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2023年5月6日 星期六

112年全國教甄聯招-數學詳解

教育部受託辦理112學年度公立高級中等學校教師甄選

第一部分:選擇題( 共40分)
一、單選題( 每題3分, 共24分)

解答|202020212022202120222023202220232025|R1+R2,R1+R3|202020212022111223|2R2+R3|202020212022111001|=20202021=1(A)
解答n3+103n+11=n211n+1211228n+111228=n+11n=1217(A)
解答x2+xy+y2=32x+y+xy+2yy=0y=2x+yx+2yy(1,1)=1(1,1)1:y=(x1)+1x+y=2(B)
解答k=012kcos(120k)=k=0123kcos(1203k)+123k+1cos(120(3k+1))+123k+2cos(120(3k+2))=11181411818118=872717=57(B)
解答S=13910k=2k(k1)C10k2k2f(x)=(1+x)10=10k=0C10kxkf(x)=10(1+x)9=10k=1kC10kxk1f(x)=90(1+x)8=10k=2k(k1)C10kxk2S=f(2)39=903839=30(A)
解答ω=1+3i2=cos2π3+isin2π3=e2πi/3ω3=1(ω1)(ω2+ω+1)=0{ω3=1ω2+ω+1=0(ω5+ω4+5)(3ω2+2ω+2)(3ω2ω+3)=(ω2+ω+5)(3(ω1)+2ω+2)(3(ω1)ω+3)=4(5ω+5)(4ω)=4(5(ω2))(4ω)=80ω3=80(B)
解答{:C31=3:C32=310610(B)
解答log2n12n12log2=120.30139.8n=40(A)

二、複選題(每題 4 分,共 16 分,全對才給分)

解答


11:P(A2WB2WA)+P(A1R1WB1R1WA)=13+49=7912:P(A1R1WB2WA)=2313=29(A)×:=79(B):=29(C)×:P(S1S1S1)+P(S1S2S1)=(79)2+2923>2081(D):P(S1S1S1S1)+P(S1S1S2S1)+P(S1S2S2S1)+P(S1S2S1S1)=(79)3+792923+291323+292379=547729(BD)
解答f(x)=(xα)(xβ)(xγ)(xδ)=(x2(α+β)x+αβ)(x2(γ+δ)x+γδ)=(x2(32i)x+4i)(x2(3+2i)x+4+i)=x46x3+21x2+28x+17(A):a=6(B)×:b=21(C):c=28(D):d=17(ACD)

解答(A):x+1503<xx>7575調(B)×:1(C)×:z2i1(D):,,(AD)
解答{a=OAb=OBc=OCd=OD(A):O,A,Dda=ka+(2k)ba=k=2(B):CD=c+d=ab+ka+(2k)b=(1k)a+(1k)b=(1k)(a+b)=(1k)ABCDAB(C):{AB=(b1a1,b2a2)AD=((k1)a1+(2k)b1,(k1)a2+(2k)b2)ABD=|b1a1b2a2(k1)a1+(2k)b1(k1)a2+(2k)b2)|=|a2b1a1b2|k(D)×:{AC=OB=(b1,b2)AD=((k1)a1+(2k)b1,(k1)a2+(2k)b2)ACD=|b1b2(k1)a1+(2k)b1(k1)a2+(2k)b2)|=|(k1)(a2b1a1b2)|k(ABC)
解答[cos2θsin2θsin2θcos2θ],{L1:x+y=0θ=135L2:x=3yθ=30{A=[0110]B=[1/23/23/21/2]Q=AP=[11]R=BQ=[(13)/2(13)/2]R=(132,132)
解答A=[1a12]A3=[1a2+3aa33a8]=Ia=3A=[1312]A2=[2311]A2+A+I=0B=A+2A2++20A20AB=A2+2A3++20A21BAB=A+A2++A2020A21=A19+A2020A21(IA)B=A+A220I=21IB=21I(IA)1=[210021][0313]=[210021][111/30]=[212170]
解答Xb(n=10,p=12){μ=E(X)=np=5σ=np(1p)=10/21.58P(μσXμ+σ)=P(51.58X5+1.58)=P(3.42X6.58)=P(X=4)+P(X=5)+P(X=6)=1210(C104+C105+C106)=6721024=2132
解答6×24x2×27x+3×16x18x=233x23x233x+324x2x32x=23x+13x+1233x+324x2x32x=(23x+13x+1+324x)(233x+2x32x)=623x(3x+2x1)232x(3x+2x1)=(623x232x)(3x+2x1)=0623x=232xlog6+3xlog2=log2+2xlog3log3=(2log33log2)xx=log32log33log22x13x=3log22log33log2×2log33log23log3=log2log3=log32
解答an=nk=1nn2+k2=nk=11/n1+(k/n)2lim
解答令f(k)={x\over k+4^2} +{y\over k+5^2} +{z\over k+6^2} -1,依題意:f(1^2)=f(2^2)=f(3^2)=0 \\ 因此(k-1)(k-4) (k-9) \\ \qquad =(k+4^2)(k+5^2)(k+6^2)-x(k+5^2)(k+6^2)-y(k+4^2)(k+6^2)-z(k+4^2)(k+5^2) \\ \Rightarrow \cases{左式k^2係數=-1-4-9=-14\\ 右式k^2係數=4^2+5^2+6^2-(x+y+z)} \Rightarrow x+y+z=77+14=\bbox[red, 2pt]{91}
解答x^2f(x)=(3x-2)p(x)+r \Rightarrow x^3f(x) =(3x-2)xp(x)+rx =(3x-2)xp(x)+{r\over 3}(3x-2)+{2r\over 3} \\ \Rightarrow {2r\over 3}=6 \Rightarrow r=\bbox[red, 2pt]9
解答Q=a^2-2a =(a-1)^2-1為有理數\Rightarrow a=1+ \sqrt q,其q為有理數,但\sqrt q為無理數\\ 又{P\over Q} ={a^3+3a^2-16a+6\over a^2-2a } =a+5+{-6a+6\over a^2-2a}=6+ \sqrt q+{ - 6\sqrt q\over q-1} 為有理數\\ \Rightarrow q-1=6 \Rightarrow q=7 \Rightarrow a=\bbox[red, 2pt]{1+\sqrt 7}
解答\cases{A(2,2,1)\\ B(1,0,2)\\ C(0,1,2)} \Rightarrow \cases{\overrightarrow{AB} =(-1,-2,1)\\ \overrightarrow{AC} =(-2,-1,1)} \Rightarrow \vec n=\overrightarrow{AB} \times \overrightarrow{AC} =(-1,-1, -3) \\ \Rightarrow E=\triangle ABC:x+y+3z =7 \Rightarrow \cases{E\cap 直線(2,0,s) =\bbox[red,2pt]{(2,0,5/3)}\\ E\cap 直線(0,2,t)=\bbox[red,2pt]{(0,2,5/3)}} 

