110年台北市立高中聯合教甄-數學詳解

臺北市110學年度市立普通型高級中等學校正式教師聯合甄選

壹、 多選題

解答：$$(A)\bigcirc:2,2,2,\dots ，公差0，公比1，既是等差也是等比數列 \\(B)\bigcirc: 例如:將所有質數排成一列，既不是等差也不是等比數列\\ (C)\times: a_n=\cases{n,n是奇數\\ {1\over 2^n},n是偶數} \quad\Rightarrow \cases{\sum_{n=1}^\infty a_n 發散\\ \sum_{n=1}^\infty a_{2n}={1\over 2^2} +{1\over 2^4} + \cdots={1\over 3}收斂} \\(D)\times: a_n=\begin{cases}{1\over 3^n}-{1\over n} & n是奇數\\ {1\over n-1} & n是偶數 \end{cases} \quad\Rightarrow \cases{\sum_{n=1}^\infty a_n =1/3+1/3^3+\cdots = 3/8 收斂\\ \sum_{n=1}^\infty a_{2n} =1+1/3+1/5 發散} \\(E)\times: 理由同(D)\\故選\bbox[red, 2pt]{(AB)}$$

解答：$$令\cases{\omega_1={z^3-z^2\over z-z^2} = -z\\ \omega_2= {z^3-z\over z^2-z} = z+1\\ \omega_3= {z-z^3 \over z^2-z^3}=1+1/z }，則\\(A) \times:z=2e^{i\pi} \Rightarrow  \cases{\omega_1  =2 \\ \omega_2=-1 \\ \omega_3= 1/2} \Rightarrow Re(\omega_i) \ne  0，無直角\\ (B)\bigcirc: z=2e^{i2\pi/3} \Rightarrow \omega_2=z+1= \sqrt 3i \Rightarrow Re(\omega_2)=0 \\(C)\bigcirc: z=2e^{i3\pi/2} \Rightarrow \omega_1=-z= 2i \Rightarrow Re(\omega_1)=0 \\(D)\bigcirc: z=2e^{i4\pi/3} \Rightarrow \omega_3=1+1/z= \sqrt 3i \Rightarrow Re(\omega_3)=0 \\(E)\times:z=2e^{i5\pi/3} \Rightarrow  \cases{\omega_1  =-1+\sqrt 3i \\ \omega_2=2-\sqrt 3i \\ \omega_3= 2+\sqrt 3i} \Rightarrow Re(\omega_i) \ne  0，無直角\\故選\bbox[red, 2pt]{(BCD)}$$

解答：$$(A)\bigcirc:f(x)=\ln x \Rightarrow f'(x)=1/x \Rightarrow f'(e)=1/e \Rightarrow 斜率=1/e \\(B)\times: \overleftrightarrow{AQ} 與切線垂直，其斜率為-e(不是e) \Rightarrow \overleftrightarrow{AQ}: y-1=-e(x-e) \\(C)\bigcirc: Q在\overleftrightarrow{AQ}上 \Rightarrow Q(t,1-e(t-e)),又\overline{AQ}=Q至x軸矩離\\ \qquad \Rightarrow (t-e)^2+(-e(t-e))^2 = (1-e(t-e))^2 \Rightarrow (t-e)^2=1-2e(t-e)\\ \qquad \Rightarrow t=\sqrt{1+e^2}(註:-\sqrt{1+e^2}應該也對!!)\\\qquad \Rightarrow Q(\sqrt{1+e^2},1-e(\sqrt{1+e^2}-e)) 符合y={1\over 2}(x-e)^2+{1\over 2} \\(D)\bigcirc: 圓心當然在弦的中垂線上\\(E) \bigcirc: \overline{QB} \bot x軸 \Rightarrow B(\sqrt{1+e^2},0)(註:(-\sqrt{1+e^2},0)應該也對)\\ 故選\bbox[red, 2pt]{(BCD)}$$

