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：
$$橢圓\Gamma_1:{x^2\over 33} +{y^2\over 49}=1 \Rightarrow \cases{a_1=7\\ b_1=\sqrt{33}} \Rightarrow a_1^2=b_1^2+c_1^2 \Rightarrow c_1=4 \Rightarrow \cases{F_1(0,4)\\ F_2(0,-4)}\\P在\Gamma_1上 \Rightarrow \overline{PF_1}+ \overline{PF_2}= 2a_1 =14; \\ 餘弦定理: \cos \angle F_1PF_2= {\overline{PF_1}^2+ \overline{PF_2}^2 -\overline{F_1F_2}^2 \over 2\overline{PF_1}\times \overline{PF_2}} \Rightarrow \cos 60^\circ = {14^2-2\overline{PF_1}\times \overline{PF_2}-8^2 \over 2\overline{PF_1}\times \overline{PF_2}}\\ \Rightarrow {1\over 2}= {132-2\overline{PF_1}\times \overline{PF_2} \over 2\overline{PF_1} \times \overline{PF_2}} \\ \Rightarrow \overline{PF_1} \times \overline{PF_2} =44 \Rightarrow (\overline{PF_1}- \overline{PF_2})^2 =(\overline{PF_1}+ \overline{PF_2})^2-4\overline{PF_1}\times \overline{PF_2}= 14^2-4\times 44 = 20 \\ \Rightarrow |\overline{PF_1}- \overline{PF_2}|= \sqrt{20}= 2\sqrt 5 =2a_2 \Rightarrow a_2=\sqrt 5 \Rightarrow c_2^2=c_1^2 =a_2^2+b_2^2 \Rightarrow 16=5+b_2^2 \\ \Rightarrow b_2=\sqrt{11} \Rightarrow 共軛軸長=2b_2 = \bbox[red,2pt]{2\sqrt{11}}$$
$$|z_1-3+4i|=1為一圓，其中\cases{圓心C(3,-4)\\ 半徑r=1};\\ {z_2\over z_1} =1+\sqrt 3i= 2(\cos 60^\circ+i\sin 60^\circ) \Rightarrow \cases{|z_2|=2|z_1|\\ \angle AOB =60^\circ }\\ 因此\triangle OAB面積= {1\over 2}\overline{OA}\times \overline{OB} \sin \angle AOB ={1\over 2}\overline{OA}\times (2\overline{OA} )\sin \angle 60^\circ = {\sqrt 3\over 2}\overline{OA}^2;\\最大的\overline{OA} =\overline{OC}+r=5+1=6 \Rightarrow 最大的\triangle OAB面積={\sqrt 3\over 2} \times 6^2= \bbox[red,2pt]{18\sqrt 3}$$
解：
$$1人整季(3個月)都不達標的機率為(1-0.2)^3=0.8^3 \Rightarrow n人整季都不達標的機率為 0.8^{3n} \\ \Rightarrow 1-0.8^{3n} > 0.999 \Rightarrow 0.001 > 0.8^{3n} \Rightarrow \log 0.001 > \log 0.8^{3n} \Rightarrow -3 > 3n(3\log 2-1) \\ \Rightarrow -3 > 3n(3\times 0.301-1) =-0.291n \Rightarrow n > {3\over 0.291} \approx 10.3 \Rightarrow n=\bbox[red,2pt]{11}$$
解：
$$三次函數的對稱中心為(x_0,y_0)，則該函數可寫成 f(x)=a(x-x_0)^3+b(x-x_0)+y_0;\\因此本題函數可寫成y=f(x)=a(x+2)^3+b(x+2) +5 \Rightarrow f'(x)=3a(x+2)^2+b\\在 x=1附近的局部圖形近似直線y=7x+1 \Rightarrow \cases{f'(1)=7\\ f(1)=8} \Rightarrow \cases{27a+b=7 \\ 27a+3b+5=8} \\ \Rightarrow \cases{a=1/3\\ b=-2} \Rightarrow \bbox[red,2pt]{y={1\over 3}(x+2)^3-2(x+2)+5}$$
