115年南港高中教甄2-數學詳解

 臺北市立南港高級中學115學年度第2次正式教師甄試

共 20 題, 每題 5 分, 滿分 100 分

解答:$$依題意最大數9需擺在中間，剩下8個數任取4個數放左邊，剩下4個數放右邊\\因此共有C^8_4= \bbox[red, 2pt]{70}個$$

解答:$$f(x)=  \sum_{k=1}^{10}(11-k)(x-k)^2 \Rightarrow f'(x) =2\sum_{k=1}^{10}(11-k)(x-k)=0 \\\Rightarrow x_1 \sum_{k=1}^{10}=(11-k)  = \sum_{k=1}^{10}k(11-1) \Rightarrow 55x_1=220 \Rightarrow x_1=4\\ g(x)=f(x) \Rightarrow x_2=4 \Rightarrow (x_1,x_2)= \bbox[red, 2pt]{(4,4)}$$

解答:$$(a,b)=k(b,-1) \Rightarrow \cases{a=kb\\ b=-k} \Rightarrow a=-b^2 \Rightarrow a-b=-b^2-b=- \left( b+{1\over 2} \right)^2+{1\over 4} \\ \Rightarrow 當b=-{1\over 2}時, a-b有最大值\bbox[red, 2pt]{1\over 4}, 此時a=-{1\over 4}, 即 \bbox[red, 2pt]{\cases{a=-1/4\\ b=-1/2}  }$$

解答:$$f(x)=x^4-4x^3-x^2+x+ \int_k^x f(t)\,dt \Rightarrow f'(x)=4x^3-12x^2-2x+1+f(x) \\ \Rightarrow f(x)-f'(x)=-4x^3+12x^2+2x-1 \Rightarrow f(x)為三次式\Rightarrow f(x)=Ax^3+ Bx^2+Cx+D \\ \Rightarrow f'(x)=3Ax^2+2Bx+C \Rightarrow f(x)-f'(x)= Ax^3+(B-3A)x^2+(C-2B)x+D-C \\ \Rightarrow \cases{A=-4\\ B-3A=12\\ C-2B=2\\ D-C=-1} \Rightarrow \cases{B=0\\ C=2\\ D=1} \Rightarrow f(x)=-4x^3+2x+1 \Rightarrow \int_{-3}^2 f(x)\,dx  \\= \left. \left[ -x^4+x^2+x \right] \right|_{-3}^2=-10-(-75) =\bbox[red, 2pt]{65}$$

解答:$$E:2x+y-z=4 \Rightarrow 法向量\vec n=(2,1,-1) \Rightarrow \vec n\bot \overrightarrow{PQ} \Rightarrow (2,1,-1) \cdot (1,c,5)=0 \\ \Rightarrow 2+c-5=0 \Rightarrow c=3 \\ 又\overrightarrow{AB}=(7-a,b+2,3)=\overrightarrow{PQ}+t\vec n=(1,3,5)+t(2,1,-1) =(2t+1,t+3,-t+5) \\ \Rightarrow \cases{7-a=2t+1\\ b+2=t+3\\ 3=-t+5} \Rightarrow \cases{t=2\\ a=2\\ b=3} \Rightarrow (a,b,c)= \bbox[red, 2pt]{(2,3,3)}$$

解答:$$假設\log_4 A=\log_6 B= \log_9 (A+B)=t \Rightarrow \cases{A=4^t\\ B=6^t\\ A+B=9^t} \Rightarrow 4^t+6^t=9^t \Rightarrow  \left( {4\over 6} \right)^t+1= \left( {9\over 6} \right)^t \\ \Rightarrow  \left( {2\over 3} \right)^t+1= \left( {3\over 2} \right)^t ,取{A\over B}=x \Rightarrow x+1={1\over x} \Rightarrow x^2+x-1=0 \Rightarrow x=\bbox[red, 2pt]{-1+\sqrt 5\over 2}$$

解答:$$振帳a=10, 又\cases{波峰在t=1/300\\ 波谷在t=4/300} \Rightarrow 半週期T/2={4\over 300}-{1\over 300} ={ 1\over 100}\Rightarrow T={1\over 50} \\\Rightarrow b={2\pi\over T} =100\pi \Rightarrow 峰值點10\sin \left( 100\pi\cdot {1\over 300}+c \right)=10 \Rightarrow {\pi\over 3}+c={\pi\over 2} \Rightarrow c={\pi\over 6} \\ \Rightarrow (a,b,c)= \bbox[red, 2pt]{\left( 10, 100\pi, {\pi\over 6} \right)}$$