解答：$$\cases{x\ge 1\\ y\ge 1\\ x+2y \le 11} \Rightarrow 頂點坐標\cases{A(1,1)\\ B(1,5) \\ C(9,1)} \Rightarrow \cases{P(A)={1\over a}+ {1\over b}\\ P(B)={1\over a}+ {5\over b}\\ P(C)={9\over a}+ {1\over b}}\\(A)\times: a\ge 2b\gt 0 \Rightarrow 1\ge 2\cdot {b\over a} \Rightarrow {1\over 2} \ge {b\over a} \\ (B)\bigcirc: a\ge 2b \gt 0 \Rightarrow 0\lt {1\over a} \le {1\over 2b} \Rightarrow ({1\over a}+{5\over b})-({9\over a}+{1\over b}) ={4\over b}-{8\over a} \ge {4\over b}-4\cdot {1\over b}=0\\ \qquad\Rightarrow 最大值為{1\over a}+{5\over b}\\(C)\times :{9\over a}+{1\over b}不是最大值，理由如(B)\\(D) \times:{1\over a} +{5\over b}={5a+b\over ab}=1 \Rightarrow 5a+b=ab \Rightarrow 10b+b=2b^2 ( \because a=2b) \Rightarrow b(2b-11)=0 \\\qquad \Rightarrow b=11/2 \Rightarrow a=11 \Rightarrow 5a+b=55+11/2=60{1\over 2} \ne 60 \\ (E)\bigcirc: a\ge 2b\gt 0 \Rightarrow 極值發生在a=2b \\故選\bbox[red, 2pt]{(BE)}$$

解答：

$$\int_0^1 (\sqrt x-(-\sqrt x))\;dx +\int_1^4 \sqrt x-(x-2)\;dx =\int_0^1 2\sqrt x \;dx +\int_1^4 \sqrt x-x+2\;dx \\= \left. \left[{4\over 3}x^{3/2} \right] \right|_0^1 +\left. \left[ {2\over 3}x^{3/2} -{1\over 2}x^2+2x \right] \right|_1^4 ={4\over 3} +{19\over 6} =\bbox[red, 2pt]{9\over 2}$$

解答：$$令\cases{a=\sqrt[3]{10+x} \\b=\sqrt[3]{-3-x}} \Rightarrow a^3+b^3= 10+x-3-x =7\\，依題意:a^2+b^2 = ab+7 \Rightarrow a^2-ab+b^2=7  \Rightarrow (a+b)(a^2-ab +b^2)= 7(a+b) \\\Rightarrow a^3+b^3=7(a+b) \Rightarrow 7=7(a+b) \Rightarrow a+b=1 \Rightarrow (a+b)^3=1 \\ \Rightarrow a^3+b^3 +3ab(a+b)=1 \Rightarrow 7+3ab=1 \Rightarrow ab=-2 \Rightarrow \sqrt[3]{(10+x)(-3-x)} =-2 \\ \Rightarrow (x+10)(x+3)=8 \Rightarrow x^2+13x+22=0 \Rightarrow (x+11)(x+2)=0 \Rightarrow x=\bbox[red, 2pt]{-2,-11}$$

解答：$$先將4個a排好，在中間3個位置一定要塞入數字，\\\square a \bigcirc a \bigcirc a \bigcirc a \square : 中間3個\bigcirc 填入bbc，bbc排列數為3，最後有6個位置可放入最後1個c，共18種;\\同理中間3個\bigcirc 填入bcc，bcc排列數為3，最後有6個位置可放入最後1個b，也是18種；\\兩者合計36種排法，但其中有12種是重複的，因此有36-12=\bbox[red, 2pt]{24}種排法。\\重複的地方: bcc 排列後，將最後一個b填入時，若放在c旳左或右皆與前次的bbc插入c有相同排列；\\例如:abacaca，插入b\to ababcaca，與ababaca, 插入c\to ababcaca相同；\\因此要扣掉4\times 3=12個。$$

解答：$$令g(x)=xf(x)-1 \Rightarrow g(1)=g(2)=\cdots=g(11)=0 \Rightarrow 1,2,\dots,11是g(x)=0的11個根\\ \Rightarrow g(x)=a(x-1)(x-2) \cdots (x-11) \Rightarrow f(x) = {g(x)+1\over x} ={a(x-1)(x-2) \cdots (x-11) +1\over x} \\由於\cases{f(x)是10次\\ g(x)是11次} \Rightarrow x 是a(x-1)(x-2) \cdots (x-11) +1的因式 \Rightarrow -11!a+1=0 \Rightarrow a={1\over 11!} \\ \Rightarrow f(12) ={{1\over 11!}\cdot 11!+1\over 12} ={2\over 12} =\bbox[red, 2pt]{1\over 6}$$

解答：$$將1-110拆成六個子集合:\cases{A=\{6k,k=1-18\}\\ B=\{6k+1,k=0-18\} \\ C=\{6k+2,k=0-18\} \\ D=\{6k+3,k=0-17\} \\ E=\{6k+4,k=0-17\} \\ F=\{6k+5,k=0-17\} } \\ \Rightarrow \cases{不能取A\\ 取B就不能取F，或取F就不能取B \\ 取C就不能取E，或取E就不能取C\\ 不能取D} \\\Rightarrow 取B全部，C全部及A、D各1個數字，共有19+19+1+1=\bbox[red, 2pt]{40}$$

