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2022年6月10日 星期五

111年彰化女中教甄-數學詳解

國立彰化女子高級中學111 學年度第二次教師甄選

一、填充題 (每題5 分,共計75 分)

1.某冰淇淋店最少需準備n 桶不同口味的冰淇淋,才能滿足廣告所稱「任選兩球不同口味冰淇淋的組合數超過500 種」。試問來店顧客從n 桶中任選兩球(可為同一口味)共有___________種方法。
解答$$C^n_2 \gt 500 \Rightarrow {n(n-1)\over 2}\gt 500 \Rightarrow n^2-n-1000\gt 0 \Rightarrow n \gt {1+\sqrt{4001}\over 2} \approx 32.1 \Rightarrow n=33\\ 當n=33時,至少提供C^{33}_1+C^{33}_2 =33+{33\times 32\over 2} =\bbox[red,2pt]{561}$$
2.彰化女中籃球校隊想招收隊員,某參加甄選的學生聲稱自身的投籃命中率\(p \ge 0.4\),校方想透過檢定的方式來決定她的聲稱是否採信。假設「此學生的投籃命中率\(p\ge 0.4\)」且「投籃直到第一次進球共需X 次」,在顯著水準為0.05的條件之下,求隨機變數X 的拒絕域為__________。( \(\log 2\approx 0.3010、\log 3\approx 0.4771\)) 

