111年台南家齊高中教甄-數學詳解

國立臺南家齊高級中學 111 學年度第一次教師甄選

一、 填充題(每題 4 分，共 60 分)

解答：

$$C:\cases{\Gamma:x^2+ (y-1)^2 +(z-5)^2=13 \\ E:x+2y+2z=3} \Rightarrow \Gamma為一圓球，球心O(0,1,5)，球半徑R=\sqrt{13} \\ 通過球心O且方向向量為E的法向量(1,2,2)的直線L:{x\over 1} ={y-1\over 2} ={z-5\over 2}；\\ 若Q\in L \Rightarrow Q(t,2t+1,2t+5), t\in \mathbb{R}； Q在E上\Rightarrow t+2(2t+1)+2(2t+5) = 3 \Rightarrow 9t+9=0 \Rightarrow t=-1\\ \Rightarrow Q(-1,-1, 3) 即為截圓圓心；又d(O,E)={2+10-3\over 3} =3 \Rightarrow 截圓半徑r= \sqrt{R^2-3^2} = 2 \\ d(P,E)={4+6+2-3\over 3}=3\\過P且方向向量為E的法向量(1,2,2)的直線L':{x-4\over 1} ={y-3\over 2} ={z-1\over 2};\\  若R\in L'\Rightarrow R(t+4,2t+3,2t+1),t\in \mathbb{R} ；又R在E上\Rightarrow (t+4)+2(2t+3)+2(2t+1)=3\\  \Rightarrow 9t+9=0 \Rightarrow t=-1 \Rightarrow R'(3,1,-1) \Rightarrow \overline{RQ} = \sqrt{16+4+16} =6 \\\Rightarrow 最遠距離\overline{PA}(見上圖)= \sqrt{(\overline{RQ}+r)^2 +(d(P,E))^2} =\sqrt{8^2+3^2} = \bbox[red, 2pt]{\sqrt{73}}$$

解答：

$$正弦定理: \cases{\triangle ABP \Rightarrow \cfrac{\overline{AP}}{\sin(60^\circ-\theta)} =\cfrac{\overline{BP}}{\sin 30^\circ } \\ \triangle APC \Rightarrow \cfrac{\overline{AC}}{\sin(90^\circ-\theta)} =\cfrac{\overline{AP}}{\sin \theta } } \Rightarrow \cases{  \cfrac{\overline{AP}}{\sin(60^\circ-\theta)} =\cfrac{a}{1/2 } \\  \cfrac{a}{\cos \theta} =\cfrac{\overline{AP}}{\sin \theta } } \\ \Rightarrow \cases{\cfrac{\overline{AP}}{a}= 2\sin(60^\circ-\theta)\\ \cfrac{\overline{AP}}{a}= \cfrac{\sin \theta}{\cos \theta}} \Rightarrow 2\sin(60^\circ-\theta)= \sqrt 3\cos \theta-\sin \theta =\cfrac{\sin \theta}{\cos \theta} \\\Rightarrow \sin \theta= \cfrac{\sqrt 3\cos^2\theta}{\cos\theta+1} \Rightarrow \left(  \cfrac{\sqrt 3\cos^2\theta}{\cos\theta+1} \right)^2+ \cos^2\theta =1 \Rightarrow 4\cos^4\theta +2\cos^3\theta -2\cos\theta -1=0\\ \Rightarrow (2\cos^3\theta-1)(2\cos\theta+1)=0 \Rightarrow \cos\theta =\sqrt[3]{1\over 2} \Rightarrow \overline{PC} = {a\over \cos\theta } =\bbox[red, 2pt]{ a\sqrt[3]{2}}$$

解答：

$$本題相當於求圖形y=1-x^4與x軸所圍面積在區間[a,b]內的最大值\\ 由於圖形對稱y軸，因此最大面積=\int_{-1/2}^{1/2} 1-x^4\,dx = \left.\left[x-{1\over 5}x^5 \right] \right|_{-1/2}^{1/2} \\=2({1\over 2}-{1\over 160})=1-{1\over 80} =\bbox[red, 2pt]{79 \over 80}$$

