114年高師大附中教甄-數學詳解

 國立高雄師範大學附屬高級中學 114 學年度教師甄試

計算證明題:(每題 10 分，共 100 分，每題須詳述計算過程，否則不予計分)

解答:

$$取\cases{C(0,0)\\ A(0,2) \\B(-2,0)\\ C(2,0) \\G(a,b)} \Rightarrow E=G旋轉90^\circ =(-b,a)  \Rightarrow {E+G\over 2}={D+F\over 2} \Rightarrow F=(a-b,a+b)\\ 又\overleftrightarrow{AC}:x+y=2且F\in \overleftrightarrow{AC} \Rightarrow a-b+a+b=2 \Rightarrow a=1 \Rightarrow \cases{G(1,b)\\E(-b,1) } \\ \overline{BE} =\sqrt 3 \cdot\overline{CG} \Rightarrow (b-2)^2+1= 3(1+b^2) \Rightarrow b=\sqrt 2-1 \\\Rightarrow DEFG面積=\overline{DG}^2 =1+(\sqrt 2-1)^2= \bbox[red, 2pt]{4-2\sqrt 2}$$

解答:$$z^7=1 \Rightarrow (1-z)(1+z+z^2+\dots+z^6)=0 \Rightarrow 1+z+\cdots+z^6=0 \Rightarrow z+z^2+\cdots +z^6=-1\\假設\cases{\alpha=z+z^2+z^4\\ \beta=z^3+z^5+z^6} \Rightarrow \alpha+\beta=z+ z^2+ \cdots+ z^6=-1 \\ 又\alpha \beta =(z+z^2+z^4) (z^3+z^5+z^6) = z^4(1+z+z^2+ \cdots+z^6+2z^3) =z^4(0+2z^3)=2z^7=2\\ 因此我們有\cases{\alpha+\beta=-1\\ \alpha\beta=2} \Rightarrow \alpha,\beta 是x^2+x +2=0的兩根 \Rightarrow \alpha=z+z^2+z^4 =\bbox[red, 2pt]{-1\pm \sqrt 7i\over 2}$$

解答:$$g(x)=\int xf(x)\,dx \Rightarrow g'(x)=xf(x) \Rightarrow {d\over dx}[f(x)+g(x)=x^4-4x^2+x-7\\ \Rightarrow f'(x)+g'(x) =f'(x)+xf(x)= x^4-4x^2+x-7 一階微分方程 \\ 積分因子I(x)=e^{\int x\,dx } =e^{x^2/2} \Rightarrow e^{x^2/2}f'(x)+ xe^{x^2/2}f(x)=e^{x^2/2}(x^4-4x^2+x-7) \\ \Rightarrow \left( e^{x^2/2}f(x) \right)'=e^{x^2/2}(x^4-4x^2+x-7)  \Rightarrow e^{x^2/2}f(x) =\int e^{x^2/2}(x^4-4x^2+x-7)\,dx \\ \Rightarrow e^{x^2/2}f(x) =e^{x^2/2}(x^3-7x+1)+C \Rightarrow f(x)=x^3-7x+1+Ce^{-x^2/2} \\ f(x)為整係數多項\Rightarrow 常數C=0 \Rightarrow \bbox[red, 2pt]{f(x)=x^3-7x+1}$$

解答:

$$\textbf{Case I }2x-y\ge 0 \Rightarrow 2x\ge y 也就是\triangle OQR \Rightarrow A(a,b)\in \triangle OQR \Rightarrow \cases{2a-b\le a\le 1-(2a-b)\\ 2a-b\le b\le 2-(2a-b)} \\\qquad \Rightarrow (a,b) \in \triangle OQS, 其中S=(1/2,1/2) \\ \textbf{Case II }2x-y\le 0 \Rightarrow 2x\le y也就是\triangle PQO \Rightarrow A(a,b) \in \triangle PQO  \\ \qquad \Rightarrow (a,b)\in \triangle OQT,其中T=(1/2,3/2) \\ \triangle OQS\cup\triangle OQT =平行四邊形OTQS \Rightarrow 面積=\overline{OS}\cdot d(y=x,y=x+1) ={\sqrt 2\over 2}\cdot {1\over \sqrt 2} =\bbox[red, 2pt]{1\over 2}$$

