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2020年9月27日 星期日

109年台中文華高中第1次教甄-數學科詳解

 臺中市立文華高級中等學校 109 學年度
第1次教師甄選-數學科試題



解:
Γ1:x233+y249=1{a1=7b1=33a21=b21+c21c1=4{F1(0,4)F2(0,4)PΓ1¯PF1+¯PF2=2a1=14;:cosF1PF2=¯PF12+¯PF22¯F1F222¯PF1ׯPF2cos60=1422¯PF1ׯPF2822¯PF1ׯPF212=1322¯PF1ׯPF22¯PF1ׯPF2¯PF1ׯPF2=44(¯PF1¯PF2)2=(¯PF1+¯PF2)24¯PF1ׯPF2=1424×44=20|¯PF1¯PF2|=20=25=2a2a2=5c22=c21=a22+b2216=5+b22b2=11=2b2=211


解:
|z13+4i|=1{C(3,4)r=1;z2z1=1+3i=2(cos60+isin60){|z2|=2|z1|AOB=60OAB=12¯OAׯOBsinAOB=12¯OA×(2¯OA)sin60=32¯OA2;¯OA=¯OC+r=5+1=6OAB=32×62=183



解:
1(3)(10.2)3=0.83n0.83n10.83n>0.9990.001>0.83nlog0.001>log0.83n3>3n(3log21)3>3n(3×0.3011)=0.291nn>30.29110.3n=11



解:
(x0,y0)f(x)=a(xx0)3+b(xx0)+y0;y=f(x)=a(x+2)3+b(x+2)+5f(x)=3a(x+2)2+bx=1y=7x+1{f(1)=7f(1)=8{27a+b=727a+3b+5=8{a=1/3b=2y=13(x+2)32(x+2)+5



解:
{x+y+z=1(1)x2+y2+z2=3(2)x3+y3+z3=5(3)(x+y+z)2=(x2+y2+z2)+2(xy+yz+zx)1=3+2(xy+yz+zx)xy+yz+zx=1(1)×(2)=(x+y+z)(x2+y2+z2)=(x3+y3+z3)+xy(x+y)+yz(y+z)+zx(z+x)3=5+xy(1z)+yz(1x)+zx(1y)=5+(xy+yz+zx)3xyz=513xyzxyz=1/3(1)×(3)=(x+y+z)(x3+y3+z3)=x4+y4+z4+xy(x2+y2)+yz(y2+z2)+zx(z2+x2)5=x4+y4+z4+xy(3z2)+yz(3x2)+zx(3y2)=x4+y4+z4+3(xy+yz+zx)xyz3x2yzxy2z=x4+y4+z43xyz(x+y+z)=x4+y4+z431/3x4+y4+z4=5+3+1/3=253



解:
f(x)=x3+(1a)x2+(a+5)x+2a+5f(1)=01;f(x)=01,1+d,1+2d{=3+3d=a1=(1+d)(1+2d)=(2a+5){a=3d2(1)2d23d4=2a(2)(1)(2)2d23d4=6d4d(2d9)=0d=9/2(d=02)a=3d2=27/22=232



解:
f(x+2)=1+f(x)1f(x)f(3)=1+f(1)1f(1)=222=21f(5)=222=2+1f(7)=2+22=21f(9)=22+2=12=f(1):f(2k1)={12if (kmod4)=121if (kmod4)=22+1if (kmod4)=312if (kmod4)=0{1021=2×5111(k=511mod4=3)2025=2×10131(k=1013mod4=1){f(1021)=2+1f(2025)=12f(1021)+f(2025)=2