解答


令\overline{MF} =\overline{FC}=a \Rightarrow 在直角\triangle BMF中,\overline{MF}^2 = \overline{MB}^2 + \overline{BF}^2 \Rightarrow a^2=120^2+(288-a)^2 \Rightarrow a=169\\ 令\overline{DE}= \overline{EG} =b \Rightarrow \overline{AE}=288-b \Rightarrow \cases{直角\triangle MAE: \overline{ME}^2 = \overline{MA}^2 + \overline{AE}^2 =120^2+(288-b)^2 \\ 直角\triangle EDC: \overline{EC}^2 = \overline{ED}^2 + \overline{DC}^2 =b^2+240^2}\\ 由於\overline{EM} =\overline{EC} \Rightarrow 120^2+(288-b)^2 =b^2+240^2 \Rightarrow b=69 \\ 直角\triangle FHE: \overline{EF}^2= \overline{EH}^2 + \overline{HF}^2 =240^2+ (a-b)^2=240^2+ (169-69)^2=260^2 \Rightarrow \overline{EF} =\bbox[red, 2pt]{260}

解答\cases{u=\ln x\\ dv =xdx} \Rightarrow \cases{du={dx\over x}\\ v={1\over 2}x^2} \Rightarrow \int x\ln x\,dx ={1\over 2}x^2\ln x -{1\over 2}\int xdx ={1\over 2}x^2 \ln x-{1\over 4}x^2+C\\ \Rightarrow \int_1^e x\ln x\,dx =\left. \left[{1\over 2}x^2 \ln x-{1\over 4}x^2 \right]\right|_1^e ={1\over 2}e^2-{1\over 4}e^2+{1\over 4} =\bbox[red, 2pt]{{1\over 4}e^2+{1\over 4}}
解答
假設\overline{AB} \bot \overline{BO} 且\overline{OB}\bot \overline{BC},則欲求之|\cos \theta | = |\cos \angle ABC|\\ 在直角\triangle OAB中,假設\overline{AB}=1再加上\angle AOB=30^\circ \Rightarrow \cases{\overline{OA}=2\\ \overline{OB}=\sqrt 3}\\ 同理,在直角\triangle OBC中,\angle BOC=45^\circ \Rightarrow \overline{BC}=\overline{OB}=\sqrt 3,且\overline{OC}=\sqrt 6\\ 在\triangle OAC中,\cos \angle AOC =\cos 60^\circ = {1\over 2} ={2^2+(\sqrt 6)^2 -\overline{AC}^2 \over 2\cdot 2\cdot \sqrt 6} \Rightarrow \overline{AC}^2 =10-2\sqrt 6\\ 在\triangle ABC中,\cos \angle ABC={1^2+(\sqrt 3)^2-(10-2\sqrt 6) \over 2\cdot 1\cdot \sqrt 3} = {2\sqrt 6-6\over 2\sqrt 3}=\sqrt 2-\sqrt 3\\ \Rightarrow |\cos \theta| =|\sqrt 2-\sqrt 3|= \bbox[red, 2pt]{\sqrt 3-\sqrt 2}


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解題僅供參考,其他試題及詳解

5 則留言:

  1. 多選九D選項第一項應該都是轉移到S_1機率

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  2. 請問複選題
    9.
    狀態1---->狀態1與狀態1---->狀態2的機率怎麼得到

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    1. 狀態1->狀態1有兩種情形,情形1:A箱抽到2W(p=1/3)丟到B箱,再從B箱抽2W(p=1)丟回A箱,機率=1/3; 情形2:A箱抽到1R1W(p=2/3)丟到B箱,再從B箱抽1R1W(p=2/3)丟回A箱,機率=4/9; 兩情況機率和=1/3+4/9=7/9

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