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2023年5月30日 星期二

112年關西高中教甄-數學詳解(待續)

國立關西高級中學112學年度第1次教師甄選

一、多重選擇題(全對得6分,答錯一個選項得4分,答錯二個選項得2分,答錯三個選項以上不給分

解答(A):{(111)2+(01)2=100+1=101(01)2+(111)2=1+100=101{(11,0)=xΓ(0,11)=yΓ(B)×:(1,1)Ly=1Γ(1101,1),1101<1100=9x(1101,1)(C):CL:y=xAB(AB¯OB=¯OC=21011012=1012(D)×:(E):xy(ACE)
解答(A):{7=7!/(2!2!3!)=2102=6!/(2!3!)=602=21060=150P(B)>P(A)(B):22=C42=6,2253C53=10P(C)=6×10/210=2/7(C)×:{#(C)=60#(AC)=3C43=12#(BC)=6012=48{P(CA)=12/210P(CB)=48/2102P(CA)+5P(CB)=21260+548150=22(D)×:P(CA)=1260=0.20.2(E):P(CB)=48150=0.32>0.3(ABE)
解答(A):ϕcot2ϕ=584=34tan2ϕ=43tan2θ=43(B)×:tan2θ=42cos2θ=3535(C)×:{sin2θ=4/5cos2θ=3/5{sinθ=1/5cosθ=2/5R=[cosθsinθsinθcosθ][xy]=R[xy][xy]=R1[xy]=[cosθsinθsinθcosθ][xy]=[(2x+y)/5(x+2y)/5]4x2+9y2=36x29+y24=1{=2a=6=2b=4(D):=abπ=32π=6π(E)×:2b2a=8343(AD)
解答(A):a0f(t)dt=[F(t)]|a0=F(a)F(0)(B):F(x)=xQ(x)+cF(x)=f(x)=Q(x)+xQ(x)f(0)=Q(0)(C)×:f(x)=x+2F(x)=f(x)dx=12x2+2x+CF(x)F(0)=12x2+2x(x+2)2(D)×:F(x)=x2滿F(x)x22f(x)=F(x)=2x<x,x<0(E)×:f(x)={x0<x1x2x>1F(x)={x2/20<x1x3/3x>1{F(1.2)=0.5761.22/2=0.72x=1.2F(x)<12x2(AB)

二、填充題(每題5分)

解答(1+12+122+123+)(1+13+132+133+)(1+15+152+153+)(1+17+172+173+)=2×32×54×76=358
解答|z|=1z=cosθ+isinθz2z+2=(cosθ+isinθ)2(cosθ+isinθ)+2=cos2θcosθ+2+(sin2θsinθ)i|z2z+2|2=(cos2θcosθ+2)2+(sin2θsinθ)2=8cos2θ6cosθ+2cosθ=38|z2z+2|78=144
解答

{A(0,0)B(8,0)C(x,y){¯AC=9¯BC=7{x2+y2=81(x8)2+y2=49C(6,35)L1=AC:y=52xP(a,5a/2),ACBP=0a=329P(329,1695)L2=BP:25x+5y=165L2¯ABQ(4,855){AQ=(4,855)AB=(8,0)AC=(6,35)AQ=xAB+yAC{x=1/10y=8/15
解答x=11f(1)=32+4+0=5f(1)=5ddx(xf(x))=f(x)+xf(x)=12x36x2+8x+f(x)f(x)=12x26x+8f(x)=4x33x2+8x+Cf(1)=543+8+C=5C=4f(x)=4x33x2+8x4
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解答

{CDB=CAE()=1E=EACEDBE(AAA)¯CD=¯CBCDB=CBD=CAD=2CDB=CAB1=2CDBCAE(AAA)CABBDECDB¯BC¯BD=¯BE¯DEak30k=30kak+11ka2+11a900=0(a25)(a+36)=0a=25¯BC:¯CD:¯BD=25:25:30=5:5:6cosBCD=52+5262255=725sinBCD=2425¯BDsinBCD=2¯BD=22425=4825
解答a,b,cx35x23x+7=0{a+b+c=5ab+bc+ca=3abc=7|abc2a2a2bbca2b2c2ccab|=|2a52a2a2b2b52b2c2c2c5|=|2(a+b+c)52(a+b+c)52(a+b+c)52b2b52b2c2c2c5|=|5552b2b52b2c2c2c5|=5|1112b2b52b2c2c2c5|=5|111050005|=525=125
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解答174:{A:{a,d,e,,o,s,u,z}B:{i,i,i,i}C:{n,n}D:{t,t,t}9539


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三、計算證明題(每題7分,請詳細寫出計算過程,否則不予給分)
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cosPAB=152+35225221535=1114sinPAB=5314cosPAC=cos(60PAB)=cos60cosPAB+sin60sinPAB=121114+325314=1314=152+562¯PC221556¯PC=1801
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解題僅供參考,其他教甄試題及詳解

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