解:
A(0,0)至B(6,5)的捷徑走法,相當於6個→(向右)及5個↑(向上)的排列數;\\我們先將6個→先擺好,並在間隔放上待填符號 \square及頭尾處放置待填符號\bigcirc,如下:\\\square→\bigcirc→\bigcirc→\bigcirc→\bigcirc→\bigcirc→\square\\對任意\bigcirc而言,無論放入多少個↑,都會發生2次轉彎(1次右轉上,另1次上轉右);\\對任意\square而言,無論放入多少個↑,都會發生1次轉彎(上轉右,或右轉上);\\現在只能發生3次轉彎,也就是在1個\bigcirc及1個\square填入↑,其它的\bigcirc及\square都不填入↑;\\2個\square選1個,有2種選法;5個\bigcirc選1個,有5種選法;共有2\times 5=10種選法;\\假設\square有x個↑,\bigcirc有y個↑,則 x+y=5, x,y\in N共有H^2_3=C^4_3=4組解;\\因此轉3次彎有10\times 4= \bbox[red,2pt]{40}種走法





解:
\cases{\int_0^k f(x)-g(x)\;dx = {2\over 5}k^5-k^4+k^2\\ \pi \int_0^k g^2(x)-f^2(x)\;dx = (-{4\over 9}k^9+k^8+ {2\over 3}k^6-{16\over 5}k^5 +{8\over 3}k^3)\pi} \\ \Rightarrow \cases{{d\over dk}\int_0^k f(x)-g(x)\;dx = {d\over dk}({2\over 5}k^5-k^4+k^2)\\ {d\over dk} \int_0^k g^2(x)-f^2(x)\;dx = {d\over dk}(-{4\over 9}k^9+k^8+ {2\over 3}k^6-{16\over 5}k^5 +{8\over 3}k^3)} \\ \Rightarrow \cases{f(k)-g(k)= 2k^4-4k^3+2k\cdots(1)\\ g^2(k)-f^2(k) =-4k^8+8k^7 +4k^5-16k^4+8k^2} \\ \Rightarrow {g^2(k)-f^2(k) \over f(k)-g(k)} ={-4k^8+8k^7 +4k^5-16k^4+8k^2 \over 2k^4-4k^3+2k}= -2k^4+4k \Rightarrow f(k)+g(k) =2k^4-4k \cdots(2) \\ \Rightarrow (2)-(1) \Rightarrow 2g(k)=4k^3-6k \Rightarrow g(k)=2k^3-3k \\ \Rightarrow \bbox[red,2pt]{g(x)=2x^3-3x}



解:


假設D為坐標原點,則各點坐標如上圖;又\cases{P為\overline{GH}中點\\ \overline{BQ}:\overline{QC} =2:1\\ R在\overline{AB}上} \Rightarrow \cases{P(2,0,2) \\ Q(4,1,0) \\R(a,3,0)};\\ \overline{EQ}直線方程式: {x\over 4} ={y-3\over -2} = {z-2\over -2} \Rightarrow S(4t,-2t+3,-2t+2) \Rightarrow \cases{\overrightarrow{PS}=(4t-2,-2t+3,-2t) \\ \overrightarrow{PR}=(a-2,3,-2)}\\ 由於\overrightarrow{PS} \parallel \overrightarrow{PR} \Rightarrow {-2t+3 \over 3} ={-2t\over -2} \Rightarrow t=3/5 \Rightarrow S({12\over 5},{9 \over 5},{4 \over 5})\\ \Rightarrow \cases{\overline{SP}=\sqrt{{4\over 25} +{81\over 25} +{36 \over 25} } ={11\over 5} \\ \overline{SE}=\sqrt{{144\over 25} +{36\over 25} +{36 \over 25} } ={6\sqrt 6\over 5} \\ \overline{EP}= \sqrt{4+9}=\sqrt 13} \\ \Rightarrow \cos \angle PSE ={\overline{SE}^2 +\overline{SP}^2 -\overline{EP}^2 \over 2\times \overline{SE}\times \overline{SP}} ={{12\over 25} \over 2\times {66\over 25}\sqrt 6}= \bbox[red,2pt]{\sqrt 6\over 66}




解:


令直線L與拋物線交於A、B兩點,其中\cases{A({1\over 4}y_1^2,y_1)\\ B({1\over 4}y_2^2,y_2)} \Rightarrow L的斜率m_L={y_1-y_2\over (y_1^2-y_2^2)/4} ={4 \over y_1+y_2}  \\ \Rightarrow L: y-y_1= {4 \over y_1+y_2} (x-{1\over 4}y_1^2) \Rightarrow x={(y-y_1)(y_1+y_2) \over 4} +{1\over 4}y_1^2 ={1\over 4}(y_1+y_2)y- {1\over 4}y_1y_2 \\ \Rightarrow 所圍面積={9\over 8} =\int_{y_2}^{y_1}{1\over 4}(y_1+y_2)y- {1\over 4}y_1y_2-{1\over 4}y^2\;dy = \left. \left[ {1\over 8}(y_1+y_2)y^2 -{1\over 4} y_1y_2y -{1\over 12}y^3\right] \right|_{y_2}^{y_1} \\ =({1\over 8}(y_1+y_2)y_1^2 -{1\over 4} y_1^2y_2 -{1\over 12}y_1^3)- ({1\over 8}(y_1+y_2)y_2^2 -{1\over 4} y_1y_2^2 -{1\over 12}y_2^3) \\ ={1\over 24}(y_1^3-y_2^3)-{1\over 8}(y_1^2y_2-y_1y_2^2) = {1\over 24}(y_1-y_2)^3 \Rightarrow (y_1-y_2)^3 =27 \Rightarrow y_1-y_2=3 \\ \Rightarrow (y_1-y_2)^2 =9 \Rightarrow y_2y_2= {1\over 2}(y_1^2+y_2^2- 9)\\ 令P為\overline{AB}中點\Rightarrow P({1\over 8}(y_1^2+y_2^2),{1\over 2}(y_1+y_2)) \equiv (x,y) \Rightarrow y^2 ={1\over 4}(y_1+y_2)^2 ={1\over 4}(y_1^2+y_2^2)+ {1\over 2}y_1y_2 \\ ={1\over 4}(y_1^2+y_2^2)+ {1\over 2}\cdot {1\over 2}(y_1^2+y_2^2- 9) ={1\over 2}(y_1^2+y_2^2)-{9\over 4} =4x-{9\over 4} \Rightarrow \bbox[red,2pt]{y^2=4x-{9\over 4}}



解:
f(x)=x^3-kx^2+4k \Rightarrow f'(x)=3x^2-2kx=0 \Rightarrow x=0,2k/3;\\ f(0)f(2k/3) < 0 \Rightarrow 4k(4k-{4\over 27}k^3) < 0 \Rightarrow 16k^2(1-{1\over 27}k^2)< 0 \Rightarrow k^2 > 27 \\ \Rightarrow k > 3\sqrt 3 或k< -3\sqrt 3\\ 由於f(x)=0有兩負根一正根\Rightarrow f(0) < 0 \Rightarrow 4k< 0 \Rightarrow k<0 \Rightarrow \bbox[red,2pt]{k < -3\sqrt 3}


解:
\cases{y-x> 2\\ z-y> 3} \Rightarrow \cases{y\ge x+3\\ z\ge y+4} \\ \Rightarrow \begin{array}{}x & y & z & 數量\\\hline 1 & 4 & 8-15 & 8\\ & 5 & 9-15 & 7\\ & \cdots & \cdots & \cdots \\ & 11 & 15 &  1 \\\hdashline 2 & 5 & 9-15 & 7 \\ \cdots \\\hdashline 7 & 10 & 14-15 & 2\\\hdashline 8 & 11 & 15 &1\\\hline \end{array} \Rightarrow 共有\sum_{k=1}^8 {1\over 2}k(k+1) =120\\ 1~15選出三個數x,y,z滿足x\le y\le z共有H^{15}_3=C^{17}_3= 680 \Rightarrow 機率為{120\over 680} = \bbox[red,2pt]{3\over 17}