解答:$$P(X=k)=(1-p)^{k-1}p \Rightarrow X 為幾何分配 \Rightarrow P(X\le k)= 1-(1-p)^k \ge 0.95 \\ \Rightarrow 1-0.6^k \ge 0.95 \Rightarrow 0.6^k \le 0.05 \Rightarrow k(\log 6-\log 10) \le \log 5-\log 100 \\ \Rightarrow k(-0.2219) \le -1.301 \Rightarrow k\ge 5.863 \Rightarrow \bbox[red,2pt]{k\ge 6}\\公布的答案是\bbox[blue,2pt]{k\ge 7}$$
解答:$$\sqrt x+\sqrt{2022} =\sqrt{xy+2022} \Rightarrow (\sqrt x+\sqrt{2022})^2 =(\sqrt{xy+2022}  )^2 \\ \Rightarrow x+2022 +2\sqrt{2022x} =xy+2022 \Rightarrow 2\sqrt{2022x}=x(y-1) \Rightarrow \sqrt{2022\over x}={y-1\over 2}\gt 0\\ {y-1\over 2}是有理數 \Rightarrow {2022\over x} 是完全平方數 \Rightarrow x=2022 或2022\cdot 2^2 \Rightarrow (x,y)=\bbox[red, 2pt]{(2022,3)或(8088,2)}$$
解答:$$\cfrac{n+4}{n^2+7} =\cfrac{n+4}{(n^2-16)+23} =\cfrac{n+4}{(n+4)(n-4)+23} \Rightarrow 若n+4是23的倍數,則\cfrac{n+4}{n^2+7} 不是最簡分數\\ 而1-1000有\lfloor {1000\over 23} \rfloor =43個23的倍數,因此n有1000-43= \bbox[red, 2pt]{957}可能值$$
解答:$$假設a_n=兩人中間有n 格,提前結束的方法數,則\cases{a_0=1:甲走一步\\ a_1=2:甲走兩步、甲走一步乙走一步};\\ 而且每人只可走1 或2步,因此a_n= a_{n-1}+ a_{n-2}  \Rightarrow a_2=3 \Rightarrow a_3=5 \Rightarrow a_4=8\\ \Rightarrow a_5=13 \Rightarrow a_6=21 \Rightarrow a_7=34 \Rightarrow a_8= 55 \Rightarrow a_{9}=89 \Rightarrow a_{10}=\bbox[red,2pt]{144}$$
解答:$${n+2\over n!+(n+1)! +(n+2)!} ={n+2\over n!(1+(n+1) +(n+2)(n+1))} ={n+2\over n!((n+2) +(n+2)(n+1))}\\ = {n+2\over n!(n+2)^2} = {1\over n!(n+2)} ={n+1\over n!(n+1)(n+2)} ={n+1\over n+2!} ={(n+2)-1\over n+2!} = {1\over (n+1)!}-{1\over (n+2)!} \\ 因此\lim_{n\to \infty} \sum_{k=1}^n  {k+2\over k!+(k+1)!+ (k+2)!}  =\lim_{n\to \infty} \sum_{k=1}^n \left( {1\over (k+1)!}-{1\over (k+2)!}\right)  =\lim_{n\to \infty}  \left( {1\over 2!}-{1\over (n+2)!}\right)\\ =\bbox[red, 2pt]{1\over 2}$$
解答
$$延長\overline{BP}交\overline{AC}於D,如上圖,並令\cases{\angle PAD=\theta\\ \overline{PD}=a},則\angle PDA= 180^\circ - \angle ABD-\angle A= 60^\circ \\\Rightarrow \angle DPC=30^\circ \Rightarrow \overline{DC} =\overline{DP}=a \Rightarrow \cos \angle PDC = \cos 120^\circ = -{1\over 2}={a^2 +a^2-\overline{CP}^2 \over 2a^2}\\ \Rightarrow \overline{CP}= \sqrt 3a\\ \cases{\triangle ADP: {a\over \sin \theta}={\overline{AP} \over \sin 60^\circ} \Rightarrow \overline{AP}={\sqrt 3a \over 2\sin \theta} \cdots(1)\\[1ex] \triangle ABP: {\overline{AP} \over \sin 80^\circ} ={\overline{BP} \over \sin(40^\circ-\theta)} \Rightarrow \overline{AP} ={\sin 80^\circ \over \sin(40^\circ-\theta)}\overline{BP} \cdots(2)\\[1ex] \triangle BCP: {\sqrt 3a \over \sin 20^\circ}={\overline{BP} \over \sin 10^\circ} \Rightarrow \overline{BP}= {\sin10^\circ \over \sin 20^\circ}\sqrt 3a \cdots(3) }\\ 將(3)代入(2) \Rightarrow \overline{AP}= {\sin 80^\circ \over \sin(40^\circ-\theta)} \times {\sin10^\circ \over \sin 20^\circ}\sqrt 3a ={\sqrt 3a \over 2\sin \theta}\\ \Rightarrow \sin \theta = {\sin 20^\circ \over 2\sin 10^\circ \sin 80^\circ}\sin(40^\circ-\theta)  ={\sin 20^\circ \over \cos 70^\circ -\cos 90^\circ }\sin(40^\circ-\theta) ={\sin 20^\circ \over \cos 70^\circ  }\sin(40^\circ-\theta)\\ ={\sin 20^\circ \over \sin 20^\circ  }\sin(40^\circ-\theta) =\sin(40^\circ-\theta)  \Rightarrow \sin \theta = \sin(40^\circ-\theta) \Rightarrow \theta=\bbox[red,2pt]{20^\circ }$$
解答:$$y=2^{2n+1}x^2 -3\cdot 2^nx+1 = (2^n x-1)(2^{n+1}x-1) \Rightarrow \cases{P_n(1/2^{n+1},0)\\ Q_n(1/2^{n},0)\\ R_n(0,1)} \\ \Rightarrow a_n= \triangle P_nQ_nR_n面積={1\over 2}\cdot ({1\over 2^n}-{1\over 2^{n+1}})\cdot 1 = {1\over 2^{n+2}} \Rightarrow \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty {1\over 2^{n+2}} = \bbox[red, 2pt]{1\over 4}$$
解答:$$本題\bbox[blue,2pt]{送分}$$
解答



