114年蘭陽女中教甄-數學詳解

 國立蘭陽女子高級中學 114 學年度第一次正式教師甄選

一、 多選題(請在答案卷第一頁作答，並依序標明題號，每一題 8 分，只錯一個選項者得 5 分，只錯二個選項者得 2 分，錯三個以上選項者得 0 分，共 16 分)

解答: $$\cases{f(x)=(x-1)^3p(x)-1\\ f(x)=(x+1)^3q(x)+1} \Rightarrow \cases{f(1)=-1\\ f(-1)=1}\\ (1) \bigcirc: f(x)=(x-1)(x+1)g(x)+ax+b \Rightarrow \cases{f(1)=a+b=-1\\ f(-1)=-a+b=1} \Rightarrow \cases{a=-1\\ b=0} \Rightarrow 餘式為-x \\(2)\times: f(x)=(x-1)(x+1)^2g(x)+{1 \over 2}x^2+x+{3\over 2} \Rightarrow f(1)=3 \ne -1\\ (3)\times: f(x)=(x-1)(x+1)^3g(x)+ {1\over 4}x^3+{3 \over 4}x^2+{3\over 4}x +{5\over 4} \Rightarrow f(1)=3 \ne -1 \\(4)\times: f(x)=-{3\over 8}(x-1)^5+{15\over 8}(x-1)^4-{5\over 2}(x-1)^3-1 \Rightarrow f(1)=61\ne -1\\ (5) \bigcirc: f(x)=a(x-1)^5+b(x-1)^4+c(x-1)^3-1 =(x+1)^3 p(x)+1 \\\qquad \Rightarrow \cases{f'(x)=5a(x-1)^4 +4b(x-1)^3+ 3c(x-1)^2 \\f''(x)=20a(x-1)^3+12b(x-1)^2+6c(x-1)  } \Rightarrow \cases{f(-1)=-32a+16b -8c-1=1 \\ f'(-1)=80a-32b+12c=0\\ f''(-1)=-160+48-12c=0} \\\qquad \Rightarrow \cases{a=-3/8\\ b=-15/8\\ c=-5/2} \Rightarrow f(x)=-{3\over 8}(x-1)^5-{15\over 8}(x-1)^4-{5\over 2}(x-1)^3-1 \Rightarrow f(-3)=63\\ 故選\bbox[red, 2pt]{(15)}$$

解答: $$(1)\bigcirc: y=3x-x^2=0 \Rightarrow x(3-x)=0 \Rightarrow x=0,3 \Rightarrow 與x軸交於(0,0),(3,0) \\  (2) \bigcirc: \int_0^3 (3x-x^2)\,dx =\left. \left[ {3\over 2}x^2-{1\over 3}x^3\right] \right|_0^3 ={27\over 6}={9\over 2} \\(3) \bigcirc: y=3x-x^2=-(x-{3\over 2})^2+{9\over 4} \Rightarrow 對稱軸:x={3\over 2} \Rightarrow a={3\over 2} \\(4) \bigcirc: 3x-x^2=mx \Rightarrow x^2+(m-3)x=0 \Rightarrow x(x+m-3)=0 \Rightarrow x=0,3-m \\\qquad \Rightarrow 交點(0,0),(3-m,3m-m^2) \\(5)\bigcirc: \int_0^{3- m} [(3x-x^2)-mx] \,dx =\left. \left[{3\over 2}x^2-{1\over 3}x^3-{m\over 2}x^2 \right] \right|_0^{3-m} \\\qquad = \left( {3\over 2}-{m\over 2}\right)(3-m)^2-{1\over 3}(3-m)^3 ={1\over 6}(3-m)^3={9\over 2}\times {1\over 2} \Rightarrow 3-m={3\over \sqrt[3]2} \\\qquad \Rightarrow m=3-{3\over \sqrt[3]2}\\ 故選\bbox[red, 2pt]{(12345)}$$

二、 填充題(請在答案卷第二頁作答，並依序標明題號，不須計算過程，僅須寫出最後的答案，每一題 5 分，共 65 分)