$$\cases{x+y+z=1 \cdots(1)\\ x^2+y^2+z^2=3\cdots(2) \\ x^3+y^3+z^3=5 \cdots(3)} \Rightarrow (x+y+z)^2 = (x^2+y^2+z^2) +2 (xy+yz+zx) \\ \Rightarrow 1=3+2 (xy+yz+zx) \Rightarrow xy+yz+zx=-1\\ 又(1)\times (2) = (x+y+z) (x^2+y^2+z^2)=(x^3 +y^3+z^3) + xy(x+y) +yz(y+z) +zx(z+x) \\ \Rightarrow 3=5+xy(1-z) +yz(1-x) +zx(1-y) =5+ (xy+yz+zx)-3xyz =5-1-3xyz \\\Rightarrow \color{blue}{xyz=1/3}\\ 同理(1)\times (3)=(x+y+z)(x^3+y^3+z^3) = x^4+y^4+z^4 +xy(x^2+y^2)+yz(y^2+z^2) +zx(z^2+x^2) \\ \Rightarrow 5=x^4+y^4+z^4 +xy(3-z^2) +yz(3-x^2) +zx(3-y^2) \\ =x^4+y^4+z^4 + 3(xy+yz+zx)-xyz^3-x^2yz-xy^2z = x^4+y^4+z^4-3-xyz(x+y+z) \\ =x^4+y^4+z^4-3-1/3 \Rightarrow x^4+y^4+z^4 =5+3+1/3= \bbox[red,2pt]{25\over 3}$$
$$f(x)=x^3+(1-a)x^2+ (a+5)x+2a+5 \Rightarrow f(-1)=0 \Rightarrow 二正根外，另一根為-1;\\ 三數中任二數之和成等差，該三數也成等差；因此f(x)=0的三根可假設為-1,-1+d,-1+2d\\ \Rightarrow \cases{三根之和=-3+3d= a-1 \\ 三根之積=-(-1+d)(-1+2d)=-(2a+5)} \Rightarrow \cases{a=3d-2 \cdots(1) \\ 2d^2-3d-4=2a\cdots(2)}\\ 將(1)代入(2) \Rightarrow 2d^2-3d-4=6d-4 \Rightarrow d(2d-9)=0 \Rightarrow d=9/2(d=0違反有2正根的存在)\\ \Rightarrow a=3d-2=27/2-2 =\bbox[red,2pt]{23\over 2}$$
$$ f(x+2)={1+f(x) \over 1-f(x)} \Rightarrow f(3)={1+f(1)\over 1-f(1)} ={2-\sqrt 2\over \sqrt 2 } =\sqrt 2-1 \Rightarrow f(5)={\sqrt 2\over 2-\sqrt 2} =\sqrt 2+1 \\ \Rightarrow f(7)={2+\sqrt 2\over -\sqrt 2} = -\sqrt 2-1 \Rightarrow f(9) ={-\sqrt 2 \over 2+\sqrt 2} =1-\sqrt 2=f(1)\\ 由以上可知: f(2k-1)=\begin{cases} 1-\sqrt 2 & \text{if }(k \mod 4)=1 \\\sqrt 2-1 & \text{if }(k \mod 4)=2 \\\sqrt 2+1 & \text{if }(k \mod 4)=3 \\-1-\sqrt 2 & \text{if }(k \mod 4)=0 \end{cases} \\ 因此\cases{1021 = 2\times 511-1 (k=511 \mod 4 = 3) \\ 2025 = 2\times 1013-1(k=1013 \mod 4=1)}\Rightarrow \cases{f(1021)=\sqrt 2+1 \\ f(2025)=1-\sqrt 2} \\ \Rightarrow f(1021)+f(2025) =\bbox[red,2pt]{2}$$
$$A(0,0)至B(6,5)的捷徑走法，相當於6個→(向右)及5個↑(向上)的排列數;\\我們先將6個→先擺好，並在間隔放上待填符號 \square及頭尾處放置待填符號\bigcirc，如下：\\\square→\bigcirc→\bigcirc→\bigcirc→\bigcirc→\bigcirc→\square\\對任意\bigcirc而言，無論放入多少個↑，都會發生2次轉彎(1次右轉上，另1次上轉右);\\對任意\square而言，無論放入多少個↑，都會發生1次轉彎(上轉右,或右轉上);\\現在只能發生3次轉彎，也就是在1個\bigcirc及1個\square填入↑，其它的\bigcirc及\square都不填入↑；\\2個\square選1個，有2種選法;5個\bigcirc選1個,有5種選法;共有2\times 5=10種選法;\\假設\square有x個↑,\bigcirc有y個↑，則 x+y=5, x,y\in N共有H^2_3=C^4_3=4組解;\\因此轉3次彎有10\times 4= \bbox[red,2pt]{40}種走法$$
解：
$$\cases{\int_0^k f(x)-g(x)\;dx = {2\over 5}k^5-k^4+k^2\\ \pi \int_0^k g^2(x)-f^2(x)\;dx = (-{4\over 9}k^9+k^8+ {2\over 3}k^6-{16\over 5}k^5 +{8\over 3}k^3)\pi} \\ \Rightarrow \cases{{d\over dk}\int_0^k f(x)-g(x)\;dx = {d\over dk}({2\over 5}k^5-k^4+k^2)\\ {d\over dk} \int_0^k g^2(x)-f^2(x)\;dx = {d\over dk}(-{4\over 9}k^9+k^8+ {2\over 3}k^6-{16\over 5}k^5 +{8\over 3}k^3)} \\ \Rightarrow \cases{f(k)-g(k)= 2k^4-4k^3+2k\cdots(1)\\ g^2(k)-f^2(k) =-4k^8+8k^7 +4k^5-16k^4+8k^2} \\ \Rightarrow {g^2(k)-f^2(k) \over f(k)-g(k)} ={-4k^8+8k^7 +4k^5-16k^4+8k^2 \over 2k^4-4k^3+2k}= -2k^4+4k \Rightarrow f(k)+g(k) =2k^4-4k \cdots(2) \\ \Rightarrow (2)-(1) \Rightarrow 2g(k)=4k^3-6k \Rightarrow g(k)=2k^3-3k \\ \Rightarrow \bbox[red,2pt]{g(x)=2x^3-3x}$$
