \\ xy=4k} \Rightarrow \cases{y^2=2k^2-x^2\\ y^2={16k^2\over x^2}} \stackrel{求交點}{\Longrightarrow } 2k^2-x^2 ={16k^2\over x^2} \Rightarrow x^4-2k^2x^2+16k^2=0 \\ \Rightarrow x^2 =\cfrac{2k^2\pm \sqrt{4k^4-64k^2}}{2}，由於是無交點，所以4k^4-64k^2 < 0 \Rightarrow 4k^2(k^2-16)< 0 \\ \Rightarrow \cases{k^2 < 16\\ k\ne 0} \Rightarrow \cases {-4< k < 4\\ k\ne 0} \Rightarrow k=\bbox[red,2pt]{\pm 1,\pm 2,\pm 3}$$

解：$$令a=10^{21}+3 \Rightarrow 10^{21}=a-3 \Rightarrow 10^{63}= (10^{21})^3= (a-3)^3 \\ \Rightarrow \cfrac{10^{63}}{10^{21}+3} =\cfrac{(a-3)^3}{ a} =\cfrac{a^3-9a^2+27a-27}{a} =a^2-9a+26+{a-27\over a} \\ \Rightarrow \left\lfloor \cfrac{10^{63}}{10^{21}+3}\right\rfloor =\left\lfloor a^2-9a+26+{a-27\over a} \right\rfloor =a^2-9a+26 (\because 0<{a-27\over a}<1)\\ 只考慮末兩位數字，即求除以100的餘數；a=10^{21}+3 \equiv 3\mod 100 \\\Rightarrow (a^2-9a+26) \mod 100 \equiv (9-27+26) \mod 100 \equiv 8 \Rightarrow 末兩位數字=\bbox[red,2pt]{08}$$

解：$$\angle A=90^\circ \Rightarrow \cfrac{z^3-z}{z^2-z}=k(\cos 90^\circ+i\sin 90^\circ) \Rightarrow \cfrac{z(z+1)(z-1)}{z(z-1) }=ki \Rightarrow z+1=ki\\ \Rightarrow z=-1+ki \Rightarrow |z|=\sqrt{1+k^2} =2 \Rightarrow k=\pm \sqrt 3 \Rightarrow z= \bbox[red,2pt]{-1\pm \sqrt 3i}$$

解：$$\cases{\tan A=1\\ \tan B=2} \Rightarrow \tan C= \tan (\pi-(A+B))= -\tan (A+B)=-\cfrac{\tan A+\tan B}{1-\tan A\tan B}=3\\ \Rightarrow \cases{\sin A=1/\sqrt 2\\ \sin B=2\sqrt 5\\ \sin C=3\sqrt{10}} \Rightarrow \cases{h_a=c\sin B\\ h_b=a\sin C\\ h_c=b\sin A} \Rightarrow \cfrac{abc}{h_ah_bh_c}= \cfrac{1}{\sin A\sin B\sin C} =\cfrac{10}{6}=\bbox[red,2pt]{\cfrac{5}{3}}$$

解：$$假設最後出現的數字為N，N不可能為4,5,6，因為出現3次N就超過10了；\\\begin{array}{} N & 樣本 & 排列數 & 機率\\\hline 3 & 3313 & 3 & 3/6^4 \\\hdashline 2 & 2242 & 3 & 3/6^4\\ & 22132 & 4!/2=12 & 12/6^5 \\\hdashline 1 & 11251 & 4/2  =12 & 12/6^5\\ & 11341 &  4/2  =12 & 12/6^5\\ &112231 & 5!/4= 30 & 30/6^6\\\hline\end{array} \\ \Rightarrow \cfrac{總和為10且最後出現1的機率}{總和為10的機率} =\cfrac{24/6^5+30/6^6}{36/6^5+30/6^6 + 6/6^4} =\cfrac{24\times 6+30}{36\times 6+30 +6\times 36}= \bbox[red,2pt]{29\over 77}$$