解答: $$\cases{F_1(0,0) \\F_2(2,2) \\ P(2+1/\sqrt 2, 2-1/\sqrt 2)} \Rightarrow \cases{ \overline{PF_1} =3\\ \overline{PF_2} =1} \Rightarrow 2a=\overline{PF_1}-\overline{PF_2}=2 \\ \Rightarrow \Gamma: \sqrt{(x-2)^2+(y-2)^2}- \sqrt{x^2+ y^2} =2 \Rightarrow \bbox[red, 2pt]{2xy-2x-2y+1=0}$$

解答: $$c=(a+bi)^3-47i \Rightarrow c+47i=(a+bi)^3= a^3-3ab^2+(3a^2b-b^3)i\\ \Rightarrow 3a^2b-b^3=b(3a^2-b^2) =47 \Rightarrow \cases{3a^2-b^2=47\\b=1} \Rightarrow a=4 \\ \Rightarrow c=a^3-3ab^2=64-12= \bbox[red, 2pt]{52}$$

解答:

$$假設\overline{CD}=x \Rightarrow \overline{BD}=4-x \Rightarrow \overline{AD}^2= \overline{AC}^2- \overline{CD}^2 =\overline{AB}^2-\overline{BD}^2 \\ \Rightarrow 25-x^2=36-(4-x)^2 \Rightarrow x={5\over 8} \Rightarrow \overline{AD}={15\over 8}\sqrt 7 \Rightarrow \cases{A(0,{15\over 8}\sqrt 7) \\B(-{27\over 8},0)\\ C({5\over 8},0)\\ D(0,0)}\\ \Rightarrow \cases{\overrightarrow{AB} =(-{27\over 8},-{15\over 8}\sqrt 7) \\ \overrightarrow{AD} =(0,-{15\over 8}\sqrt 7) \\ \overrightarrow{AC} =({5\over 8},-{15\over 8}\sqrt 7)} \Rightarrow (0,-{15\over 8}\sqrt 7)=x(-{27\over 8},-{15\over 8}\sqrt 7)+y ({5\over 8},-{15\over 8}\sqrt 7) \\ \Rightarrow \cases{x=5/32\\ y=27/32} \Rightarrow (x,y)= \bbox[red, 2pt]{\left( {5\over 32}, {27\over 32}\right)}$$

解答: $$\lim_{n\to \infty} \sum_{k=1}^n {n\over (n+(2k-2))^2} =\lim_{n\to \infty} \sum_{k=1}^n {n\over n^2(1+({2k-2\over n}))^2} = \lim_{n\to \infty} \sum_{k=1}^n {1/n\over (1+{2(k-1)\over n})^2} \\ =\int_0^1 {1\over (1+2x)^2} \,dx = \left. \left[ -{1\over 2(1+2x)}\right] \right|_0^1 ={1\over 2}-{1\over 6} =\bbox[red, 2pt]{1\over 3}$$

解答: $$\cases{P(x_1,y_1) \\Q(x_2,y_2)} \;在\Gamma:x^2+2y^2=3上\Rightarrow \cases{\overline{PQ}中點M=((x_1+x_2)/2, (y_1+y_2)/2)\\ \overline{PQ}\bot L \Rightarrow \overline{PQ}斜率={y_2-y_1\over x_2-x_1} = -1 \\ x_1^2+2y_1^2 =3 \cdots(1) \\ x_2^2 +2y_2^2 =3\cdots(2)} \\ \Rightarrow (2)-(1) =(x_2^2-x_1^2)+ 2(y_2^2 -y_1^2) =0 \Rightarrow -{1\over 2} ={y_2^2-y_1^2\over x_2^2-x_1^2} ={y_2-y_1\over x_2-x_1} \cdot {y_2+y_1\over x_2+x_1} =-{y_2+y_1\over x_2+x_1} \\ \Rightarrow y_2+y_1= {1\over 2}(x_2+x_1) \Rightarrow M=(x_0,x_0/2) \in L \Rightarrow {1\over 2}x_0=x_0+m \Rightarrow x_0=-2m \\ \Rightarrow M(-2m,- m) \in \Gamma內部\Rightarrow 4m^2+2m^2\le 2 \Rightarrow m^2\le {1\over 3} \Rightarrow \bbox[red, 2pt]{-{\sqrt 3\over 3}\le m\le {\sqrt 3\over 3}}$$