$$假設D為坐標原點，則各點坐標如上圖；又\cases{P為\overline{GH}中點\\ \overline{BQ}:\overline{QC} =2:1\\ R在\overline{AB}上} \Rightarrow \cases{P(2,0,2) \\ Q(4,1,0) \\R(a,3,0)};\\ \overline{EQ}直線方程式: {x\over 4} ={y-3\over -2} = {z-2\over -2} \Rightarrow S(4t,-2t+3,-2t+2) \Rightarrow \cases{\overrightarrow{PS}=(4t-2,-2t+3,-2t) \\ \overrightarrow{PR}=(a-2,3,-2)}\\ 由於\overrightarrow{PS} \parallel \overrightarrow{PR} \Rightarrow {-2t+3 \over 3} ={-2t\over -2} \Rightarrow t=3/5 \Rightarrow S({12\over 5},{9 \over 5},{4 \over 5})\\ \Rightarrow \cases{\overline{SP}=\sqrt{{4\over 25} +{81\over 25} +{36 \over 25} } ={11\over 5} \\ \overline{SE}=\sqrt{{144\over 25} +{36\over 25} +{36 \over 25} } ={6\sqrt 6\over 5} \\ \overline{EP}= \sqrt{4+9}=\sqrt 13} \\ \Rightarrow \cos \angle PSE ={\overline{SE}^2 +\overline{SP}^2 -\overline{EP}^2 \over 2\times \overline{SE}\times \overline{SP}} ={{12\over 25} \over 2\times {66\over 25}\sqrt 6}= \bbox[red,2pt]{\sqrt 6\over 66}$$
解：
$$令直線L與拋物線交於A、B兩點，其中\cases{A({1\over 4}y_1^2,y_1)\\ B({1\over 4}y_2^2,y_2)} \Rightarrow L的斜率m_L={y_1-y_2\over (y_1^2-y_2^2)/4} ={4 \over y_1+y_2} \\ \Rightarrow L: y-y_1= {4 \over y_1+y_2} (x-{1\over 4}y_1^2) \Rightarrow x={(y-y_1)(y_1+y_2) \over 4} +{1\over 4}y_1^2 ={1\over 4}(y_1+y_2)y- {1\over 4}y_1y_2 \\ \Rightarrow 所圍面積={9\over 8} =\int_{y_2}^{y_1}{1\over 4}(y_1+y_2)y- {1\over 4}y_1y_2-{1\over 4}y^2\;dy = \left. \left[ {1\over 8}(y_1+y_2)y^2 -{1\over 4} y_1y_2y -{1\over 12}y^3\right] \right|_{y_2}^{y_1} \\ =({1\over 8}(y_1+y_2)y_1^2 -{1\over 4} y_1^2y_2 -{1\over 12}y_1^3)- ({1\over 8}(y_1+y_2)y_2^2 -{1\over 4} y_1y_2^2 -{1\over 12}y_2^3) \\ ={1\over 24}(y_1^3-y_2^3)-{1\over 8}(y_1^2y_2-y_1y_2^2) = {1\over 24}(y_1-y_2)^3 \Rightarrow (y_1-y_2)^3 =27 \Rightarrow y_1-y_2=3 \\ \Rightarrow (y_1-y_2)^2 =9 \Rightarrow y_2y_2= {1\over 2}(y_1^2+y_2^2- 9)\\ 令P為\overline{AB}中點\Rightarrow P({1\over 8}(y_1^2+y_2^2),{1\over 2}(y_1+y_2)) \equiv (x,y) \Rightarrow y^2 ={1\over 4}(y_1+y_2)^2 ={1\over 4}(y_1^2+y_2^2)+ {1\over 2}y_1y_2 \\ ={1\over 4}(y_1^2+y_2^2)+ {1\over 2}\cdot {1\over 2}(y_1^2+y_2^2- 9) ={1\over 2}(y_1^2+y_2^2)-{9\over 4} =4x-{9\over 4} \Rightarrow \bbox[red,2pt]{y^2=4x-{9\over 4}}$$
