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2021年4月24日 星期六

110年台南女中第1次教甄-數學詳解

國立臺南女中 110 學年度第一次教師甄選

一、填充題

解答kNf(k){f(k)f(k+1)f(k)f(k1){f(k)/f(k+1)1f(k)/f(k1)1{(kk+1)21.0521(kk1)211.0521{(11k+1)1.051(1+1k1)11.051{k201k21k=2021

解答y=x2y=2xy(1)=2Q1P2:y=2x1Q2(12,0)P2(12,14){P3(14,0),Q3(14,116)P4(18,0),Q4(18,164){¯P1Q1=1¯P2Q2=1/4¯P3Q3=1/42¯P4Q4=1/43k=1¯PkQk=111/4=43

解答

¯AC¯OBD¯CP¯OB¯AQ¯AD¯DC=OAQBOBPC=4012=103=OADOCD{OAD=1013OAC=18013OCD=313OAC=5413ABD=12OAQBOAD=2018013=8013;¯OD¯DB=OADABD=180/1380/13=94OB=134OD=139(313OA+1013OC)=13OA+109OC(a,b)=(13,109)
解答f(x)=(xa1)(xa2)(xa2011){x2020=aix2019=aiajx2018=aiajak=a1a2a2011b2010=1f(1)=1(11112)(11122)(11132)(1120212)=11012112×1113122×1214132××2020×202220212=11011×20222021=12022022231=201122231
解答{7x1y=16xy+1xy=30(7x1y)(y7x)=7(xy+1xy)5016(y7x)=73050y7x=102xy20x+9=2x(y10)+9=2x(y(y7x))+9=14+9=23


解答rnrn+1x2anx+(15)n=0{an=rn+rn+1rnrn+1=15n{r1=2rnrn+1=15n{r1=2r2k+1=25kr2k=125k,kNn=1an=r1+2(r2+r3++rn+)=2+2k=1(125k+25k)=2+252k=115k=2+5×14=134
解答{N=::xi:yi{54=C94=126xi=5yi=4xi,yjNN1()5()41()4()512()x1()4()x2H23=4()y1()5()y2H22=33()x1()y1()x2()y2H23×H22=12()y1()x1()y2()x2H22×H23=124()x1()y1()x2()y2()x3H32×H22=18()y1()x1()y2()x2()y3H31×H23=125()x1()y1()x2()y2()x3()y3H32×H31=18()y1()x1()y2()x2()y3()x3H32×H31=186()x1()y1()x2()y2()x3()y3()x4H41×H31=12()y1()x1()y2()x2()y3()x3()y4H32=67()x1()y1()x2()y2()x3()y3()x4()y4H41=4()y1()x1()y2()x2()y3()x3()y4()x4H41=481=(1×2+2×7+3×24+4×30+5×36+6×18+7×8+8)÷126=560126=409
解答{|z1|=2|z2|=3{z1=2(cosA+isinA)z2=3(cosB+isinB)3z12z2=32i{6(cosAcosB)=326(sinAsinB)=1{12sinA+B2sinAB2=32(1)12cosA+B2sinAB2=1(2)(1)(2)tanA+B2=32tanA+B2=32{sinA+B2=313cosA+B2=213{sin(A+B)=2sinA+B2cosA+B2=1213cos(A+B)=cos2A+B2sin2A+B2=513z1z2=6(cos(A+B)+isin(A+B))=3013+7213i

解答¯PB=a{cosA=4a223a2=423cosAcosQ=2a22a2=22cosQ423cosA=22cosQcosQ=3cosA1{S=123sinAT=12sinQS2+T2=34sin2A+14sin2Q=34(1cos2A)+14(1cos2Q)=134cos2A14(3cosA1)2=32cos2A+32cosA+34cosA=36S2+T278

解答


=JABKJDGHKBCI,{G,H,IHJHIzKHIxy{{AB=(2,1,0)AK=(0,3,0)AJ=(0,0,3)JABK=16210030003=3{JD=(0,0,1)AK=(2/3,1/3,1)JH=(0,1,1)JDGH=160012/31/31011=19{KI=(0,1,1)KC=(0,1,0)KB=(2,2,0)KBCI=16011010220=13=31913=239




解答
    由題意知: 每抓一次,箱子(唉!題目的箱子就是盒子啦!)就少一個白球或2個黑球;也就是說箱子內的黑球永遠都是偶數,如果最後剩下一球,那一定是白球,所以最後一球是黑色的機率為0