解答: $$假設P(x,y,z) \Rightarrow \overline{PA}^2 +\overline{PB}^2 +\overline{PC}^2 =f(x,y,z)\\= (x-1)^2+(y-1)^2+ (z-1)^2 +(x-2)^2 +(y-4)^2+z^2 +(x-3)^2+(y-2)^2+(z-1)^2\\ 並令g(x,y,z)=x+y+z-6, 使用\text{ Lagrange's }算子求極值 \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ f_z= \lambda g_z\\ g=0} \\ \Rightarrow \cases{2(x-1)+2(x-2)+2(x-3)=6x-12= \lambda \\ 2(y-1)+ 2(y-4) +2(y-2)= 6y-14= \lambda \\ 2(z-1) +2z+2(z-1) = 6z-4=\lambda} \Rightarrow \cases{x=y-2/6\\ z=y-10/6} \\ \Rightarrow x+y+z-6=3y-2-6=0 \Rightarrow y={8\over 3} \Rightarrow \cases{x=7/3\\ z=1} \Rightarrow P= \bbox[red, 2pt]{({7\over 3},{8\over 3},1)}$$

解答: $$假設\triangle ABC中外接圓半徑R={1\over 2}, 且\cases{\angle A=135^\circ \\ \angle B=37^\circ \\ \angle C=8^\circ}, 則{\overline{BC} \over \sin \angle A} ={\overline{AC} \over \sin \angle B} = {\overline{AB} \over \sin \angle C} =2R=1 \\ \Rightarrow \cases{\overline{BC} =\sin 135^\circ =1/\sqrt 2 \\ \overline{AB}= \sin 8^\circ \\ \overline{AC} = \sin 37^\circ} \Rightarrow \cos \angle A=-{\sqrt 2\over  2}={\sin^2 37^\circ +\sin ^28-\sin^2 135^\circ \over 2\cdot \sin 37^\circ \sin 8^\circ} \\ \Rightarrow \sin^2 37^\circ +\sin ^28+ \sqrt 2\sin 37^\circ \sin 8^\circ= \sin^2 135^\circ = \bbox[red, 2pt]{1\over 2}$$

解答: $$\sqrt 3=\cos \alpha+\sin(\alpha +\beta) +\cos(\alpha+\beta) =\cos \alpha+\sin \alpha \cos \beta + \sin \beta \cos \alpha+\cos \alpha \cos\beta- \sin \alpha \sin \beta \\= \cos \alpha(1+\sin \beta+ \cos \beta) +\sin \alpha(\cos \beta-\sin \beta) =\sqrt{(1+\sin \beta+ \cos \beta)^2+ (\cos \beta-\sin \beta)^2} \sin(\alpha+ \gamma) \\ \le \sqrt{(1+\sin \beta+ \cos \beta)^2+ (\cos \beta-\sin \beta)^2} =\sqrt{3+2\sin \beta +2\cos \beta} \\ \Rightarrow 3\le 3+2\sin \beta +2\cos \beta \Rightarrow 2(\sin \beta+ \cos \beta)\ge 0 \Rightarrow \sin \beta+\cos \beta \ge 0 \Rightarrow \sin(\beta+{\pi\over 4})\ge 0 \\ \Rightarrow \beta= \bbox[red, 2pt]{3\pi\over 4}$$

