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2024年7月4日 星期四

113年台南市國中教甄聯招-數學詳解

臺南市 113 學年度市立國民中學正式教師聯合甄選 

以下題目共 100 題,為四選一單選選擇題(每題 1 分,共 100 分)

解答:{L1:2x+y=qL2:y=xp{4+k=qk=2pp+q=(2k)+(4+k)=6(D)

解答::23,2447;53,54107(D)


解答::11,13,17,19,31,33,37,39,71,73,77,79,91,93,97,99,1633,39,77,91,93,99,,101016=58(B)
解答:×::y=9x12:×:(1,2)xy(C)

解答:{¯AB=2a¯BC=2b¯AC=2c{aπ=4π12c2π=2898π{a=4c=172(2a)2+(2b)2=(2c)264+4b2=289b2=2254b=152(B)


解答:=108÷4(10+4)×(8+3)=20154=1077(D)

解答:0+1=C83+C82C21C103=112120=1415(D)
解答:{=2=+1={α=2β=γ=1α>β=γ(A)
解答:f(x)=2x2+x+3=ax(x1)+bx(x3)+c(x1)(x3){f(0)=3=3cf(1)=6=2bf(3)=24=6a{a=4b=3c=1a+b+c=2(B)
解答:(0,b)(C)(C)
解答:xiT0 100(C)
解答:3(D)
解答:z(C)
解答:f(x)f(3)=f(3)=4(C)
解答:f(x)=x(2x1)(13x+2)427x+9f(x)=(2x1)(13x+2)427x+9+xg(x)f(0)=249=163(A)
解答:limn(2n2+1n2n2+n+2n+2)=limn((2n2+1)(n+2)n(2n2+n+2)n(n+2))=limn3n2n2=3()(D)
解答:L(9,5)(3,1)y=f(x)=23x1f(x)=23limh0f(3+h)f(3)h=23(B)
解答:31|2x1|dx=1/21(12x)dx+31/2(2x1)dx=[xx2]|1/21+[x2x]|31/2=94+254=172(B)
解答:limx0(x1x)sinx=limx0sinxxx21=limx0(sinx)(xx21)=limx0cosx1x212x2(x21)2=11+0=1(A)
解答:{f1(x)=x2/3f1(x)=231x1/3f1(0)f2(x)=|x22|x=±2,f2(x)(D)


解答:(B)×:{A=[1000]B=[0100]{AB=[0100]BA=[0000]ABBA(C)×:(A+B)2=A2+AB+BA+B2,ABBA(D)×:C=0AC=BCA=B(A)

 
解答:{x+2y+(3a)=0(1)x+(2a)y+3=0(2)(1a)x+2y+3=0(3),(1)(2){x=a5y=1(3)(1a)(a5)+5=0a26a=0a=0,6(B)

解答:limx6f(x)=±1(C)

解答:f(x)=x5x29x+20=x5(x4)(x5)(D)

解答:f(x)=(1x+1)(2x+1)f(x)=(1x2)(2x+1)+(1x+1)2=2x1x2+2x+2=21x2(D)

解答:limx4x+53x4=limx4(x+53)(x4)=limx412x+51=16(B)
解答:f(x)=x39x2+27x27f(x)=3x218x+27f(x)=6x18=0x=3(C)


解答:(A):rref([101251043])=[100010001]=I3 rank =3(B)×:rref([241031601])=[10160113000] rank =2(C):rref([121102211])=I3 rank =3(D):rref([131243382])=I3 rank =3(B)

解答:(C)×:{v1=(1,0),v2=(0,1),v3=(2,0)},v3=2v1,,{v1,v2}(C)


解答:A=[110211111]rref(A)=[100010001]=I3rank(A)=3(A)

解答:|2i346i|=1212=24(C)

解答:4x213x+7=0{a1=13/4a2=7/4a1a2=64=32(D)


解答:(B):39=13=721(B)

解答:2x=5log22x=log25x=log25(B)

解答:|2+2k|5=|4k|5=154k=±1k=3,5(A)


解答:(A)×:x2+y2+z2=321(B)×:x2+y2+z2=21(C):x2+y2+z2=1(D)×:x2+y2+z2=541(C)

解答:{2200/2=1003200/3=665200/5=406200/6=3310200/10=2015200/15=1330200/30=6100+66+40332013+6=146(B)