解：$$\sum_{k=1}^n \cfrac{k^2+3k+1}{(k+2)!} =\sum_{k=1}^n \cfrac{(k+2)(k+1)-1}{(k+2)!} =\sum_{k=1}^n\left( \cfrac{1}{k!}-\cfrac{1}{(k+2)!} \right)\\ =(1-{1\over 3!})+ ({1\over 2!}-{1\over 4!})+ ({1\over 3!}-{1\over 5!})+\cdots + ({1\over (n-1)!}-{1\over (n+1)!})+ ({1\over n!}-{1\over (n+2)!})\\ =1+{1\over 2!}-{1\over (n+1)!}-{1\over (n+2!)} =\bbox[red,2pt]{{3\over 2}-{1\over (n+1)!}-{1\over (n+2)!}}$$

解：$$x^3+3x-2=(x-a)(x-b)(x-c) \Rightarrow \cases{a+b+c=0\\ ab+bc+ca =3\\ abc=2\\ a^3+3a= b^3+3b= c^3+3c=2}\\ 由於a,b皆為x^3+3x-2=0之二根\Rightarrow \cases{a^3+3a=2\\ b^3+3b=2} \Rightarrow a^3+3a=b^3+3b \\\Rightarrow a^3-b^3+3a-3b=0  \Rightarrow (a-b)(a^2+ab+b^2)+3(a-b)=0 \Rightarrow (a-b)(a^2+ab+b^2+3)=0\\ \Rightarrow a^2+ab+b^2=-3 \Rightarrow (a-b)^2=-3-3ab=-3(ab+1) =-3({2\over c}+1) =-3({c+2\over c})\\ 同理，我們有\cases{(a-b)^2=-3(ab+1) =-3({c+2\over c})\\(b-c)^2=-3(bc+1)=-3({a+2\over a}) \\ (c-a)^2=-3(ca+1) =-3({b+2\over b})}\\  因此三根之和=(a-b)^2+ (b-c)^2+(c-a)^2 =-3(1+{2\over c}+1+{2\over a}+1+ {2\over b}) \\=-3(3+{2(ab+bc +ca)\over abc}) =-3(3+{2\cdot 3\over 2})= \color{blue}{-18};\\ 三根兩兩相乘之和=(a-b)^2 (b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2\\ =9\left( (ab+1)(bc+1) + (bc+1)(ca+1) +(ca+1)(ab+1)\right) \\ =9((2b+ab+bc+1)+(2c+bc+ca+1) +(2a+ca+ab+1)) \\ =9(2(a+b+c) +2(ab+bc+ca)+3) =9(0+6+3)=\color{blue}{81}\\三根之積= (a-b)^2(b-c)^2 (c-a)^2 =-3({c+2\over c})\cdot -3({a+2\over a})\cdot -3({b+2\over b})\\ =-27{(a+2)(b+2)(c+2) \over abc} =-{27\over 2}(a+2)(b+2)(c+2)\\=-{27\over 2}\cdot (-f(-2))={27\over 2}\cdot f(-2)={27\over 2}(-8-6-2)= \color{blue}{-216}\\ 因此該多項式為\bbox[red,2pt]{x^3+18x^2+81x +216}$$

解：$$柯西不等式: (x^2+y^2)(t^2+(-z)^2) \ge (xt-yz)^2，已知\cases{x^2+y^2=16 \\ z^2+t^2=25 \\ xt-yz=20}\\ \Rightarrow (x^2+y^2)(t^2+(-z)^2)=16\times 25= 400=20^2 \Rightarrow 柯西不等式的等號成立\\ \Rightarrow {x\over t} ={y\over -z} ={4\over 5} \Rightarrow z= -{5\over 4}y \Rightarrow xz=-{5\over 4}xy\\ 由於{x^2+y^2 \over 2} \ge \sqrt{x^2y^2} \Rightarrow 8\ge xy \ge -8 \Rightarrow -10\le -{5\over 4}xy \le 10\Rightarrow xz最大值為\bbox[red,2pt]{10}$$