解答: $$\alpha+ \beta +\gamma+ \delta ={5\over 4} \Rightarrow \cases{\alpha/2 +\beta/4+ \gamma/5+ \delta/8=1\\  (\alpha/2) \cdot (\beta/4) \cdot (\gamma/5) \cdot (\delta/8) =1/256} \\算幾不等式: \alpha/2 +\beta/4+ \gamma/5+ \delta/8 \ge 4\sqrt[4]{(\alpha/2) \cdot (\beta/4) \cdot (\gamma/5) \cdot (\delta/8)} \\ 等號剛好成立，即 \alpha/2 =\beta/4= \gamma/5= \delta/8=1/4 \Rightarrow \cases{\alpha=1/2\\ \beta=1\\ \gamma =5/4\\ \delta=2} \Rightarrow \alpha+\beta +\gamma+\delta={19\over 4} \\ \Rightarrow -{b\over 4}={19\over 4} \Rightarrow b= \bbox[red, 2pt]{-19}$$

解答: $$\cases{\log(2000xy) =3+\log 2+\log x+\log y=4+\log x\cdot \log y\\ \log(2yz) =\log 2+\log y+\log z=1+\log y\cdot \log z\\ \log zx=\log z+\log x=\log z\cdot \log x}\\ \Rightarrow \cases{\log x+\log y=\log 5+\log x\cdot \log y \cdots(1)\\ \log y+\log z=\log 5+\log y\cdot \log z \cdots(2)\\ \log z+\log x=\log z\cdot \log x \cdots(3)} \\ (1)-(2) =\log x-\log z=\log y(\log x-\log z) \Rightarrow (\log y-1)(\log x-\log z) =0\\ 若\log y=1 代入(1)\Rightarrow  \log x+1=\log 5+\log x 矛盾,無解 \\ \log x=\log z代入(3) \Rightarrow 2\log x=(\log x)^2 \Rightarrow \log x(\log x-2)=0 \\\Rightarrow \cases{\log x=0 \Rightarrow x=1 \Rightarrow z=1 \Rightarrow y=5 \Rightarrow x+y+z=7\\ \log x=2 \Rightarrow x=100\Rightarrow z=100 \Rightarrow y=20 \Rightarrow x+y+z=220} \Rightarrow \bbox[red, 2pt]{7或200}$$

解答: $$\begin{array}{r} 1的個數& 4的個數& 數量\\\hline 0& 0 & 2^5=32\\ 1& 0&2^4\cdot 5= 80 \\ 2& 0& 2^3\cdot C^5_3=80 \\ 3& 0 & 2^2\cdot C^5_3=40 \\ 4& 0& 2\cdot 5=10 \\ 5& 0& 1\\\hdashline 0&1 & 80\\ 0& 2& 80\\ 0&3& 40\\ 0& 4&10\\ 0& 5& 1\\\hdashline 1& 1 & 80\\  1& 2& 40\\ 1& 3&  10\\ 1& 4& 1 \\\hdashline  2& 1& 40\\ 2& 2& 10\\ 2& 3& 1 \\\hdashline   3& 1& 10\\ 3& 2& 1\\ \hdashline   4& 1& 1 \end{array} \\ \Rightarrow 合計\bbox[red, 2pt]{648}$$

解答: $$A箱點數不是0 \Rightarrow A箱沒有5或沒有偶數\Rightarrow 有C^8_5+C^6_5=62種情形 \\ \Rightarrow A箱點數是0的情形有C^{10}_5-62=190 \\ A箱點數是0且B箱點數是0 \Rightarrow 兩箱皆有5及1個偶數\Rightarrow 有(C^8_4-C^4_4-C^4_4)\times 2=136 \\ \Rightarrow 條件機率={136\over 190} = \bbox[red, 2pt]{68\over 95}$$

解答: $$\cases{A={ab\over bc} \\ B={a\over c}} \Rightarrow {10a+b\over 10b+c}={a\over c} 且符合\cases{a\lt c\\ a,c 互質\\ 10a+b\lt 10b+c\\ 10a+b與10b+c互質\\ 1\le a,b,c\le 9}\\ 只能從b=1-9一個一個試, 只有b=6或9有可能\\ \cases{b=6  \Rightarrow {a6\over 6c}={a\over c} \Rightarrow \cases{a=1, c=4\\ a=2, c=5} \\ b=9\Rightarrow {a9\over 9c}= {a\over c}\cases{a=1,c=5\\ a=4,c=8}} \Rightarrow A= \bbox[red, 2pt]{{16\over 64}, {26\over 65},{19\over 95}, {49\over 98}}$$