解答:$$A(1,1)為反曲點\Rightarrow f(x)=a(x-1)^3+p(x-1)+1, 通過(3,5) \Rightarrow f(3)=5\\ \Rightarrow 8a+2p=4 \Rightarrow p=2-4a \Rightarrow f(x)-L=a(x-1)^3+(2-4a)(x-1)+1-(2x-1)\\ =a(x-1)^3-4a(x-1) =a(x-1)(x-3)(x+1) \Rightarrow 交點在x=-1,1,3 \\ \Rightarrow 面積=2 \int_1^3 |a(x-1)(x-3)(x+1)|\,dx =24 \Rightarrow -2a \int_1^3 (x-1)(x-3)(x+1) \,dx=24 \\ \Rightarrow a=3 \Rightarrow p=2-12=-10 \Rightarrow f(x)=3(x-1)^3-10(x-1)+1= 3x^3-9x^2-x+8\\ \Rightarrow (a,b,c,d)= \bbox[red, 2pt]{(3,-9,-1,8)}$$

解答:$$M= \begin{bmatrix}5&2\\3& 4 \end{bmatrix} \Rightarrow \det(M-\lambda I)= \lambda^2-9 \lambda+14=0 \Rightarrow (\lambda-2)(\lambda-7)=0 \Rightarrow \lambda=2,7\\ \lambda_1=2 \Rightarrow (M-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}3& 2\\3&2  \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix}=0 \Rightarrow 3x+2y=0 \\ \lambda_2=7 \Rightarrow (M-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}-2&2\\3&-3 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix}=0 \Rightarrow  x-y=0 \\L: \bbox[red, 2pt]{3x+2y=0, x-y=0}$$

解答:$$三個數乘積為一完全平方數:(1,4,9),(1,2,8), (2,4,8),(2,8,9),(2,3,6),(3,6,8)，共6種\\ \Rightarrow 機率={6\over C^9_3} ={6\over 84}= \bbox[red, 2pt]{1\over 14}$$

解答:

$$\triangle ABD面積= {1\over 2} \begin{Vmatrix}2&-6\\ 1& 5  \end{Vmatrix} =8 \Rightarrow \triangle ADE面積= \triangle ABD-\triangle ABE= 8-3 =5 \Rightarrow \triangle BCE=5 \\ \Rightarrow {\overline{BE} \over \overline{DE}} ={\triangle ABE\over \triangle ADE} ={3\over 5} \Rightarrow {\triangle ABE\over \triangle CDE} ={\overline{BE}^2\over \overline{DE}^2} ={9\over 25} \Rightarrow \triangle CDE={25\over 9} \cdot \triangle ABE={25\over 3} \\ \Rightarrow ABCD面積= \triangle ABE+ \triangle BCE+\triangle CDE+ \triangle ADE =3+5+{25\over 3}+5 = \bbox[red, 2pt]{64\over 3}$$

解答:

$$z=3+yi \Rightarrow (3+yi)^3=-117-44i \Rightarrow y=-4 \Rightarrow z=3-4i \Rightarrow |z|=5 \\\Rightarrow A(3,-4) \Rightarrow \cases{B=A逆時針旋轉120度 \\C=A順時針旋轉120度} \Rightarrow \angle BOC= 120^\circ \Rightarrow \angle OPB= 30^\circ \Rightarrow \overline{OP}=5\cdot 2=10\\ P\in \overleftrightarrow{OA}: 4x+3y=0 \Rightarrow P(3t,-4t) \Rightarrow \overline{OP}^2=25t^2=10^2 \Rightarrow t^2=4\Rightarrow t=-2 \Rightarrow P=\bbox[red, 2pt]{(-6,8)}$$

解答:$$若第 n − 1 次交換後手中為紅球(發生機率P_{n-1})，此時袋中有 3 紅 2 白。下一次抽到紅球的機率為{3\over 5} \\若第 n − 1 次交換後手中為白球(發生機率1-P_{n-1})，此時袋中有 4 紅 1 白。下一次抽到紅球的機率為{4\over 5}\\因此 P_n= {3\over 5}P_{n-1}+{4\over 5}(1-P_{n-1}) =-{1\over 5}P_{n-1}+ {4\over 5} \Rightarrow (r,s) = \bbox[red, 2pt]{\left( -{1\over 5},{4\over 5} \right)}$$

解答:$$全部組合數=5(衣料)\times 4(模式)\times 2^3(附加功能) =160種\\ 違反限制的組合數=1(衣料1)\times 4(模式)\times 1(A開啟)\times 2^2(B,C任意)=16\\ \Rightarrow 可行的行程數=160-16= \bbox[red, 2pt]{144}$$

