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2022年6月13日 星期一

111年香山高中教甄-數學詳解

新竹市立香山高級中學111學年度教師甄選

一、單選題(每題4分,共計60分)

解答1+2++9=4532aba+b373(a,b)=(1,2),(1,5),(1,8),(2,4),(2,7),(3,6),(3,9),(4,5),(4,8),(5,7),(6,9),(7,8)12(D)
解答:27sinA=48sin3A9sinA=163sinA4sin3A27sinA36sin3A=16sinA36sin3A11sinA=0sinA(36sin2A11)=0sinA=116cosA=5656=482+¯AC227296¯AC¯AC280¯AC+1575=0(¯AC35)(¯AC45)=0¯AC=3545(B)
解答

{sin2x+cos2y=y2(1)sin2y+cos2x=x2(2){(1)+(2)(2)(1){x2+y2=2:x2y2=cos(2x)cos(2y):{A(1,1)B(1,1)C(1,1)D(1,1)(B)
解答{z1=1+i32=e2πi/3z2=1i32=e4πi/3{z1011=e202πi/3=e66π+4πi/3=e4πi/3=z2z1012=e404πi/3=e134π+2πi/3=e2πi/3=z1z1011+z1012=z2+z1=1(A)
解答cosx=112x2+14!x4cosx2=112x4+14!x8{1cosx=12x214!x4+1cosx2=12x414!x8+limx01cosx21cosx=(1/2)1/2=2(D)
解答x2+18x+30=2(x+9)236(x+9)23615=2(x+9)236a15=2a, where a=(x+9)236a234a+225=0(a25)(a9)=0{(x+9)236=a=25(x+9)236=a=9(a15=2a<0)(x+9)2=61x=9±61=8161=20(A)
解答5n23n7=5+12n7n7=±1,±2,±3,±4,±6,±1212(C)
解答{a=3b=4c=5滿a,b,c{tanA=3/4tan(A/2)=1/3tan(C/2)=tan45=1tanA2tanC2=13(B)
解答{3a2+2022a+8=08b2+2022b+3=0{a=2022±k6b=2022±k16,where k=20222438(a,b)=(2022+k6,2022+k16)(2022k6,2022k16)ab=166=83ab=1(C)
解答limx0x3sin1xsinx=limx0(x3sin1x)(sinx)=limx03x2sin1xxcos1xcosx=01=0(B)
解答limn1n(1+12+13++1n)=limnnk=11n1k=limnnk=11n1(k/n)=101xdx=[2x]|10=2(C)
解答¯AB=acosC=32=(2+23)2+16a28(2+23)a2=8¯AB=22cosB=42+(22)2(2+23)2162=264<0B=18075=105(D)
解答y=8nx22n(2n+1)x+1=(22nx1)(2nx1)Ln=¯AnBn=12n122nn=1Ln=n=1(12n122n)=1/211/21/411/4=113=23(D)
解答{f(x,y,z)=cos2x+cos2y+cos2zg(x,y,z)=cosx+cosy+cosz3f,g,3af+2(1a)g=9a6a(f2g)+2g=9a6a=6+2g9f+2ga=1,0{a=02g=6g=cosx+cosy+cosz=3(x,y,z)=(π,π,π)a=1f=cos2x+cos2y+cos2z=3(x,y,z)=(0,0,0),(π,π,π),(2π,2π,2π)(C)a=1(x,y,z)=(0,0,2π),(0,2π,0),...23+1(π,π,π)=9
解答m>n+14m>4n+4m24m<m24n4(m2)2<m24nm24n<m2(m2)2<m24n<m2m24n=(m1)2=m22m+12m=4n+1=mn+1;m=n+1,m=n;m=n+1{m24n=(n+1)24n=(n1)2n24m=n24n4=(n2)28n=5m=6m=nn24m=m24n=n24nm=n=4n>m+1n=m+1n=m(m,n)=(4,4),(5,6),(6,5)(C)

二、多選題(每題8分,共計40分;每題有5個選項,至少有一個是正確的選項)

解答
(A):EAF=ABAEDAF=903015=45(B)×:{¯DF=23tan15=436¯CF=23(436)=623¯BE=23tan30=2¯CE=232¯CF/¯CE=3FEC=60AEF=6075(C):CFE=180AEFC=30(D):(B):¯CE=232(E):¯EF=2¯CE=434(ACDE)
解答sinAsinB(m+5)x2(2m5)x+12=0{sinA+sinB=(2m5)/(m+5)sinAsinB=12/(m+5)(sinA+sinB)2=sin2A+sin2B+2sinAsinB=1+2sinAsinB(2m5m+5)2=1+212m+5=m+29m+5(2m5)2=(m+29)(m+5)m218m40=0(m20)(m+2)=0m=20(m>0,m=2)25x235x+12=0(5x3)(5x4)=0{sinA=3/5sinB=4/5asinA=bsinB=2R=4{a=12/5b=16/5{a2+b2=c2=40025c=205=4a+b=28/5ab=192/25(BCD)
解答:{a2+b2=c2a+b+c=126ab=630×2=1260{(a+b)2=(126c)2=c2252c+1262(a+b)2=a2+b2+2ab=c2+2ab=c2+2520c2252c+1262=c2+2520252c=13356c=53a+b=12653=73ab=a(73a)=1260a273a+1260=0(a45)(a28)=0{(a,b)=(28,45)(a,b)=(45,28)b>a{ba=4528=1715cb=5345=8ca=5328=25a+b+c==126120(BCD)
解答(D)×:{g(x)=2/|x|f(x)=1/|x|滿g(x)>f(x)x0{limx0f(x)=limx0g(x)=limx0f(x)limx0g(x)(E)×:{g(x)=2/xf(x)=xlimx0(f(x)×g(x))=limx02=2limx0g(x)(ABC)
解答a=2+b2ab+22c2+1=02(2+b)b+22c2+1=02b2+22b+22c2+1=0(22)242(2c2+1)016c20c=02b2+22b+1=0b=22a=222=22{a=2/2b=2/2c=0a+b+c=0(BCE)

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解題僅供參考,其他教甄歷年試題及詳解

2 則留言:

  1. 感謝老師的詳解,獲益良多
    但發現一點小筆誤,第15題最後答案應該是(4,4)、(6,5)、(5,6)

    回覆刪除