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2021年9月16日 星期四

108年松山工農教甄-數學詳解

臺北市立松山高級工農職業學校108學年度正式教師甄選

一、填充題

解答30=2×3×5=(1+1)×(2+1)×(4+1)N=24325=720
解答f(x)x=25f(2)=0(5,f(5))7f(5)=752f(x)dx=f(x)|52=f(5)f(2)=70=7
解答cosDAF=ADAF¯AD¯AF=1125cosCBE=cosDAF=1125=¯BC2+¯BE2¯CE22¯BE¯BE=50¯CE250¯CE2=28cosCAE=¯AC2+¯AE2¯CE22¯AC¯AE=50+502825252=72100ACAE=¯AC¯AEcosCAE=525272100=36
解答
x+y=1xyABCD{A,Dx=yAB,Cx=yPx=yP(t,t),tRd(P,x+y=1)=t|2t1|2=t2t24t+1=0t=2±2{A((22)/2,(22)/2)D((2+2)/2,(2+2)/2)Qx=yQ(t,t)d(Q,x+y=1)=|t|12=|t|t=±12{B(1/2,1/2)C(1/2,1/2)2±22,22
解答z=x+iy,x,yRz22z2+2=14z2+2=i4(x+iy)2+2=4x2y2+2+2xyi=1i((x2y2+2)+2xyi)(1i)=4(x2y2+2+2xy)(x2y2+22xy)i=4{x2y2+2+2xy=4x2y2+22xy=0{xy=1x2y2=0(x,y)=(1,1),(1,1)z=1+i1i
解答k3k171782622×19=38276333×37=1116412444×61=244125557+38+11+244+5=405
解答
解答
{tanA=3/41/41+3/16=819tanB=1/4(3/4)13/16=1613{¯BD¯AC¯BD=h¯AD=a{ha=819h51a=1613h=819a=1613(51a)a=38h=16ABC=125116=408
解答3611263×5=1516163×52=75912313!=697232310023331032343!=61092353!=61153413!=6121343312434431273453!=613313363=133216
解答{y1nx+nx,y0,nN,x,yZ;yx00n2nn+110n(n1)n(n1)+120n(n2)n(n2)+1n10nn1+1n0n0+1=n(n+(n1)+(n2)++1+0)+n+1=nn(n+1)2+n+1=n3+n22+n+1=n3+n2+2n+22
解答
y=x3/2+1x=(y1)2/3y=91x2πdy=91x2πdy=91(y1)4/3πdy=[37(y1)7/3]|91=3847π
解答g(x)=3ax+b2lim
解答\cases{{1\over x}+ {1\over y}+{1\over z}=1 \Rightarrow yz+xz+ xy= xyz\cdots(1)\\ x^2+y^2+z^2 =3/2\cdots(2) \\ x^3+y^3 +z^3=1 \cdots(3)} \\ 因此 \cases{(x+y+z)^2 =x^2 +y^2 +z^2 +2(xy+yz +zx)\\ x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^z+z^2 -(xy+ yz+zx)}\\ \Rightarrow \cases{(x+y+z)^2 =3/2+ 2xyz \\ 1-3xyz=(x+y +z) (3/2-xyz)} \Rightarrow \left({1-3xyz\over 3/2-xyz} \right)^2 ={3\over 2}+2xyz \\\Rightarrow ({1-3a\over 3/2-a})^2={3\over 2}+2a,其中a=xyz \Rightarrow 16a^3-108a^2+48a+19=0 \Rightarrow (4a+1)(4a^2-28a+19)=0\\ xyz=a=-{1\over 4}(\because xyz\in \mathbb{Z}) \Rightarrow x+y+z ={1-3xyz\over 3/2-xyz} ={7/4\over 7/4} =\bbox[red,2pt]{1}
解答

A=\begin{bmatrix} 4 & -3\\ 3 & 4\end{bmatrix} =5\begin{bmatrix} 4/5 & -3/5\\ 3/5 & 4/5\end{bmatrix} =5\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{bmatrix} ,其中\cases{\cos\theta =4/5\\ \sin \theta = 3/5} \\\Rightarrow \cases{\angle P_2OP_1= \angle P_3OP_2 = \theta\\ \overline{OP_1}=a\\ \overline{OP_2}=5\cdot \overline{OP_1}= 5a\\ \overline{OP_3}= 5\cdot \overline{OP_2}=25a} \Rightarrow \triangle P_1P_2P_3 =\triangle OP_1P_2 +\triangle OP_2P_3- \triangle OP_1P_3 \\ ={1\over 2}\left(\overline{OP_1}\cdot \overline{OP_2}\sin \theta+ \overline{OP_2}\cdot \overline{OP_3}\sin \theta -\overline{OP_1}\cdot \overline{OP_3}\sin 2\theta \right)\\ ={1\over 2}\left(5a^2\cdot{3\over 5}+125a^2\cdot {3\over 5}-25a^2\cdot 2\cdot {3\over 5}\cdot {4\over 5} \right)={1\over 2}\cdot 54a^2= \bbox[red,2pt]{27a^2}
解答{1\over 1!}+{2\over 3!}+{3\over 5!}+{4\over 7!}+\cdots =\sum_{k=0}^\infty {k+1\over (2k+1)!} ={1\over 2}\sum_{k=0}^\infty {(2k+1)+1\over (2k+1)!}\\ ={1\over 2}\sum_{k=0}^\infty \left({1\over (2k)!}+{ 1\over (2k+1)!} \right)=\bbox[red,2pt]{ {1\over 2}e}
解答

\cases{A(0,0,6)\\ B(0,0,20)\\ P(x,y,0)} \Rightarrow \cases{\overline{AB}=14\\ \overline{AP}=\sqrt{x^2+y^2+36} =\sqrt{a+ 36}\\ \overline{BP}= \sqrt{x^2+y^2+400}=\sqrt{a+400}},其中a=x^2+y^2\\ \Rightarrow \cos \angle APB= {a+36+a+400-14^2\over 2\sqrt{a+36} \cdot \sqrt{a+400}} \le \cos 30^\circ ={\sqrt 3\over 2} \Rightarrow a^2-348a+14400 \le 0\\ \Rightarrow (a-48)(a-300)\le 0 \Rightarrow 48\le a\le 300\\ 本題相當於求聯立方程式\cases{48\le x^2+y^2\le 300\\ 0\le x\le 15\\ 0\le y\le 15}所圍面積=2\left({75\over 2}\sqrt 3-4\pi\right)+{1\over 12}(300-48)\pi\\ =\bbox[red,2pt]{75\sqrt 3+13\pi}
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解題僅供參考,其他教甄試題及詳解


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