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2021年5月4日 星期二

110年臺中一中教甄-數學詳解

 臺中市立臺中第一高級中等學校110學年度第1次教師甄選

壹、填充題第一部分

解答{2k4m8n=5124p3q6r=211616{2k+2m+3n=2922p+r3q+r=227316{k+2m+3n=92p+r=27q+r=16{(k,m,n)=(4,1,1),(2,2,1),(1,1,2)(p,q,r)=(14k,172k,2k1),k=18{a=3b=8(a,b)=(3,8)
解答A1B=CB=AC[1412]=[xyzu][426318]=[4x3y26x+18y4z3u26z+18u]{4x3y=126x+18y=44z3u=126z+18u=2(x,y,z,u)=(5,7,2,3)
解答log4xlogx8+2=012log2x3log2x+2=0(log2x)2+4(log2x)6=0log2x=2+10(210,x>1log2x>0)2(log2x)3+9(log2x)27(log2x)3=2(2+10)3+9(2+10)27(2+10)3=2(68+2210)+9(14410)7(2+10)3=1+10
解答

O¯BCO{B(3,0,0)C(3,0,0)A(0,33,0),¯CD=¯BDDx=0D(0,y,z)=13ABCz=1831393z=18z=6;ABCDBC60tan60=zyy=33z=23D(0,23,6)¯AD=(3)2+62=39
解答{PxQy{P(a,0)Q(0,b){¯PQ=10¯PR:¯RQ=3:2{a2+b2=102R=25P+35Q{b2=100a2R(25a,35b)R(x,y)259y2=100254x2x216+y236=1RR(4cosθ,6sinθ)d=R(2,0)=(4cosθ2)2+36sin2θ=20sin2θ16cosθ+20=20cos2θ16cosθ+40{cosθ=2/5M=630/5cosθ=1m=2(M,m)=(6305,2)

貳、填充題第二部分

解答an=x,y,x+y,x+2y,2x+3y,3x+5y,5x+8y,8x+13y,13x+21y,21x+34y,34x+55y,...x,yN,an=11513x+21y=115{x=4y=3x+y=7
解答{limx1f(x)x1=36limx1f(x)x+1=36limx2f(x)x2=0limx2f(x)x+2=0{f(x)=p(x)(x1)p(1)=36f(x)=q(x)(x+1)q(1)=36f(x)=r(x)(x2)r(2)=0f(x)=s(x)(x+2)s(2)=0f(x)=(ax+b)(x1)(x+1)(x2)2(x+2)2{limx1f(x)x1=36limx1f(x)x+1=36{(a+b)2(1)232=36(a+b)(2)(3)212=36{a+b=2a+b=2{a=0b=2f(x)=2(x21)(x2)2(x+2)2f(3)=28125=400
解答{E(X)=1E(X2)=3E(Y)=2E(Y2)=5E(XY)=3{Var(X)=E(X2)(E(X))2=31=2Var(Y)=E(Y2)(E(Y))2=54=1Cov(X,Y)=E(XY)E(X)E(Y)=32=1Var(3X2Y+7)=9Var(X)+4Var(Y)12Cov(X,Y)=18+412=10
解答f(x)=x3+a2x2+a1x+a0f(x)=0b,c,dabcd=5bcd=5/aa0=5/af(x)=x3+a2x2+a1x+5af(x)=0b,c,d{f(x+1)=0b1,c1,d1f(x+2)=0b2,c2,d2f(x+3)=0b3,c3,d3{f(x+1)=1+a2+a1+5/a=(b1)(c1)(d1)=11/af(x+2)=8+4a2+2a1+5/a=(b2)(c2)(d2)=33/af(x+3)=27+9a2+3a1+5/a=(b3)(c3)(d3)=73/a{a1=17/2a2=3/2a=2f(x)=x332x2+172x52f(x1)=0b+1,c+1,d+1(b+1)(c+1)(d+1)=(1)×f(x1)=(1)×(13217252)=272a(b+1)(c+1)(d+1)=2×272=27
解答f(x)=ax3+bx2+cx+d(b3a,f(b3a)){b3a=1b=3a(1)f(b3a)=f(1)=a+b+c+d=2(2)f(2)=412a+4b+c=4(3)(1)(3)c=4(2)a3a4+d=2d=2a+6k=|b2+c2+d2a|=|9a2+16+4a2+24a+36a|=|13a2+24a+52a|=|13a+24+52a|a>0,13a+52a213a52a=226=52;a<013a+52a52;k13a+52a=52k=|52+24|=|28|=28
解答A1:126,11526×15=115A23115A115×3=15(15)3=112511125=124125124125×12=62125
解答{a=2xb=3x6x+138x+227x36x=6ab3a3+2b3a2b2=3a(2ba2)+b2(2ba2)=(3a+b2)(2ba2)=0a2=2b22x=23x2xlog2=log2+xlog3x(2log2log3)=log2x=log22log2log3x2x1=log2/(2log2log3)log3/(2log2log3)=log2log3=log32
解答
ABCDA¯ACx¯AC=¯AD2+¯DC2=5¯ADׯDC=¯ACׯDE¯DE=12/5,CDE¯EC=9/5ABCDxyDxθD(165,125cosθ,125sinθ)¯BD=(75)2+(125)2(1cosθ)2+(125)2sin2θ=4925+14425(22cosθ)=337288cosθ5

參、計算證明題(第一題 9 分,第二題 10 分,合計 19 分)


解答

A(6,9)LL:4x3y=51BLL
解答\mathbf{(1)}\; f(x)=x^4+x^3+x^2+x+1 = (x-\alpha) (x-\beta) (x-\gamma) (x-\phi) \\ \Rightarrow f'(x)=4x^3+3x^2+2x+1 \\=  (x-\beta) (x-\gamma) (x-\phi) +(x-\alpha)  (x-\gamma) (x-\phi) +(x-\alpha) (x-\beta)  (x-\phi)  +(x-\alpha) (x-\beta) (x-\gamma)  \\ 因此{1\over 1-\alpha}+{1\over 1-\beta} +{1\over 1-\gamma}+{1\over 1-\phi} \\={ (1-\beta)(1-\gamma )(1-\phi)+ (1-\alpha) (1-\gamma )(1-\phi) +(1-\alpha) (1-\beta)(1-\phi)+ (1-\alpha) (1-\beta)(1-\gamma ) \over (1-\alpha) (1-\beta)(1-\gamma )(1-\phi)}\\ ={f'(1)\over f(1)} ={10\over 5} =\bbox[red, 2pt]2 \\\mathbf{(2)}\; \overline{AP} \times \overline{AQ} \times \overline{AR} \times \overline{AS} =|1+i-\alpha| \times|1+i-\beta| \times|1+i-\gamma| \times|1+i-\phi| \\ =|(1+i-\alpha) \times(1+i-\beta) \times(1+i-\gamma) \times(1+i-\phi)| =|f(1+i)| \\=|-4+(2i-2)+ 2i+(1+i)+1|=|5i-4|= \bbox[red, 2pt]{\sqrt{41}}


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解題僅供參考,學校未公布計算題題目,其它教甄試題及詳解

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