三、 計算證明題(請從答案卷第三頁開始作答，依序標明題號，並詳細寫出計算或證明過程，每一題 10 分，共 30 分)

解答: $$x=\sqrt 2+\sqrt[3]3 \Rightarrow (x-\sqrt 2)^3 =(\sqrt[3]3)^3=x^3-3\sqrt 2x^2+6x-2\sqrt 2=3 \\ \Rightarrow (x^3+6x-3)^2=(3\sqrt 2x^2+2\sqrt 2)^2 \Rightarrow x^6+36x^2+9+12x^4-36x-6x^3=18x^4+24x^2+8 \\ \Rightarrow x^6-18x^4-6x^3+12x^2-36x+1=0\\ 依勘根原理, x^6-18x^4-6x^3+12x^2-36x+1=0的有理根僅可能是\pm 1,\\ 因此\sqrt 2+\sqrt[3] 3不是有理根, 也就是\sqrt 2+\sqrt[3] 3是無理數\qquad \bbox[red, 2pt]{QED}$$

解答:

$$\textbf{(1) }假設\overline{AB}中點O(0,0)及 \angle OBC=\angle OCB=\theta \Rightarrow \angle BOC=\pi-2\theta \Rightarrow \cases{B(R,0)\\ C(R\cos(\pi-2\theta),R\sin(\pi-2\theta))} \\ \Rightarrow C(-R\cos 2\theta, R\sin 2\theta) \Rightarrow 梯形周長)= \overline{AB}+ \overline{CD}+ 2\overline{BC}\\=f(\theta) =2R-2R\cos 2\theta+ 2 R\sqrt{2+2\cos 2\theta} \Rightarrow f'(\theta) =4R\sin 2\theta-{4R\sin 2\theta\over \sqrt{2+2\cos 2\theta}} =0\\ \Rightarrow 2+2\cos 2\theta=1 \Rightarrow \cos 2\theta=-{1\over 2} \Rightarrow \theta= {\pi\over 3} \Rightarrow f( {\pi\over 3}) =2R+R+2R= \bbox[red, 2pt]{5R} \\ \textbf{(2) }當C\to B且C\to A時，面積趨近0，因此最小值\bbox[red, 2pt]{不存在}$$

解答: $$\textbf{(1) }X\sim G(p) \Rightarrow p(x) =P(X=x) =q^{x-1}p,其中q=1-p, x=1,2,\dots\\\qquad  \Rightarrow E(X)= \sum_{x=1}^{\infty} x q^{x-1}p = p \sum_{x=1}^{\infty} x q^{x-1}= p{d\over dq}\sum_{x=0}^{\infty}  q^{x} =p{d\over dq}\left( {1\over 1-q}\right) =p\cdot {1\over (1-q)^2} \\ \qquad =p\cdot {1\over p^2} ={1\over p} \Rightarrow E(X)={1\over p} \quad \bbox[red, 2pt]{QED} \\\textbf{(2 )} E(X(X-1)) =\sum_{x=1}^\infty x(x-1)q^{x-1}p =pq \sum_{x=1}^\infty x(x-1)q^{x-2}  =pq {d^2 \over dq^2}\sum_{x=0}^\infty q^x  =pq {d^2 \over dq^2} \left({1\over 1-q} \right) \\\qquad =pq\cdot {2\over (1-q)^3} ={2q\over p^2} \Rightarrow E(X^2)=E(X(X-1))+E(X)={2q\over p^2}+{1\over p} \\\qquad \Rightarrow Var(X) =E(X^2)-(E(X))^2 ={2q\over p^2}+{1\over p}-{1\over p^2} ={1-p\over p^2} \quad \bbox[red, 2pt]{QED}$$

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