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2021年6月22日 星期二

105年鳳山高中教甄-數學詳解

 國立鳳山高級中學105學年度第1次專任教師甄選

一、填充題

解答{A(0,0)B(b,11)C(c,37){C=B60C=B60{[1/23/23/21/2][b11]=[c37][1/23/23/21/2][b11]=[c37]{{b113=2c3b+11=74{b=213c=53{b+113=2c3b+11=74{b=213c=53bc=105×3=315
解答
¯AB{¯BC=8222=215¯AD=8272=15{tanθ1=tanABC=2/215=1/15tan(θ1+θ2)=tanABD=15/7tan(θ1+θ2)=tanθ1+tanθ21tanθ1tanθ2157=115+tanθ21115tanθ2tanθ2=115cosθ2=154=(215)2+72¯CD22815¯CD=2
解答{Pn=(xn,yn),1n179P0=A=(1,0)P180=B=(1,0)Pn+1OPn=1,0n179xn=cosn179n=1x2n=179n=1cos2ncos21+cos22++cos290=cos21+cos22++cos288+cos289+0=cos21+cos22++cos2(902)+cos2(901)=cos21+cos22++sin22+sin21=(cos21+sin21)+(cos22+sin22)++(cos244+sin244)+cos245=1+1++1+12=44+12cos291+cos292++cos2179=44+12179n=1x2n=2(44+12)=89

解答

xπ63236x2dx=π[36x13x3]|632=π(144902)=23π63=144π=144π(144902)π=902π
解答limn(12n)p+(22n)p++(2n2n)p(12+12n)p+(12+22n)p++(12+n2n)p=limn1n((12n)p+(22n)p++(2n2n)p)1n((12+12n)p+(12+22n)p++(12+n2n)p)=limn1n((12n)p+(22n)p++(2n2n)p)limn1n((12+12n)p+(12+22n)p++(12+n2n)p)=limn1n2nk=1(k2n)plimn1nnk=1(12+k2n)p=20(x2)pdx10(12+x2)pdx=2p+1/2p(p+1)(2p+11)/(p+1)2p=2p+12p+11
解答2a+b+2a+18ab=a2+a2+a2+a2+b3+b3+b3+2a+6ab+6ab+6ab1111a424b3332a63a3b3=11
解答M=[01/31/31/31/301/31/31/31/301/31/31/31/30]M=PAP1=[1111110010101001][100001/300001/300001/3][1/41/41/41/41/43/41/41/41/41/43/41/41/41/41/43/4]M60=PA60P1=P[100001/36000001/36000001/360]P1=[11/3601/3601/36011/36000101/36001001/360][1/41/41/41/41/43/41/41/41/41/43/41/41/41/41/43/4]=[14+141359]APA60P1[1000]14+141359
解答a,b,c{a=3b=4c=5C=90{tanC2=tan45=1tanA=34=2tan(A/2)1tan2(A/2)tanA2=13tanA2tanC2=131=13
解答{L1:x+3y=4L2:5x+3y=16A(4,0)B(4,12){L1L2P(5,3)¯ABQ(4,6){PQ:x+y=2¯PQC(12,32)ΓPQΓ45ΓΓx(x,y)45(x,y){x=xyy=x+y{A(4,0)A(4,4)B(4,12)B(16,8)C(1/2,3/2)C(1,2)Γ:x=ay2+by+cA,B,CΓ{16a+4b+c=464a8b+c=164a2b+c=1{a=1/4b=c=0x=14y2(xy)=14(x+y)2Γ:x2+2xy+y24x+4y=0