解答：

$$拋物線y^2=8x對稱x軸，\triangle OAB垂心也在x軸上，因此A、B亦對稱x軸；\\因此令\cases{A(2t^2,4t),t\gt 0\\ B(2t^2,-4t) \\P為\triangle OAB 外心}\quad \Rightarrow \cases{\overleftrightarrow{AF}斜率m_1=4t/(2t^2-2)\\ \overleftrightarrow{BF}斜率m_2=-4t/(2t^2-2)\\ \overline{OA}中點A'(t^2, 2t)\\ \overline{OB} 中點B'(t^2,-2t)} \\\Rightarrow \cases{\overleftrightarrow{PA'}: y=m_2(x-t^2)+2t \\ \overleftrightarrow{PB'}: y=m_1(x-t^2)-2t }，求交點\Rightarrow m_2(x-t^2)+2t=m_1(x-t^2)-2t \\ \Rightarrow x-t^2={4t \over m_1-m_2} ={2t^2-2\over 2} \Rightarrow x=2t^2-1 \Rightarrow P(2t^2-1,0) \\ 又半徑= \overline{OP} = \overline{AP} \Rightarrow (2t^2-1)^2 = 1+16t^2 \Rightarrow 4t^2(t^2-5)=0 \Rightarrow t=\sqrt 5 \\ \Rightarrow P(2\cdot 5-1,0)= \bbox[red, 2pt]{(9,0)}$$

解答：$$x^{10}+ (nx-1)^{10}=0 \Rightarrow \left({nx-1\over x} \right)^{10} =-1 \Rightarrow \left( n-{1\over x}\right)^{10} =e^{i\pi} \Rightarrow n-{1\over x}=\omega = e^{{2k+1\over 10}\pi i},k=0-9 \\ {1\over x}= n-\omega \Rightarrow {1\over x\bar x} =(n-\omega)(n-\bar \omega) = n^2+1-n(\omega +\bar \omega) \\ \Rightarrow \sum_{k=1}^5 {1\over z_k \bar z_k} =5(n^2+1) -n \sum_{k=0}^9 e^{{2k+1\over 10}\pi i} =5(n^2+1)-0 = \bbox[red, 2pt]{5n^2+5}$$

解答：

$$令正五邊形邊長為a \Rightarrow \cases{\overline{DF} =a\sin 72^\circ \\ \overline{AG} =a\cos 54^\circ =a\sin 36^\circ} \Rightarrow {\triangle DEB \over \triangle ABE} ={\overline{DF} \over \overline{AG}} ={a\sin 72^\circ \over a\sin 36^\circ}\\  ={2a\sin 36^\circ \cos 36^\circ\over a\sin 36^\circ} =2\cos 36^\circ  \Rightarrow \triangle DEB= 2\cos 36^\circ \triangle ABE \\\Rightarrow {正五邊形\over \triangle ABE} ={\triangle DEB+\triangle ABE+ \triangle BCD \over \triangle ABE} ={\triangle ABE(2\cos 36^\circ +1+1)\over \triangle ABE} \\ =2+2\cos 36^\circ =2 + 2\cdot {\sqrt 5+1\over 4} =\bbox[red, 2pt]{5+ \sqrt 5\over 2}$$

解答：$$五人:A,B,C,D,E\\ A贏:A出剪刀，其他人出布，或A出石頭，其他人出剪刀，或A出布，其他人石頭；\\因此A贏的機率=3/3^5=1/3^4 \Rightarrow P(1)=1人贏的機率=C^5_1\cdot {1\over 3^4}；\\P(2)=2人贏的機率= C^5_2\cdot {1\over 3^4}；P(3)=3人贏的機率= C^5_3\cdot {1\over 3^4}；\\P(4)=4人贏的機率= C^5_4\cdot {1\over 3^4}；\\ \Rightarrow 有人贏的機率 (分出勝負)= {1\over 3^4}\sum_{k=1}^4C^5_k = {10\over 27}；\\ 因此期望值= 玩1次分出勝負的機率+ 不分勝負的機率(再玩一次)\\ EX= {10\over 27}+(1-{10\over 27})(EX+1) \Rightarrow {10\over 27}EX =1 \Rightarrow EX= \bbox[red, 2pt]{27 \over 10}$$