解答：

$$\cases{\overline{AB}=8\\ \overline{AD}=4} \Rightarrow \overline{BD} =\sqrt{8^2+4^2} = 4\sqrt 5 \Rightarrow \overline{MB}= 4\sqrt 5\div 2= 2\sqrt 5;\\ \triangle EMB \sim \triangle BCD(AAA) \Rightarrow \cfrac{\overline{EM} }{\overline{MB}} =\cfrac{\overline{BC} }{\overline{CD}} \Rightarrow \cfrac{\overline{EM} }{2\sqrt 5} =\cfrac{4 }{8} \Rightarrow \overline{EM} =\sqrt 5 \Rightarrow \overline{BE}=5\\ 在延長\overleftrightarrow{MF}線上找一點G，符合\overline{FG} \bot \overline{GC}，如上圖；假設\cases{\overline{FG}=a\\ \overline{CG}=b} \Rightarrow \cases{a^2+b^2 = 3^2 \\ (a+\sqrt 5)^2 +b^2 = (2\sqrt 5)^2} \\ \Rightarrow \cases{a=3\sqrt 5/5\\ b=6\sqrt 5/5} \Rightarrow \cases{M(0,0,0)\\ B(2\sqrt 5,0,0)\\ D(0,0, 2\sqrt 5)\\ F(0,-\sqrt 5,0) \\ C(b,-(a+\sqrt 5),0)= (6\sqrt 5/5,-8\sqrt 5/5,0)} \Rightarrow \cases{\overrightarrow{FC} = (6\sqrt 5/5,-3\sqrt 5/5,0)\\\overrightarrow{FD} =(0, \sqrt 5, 2\sqrt 5)} \\ \Rightarrow \cos \angle CFD = \overrightarrow{FC}\cdot \overrightarrow{FD}/|\overrightarrow{FC}| | \overrightarrow{FD}| =\cfrac{-3}{3\cdot 5} =\bbox[red, 2pt]{-{1\over 5}}$$

解答： $$\begin{vmatrix} x-\sin \theta & \cos \theta \\ -\cos\theta & x-\sin \theta\end{vmatrix} =0 \Rightarrow (x-\sin\theta)^2+\cos^ \theta=0 \Rightarrow x^2-2\sin \theta x+1=0 \\ \Rightarrow x=\sin\theta \pm \sqrt{-\cos^2\theta} =\sin \theta \pm i\cos\theta = \cos({\pi\over 2}-\theta) \pm i\sin ({\pi\over 2}-\theta)\\ 令\cases{\alpha = \cos({\pi\over 2}-\theta) + i\sin ({\pi\over 2}-\theta)\\ \beta =\cos({\pi\over 2}-\theta) - i\sin ({\pi\over 2}-\theta)} \Rightarrow \cases{\alpha^n = \cos({n\pi\over 2}-n\theta)+ i\sin ({n\pi\over 2}-n\theta) \\ \beta^n = \cos({n\pi\over 2}-n\theta) - i\sin ({n\pi\over 2}-n\theta)} \\ \Rightarrow \alpha^n +\beta^n = 2\cos({n\pi\over 2}-n\theta) =\bbox[red, 2pt]{2\cos(n\theta-{n\pi\over 2})}$$

解答： $$x=2+i \Rightarrow x^2-4x+5=0 \Rightarrow g(x)=x^2-4x+5 是f(x)的因式\\ \Rightarrow f(x)=g(x)(x^2+2kx+k+2)+ (-6k+p+8)x +(q-5k-10)\\ \Rightarrow \cases{-6k+p+8=0 \\ q-5k-10=0} \Rightarrow \cases{p=6k-8\\ q=5k+10} \Rightarrow pq = 30(k +{1\over 3})^2-{250 \over 3} \cdots(1)\\ x^2+2kx+k+2=0有另二實根\Rightarrow 4k^2-4(k+2)\ge 0 \Rightarrow k^2-k-2\ge 0 \Rightarrow (k-2)(k+1)\ge 0 \\ \Rightarrow \cases{k\ge 2\\ k \le -1}，要使(1)有最小值，故取k=-1 代回(1) \Rightarrow pq = 30\cdot (-{2\over 3})^2-{250\over 3} =\bbox[red, 2pt]{-70}$$

解答：

$$在\overline{AC}上找一點P，滿足\angle PBC=\angle C=\theta，並令\overline{PB} =\overline{PC}=a，則\overline{AP}=6-a\\ 因此\cos (B-C) =\cos \angle ABP = {4^2+a^2-(6-a)^2\over 2\cdot a\cdot 4} ={2\over 3} \Rightarrow a=3;\\ 令\overline{BC}=b \Rightarrow \cases{\cos \angle PBC = \cos \theta = {a^2+b^2-a^2\over 2ab} ={b^2\over 6b}\\[1ex] \cos ACB= \cos \theta = {b^2+6^2-4^2 \over 12b} ={20+b^2 \over 12b}} \Rightarrow {b^2\over 6b} ={20+b^2\over 12b} \Rightarrow b^2=20 \Rightarrow b=\bbox[red, 2pt]{2\sqrt 5}$$