$$改變視角,從原圖(上圖)\overleftrightarrow{PQ}視角望去,讓P、Q兩點重疊(命名為T),則T、R、S為一三角形\\,且\angle RTS=120^\circ;因此兩歪斜線距離就是\overline{RS}上的高(h),先利用餘弦定理求\overline{RS},\\\cos 120^\circ=\cfrac{3^2+5^2-\overline{RS}^2}{2\cdot 3\cdot 5} \Rightarrow \overline{RS}=7;再利用2\triangle TRS面積= 3\cdot 5\cdot \sin 120^\circ = 7h \Rightarrow h=\bbox[red,2pt]{{15\over 14}\sqrt 3}$$
解答
$$由於\overline{AB}是直角\triangle APB與直角\triangle AOB的共同斜邊,因此A、B、O、P共圓;\\ 假設圓半徑=r,則\cases{\overline{PB}^2 = \overline{AB}^2-\overline{AP}^2 = 4r^2-8 \Rightarrow \overline{PB} =\sqrt{4r^2-8}\\ \overline{OA}^2+\overline{OB}^2 =\overline{AB}^2 \Rightarrow \overline{OA}=\overline{OB} =\sqrt 2r}\\ 依\href{https://chu246.blogspot.com/2020/11/ptolemys-theorem.html}{托勒密定理}: \overline{OA}\times \overline{PB} = \overline{OP}\times \overline{AB}+ \overline{PA}\times \overline{OB} \Rightarrow \sqrt 2r \cdot \sqrt{4r^2-8} =6r+ 4r\\ \Rightarrow 2r^2(4r^2-8)=100r^2 \Rightarrow 4r^2-8=50 \Rightarrow \overline{PB}= \sqrt{4r^2-8} =\bbox[red,2pt]{5\sqrt 2}$$
解答:$$f(1-{1\over x})+2f({1\over 1-x}) +3f(x)=12x \Rightarrow \cases{x=-1 \Rightarrow f(2)+2 f({1\over 2})+3f(-1)= -12\\ x={1\over 2} \Rightarrow f(-1)+2f(2) +3f({1\over 2}) =6 \\ x= 2 \Rightarrow f({1\over 2}) +2 f(-1)+3f(2) = 24} \\ \Rightarrow \cases{f(-1)=f({1\over 2})= -5\\ f(2)=\bbox[red,2pt]{13}}$$
解答:$$即然兩焦點至任一切線的距離為定值,因此假設橢圓為左右形,即a\gt b\gt 0,因此\cases{F_1(-c,0)\\ F_2(c,0)};\\ 並取右端點切線L:x=a \Rightarrow \cases{d(F_1,L)=a+c\\ d(F_2,L)=a-c} \Rightarrow d(F_1,L)\times d(F_2,L)= a^2-c^2= b^2\\ 若橢圓為上下形,即b\gt a\gt 0,此時\cases{F_1(0,c)\\ F_2(0,-c)},右端點切線L:x=a \Rightarrow \cases{d(F_1,L)=a\\ d(F_2,L)=a} \\ \Rightarrow d(F_1,L)\times d(F_2,L)=a^2;因此兩焦點至任一切線的距離為\bbox[red,2pt]{\begin{cases}b^2 & \text{if }a\gt b\gt 0\\ a^2 & \text{if } b\gt a\gt 0 \end{cases}}$$
解答:$$兩圖形\cases{\Gamma: y=f(x)=x^4-4x^3+10\\ L:y=mx+a} 的交點有兩個\Rightarrow f(x)=mx+a 有二重根\\ \Rightarrow x^4-4x^3-mx+(10-a)=0的二重根\alpha,\beta,滿足\cases{\alpha+\alpha+\beta+\beta =2(\alpha+\beta)= 4\\ \alpha^2+4\alpha\beta+ \beta^2= (\alpha+\beta)^2 +2\alpha\beta =0\\ 2\alpha\beta(\alpha+\beta)=m \\ \alpha^2\beta^2=10-a} \\ \Rightarrow \cases{\alpha+\beta =2 \\ \alpha\beta =-2 \\ m=-8\\ a=6} \Rightarrow L:\bbox[red, 2pt]{y=-8x+6}$$
解答:$$邊長為6的正方體其四頂點可假設為\cases{O(0,0,0)\\ A(a,0 ,a) \\ B(0, a,a)\\ C(a,a,0)},其中a=3\sqrt 2\\\Rightarrow P={1\over 2}A+{1\over 3}B+ {1\over 6}C = ({2\over 3}a , {1\over 2}a , {5\over 6}a); \\ 令\cases{A'(s,0,s) \\ B'(0,t,t)\\ C'(u,u,0)},P為\triangle A'B'C'重心 \Rightarrow \cases{s+u =   2a\\ t+u= 3a/2\\ s+t=  5a/2} ,其中s,t,u\in \mathbb{R} \Rightarrow \cases{s =3a/2 \\ t=a \\ u=a/2} \\ \Rightarrow \cases{A'(3a/2,0,3a/2)\\ B'(0,a, a)\\ C'(a/2,a/2,0)} \Rightarrow OA'B'C'體積={1\over 6}\begin{Vmatrix} 3a/2 & 0 & 3a/2 \\ 0& a & a \\ a/2 & a/2 & 0\end{Vmatrix} ={3\over 2}a^3 =\bbox[red,2pt]{{27\over 2}\sqrt 2}$$