解：$$\cases{\overrightarrow{OA}=(1,3,-2)\\ \overrightarrow{OB}=(-1,2,1)\\ \overrightarrow{OC}=(1,-1,0)} \Rightarrow \overrightarrow{OA}-x\overrightarrow{OB}-y\overrightarrow{OC} =(1+x-y,3-2x+y,-2-x) \\ \Rightarrow \left| \overrightarrow{OA}-x\overrightarrow{OB}-y\overrightarrow{OC}\right| ^2 =(1+x-y)^2+ (3-2x+y)^2+ (-2-x)^2\\ 柯西不等式: ((1+x-y)^2+ (3-2x+y)^2+ (-2-x)^2)(1^2+1^2+(-1)^2)\\\qquad\qquad \ge (1+x-y+3-2x+y+2+x)^2 =6^2=36 \\ \Rightarrow \left| \overrightarrow{OA}-x\overrightarrow{OB}-y\overrightarrow{OC}\right|\ge \sqrt{36\over 3}= 2\sqrt{3} =m\\ 當\left| \overrightarrow{OA}-x\overrightarrow{OB}-y\overrightarrow{OC}\right|=m時，1+x-y=3-2x+y=2+x \Rightarrow \cases{x=0\\ y=-1} \\ \Rightarrow (x,y,m)= \bbox[red,2pt]{(0,-1,2\sqrt 3)}$$

二、計算證明題：5 題共 50 分

解：

$$\cases{直角\triangle ADE: \overline{AE}^2= \overline{AD}^2 +\overline{DE}^2 = 6^2+3^2=45 \\ 直角\triangle ADB: \overline{AB}^2= \overline{AD}^2 +\overline{DB}^2=6^2+6^2=72} \Rightarrow \cases{\overline{AE}=3\sqrt 5\\ \overline{AB}=6\sqrt 2}\\ \triangle AEB: \cos \angle AEB={45+9-72 \over 18\sqrt 5} = -{1\over \sqrt 5} \Rightarrow \sin \angle AEB= {2\over \sqrt 5} \\ \triangle CDE: {\overline{DE}\over \sin \angle DCE} = {\overline{DC} \over \sin DEC} \Rightarrow {3\over 1/\sqrt 2} = {\overline{CD} \over 2/\sqrt 5} \Rightarrow \overline{CD}=6\sqrt{2\over 5}\\ 又\sin \angle BDC= \sin (180^\circ-(45^\circ+\angle DEC)) = \sin(45^\circ+\angle DEC) \\ = \sin 45^\circ \cos \angle DEC+ \sin \angle DEC \cos 45^\circ = {1\over \sqrt 2}\cdot (-{1\over \sqrt 5})+{2\over \sqrt 5}\cdot {1\over \sqrt 2} = {1\over \sqrt{10}} \\ \Rightarrow \triangle BCD面積={1\over 2}\overline{CD}\times \overline{BD} \sin \angle BDC = {1\over 2}\cdot 6\sqrt{2\over 5}\cdot 6\cdot {1\over \sqrt{10}} =\bbox[red,2pt]{18\over 5}$$

解：$$假設f(x)=x^3+ax+b=0的三根為\alpha,\beta,\gamma \Rightarrow \cases{\alpha+\beta+\gamma =0 \\ \alpha\beta +\beta\gamma +\gamma\alpha =a\\ \alpha\beta\gamma =-b}\\ \Rightarrow \triangle(f)=(\alpha-\beta)^2(\beta-\gamma)^2 (\gamma-\alpha)^2 = -(4a^3+27b^2)\\ f(x)為三次式，因此三根為(1)三實根,或(2)一實根二相異之共軛複數根;\\(1)三實根 \Rightarrow \triangle(f) \ge 0 (\because 實數的平方一定大於等於0)\\ (2)一實根二相異之共軛複數根 \Rightarrow \triangle(f) < 0 \Rightarrow 4a^3+27b^2 > 0\\ 因此僅有一實根的條件為\bbox[red,2pt]{4a^3+27b^2 > 0}$$