解答： $$\cases{Q(\cos\theta, \sin\theta)\\ A(-2,0)} \Rightarrow P= \begin{bmatrix} \cos 90^\circ & -\sin 90^\circ\\ \sin 90^\circ & \cos 90^\circ\end{bmatrix} \begin{bmatrix} -2-\cos\theta \\ 0-\sin \theta\end{bmatrix} +\begin{bmatrix}  \cos\theta \\  \sin \theta\end{bmatrix}=\begin{bmatrix} \sin\theta +\cos\theta \\ -2-\cos\theta +\sin \theta\end{bmatrix} \\ \Rightarrow \cases{x=\sin\theta +\cos\theta  \\ y= -2-\cos\theta +\sin \theta} \Rightarrow \cases{\sin \theta =(x+y+2)/2\\ \cos\theta = (x-y-2)/2} \\\Rightarrow \sin^2\theta +\cos^2\theta  = {1\over 4}((x+y+2)^2 +(x-y-2)^2 )=1 \Rightarrow \bbox[red, 2pt]{x^2+(y+2)^2 =2}$$

解答：

$${x^2\over 4}-{y^2\over 5}=1 \Rightarrow \cases{a=2\\ b=\sqrt 5} \Rightarrow c=3 \Rightarrow 兩焦點\cases{F_1(-3,0)\\ F_2(3,0) } \\\Rightarrow \cases{\cases{ \overrightarrow{F_1F_2}=(6,0) \Rightarrow \vec u_1= \overrightarrow{F_1F_2}/| \overrightarrow{F_1F_2}| =(1,0)\\ \overrightarrow{F_1A}=(-1,\sqrt{15}) \Rightarrow \vec u_2= \overrightarrow{F_1A} /|\overrightarrow{F_1A}|=-1/4, \sqrt{15}/4}   \\[1ex] \cases{\overrightarrow{F_2F_1} =(-6,0) \Rightarrow \vec v_1= \overrightarrow{F_2F_1}/|\overrightarrow{F_2F_1}| =(-1,0)\\ \overrightarrow{F_2A}= (-7, \sqrt{15}) \Rightarrow \vec v_2= \overrightarrow{F_2A}/|\overrightarrow{F_2A}| = (-7/8,\sqrt{15}/8)}   } \\ \Rightarrow \cases{ \vec n_1= \vec u_1+\vec u_2=(3/4,\sqrt{15}/4) \Rightarrow \angle AF_1F_2 角平分線 L_1: \sqrt{15}x +3\sqrt{15} =3y \\\vec n_2= \vec v_1+\vec v_2= (-15/8, \sqrt{15}/8) \Rightarrow \angle AF_2F_1 角平分線L_2: \sqrt{15}x-3\sqrt{15}=15y}\\ \Rightarrow 內切圓圓心C=L_1\cap L_2 =(-2,\sqrt{15}/3) \Rightarrow x軸切點坐標\bbox[red, 2pt]{(-2,0)}\\\bbox[blue,2pt]{註}:題目的\triangle PF_1F_2應該是\triangle AF_1F_2$$

解答：

$$y=-x^2-3x+6 \Rightarrow x^2+3x-6+y=0 \Rightarrow x={-3\pm \sqrt{33-4y}\over 2} \\ \Rightarrow R_1旋轉體積= \pi \int_6^{33/4} \left( {-3-\sqrt{33-4y}\over 2}-2\right)^2 -\left( {-3+\sqrt{33-4y}\over 2}-2\right)^2 \,dy \\= \pi \int_6^{33/4} \left( {-7-\sqrt{33-4y}\over 2} \right)^2 -\left( {-7+\sqrt{33-4y}\over 2} \right)^2 \,dy = 7\pi \int_6^{33/4} \sqrt{33-4y} \,dy\\ ={7\over 4}\pi \int_0^9 \sqrt u \,du={63\over 2}\pi\\ R_2旋轉體積 = \pi \int_2^6 (3-y-2)^2-\left( {-3+\sqrt{33-4y}\over 2}-2\right)^2\,dy \\=\pi \int_2^6(1-y)^2- \left( {-7+\sqrt{33-4y}\over 2} \right)^2\,dy= {65\over 2}\pi\\ 因此欲求之旋轉體積 ={63\over 2}\pi +{65\over 2}\pi =\bbox[red, 2pt]{64\pi}$$