解答：$$將兩直線\cases{L_1:y=x+2021\\ L_2:y=2x+110}各自平移，並不會影兩直線夾角，也就是不影響\triangle ABC的面積;\\因此將兩直線平移後變成\cases{L_1':y=x\\ L_2':y=2x} \Rightarrow \cases{A(a,a)\\ B(b,2b) \\ 重心P=L_1'\cap L_2'=(0,0)} \\ P=(A+B+C)/3 \Rightarrow C(-a-b,-a-2b) \Rightarrow \cases{\overrightarrow{CA}= (2a+b,2a+2b)\\ \overrightarrow{CB}= (a+2b,a+4b)} \\ \cases{ \overline{AB}=60 \Rightarrow (a-b)^2+( a-2b)^2=60^2\\\angle C=90^\circ \Rightarrow \overrightarrow{CA}\cdot \overrightarrow{CB}=0 \Rightarrow (2a+b)(a+2b)+(2a+2b)(a+4b)=0 } \\ \Rightarrow \cases{2a^2+5b^2=3600 +6ab\\ 2a^2+5b^2 =-15ab/2} \Rightarrow {3\over 2}ab=3600 \Rightarrow ab= -{800\over 3}\\ 因此\triangle ABC = \triangle PAB+ \triangle PBC +\triangle PCA\\ ={1\over 2}\left( \begin{Vmatrix} a & a\\ b & 2b\end{Vmatrix} + \begin{Vmatrix} b & 2b\\ -a-b & -a-2b\end{Vmatrix} + \begin{Vmatrix} -a-b & -a-2b\\ a & a\end{Vmatrix} \right) \\ ={1\over 2}(|ab|+|ab| +|ab|) ={3\over 2}|ab| ={3\over 2}\times {800\over 3} =\bbox[red, 2pt]{400}$$

解答：

$$令\cases{\overline{AD}=a \\ \angle DFB=\theta} \Rightarrow \cases{\overline{DB}=1-a\\ \overline{DF}=a}，正弦定理:{\overline{DF} \over \sin \angle B} ={\overline{DB}\over \sin \angle DFB} \Rightarrow {a \over \sin 60^\circ} ={1-a\over \sin \theta} \\ \Rightarrow {1\over a}-1= {\sin \theta \over \sin 60^\circ} \Rightarrow a={\sqrt 3\over 2\sin \theta+\sqrt 3} \Rightarrow \theta=90^\circ 時，a有最小值{\sqrt 3\over 2+\sqrt 3} = \bbox[red, 2pt]{2\sqrt 3-3}$$

解答：

(1)$$令x=\sin\theta ，則dx = \cos \theta d\theta \Rightarrow a_n=\int_0^1 (1-x^2)^{n/2} \;dx = \int_0^{\pi/2} (\cos^2\theta)^{n/2}\cdot \cos \theta \;d\theta \\ =\int_0^{\pi/2} \cos ^{n+1}\theta \;d\theta = \left.\left[ \sin\theta \cos^n\theta\right] \right|_0^{\pi/2} +n\int_0^{\pi/2} \sin^2\theta \cos^{n-1}\theta \;d\theta \\ =0+n\int_0^{\pi/2} (1-\cos^2\theta) \cos^{n-1}\theta \;d\theta = n\int_0^{\pi/2} \cos^{n-1}\theta\;d\theta -n\int_0^{\pi/2}\cos^{n+1} \theta \;d\theta \\ =na_{n-2}-na_n \Rightarrow (n+1)a_n= na_{n-2} \Rightarrow a_n={n\over n+1}a_{n-2}，\bbox[red, 2pt]{故得證}$$(2)$$a_n= \int_0^{\pi/2} \cos^{n+1}\theta \;d\theta \Rightarrow a_n \ge a_{n+1} \Rightarrow {a_{n+2} \over a_n} \le {a_{n+1}\over a_n} \le {a_n\over a_n} =1; \\而{a_{n+2}\over a_n} ={n+2\over n+3}\cdot {a_n\over a_n} ={n+2\over n+3} \Rightarrow \lim_{n \to \infty}{a_{n+2}\over a_n} =\lim_{n \to \infty} {n+2\over n+3}=1\\ 因此1 ={a_{n+2} \over a_n} \le {a_{n+1}\over a_n} \le {a_n\over a_n} =1，由夾擠定理可知 \lim_{n\to \infty}{a_{n+1} \over a_n} = \bbox[red, 2pt]{1}$$
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