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2021年8月23日 星期一

110年嘉義高中教甄-數學詳解

國立嘉義高級中學 110 學年度第 1 次教師甄選

一、 填充題(共 10 題, 每題 8 分,共 80 分)

解答:6!(112!+13!14!+15!16!)=720×91144=4556!455=265
解答1314588=22×3×11×23×433N=(1+2+22)(1+3)(1+11)(1+23)(1+433)=7×4×12×24×434=28×32×72×31
解答{f(a,b,c)=abcg(a,b,c)=a+b2+c311f=λg{af=λagbf=λbgcf=λcg{bc=λ(1)ac=2bλ(2)ab=3c2λ(3){(1)/(2)(1)/(3){b/a=1/2bc/a=1/3c2{a=2b2a=3c32b2=3c3b2=32c3{a=3c3b2=3c3/2g(a,b,c)=0c3=2b2=3a=6abc=6332=24/333/2
解答1,4522,352f(x)=(x52)2f2
解答
P(0,0,h)O{OAP=30¯OA=3hOBP=60¯OB=h/3OCP=45¯OC=h{cosABO=h2/3+25003h2100h/3cosOBC=h2/3+2500h2100h/3ABO+OBC=180h2/3+25003h2=(h2/3+2500h2)h2=1500h=1015
解答
{¯AF=a¯FB=bCeva :33a24b=1ab=89
解答log7=0.84530log7=25.3573025+1=26log7(log6+log8)÷2=(4log2+log3)÷2=(4×0.301+0.4771)÷20.840.84×30=25.2730=49154815<730<501515log48<log730<15log50{15log48=15(4log2+log3)=15(4×0.301+0.4771)=25.216515log50=15(2log2)=15(20.301)=25.48525.2165<log730<25.485
解答an+2=2an+1+3an+1an+2+14=2(an+1+14)+3(an+14)bn+2=2bn+1+3bn{bn=an+1/4b0=a0+1/4=5/4b1=a1+1/4=9/4λ22λ3=0(λ3)(λ+1)=0λ=3,1bn=C13n+C2(1)nC1C2{b1=9/4=3C1C2b0=5/4=C1+C2{C1=7/8C2=3/8bn=783n+38(1)nb50=78350+38a50=78350+3814=78350+18
解答A2=AA(eigenvalue) 不是01a2+b2+c2+d2=tr(AAT)tr(A2)=tr(A)=0,1,2(A2×2)a2+b2+c2+d2=0a=b=c=d=0A=0()=1
解答PΓ:y=x2+x+1P(t,t2+t+1)PLm=y(t)=2t+1A(1,2)L:y=(2t+1)(x1)2PLt2+t+1=(2t+1)(t1)2t22t4=0t=1±5P{B(1+5,8+35)C(15,835){AB=(5,10+35)AC=(5,1035)ABC=12510+3551035=12×205=105

二、 計算證明題(共 2 題,每題 10 分,共 20 分)

解答x=2+3+5(x5)2=(2+3)2x2=2(6+5x)x4=4(6+5x2+230x)14x465x22x=30=2+3+5
解答
(1){1:1:Pm,n:O(0,0)P(n,m)(a,a)m1(m+1)!m!=m+1{11(a,a)m+12=m1Pm,1=m1m+1m2(m+2)!m!2!=(m+2)(m+1)2{m+1m1(a,a)(m+2)(m+1)2(m+1)m1=(m+1)(m2)2Pm,2=m2m+2(2)(0,0)(n,m)=(0,0)(n1,m)+(0,0)(n,m1)Pm,n=(m+n1)!(m1)!n!Pm1,n+(m+n1)!m!(n1)!Pm,n1(m+n)!m!n!=mm+nPm1,n+nm+nPm,n1(3)(1)Pm,n=mnm+n,mnk=m+nk=2{m=2,n=0P2,0=202+0=1m=n=1P1,1=111+1=0k=Nk=N+1Pm,n=mm+nPm1,n+nm+nPm,n1=mm+nmn1m+n1+nm+nmn+1m+n1=m2n2m+n(m+n)(m+n1)=(mn)(m+n1)(m+n)(m+n1)=mnm+n
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解題僅供參考,其他教甄試題及詳解


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