解答:1n+3n+5n+7n+9n+11n+13n=49nn=1,7,49(C)


解答:(A)×:xx(B)×:32xx(D)×:1x2x(C)

解答:xf(x)=(x3)p(x)+63f(3)=6f(3)=2(C)

解答:f(1)=f(3)=f(5/2)=01,3,5/2f(x)=0(x+1)(x3)(2x5)f(x)(C)

解答:115383116=18(115+38+3116)=92(B)

解答:an=2an1+1=2(2an2+1)+1=22an2+2+1=23an3+22+2+1==2n1a1+2n2+2n1++1=2n1+2n2+2n1++1=2n1(D)


解答:a2=21a1=2a3=32a2=3a4=43a3=4a50=50(D)

解答:AB={2,4}a2+a4=2a2+a2=0{a=1B={2,6,4}AB={2,4}a=2B={2,3,10}AB={2}a=1(D)

解答:(A):ana1,d4an4a1,4d,(A)


解答:=BCA=BCA(A)
解答:545×4=20(C)


解答:525=32321=31(B)

解答:Case I :5×4×3×3=120Case II :{B,D:5×4×3=60A,C:5×4×3=60120Case III :A,CB,D:5×4=20120+120+20=260(D)

解答:5,51×5×5=52(B)
解答:A2422=4(A)

解答:a=f(1)=1,b+c=aa+b+c=2a=2(D)


解答:3535!3636!35!36!=136(A)
解答:{=C33/C831=C51C32/C83=1C33C83C51C32C83=127=57(D)
解答:2=56×56=25362=12536=1136(A)
解答:(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3),(6,6)121236(C)
解答:45+4=49(A)
解答:S=p+2p2+3p3+pS=p2+2p3+3p4+(1p)S=p+p2+p3+S=p11p1p=p(1p)2(B)
解答:(C)
解答:(D)
解答:=10%×0.810%×0.8+90%×0.2=0.307(C)


解答:{AB={1}ABC={1}P(AB)=P(ABC)=14(A)
解答:{:16:56×12×16:(56×12)2×16=16(1+512+(512)2+)=1611512=27(B)