$$f(x)=x^3-kx^2+4k \Rightarrow f'(x)=3x^2-2kx=0 \Rightarrow x=0,2k/3;\\ f(0)f(2k/3) < 0 \Rightarrow 4k(4k-{4\over 27}k^3) < 0 \Rightarrow 16k^2(1-{1\over 27}k^2)< 0 \Rightarrow k^2 > 27 \\ \Rightarrow k > 3\sqrt 3 或k< -3\sqrt 3\\ 由於f(x)=0有兩負根一正根\Rightarrow f(0) < 0 \Rightarrow 4k< 0 \Rightarrow k<0 \Rightarrow \bbox[red,2pt]{k < -3\sqrt 3}$$
$$\cases{y-x> 2\\ z-y> 3} \Rightarrow \cases{y\ge x+3\\ z\ge y+4} \\ \Rightarrow \begin{array}{}x & y & z & 數量\\\hline 1 & 4 & 8-15 & 8\\ & 5 & 9-15 & 7\\ & \cdots & \cdots & \cdots \\ & 11 & 15 & 1 \\\hdashline 2 & 5 & 9-15 & 7 \\ \cdots \\\hdashline 7 & 10 & 14-15 & 2\\\hdashline 8 & 11 & 15 &1\\\hline \end{array} \Rightarrow 共有\sum_{k=1}^8 {1\over 2}k(k+1) =120\\ 1~15選出三個數x,y,z滿足x\le y\le z共有H^{15}_3=C^{17}_3= 680 \Rightarrow 機率為{120\over 680} = \bbox[red,2pt]{3\over 17}$$
解：
$$A=\begin{bmatrix} \sqrt 3& -1 \\ 1 & \sqrt 3\end{bmatrix} \Rightarrow {1\over 2}A= \begin{bmatrix} \sqrt 3/2& -1/2 \\ 1/2 & \sqrt 3/2 \end{bmatrix} =\begin{bmatrix} \cos (\pi/6) & -\sin (\pi/6) \\ \sin (\pi/6) & \cos (\pi/6) \end{bmatrix} \\ \Rightarrow ({1\over 2}A)^{30} =\begin{bmatrix} \cos (5\pi) & -\sin (5\pi) \\ \sin (5\pi) & \cos (5\pi) \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1\end{bmatrix} \Rightarrow A^{30}= \begin{bmatrix} -2^{30} & 0 \\ 0 & -2^{30} \end{bmatrix} \\又C^{n-2}_r =C^{n-2}_{r-2} \Rightarrow C^{n-2}_r = C^{n-2}_{n-2-r}=C^{n-2}_{r-2} \Rightarrow n-2-r=r-2 \Rightarrow r/n=1/2\\ 因此 A^{30}B=\begin{bmatrix} -2^{30} & 0 \\ 0 & -2^{30} \end{bmatrix}\begin{bmatrix} 1/2 \\ 1/2\end{bmatrix} = \bbox[red,2pt]{\begin{bmatrix} -2^{29} \\ -2^{29}\end{bmatrix}}$$
解：
解：$$\cases{O(0,0,0) \\ A(3,1,-2) \\ B(-1,1,4) \\ C(2,3,1)} \Rightarrow \cases{0\le x+y+z \le 6 \Rightarrow \cases{ A'(18,6,-12) \\ B'(-6,6,24) \\ C'(12,18,6)} \\ 0\le x+y+z\le 3 \Rightarrow \cases{A''(9,3,-6)\\ B''(-3,3,12) \\ C''(6,9,3)}} \\ \Rightarrow \cases{四面體OA'B'C'體積= {1\over 6}\begin{Vmatrix}18 & 6 &-12 \\ -6 & 6 & 24 \\ 12 & 18 & 6\end{Vmatrix} \\四面體OA''B''C''體積= {1\over 6}\begin{Vmatrix}9 & 3 &6 \\ -3 & 3 & 12 \\ 6 & 9 & 3\end{Vmatrix}} \\ \Rightarrow 欲求之體積={1\over 6}\begin{Vmatrix}18 & 6 &-12 \\ -6 & 6 & 24 \\ 12 & 18 & 6\end{Vmatrix}- {1\over 6}\begin{Vmatrix}9 & 3 &6 \\ -3 & 3 & 12 \\ 6 & 9 & 3\end{Vmatrix}= {1\over 6}(6^3-3^3) \begin{Vmatrix}3 & 1 &2 \\ -1 & 1 & 4 \\ 2 & 3 & 1\end{Vmatrix} \\ ={1\over 6}\times 189 \times 14 = \bbox[red,2pt]{441}$$
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