解答： $$假設{2b+c\over a} ={2c+a\over b} ={2a+b \over c} = k \Rightarrow \cases{2b+c=ak\\ 2c+a=bk \\ 2a+b=ck}，三式相加可得:3(a+b+c)= k(a+b+c)\\ 若a+b+c=0 \Rightarrow {(a+b)(b+c )( c+a)\over abc} ={(-c)(-a)(-b)\over abc} =-1;\\ 若k=3 \Rightarrow \cases{2b+c=3a\\ 2c+a=3b \\ 2a+b=3c} \Rightarrow \cases{3a-2b-c=0\\ a-3b+2c=0 \\ 2a+b-3c=0} \Rightarrow a=b=c=1 為其中一解\\ \Rightarrow {(a+b)(b+c )( c+a)\over abc}={2\cdot 2\cdot 2\over 1\cdot 1\cdot 1}=8\\ \Rightarrow {(a+b)(b+c )( c+a)\over abc}=\bbox[red, 2pt]{-1或8}$$

解答：

考慮不能連成一直線的情形，也就是四個角一定要有奇數；

情況(A)恰有一個角是奇數，只有四種情形；

情況(B)恰有三個角是奇數，只有四種情形；

情況(C)側邊兩個角是奇數，有3X4=12種情形；

情況(D)對角兩個角是奇數，有4X2=8種情形；

以上共有28種情形是無法連成一直線的，因此機率=$(C^9_5-28)/C^9_5= \bbox[red,2pt]{7\over 9}$

解答：

$$一個長方形P_1P_4P_5P_8 +兩個梯形P_1P_2P_3P_4 =4\times (4+2\times {3\over \sqrt 2}) +(4+4+2\times {3\over \sqrt 2})\times {3\over \sqrt 2} \\ =16+12\sqrt{2}+ 12\sqrt 2+9 = \bbox[red, 2pt]{25+24\sqrt 2}$$

解答：

$$A_k = \cfrac{{2k-1\over 99}+ {2k-2\over 99}}{2} =\cfrac{4k-3}{198} \Rightarrow a_n= \sum_{k=1}^n A_k= \sum_{k=1}^n \cfrac{4k-3}{198}= {4\over 198}\sum_{k=1}^n k-{3\over 198}\sum_{k=1}^n 1\\ ={n(n+1)\over 99}-{n\over 66} ={n^2\over 99} -{n\over 198}\\ 令S=\sum_{k=1}^{50} a_k \Rightarrow \cases{x= (S-a_{50})(S-a_1) =S^2-(a_1+a_{50})S+a_1a_{50}\\ y=S(S-a_1-a_{50}) =S^2-(a_1+a_{50})S} \\ \Rightarrow x-y=a_1a_{50} = \left({1\over 99}-{1\over 198}\right) \left({50^2\over 99}-{50 \over 198}\right) = \bbox[red, 2pt]{25\over 198}$$

解答： $$B_n在直線y=x-2上 \Rightarrow B_n(x_n,y_n) =(n+1,n-1), n\in \mathbb{N} \\ 因此{1\over \sqrt {x_1}+\sqrt {y_1}} ={1\over \sqrt{n+1}+\sqrt{n-1}} ={1\over 2}(\sqrt{n+1} -\sqrt{n-1})\\\Rightarrow  \lim_{n\to \infty} {1\over \sqrt n}\left( {1\over \sqrt {x_1}+\sqrt {y_1}} +\cdots + {1\over \sqrt {x_n}+\sqrt {y_n}}\right)\\ =\lim_{n\to \infty} {1\over 2\sqrt n} ( (\sqrt 2-\sqrt 0)+(\sqrt 3-\sqrt 1)+ (\sqrt 4-\sqrt 2)+\cdots+\sqrt{n+1}-\sqrt(n-1)) \\=\lim_{n\to \infty} {1\over 2\sqrt n} (\sqrt{n+1} +\sqrt n-1) =\bbox[red, 2pt]{1}$$

二、 計算題 (每題 8 分，共 40 分，請寫出詳細計算過程)