解答α,βx2+(k2+ak)x+k2+ak+127=0{α+β=k2akαβ=k2+ak+127αβ+α+β=127(α+1)(β+1)=128=32×4{α=31β=3{312+31(k2+ak)+k2+ak+127=032+3(k2+ak)+k2+ak+127=0k2+ak+34=00a2=4×34a=234(234a)
解答f(x)+f(11x)=x+1+1x1f(x)+f(x1x)=x+xx1(1)x=x1x(1)f(x1x)+f(x1x1x1x)=x1x+x1xx1x1f(x1x)+f(11x)=x1x+1x(2)x=x1x(2)f(x1x1x1x)+f(11x1x)=x1x1x1x+1x1xf(11x)+f(x)=11x+1x(3)(1)+(2)+(3)2(f(x)+f(x1x)+f(11x))=3f(x)+f(x1x)+f(11x)=32(4)(4)(2)f(x)=12x1x+x=x+1x12
解答:x3+y3+z33xyz=(x+y+z)(x2+y2+z2(xy+yz+zx))=12(x+y+z)((xy)2+(yz)2+(zx)2)x3+(y)3+(z)33x(y)(z)=12(xyz)((x+y)2+(y+z)2+(x+z)2=0xyz=0x=y+z(1)(1)x2=2(y+z)(y+z)2=2(y+z)(y+z)(y+z2)=0y+z=2(1)x=2a+b+c=x+y+z=2+2=4
解答

z=a+biz2z+2=a2+bia+2+bi=(a2+bi)(a+2bi)(a+2+bi)(a+2bi)=(a2+b24)+4bi(a+2)2+b2{a2+b2=4b0zC1:x2+y2=4()ω=z+iC2:x2+(y1)2=4(1)M=|ω+1|2+|ω1|2=¯RP2+¯RQ2{P(1,0)Q(1,0)RC2R(0,3)ω=3iM|ω|=3
解答

{x=m+2cosθy=3sinθ(xm2)2+(y3)2=1(xm)24+y23=1{a=2b=3{P(ma,0)=(m2,0)Q(m+a,0)=(m+2,0){x=t2+32y=6tx=(y6)2+32=y26+32y2=6x9=6(x32)R(32,0)xR¯PQ{P=Rm2=3/2m=7/2Q=Rm+2=3/2m=1/212m72a+b=12+72=3
解答log2x4log22x+12log23x++n(2)n1log2nx=log2x2log2x+4log2x++(2)n1log2x=log2(xx2x4x(2)n1)=(12+42n1)log2x=1(2)n3log2x:1(2)n3log2x>1(2)n3log2(x22)x<x22(1(2)n<0)x2x2>0(x2)(x+1)>0x>2(xx1)

解答

AABB{AA==2πBB==10π¯QRAOA=θ2π=rθ=3θθ=23πcosθ=¯OP2+¯OB2¯PB22ׯOPׯOB12=81+225¯PB2270¯PB=21OPB=12¯OQ¯PB=12¯OP¯OBsinPOB21¯OQ=13532¯OQ=45143¯QR=¯OQ3=451433=4534214=a3+b14{a=45b=42a+b=3:a3+b121214

二、計算題

解答
(1)f(x)=x6+x4+x2+1=(xα1)(xα2)(xα6)f(2i)=64+164+1=(2iα1)(2iα2)(2iα6)(2i+α1)(2i+α2)(2i+α6)=51(2)
x6+x4+x2+1=0(x21)(x6+x4+x2+1)=0x8=1P1,P2,,P6(1,0),(1,0)=P1P3P4P6+P1P2P3+P4P5P6=(2)2+2×(122)=2+21=2+1

解答f(x)=x0(xt)cos3tdt=[136(3(xt)(9sint+sin(3t))27costcos(3t))]|x0=136(27cosxcos(3x))136(271)=136(2827cosxcos(3x))f(x)=0136(27sinx+3sin(3x))=09sinx+sin(3x)=09sinx+3sinx3sin2xsin3x=0sin3x3sin2x+12sinx=0sinx(sin2x3sinx+12)=0sinx=0x=0,π{f(0)=0f(π)=14/9
 

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5 則留言:

  1. 您好:請問第6題可以拆成a+a+b+2/a+18/ab嗎?這樣值比較小,謝謝

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    回覆
    1. 不行, 因為等號成立的條件a=a=b=2/a=18/ab (你的方法) 無法求得相對應的a與b;

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  2. 另外,第7題的第二行PAP*-1是怎麼來的呢?謝謝

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  3. 然後計算題第2題解答的第一行積分就看不懂了,有用什麼公式或是變數變換嗎?謝謝

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