解：$$只要我們可以證明: {1\over 2} \in [f的最小值,f的最大值]，就可以找到f(x_0)={1\over 2},x_0\in[0,1]\\ 假設 0\le a\le b\le c\le 1 \Rightarrow f(b)=\cfrac{|b-a|+|c-b|}{3} =\cfrac{c-a}{3} \le \cfrac{1}{3} < \cfrac{1}{2} \Rightarrow f(b) <\cfrac{1}{2}\\ \cases{f(1)= ((1-a)+(1-b)+(1-c))\div 3=1-(a+b+c)/3\\ f(0)=(a+b+c)/3} \Rightarrow {f(1)+f(0)\over 2} ={1\over 2}\\ \Rightarrow \max\{f(0),f(1)\} \ge {1\over 2}\Rightarrow f(b) < {1\over 2} \le \max\{f(0),f(1)\} \Rightarrow \exists x_0 \in [0,1] ,f(x_0)=1/2，\bbox[red,2pt]{故得證}。$$

4. 證明：\(n\in N,1+\cfrac{2}{3n-2}\le \sqrt[n]{3} \le 1+\cfrac{2}{n}\)

解：$$\cfrac{3+\overbrace{1+\cdots + 1}^{n-1個1}}{n} \ge \sqrt[n]{3\cdot 1} \Rightarrow \cfrac{n+2}{n} \ge \sqrt[n]{3} \Rightarrow \sqrt[n]{3} \le 1+\cfrac{2}{n} \cdots(1)\\ \cfrac{1/3+\overbrace{1+\cdots + 1}^{n-1個1}}{n} \ge \sqrt[n]{1 \over 3} \Rightarrow \cfrac{1/3+n-1}{n} \ge   \cfrac{1}{\sqrt[n]{3}} \Rightarrow \cfrac{3n-2 }{3n} \ge   \cfrac{1}{\sqrt[n]{3}} \Rightarrow \cfrac{3n}{3n-2}\le  \sqrt[n]{3} \\ \Rightarrow 1+\cfrac{2}{3n-2}\le \sqrt[n]{3} \cdots(2) \\ 由(1)及(2) \Rightarrow 1+\cfrac{2}{3n-2}\le \sqrt[n]{3} \le 1+\cfrac{2}{n}，\bbox[red,2pt]{故得證}。$$

解：$$假設\cases{\triangle 三邊長為a,b,c \\ s=(a+b+c)\div 2}，則\triangle 面積= rs = \sqrt{s(s-a)(s-b)(s-c)} ={abc\over 4R};\\ 由於三邊成等差，故令\cases{a=x-d\\ b=x\\ c=x+d } \Rightarrow s=\cfrac{3}{2}x\Rightarrow \triangle 面積= rs= {3\over 2}rx \\ \Rightarrow \cases{ {3\over 2}rx= {abc\over 4R} ={x(x^2-d^2)\over 4R} \Rightarrow x^2-d^2 =6Rr\\{3\over 2}rx= \sqrt{s(s-a)(s-b)(s-c)} =\sqrt{{3x^2\over 4} ({x^2\over 4}-d^2) } } \\ \Rightarrow \cases{ x^2=d^2+6Rr\\{9r^2x^2\over 4} ={3x^2\over 4} ({x^2\over 4}-d^2) \Rightarrow x^2=4(3r^2+d^2)} \Rightarrow d^2+6Rr=12r^2+4d^2 \\\Rightarrow 3d^2=6Rr-12r^2 \Rightarrow d^2=2Rr-4r^2 \Rightarrow d=\sqrt{2Rr-4r^2}，\bbox[red,2pt]{故得證}。$$