解答：

(1) $$\cases{ L:2x-y=0 \\  M:2x+3y=6} \Rightarrow  A \in L\cap M \Rightarrow A(3/4,3/2)  \Rightarrow \begin{bmatrix} 7 & b \\ c & d\end{bmatrix}\begin{bmatrix} 3/4\\ 3/2\end{bmatrix} =\begin{bmatrix} 3/4\\ 3/2\end{bmatrix} \Rightarrow \cases{b=-3 \\c+2d =2\cdots(1)}\\ 在M上任找一點(0,2) \Rightarrow \begin{bmatrix} 7 & -3 \\ c & d\end{bmatrix} \begin{bmatrix} 0\\  2 \end{bmatrix} =\begin{bmatrix} -6\\  2d \end{bmatrix} \Rightarrow (-6,2d)也在M上\Rightarrow -12+6d=6 \Rightarrow d=3 \\將d=3代入(1) \Rightarrow c+6=2 \Rightarrow  c=-4 \Rightarrow P=\bbox[red,2pt]{ \begin{bmatrix} 7 & -3 \\ -4 &  3 \end{bmatrix}}$$ (2) $$L、M與x軸所圍三角形的頂點\cases{A(3/4,3/2)\\ B(3,0)\\ O(0,0)} \Rightarrow \triangle ABO = {1\over 2}\cdot 3\cdot {3\over 2}= {9\over 4};\\又\det(P)=21-12=9 \Rightarrow 變換後三角形面積={9\over 4}\cdot 9= \bbox[red, 2pt]{81\over 4}$$

解答：

$$f(x) =\sqrt{x^4-9x^2-6x+34} -\sqrt{x^4-3x^2+4} \\= \sqrt{(x^4-10x^2 +25)+(x^2-6x+9)} -\sqrt{(x^4-4x^2+4) +x^2}\\ = \sqrt{(x^2-5)^2 +(x-3)^2} -\sqrt{(x^2-2)^2 +(x-0)^2} \\ = \overline{PA}-\overline{PB}，其中\cases{P\in \Gamma:x=y^2\\ A(5,3)\\ B(2,0)} \\ 當P、A、B成一直線時，f(x)有最大值，即M=\overline{AB} = \sqrt{(5-2)^2+3^2} =3\sqrt 2\\此時P=\overleftrightarrow{AB} \cap \Gamma，而\overleftrightarrow{AB}:y=x-2代入\Gamma \Rightarrow x=(x-2)^2 \Rightarrow (x-4)(x-1)=0\\ 取近B點，即x=1 \Rightarrow y=\pm 1 \Rightarrow P(1,-1) =(t^2,t) \Rightarrow t=-1 \Rightarrow (t,M)= \bbox[red, 2pt]{(-1,3\sqrt 2)}$$

解答： $${a\over \sin A} ={b\over \sin B} ={c\over \sin C} =2R \Rightarrow \cases{\sin A=a/2R\\ \sin B=b/2R\\ \sin C=c/2R} \\\Rightarrow {\sin A\over \sin B\sin C}+  {\sin B\over \sin C\sin A} + {\sin C\over \sin A\sin B} = {a/2R\over bc/4R^2} + {b/2R\over ac/4R^2} +{c/2R\over ab/4R^2} = 2R\left( {a\over bc} +{b\over ac} +{c\over ab}\right) \\ ={2R\over abc}(a^2+b^2+ c^2) ={1\over 2\triangle }(a^2+b^2+ c^2) (\because R={abc\over 4\triangle}) \\ \ge {1\over 2\triangle }\cdot 4\sqrt 3\triangle =\bbox[red, 2pt]{2\sqrt 3}(\href{https://frankliou.wordpress.com/2012/04/09/%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8weitzenbocks-%E4%B8%8D%E7%AD%89%E5%BC%8F/}{外森比克(Weitzenböck’s)不等式}:a^2+b^2+c^2 \ge 4\sqrt 3\triangle)$$

解答： $$y=f(x)=x^2+ax+4通過(0,4)且凹向上，因此只需考慮f(0)至f(1)遞減的情形，\\ 也就是f(1) \ge 0 \Rightarrow a+5\ge 0 \Rightarrow \bbox[red, 2pt]{a\ge -5}$$

解答： $$顯然(0,0,0)為其中一組解；\\假設有不為0的解(x,y,z)，且三數互質，滿足7x^2+6y^2=5z^2 \Rightarrow 7x^2+6y^2 = 0 \mod 5 \\\Rightarrow 2x^2+y^2 = 0 \mod 5  \Rightarrow \cases{x^2 = 0 \mod 5\\ y^2 =0 \mod 5}\Rightarrow \cases{x = 0 \mod 5\\ y =0 \mod 5} \\\Rightarrow x,y,z皆為5的倍數，不互質，與假設矛盾，故只有一解(0,0,0);\\ 若三數有公因數k \Rightarrow \cases{x=ka \\ y=kb\\ z=kc} \Rightarrow 7\cdot k^2a^2 +6\cdot k^2b^2 = 5\cdot k^2c^2 \Rightarrow 7a^2+6b^2=5c^2，證明同上；\\ 故只有一解\bbox[red,2pt]{(0,0,0)}$$

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解題僅供參考，其他教甄試題及詳解