解答:a3+b3+c33abc=(a+b+c)(a2+b2+c2(ab+bc+ca))=12(a+b+c)2((ab)2+(bc)2+(ca)2)m3+n3+99mn=333m3+n3+(33)33mn(33)=0(m+n33)(m2+n2+(33)2(mn33n33m))=012(m+n33)((mn)2+(n+33)2+(m+33)2)=0{m+n=33m=n=33mn0m,nZ(m,n)=(33,0),(32,1),,(1,32),(0,32),(33,33)35


解答{1100:12,22,,10210616(100/6){3615101+15=24
解答f(x)=x0(t+a)(t+2)3dtf(x)=(x+a)(x+2)3f(x)=(x+2)3+3(x+a)(x+2)2y=f(x)x=2f(2)=0(2+a)42=0a=2f(x)=(x+2)3+3(x2)(x+2)2=4(x+2)2(x1)f(x)=0x=1,2(1,f(1))f(2+)f(2)>0(2,f(2))f(x)=x0(t2)(t+2)3dt=15x5+x48x216xf(1)=15+1816=1145(1,1145)
解答
:{y=f(x)=|logx|y=g(x)=ax+bk,2k,3k,kR+y=f(x){f(k)<0f(2k),f(3k)>0{logk=ak+b(1)log2k=2ak+b(2)log3k=3ak+b(3)(2)(1)log2k+logk=aka=log2k2k(1)logk=log2k2+bb=log2k3{a=log2k2kb=log2k3(3)log3k=3log2k2log2k3=log8k62k3=log4k33k=4k3k(4k23)=0k=32(a,b)=(23log32,log334)=(233log32,log439)
解答
Γ:y2=4xF(1,0){F(1,0)L:x=1PΓP(t2,2t),tR¯PF=dist(P,L)=t2+1k=|t2(t25)2+(2t3)2|=|¯PF1¯PQ|,Q(5,3)k¯PF¯PQ=¯FQk=|1¯FQ|=|15|=6P=AA,BQFΓ
解答{x=3x+yy=2x7yx25+y24152πxxyxxyxy=3127=23(3x+y)25+(2x7y)241=25π23
解答PL:x1=y31=z12PP(t,t+3,2t+1),tR¯PA+¯PB=(t1)2+(t1)2+4t2+(t3)2+(t+1)2+(2t7)2=6t2+2+6t236t+59=6(t2+13+t26t+596)=6((t0)2+(013)2+(t3)2+(0+56)2)(t,0)A(0,13)(t,0)B(3,56)¯AB(t,0)¯ABxABAB:y=5+236x+13x(325+2,0)t=325+2Pxt325+2=2(52)=102
解答a,b,c=V=6Vα,β,γ{|α|1|α+β|1|α+β+γ|18S=6V×8=48V:{|x|1|x+y|1|x+y+z|1111x1x1xy1xy1dzdydx=8
解答a_{n+3} =a_{n+2}-a_{n+1}+a_n \Rightarrow a_{n+3}+a_{n+1} =a_{n+2}+a_n \Rightarrow \cases{a_4+a_2=a_3+a_1 \\ a_5+a_3= a_4+a_2} \\ \Rightarrow a_5+a_3= a_3+a_1 \Rightarrow a_5=a_1 ,同理可得\cases{a_1=a_5\\ a_2=a_6\\ a_3=a_7\\ a_4=a_8\\ \cdots},即 a_m=a_{(m \mod 4)}\\ 因此 \sum_{k=-1}^{102}(-1)^ka_k = (-a_1+a_2-a_3+a_4) + (-a_5+a_6-a_7+a_7)+ \cdots \\\qquad +(-a_{97}+ a_{98}- a_{99}+a_{100})-a_{101}+a_{102} = 0 + 0 +\cdots +0 -a_{101}+a_{102}\\ =-a_1+a_2 =a_3-a_4 = a_{75}-a_{24} =13-71= \bbox[red, 2pt]{-58}
解答令u=25-x^2,則du=-2xdx;因此\int_3^5 x\sqrt{25-x^2}\;dx = \int_{16}^0 -{1\over 2}u^{1/2}\;du = \left. \left[ -{1\over 3} u^{3/2} \right] \right|_{16}^0\\ ={64\over 3} \Rightarrow f(x)在[3,5]的函數值平均為{\int_3^5f(x)\;dx \over 5-3} ={64\over 3} \times {1\over 2} =\bbox[red, 2pt]{32\over 3}