解答:$$算幾不等式\cases{x^2+ x^{-2} \ge 2 \\x+x^{-1} \ge 2} \Rightarrow f(x)=x^2+x^{-2} +4(x+ x^{-1})+6 \ge 2+4\cdot 2+6= \bbox[red, 2pt]{16}$$

解答:

$$f(x)=3x^3-18x+4\sqrt 2 \Rightarrow f'(x)=9x^2-18 \Rightarrow f''(x)=18x\\ f'(x)=0 \Rightarrow x=\pm \sqrt 2 \Rightarrow \cases{f(\sqrt 2)\gt 0 \Rightarrow A(\sqrt 2,f(\sqrt 2)=(\sqrt 2,-8\sqrt 2)) 為極小值坐標\\ f(-\sqrt 2)\lt 0\Rightarrow B(-\sqrt 2,f(-\sqrt 2)=(-\sqrt 2),16\sqrt 2) 為極大值坐標} \\ \Rightarrow \cases{A'(\sqrt 2,8\sqrt 2) \\ B(-\sqrt 2,16\sqrt 2)} 為y=|f(x)|極大值坐標\Rightarrow \cases{16\sqrt 2=2^{9/2} \\ 8\sqrt 2=2^{7/2}} \\ \Rightarrow \bbox[red, 2pt]{ \cases{k\gt 9/2有2個相異實根\\ k=9/2有3個相異實根\\7/2\lt k\lt 9/2有4個相異實根\\ k=7/2有5個相異實根 \\ k\lt 7/2有6個相異實根}}$$

解答:

$$餘弦定理: \overline{BD}^2 =\overline{AB}^2+ \overline{AD}^2-2\cdot \overline{AB}\cdot \overline{AD}\cos \angle BAD =2+4-4\sqrt 2\cdot \left( -{\sqrt 2\over 2} \right)=10\\ \Rightarrow \overline{BD}=\sqrt{10} \Rightarrow \triangle ABD={1\over 2}\overline{AB}\cdot \overline{AD}\sin \angle BAD={1\over 2}\cdot \sqrt 2\cdot 2\cdot {\sqrt 2\over 2}=1 \Rightarrow 箏形ABCD面積=2 \\ \Rightarrow {1\over 2} \overline{BD}\cdot \overline{AC}=2 \Rightarrow \overline{AC}={4\over \sqrt{10}}= \bbox[red, 2pt]{2\sqrt{10} \over 5}$$

解答:

$$圖形\Gamma:x=\sqrt{4-y^2} 為一右半圓:x^2+y^2=4,x\ge 0,兩端點坐標A(0,2), B(0,-2)\\ 直線L:y=3x+k斜率為正值，與\Gamma有相異兩交點代表L從右切點一直到過B之間都有兩交點\\ 圓心到切點距離=2 \Rightarrow {|k|\over \sqrt{10}}=2 \Rightarrow k=- 2\sqrt{10},(k是y截距,需小於0)\\ L通過B \Rightarrow k=-2 \Rightarrow \bbox[red, 2pt]{-2\sqrt{10}\lt k\le -2}$$

解答:$$取u=\log x ,則原式\log_2 x+\log_3 x=2(\log_2 x) (\log_3 x) \Rightarrow  {u\over \log 2}+{u\over \log 3}=2\cdot {u\over \log 2}\cdot {u\over \log 3 }\\ \Rightarrow u\log 3+u\log 2=2u^2 \Rightarrow 2u^2-(\log 3+\log 2)u=0 \Rightarrow u(2u-\log 6) =0 \\ \Rightarrow \cases{u=\log x=0 \Rightarrow x=1\\ u=\log x=\log 6/2=\log \sqrt 6 \Rightarrow x=\sqrt 6} \Rightarrow 所有實數解的和: \bbox[red, 2pt]{1+\sqrt 6}\\ 但公布的答案是\bbox[cyan, 2pt]{5}$$

解答:$$\textbf{(1) }\cases{抽到B(機率為1/3), 擲兩次得到「正反」或「反正」機率為1/2 \Rightarrow 機率為1/6\\ 抽到C(機率為1/3), 擲兩次得到「正反」或「反正」機率為2/p(1-p) \Rightarrow 機率為2/3p(1-p)} \\ \Rightarrow P(B\mid 一正一反) ={1/6 \over 1/6+2/3p(1-p)}= {1\over 1+4p(1-p)} ={1\over -4(p-1/2)^2+2} \ge {1\over 2},\bbox[red, 2pt]{故得證}\\ \textbf{(2) }{1\over 1+4p(1-p)} ={1\over 2} \Rightarrow -4(p-1/2)^2+2=2 \Rightarrow p= \bbox[red, 2pt]{1\over 2}$$

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