A=\begin{bmatrix} \sqrt 3& -1 \\ 1 & \sqrt 3\end{bmatrix} \Rightarrow {1\over 2}A= \begin{bmatrix} \sqrt 3/2& -1/2 \\ 1/2 & \sqrt 3/2 \end{bmatrix} =\begin{bmatrix} \cos (\pi/6) & -\sin (\pi/6) \\ \sin (\pi/6) & \cos (\pi/6) \end{bmatrix} \\ \Rightarrow ({1\over 2}A)^{30} =\begin{bmatrix} \cos (5\pi) & -\sin (5\pi) \\ \sin (5\pi) & \cos (5\pi) \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1\end{bmatrix} \Rightarrow A^{30}= \begin{bmatrix} -2^{30} & 0 \\ 0 & -2^{30} \end{bmatrix} \\又C^{n-2}_r =C^{n-2}_{r-2} \Rightarrow C^{n-2}_r = C^{n-2}_{n-2-r}=C^{n-2}_{r-2} \Rightarrow n-2-r=r-2 \Rightarrow r/n=1/2\\ 因此 A^{30}B=\begin{bmatrix} -2^{30} & 0 \\ 0 & -2^{30} \end{bmatrix}\begin{bmatrix} 1/2 \\ 1/2\end{bmatrix} = \bbox[red,2pt]{\begin{bmatrix} -2^{29} \\ -2^{29}\end{bmatrix}}






兩圖形\cases{x^2+(y-1)^2 =1 \\ y=x+2}交點為\cases{A(0,2) \\ D(-1,1)},兩圖形\cases{x^2+(y-1)^2 =1 \\ y=x}交點為\cases{C(0,0) \\ B(1,1)}\\  上圖左半部繞x軸旋轉之體積V_1=\pi\int_{-1}^0 (x+2)^2 - (1-\sqrt{1-x^2})^2\;dx\\\qquad = \pi\int_{-1}^0 2x^2+4x+2 +2\sqrt{1-x^2}\;dx =\pi \left.\left[ {2\over 3}x^3 +2x^2 +2x +(\sqrt{1-x^2} +\sin^{-1}x)\right] \right|_{-1}^0 \\\qquad =\pi(1-(-{2\over 3}-{1\over 2}\pi))= {1\over 2}\pi^2 +{5\over 3}\pi\\ 上圖右半部繞x軸旋轉之體積V_2=\pi\int_{0}^1 (1+\sqrt{1-x^2})^2-x^2\;dx \\ \qquad =\pi\int_0^1 2-2x^2+2 \sqrt{1-x^2}\;dx =\pi \left.\left[2x-{2\over 3}x^3+ \sqrt{1-x^2} +\sin^{-1}x\right] +\right|_0^1 \\ \qquad =\pi({4\over 3}+{1\over 2}\pi-1)= {1\over 2}\pi^2+{1\over 3}\pi \\ \Rightarrow V_1+V_2 =\bbox[red,2pt]{\pi^2 +2\pi}



\cases{O(0,0,0) \\ A(3,1,-2) \\ B(-1,1,4) \\ C(2,3,1)} \Rightarrow \cases{0\le x+y+z \le 6 \Rightarrow \cases{ A'(18,6,-12) \\ B'(-6,6,24) \\ C'(12,18,6)} \\ 0\le x+y+z\le 3 \Rightarrow \cases{A''(9,3,-6)\\ B''(-3,3,12) \\ C''(6,9,3)}} \\ \Rightarrow \cases{四面體OA'B'C'體積= {1\over 6}\begin{Vmatrix}18 & 6 &-12 \\ -6 & 6 & 24 \\ 12 & 18 & 6\end{Vmatrix} \\四面體OA''B''C''體積= {1\over 6}\begin{Vmatrix}9 & 3 &6 \\ -3 & 3 & 12 \\ 6 & 9 & 3\end{Vmatrix}} \\ \Rightarrow 欲求之體積={1\over 6}\begin{Vmatrix}18 & 6 &-12 \\ -6 & 6 & 24 \\ 12 & 18 & 6\end{Vmatrix}- {1\over 6}\begin{Vmatrix}9 & 3 &6 \\ -3 & 3 & 12 \\ 6 & 9 & 3\end{Vmatrix}= {1\over 6}(6^3-3^3) \begin{Vmatrix}3 & 1 &2 \\ -1 & 1 & 4 \\ 2 & 3 & 1\end{Vmatrix} \\ ={1\over 6}\times 189 \times 14 = \bbox[red,2pt]{441}


-- END   (解題僅供參考,學校未提供計算題資料)  --





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