解答

3\overline{AD} =2\overline{DE}=\overline{EB} \Rightarrow \overline{AD} :\overline{DE} :\overline{EB} =2:3:6 \\\Rightarrow \triangle CAD: \triangle CAD: \triangle CDE: \triangle CEB: \triangle ABC=2:3:6:11 \\ \Rightarrow {1\over 2}ab \sin \alpha: {1\over 2}bc \sin \beta: {1\over 2}cd \sin \gamma: {1\over 2}ad =2:3:6:11,其中\cases{\overline{CA}= a\\ \overline{CD}= b\\ \overline{CE}= c\\ \overline{CB}= d} \\ \Rightarrow ab \sin \alpha: bc \sin \beta: cd \sin \gamma: ad =2:3:6:11 \Rightarrow \cases{\sin\alpha = {2\over ab}k \\\sin\beta = {3\over bc}k \\\sin\gamma = {6\over cd}k \\ad=11k}\\ \Rightarrow {\sin \alpha\cdot \sin \gamma \over \sin \beta}={{12\over abcd}k^2 \over {3\over bc}k} ={4k\over ad} ={4k\over 11k}= \bbox[red, 2pt]{4\over 11}

二、計算題


解答\cases{A(a,a^3) \\ B(b,b^3)\\ C(c,c^3)} \Rightarrow {1\over 3}(\overrightarrow{OA} +\overrightarrow{OB} +\overrightarrow{OC} ) = \overrightarrow{OG}  \Rightarrow {1\over 3}(a+b+c,a^3+b^3+c^3) =(0,-2)\\ \Rightarrow \cases{a+b+c =0 \\ a^3+b^3+c^3 =-6} \Rightarrow \cases{c=-a-b\\ a^3+b^3-(a+b)^3+6 =0} \\ 令\cases{f(a,b)=-a-b\\ g(a,b)=a^3+b^3-(a+b)^3+6 =0},利用 \text{Lagrange 乘數法} \Rightarrow \cases{{\partial \over \partial a}f+ \lambda {\partial \over \partial a}g=0 \\ {\partial \over \partial b}f+ \lambda {\partial \over \partial b}g=0 \\ g=0} \\ \Rightarrow \cases{-1 +\lambda (3a^2-3(a+b)^2) =0 \cdots(1)\\ -1+ \lambda(3b^2-3(a+b)^2)=0 \cdots(2)\\ a^3+b^3-(a+b)^3+6 =0 \cdots(3)},由(1)及(2)可得 \lambda = {1\over -6ab-3b^2} = {1\over -3a^2-6ab} \\ \Rightarrow \cases{a=b\\ a=-b} 代入(3) \Rightarrow \cases{2a^3-8a^3+6=0\\ 0-0+6=0} \Rightarrow \cases{a=1=b\\ 6=0矛盾,不合} \Rightarrow c=f(1,1)=\bbox[red, 2pt]{-2}為極值(最大值)



解答\cases{n是奇數\Rightarrow 3n^2+n+1 = 奇+奇+奇=奇數\\ n是偶數\Rightarrow 3n^2+n+1= 偶+偶+奇=奇數} \qquad\Rightarrow 3n^2+n+1是奇數\\利用反證法:\\假設存在正整數n,使得f(n)=2;由於3n^2+n+1為奇數且各數字和=2\\,因此3n^2+n+1只能首尾數字是1,其它皆是0 \Rightarrow 3n^2+n+1=10^k+1,k\in \mathbb{N}\\ \Rightarrow n(3n+1)= 2^k\cdot 5^k \Rightarrow \cases{n=2^k\cdots(1)\\ 3n+1=5^k \cdots(2)},將(1)代入(2) \Rightarrow 3\cdot 2^k+1=5^k \\ \Rightarrow \cases{3\cdot 2^k+1 \gt 5^k,\text{ if }k=1\\ 3\cdot 2^k+1 \lt 5^k,\text{ if }k\ge 2} \Rightarrow 3\cdot 2^k+1\ne 5^k, \forall k\in \mathbb{N}\Rightarrow 矛盾\\ \Rightarrow f(n)\ne 2, \forall n\in \mathbb{N}


解答學校公布的作法:\\2^{234} =\sum_{k=0}^{99} \lfloor 2^kx\rfloor \le \sum_{k=0}^{99} 2^kx =(2^{100}-1)x \Rightarrow x \ge {  2^{234} \over 2^{100}-1} =2^{134}+ 2^{34} +{2^{34}\over 2^{100}-1}\\ 設最小的x=2^{134} +2^{34}+ {y\over 99} \Rightarrow \sum_{k=0}^{99}\lfloor 2^k \cdot {y\over 2^{99}}\rfloor =2^{34};\\ 取y=2^{33+1}時,可得最小的x=\bbox[red, 2pt]{2^{134}+ 2^{34} +{2^{33}+1\over 2^{99}}}

解答
學校公布的作法:


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解答僅供參考
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2 則留言:

  1. (20) m=even (am=a(m mod 5)); m=odd (am(m mod 4))

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  2. 其它數字的因數是偶數個?

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