解答:1.2n=2n=4(B)
解答:(A)×:A=[1111]A2=[0220]2<0(B)×::2>0(C):det(A2)=det(AA)=detAdetA0(D)×:A=[0010]A2=[0000]A,A2(C)
解答:{f(x,y)=xyg(x,y)=x2+y22{fx=λgxfy=λgyg=0{y=2λxx=2λyyx=xyx2=y2{x=yf(x,x)=x2x=yf(x,x)=x2{x=yx=y(C)
解答:A=[21+i1i2]det(AλI)=λ26=0λ=±6(B)
解答:(0,0)(B)(C)(5,1)(A)(A)
解答:{y=2xy=x2+1(0,1)(1,2)O(2x)>O(x2)x>1(D)
解答:{(3)6=27(34)6=16(618)6=183>618>34(B)
解答:
1ACQ,ACS,BDP,BDR,QSC,QSA,PRB,PRD28(A)
解答:(2+2i)8(1+3i)12=212(12+12i)8212(12+32i)12=(eπi/4)8(e2πi/3)12=e2πie8πi=1(A)
解答:abab=0{2x3y=a3x+y=b{x=111a+311by=311a+211b{|x|=10/11|y|=13/11xy=3/121cosθ=xy|x||y|=3130(B)
解答:(A):66n13166111113(B)×:2n132,4,8,3,6,12,11,9,5,10,7,1,,12998=12×83+22998=4 mod 13=42998+8=12 mod 1313(C):68n133,9,1,3,9,1,,333=3×116833=1 mod 136833113(D):18n135,12,8,1,5,12,,450=4×12+21850=12 mod 131850+113(B)
解答:{,bbb121C41C31×3=36C43×3!=241+36+24=61(D)
解答:{1,3,5,744,5544×5×5×4=400(C)
解答:3f(x)+f(1x)=x{3f(2)+f(12)=23f(12)+f(2)=12f(2)=1126{3f(3)+f(13)=33f(13)+f(3)=13f(3)=262424f(3)16f(2)=2611=15(D)
解答:P4x2y2=4P(4+a24,a)¯AP=4+a24+(a3)2)=54a26a+10f(a)=54a26a+10f(a)=0a=125f(125)=145=2.8¯AP=2.8(D)
解答:AP=32AQ=32(AB+BQ)=32(AB+32BR)=32AB+94BR=32AB+94(BC+CR)=32AB+94(BA+AC+43CP)=34AB+94AC+3(CA+AP)=34AB34AC+3AP2AP=34AB+34ACAP=38AB+38ACx+y=38+38=34(C)
解答:a=14+2114+16<a<14+254<18<a<19<54<a<5(B)
解答:{limx1f(x)x1=3limx2f(x)x2=5{f(x)=(x1)g(x)g(1)=3f(x)=(x2)h(x)h(2)=5f(x)=(x1)(x2)(ax+b){(a+b)=32a+b=5{a=2b=1f(x)=(x1)(x2)(2x+b)f(0)=2b=2(A)
解答:{(log3)x+logy=log2(log9)x+3logy=log16{log(3xy)=log2log(9xy3)=log16{y3x=2y=2/3xy332x=16(23x)332x=83x=163x=12y=2/3x=43a+b=12+4=92(C)
解答:f(x)=x10+2x+1f(x)=10x9+2f(1)=8limh0f(2h1)f(1)h=limh0(f(2h1)f(1))(h)=limh02f(2h1)1=2f(1)=16(A)
解答:f(x)=(x+1)p(x)1=(x2)q(x)+5=(x2x2)r(x)+ax+b=(x2)(x+1)r(x)+ax+b{f(1)=1=a+bf(2)=5=2a+b{a=2b=12x+1(B)
解答:cos60=ab|a||b|ab=1213=32(2ab)(2ab)=4|a|24ab+|b|2=46+9=7|2ab|=7(D)
解答:{log(5x)(x2+10x5)log(5x)(2x+10){5x>0x2+10x5>02x+10>0{5>x525<x<5+25x>5525<x<5(1);{5x>1x2+10x5<2x+10x<3(2)5x<1x2+10x5>2x+104<x<5(3){(1)(2)525<x<3(1)(3)4<x<5(D)
解答:|121112013|=31+006+2=8(B)
解答:{A(1,1,1)B(3,2,1)C(5,4,3){u=AB=(4,3,0)v=AC=(4,3,2)n=u×v=(6,8,0)SABC=12n=1262+82=5(C)
解答:#(ABC)=#(A)+#(B)+#(C)#(AB)#(BC)#(CA)+#(ABC)=28+29+25101112+3=52(B)
解答:z+z1=3z2+3z+1=0z=32+12i=e5πi/6z10+z10=e50πi/6+e50π/6=2cos506π=2cosπ3=1(A)
解答:limxxsinxx=limx(1sinxx)=10=1(A)
解答:30x[x]dx=100dx+21xdx+322xdx=0+32+4=112(D)
解答:x3+y39xy=03x2+3y2y9y9xy=0y=9y3x23y29xy(2,4)=36124818=2430=45(D)
解答:A5×3AT=3×5B5×n,C7×4n=7B5×7(A)
解答:{tanα+tanβ=batanαtanβ=catan(α+β)=tanα+tanβ1tanαtanβ=b/a1c/a=baccot(α+β)=cab(D)
解答:(x+1x)2=x2+1x2+22(x2+1x2)9(x+1x)+14=2(x+1x)29(x+1x)+10=(2(x+1x)5)((x+1x)2)=0{x+1x=52x+1x=2{2x25x+2=0x22x+1=0{(2x1)(x2)=0(x1)2=0x=12,1,2(C)
解答:limx01cosxx2=limx0(1cosx)(x2)=limx0sinx2x=limx0(sinx)(2x)=limx0cosx2=12(B)
解答:4x^3-24x^2+23x+18=0 \Rightarrow \cases{三根之和=6\\ 三根之積=-9/2},故選\bbox[red, 2pt]{(B)}
解答:f(x)=(x+2)(x+5)^2 \Rightarrow f'(x)=(x+5)^2+2(x+2)(x+5) =(x+5)(3x+ 9) \\ \Rightarrow f''(x)=3x+9+3x+15 =6x+24\\ f''(x)=0 \Rightarrow x=-4 \Rightarrow 反曲點(-4,f(-4)) =(-4,-2),故選\bbox[red, 2pt]{(C)}
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