解答:$$取\cases{A(0,0,0)\\ B(\sqrt 2,0,\sqrt 2)\\C(0,\sqrt 2, \sqrt 2) \\ D(\sqrt 2,\sqrt 2,0)} \Rightarrow \cases{平面E_1=\triangle BCD:x+y+z=2\sqrt 2\\ 平面E_2=\triangle ACD:-x+y-z=0} \Rightarrow \cases{A'=({4\over 3}\sqrt 2, {4\over 3}\sqrt 2, {4\over 3}\sqrt 2) \\B'=(-{\sqrt 2\over 3},{4\over 3}\sqrt 2, -{\sqrt 2\over 3})} \\ \Rightarrow \cases{\overrightarrow{A'C}=(-{4\over 3}\sqrt 2,-{1\over 3}\sqrt 2, -{1\over 3}\sqrt 2) \\ \overrightarrow{A'D} =(-{1\over 3}\sqrt 2, -{1\over 3}\sqrt 2, -{4\over 3}\sqrt 2)} \Rightarrow \vec n= \overrightarrow{A'C} \times \overrightarrow{A'D} =({2\over 3},-{10\over 3},{2\over 3}) \\\Rightarrow \cases{a\triangle A'CD={1\over 2}||\vec n||=\sqrt 3 \\ 平面E_3=\triangle A'CD: x-5y+z+4\sqrt 2=0} \Rightarrow d(B',E_3)={10\sqrt 2\over 9\sqrt 3} \\ \Rightarrow 四面體A'CB'D體積={1\over 3}\cdot \sqrt 3\cdot {10\sqrt 2\over 9\sqrt 3} = \bbox[red, 2pt]{10\sqrt 2\over 27}$$

解答:$$欲求第三次才出現第一次成功的機率，因此假設X\sim \text{geometric}(p) \Rightarrow \cases{E(X)=1/p\\ Var(X)=(1-p)/p^2} \\ \Rightarrow Var(X)=E(X^2)-(E(X))^2 \Rightarrow E(X^2)=Var(X)+(E(X))^2={1-p\over p^2}+{1\over p^2} ={2-p\over p^2} \\ 又E(X^2)={p\over 1-p}S_\infty \Rightarrow {2-p\over p^2}={180p \over 1-p} \Rightarrow 180p^3-p^2+3p-2=0 \\   \Rightarrow (5p-1)(36p^2+7p+2)=0 \Rightarrow p={1\over 5} \Rightarrow P(X=3)=(1-{1\over 5})^2\cdot {1\over 5}= \bbox[red, 2pt]{16\over 125}$$

解答:$$f(x)=|\log(x)| \Rightarrow f(1)=0 \Rightarrow a\lt 1\lt {a+b\over 2}\lt b\\ f(a)={1\over 2}f(b)=f({a+b\over 2}) \Rightarrow -\log a ={1\over 2}\log b =\log{a+b\over 2}\Rightarrow {1\over a}=\sqrt b={a+b\over 2} \\ \Rightarrow \cases{b={1\over a^2} \\ a^2+ab=2} \Rightarrow a^3-2a+1=0 \Rightarrow (a-1)(a^2+a-1)=0 \Rightarrow a={\sqrt 5-1\over 2} \\ \Rightarrow b={1\over a^2}={2 \over 3-\sqrt 5} ={\sqrt 5+3\over 2} \Rightarrow (a,b) = \bbox[red, 2pt]{\left( {\sqrt 5-1\over 2}, {\sqrt 5+3\over 2}\right)}$$

