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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Cn2>500n(n1)2>500n2n1000>0n>1+4001232.1n=33n=33C331+C332=33+33×322=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p0.4,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p0.4」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__________。( log20.3010log30.4771

解答P(X=k)=(1p)k1pXP(Xk)=1(1p)k0.9510.6k0.950.6k0.05k(log6log10)log5log100k(0.2219)1.301k5.863k6k7
解答x+2022=xy+2022(x+2022)2=(xy+2022)2x+2022+22022x=xy+202222022x=x(y1)2022x=y12>0y122022xx=2022202222(x,y)=(2022,3)(8088,2)
解答n+4n2+7=n+4(n216)+23=n+4(n+4)(n4)+23n+423n+4n2+711000100023=4323n100043=957
解答an=n{a0=1:a1=2:12an=an1+an2a2=3a3=5a4=8a5=13a6=21a7=34a8=55a9=89a10=144
解答n+2n!+(n+1)!+(n+2)!=n+2n!(1+(n+1)+(n+2)(n+1))=n+2n!((n+2)+(n+2)(n+1))=n+2n!(n+2)2=1n!(n+2)=n+1n!(n+1)(n+2)=n+1n+2!=(n+2)1n+2!=1(n+1)!1(n+2)!limnnk=1k+2k!+(k+1)!+(k+2)!=limnnk=1(1(k+1)!1(k+2)!)=limn(12!1(n+2)!)=12
解答
¯BP¯ACD{PAD=θ¯PD=aPDA=180ABDA=60DPC=30¯DC=¯DP=acosPDC=cos120=12=a2+a2¯CP22a2¯CP=3a{ADP:asinθ=¯APsin60¯AP=3a2sinθ(1)ABP:¯APsin80=¯BPsin(40θ)¯AP=sin80sin(40θ)¯BP(2)BCP:3asin20=¯BPsin10¯BP=sin10sin203a(3)(3)(2)¯AP=sin80sin(40θ)×sin10sin203a=3a2sinθsinθ=sin202sin10sin80sin(40θ)=sin20cos70cos90sin(40θ)=sin20cos70sin(40θ)=sin20sin20sin(40θ)=sin(40θ)sinθ=sin(40θ)θ=20
解答y=22n+1x232nx+1=(2nx1)(2n+1x1){Pn(1/2n+1,0)Qn(1/2n,0)Rn(0,1)an=PnQnRn=12(12n12n+1)1=12n+2n=1an=n=112n+2=14
解答
解答



()PQPQ(T)TRSRTS=120¯RS(h)¯RScos120=32+52¯RS2235¯RS=72TRS=35sin120=7hh=15143
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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