二、計算證明題:(共計25 分)

解答:$$e^x = 1+ x+ {x^2 \over 2!} +{x^3\over 3!} + \cdots \Rightarrow e^x \gt 1+x \Rightarrow  e^{\pi/e-1} \gt 1+({\pi\over e}-1) \Rightarrow e^{\pi/e-1} \gt {\pi\over e} \\ \Rightarrow e^{\pi/e} \ge \pi  \Rightarrow \bbox[red,2pt]{e^\pi \gt \pi^e,故得證}$$
解答:$$P_n在單位圓上 \Rightarrow \overline{OP_n}=1, n=1-2022 \Rightarrow a_1+a_2+ \cdots + a_{2022}\ge 1 \cdots(1)\\ 柯西不等式: (a_1^2 +a_2^2 + \cdots +a_{2022}^2)(1^2+1^2 +\cdots +1^2) \ge (a_1+a_2+\cdots +a_{2022})^2 \\ \Rightarrow {1\over 2022} \cdot 2022 \ge (a_1+a_2+\cdots +a_{2022})^2 \Rightarrow a_1+a_2+\cdots +a_{2022}\le 1 \cdots(2) \\ 由(1)及(2)可知 1\le a_1+a_2+\cdots +a_{2022}\le 1,再由夾擠定理可知a_1+a_2+\cdots +a_{2022}= \bbox[red,2pt]{1}$$
解答


$$此題相當於求P(2,0)至單位圓上正九邊形各頂點距離的平方和\\,即|2-\omega|^2 +|2-\omega^2|^2 +\cdots +|2-\omega^8|^2 = \sum_{n=1}^8 \overline{A_nP}^2\\ \triangle OA_nP: \cos \angle A_nOP = \cfrac{1^2+2^2 -\overline{A_nP}^2}{2\cdot 2\cdot 1} \Rightarrow \cos (40^\circ \cdot n) = \cfrac{5 -\overline{A_nP}^2}{ 4} \Rightarrow \overline{A_nP}^2=5-4\cos(40^\circ \cdot n)\\ \Rightarrow \sum_{n=1}^8 \overline{A_nP}^2 =\sum_{n=1}^8 (5-4\cos(40^\circ \cdot n))=40-4(\cos 40^\circ+ \cos 80^\circ +\cdots +320^\circ)\\= 40-8( \cos 40^\circ +\cos 80^\circ +\cos 120^\circ +\cos 160^\circ)=40-8(2\cos 60^\circ\cos 20^\circ-{1\over 2}+\cos 160^\circ) \\ =40-8( \cos 20^\circ+\cos 160^\circ+\cos 160^\circ-{1\over 2}) =40-8( 2\cos 90^\circ\cos 20^\circ-{1\over 2} ) \\ =40-8(-{1\over 2} )= \bbox[red,2pt]{44}$$

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解題僅供參考,其他教甄歷年試題及詳解

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