解答:$$\textbf{(1) }f(x)=x^3-3x^2+a \Rightarrow f'(x)=3x^2-6x =3x(x-2)=0  \Rightarrow x=0,x=2\\ \quad \Rightarrow \cases{f(0)=a\gt 0\\ f(2) =a-4\lt 0} \Rightarrow f(0)f(2)\lt 0 \Rightarrow 在0\lt x\lt 2之間f(x)=0恰有一實根 \quad \bbox[red, 2pt]{故得證} \\\textbf{(2) }S=\int_0^t (x^3-3x^2+a)\,dx +\int_2^t (x^3-3x^2+a)\,dx ={1\over 2}t^4-2t^3+2at+4-2a\\ f(t)=0 \Rightarrow t^3-3t^2+a=0 \Rightarrow a=-t^3+3t^2 \Rightarrow S= {1\over 2}t^4-2t^3+2t(-t^3+3t^2)+4-2(-t^3+3t^2) \\ \Rightarrow S=\bbox[red, 2pt]{-{3\over 2}t^4+6t^3-6t^2+4} \\ \textbf{(3) }f(t)= -{3\over 2}t^4+6t^3-6t^2+4 \Rightarrow f'(t)=-6t^3+18t^2-12t=0 \Rightarrow -6t(t-1)(t-2) =0\\ \\\quad \Rightarrow t=0,1,2 \Rightarrow \cases{f(0)=4\\ f(1)=5/2\\ f(2)=4} \Rightarrow 最小值為\bbox[red, 2pt]{5\over 2}$$

解答:$$\sum_{k=1}^n{4S_k\over a_k+2} =S_n \Rightarrow S_1=a_1={4a_1\over a_1+2} \Rightarrow a_1^2-2a_1=0 \Rightarrow a_1=2 (a_n為正數,a_1\ne 0) \\ a_n=S_n-S_{n-1} =\sum_{k=1}^n{4S_k\over a_k+2}-\sum_{k=1}^{n-1}{4S_k\over a_k+2} ={4S_n\over a_n+2} \Rightarrow S_n={1\over 4}a_n(a_n+2) \\ \Rightarrow   S_n=S_{n-1}+a_n={1\over 4}a_n(a_n+2) \Rightarrow S_{n-1}={1\over 4}a_n^2-{1\over 2}a_n ={1\over 4}a_{n-1}(a_{n-1}+2) \\ \Rightarrow a_n^2-2a_n=a_{n-1}^2+2a_{n-1} \Rightarrow (a_n-1)^2=(a_{n-1}+1)^2 \Rightarrow a_n-1=a_{n-1}+1\\ \Rightarrow a_n=a_{n-1}+2=a_{n-2}+4 =a_{n-3}+6=\cdots =a_1+2(n-1) =2n\\ \Rightarrow \bbox[red, 2pt]{a_n=2n} \Rightarrow S_n={1\over 4}\cdot 2n \cdot (2n+2) =n(n+1) \Rightarrow \bbox[red, 2pt]{S_n=n^2+n}$$

解答:$$\cases{B(6,0)\\ A(1,\sqrt{15}) \\ C(a,b)} \Rightarrow \cases{\vec a=\overrightarrow{OA} \\ \vec b=\overrightarrow{OB} \\\vec c=\overrightarrow{OC}} \Rightarrow \cases{\vec c-\vec a=(a-1,b-\sqrt{15}) \\ \vec c-\vec b=(a-6,b)} \\ \Rightarrow (\vec c-\vec a)\cdot (\vec c-\vec b)=(a-1)(a-6)+b(b-\sqrt {15})=0\\ 再用\text{Lagrange }算子求極值, 取\cases{f(a,b)=a^2+b^2\\ g(a,b) =(a-1)(a-6)+b(b-\sqrt {15})} \Rightarrow \cases{f_a=\lambda g_a\\ f_b= \lambda g_b} \\ \Rightarrow \cases{2a=(2a-7)\lambda\\ 2b=(2b-\sqrt{15})\lambda} \Rightarrow {a\over b}={2a-7\over 2b-\sqrt{15}} \Rightarrow b={\sqrt{15}\over 7} a 代入g(a,b)=0 \\ \Rightarrow 32a^2-224a+147=0 \Rightarrow a={28+7\sqrt{10} \over 8} \Rightarrow |\vec c|=\sqrt{f(a,b)} =\sqrt{a^2+{15\over 49}a^2} \\={8\over 7}a={8\over 7} \cdot {28+7\sqrt{10} \over 8}= \bbox[red, 2pt]{4+\sqrt{10}}$$

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