115年鳳山高中教甄-數學詳解

 115 學年度國立鳳山高級中學教師甄選

一、 填充題，每題 6 分，共 72 分

解答:$$m^3+8n^3+18-27= m^3+(2n)^3+(-3)^3-3\cdot m\cdot (2n)\cdot (-3) \\= (m+2n-3)(m^2+4n^2+9-2mn+3m+6n)=0 \Rightarrow m+2n=3 \\ 算幾不等式{(m/3)+(m/3)+(m/3)+2n\over 4} \ge \sqrt[4]{({m\over 3})^3\cdot (2n)} \Rightarrow {3\over 4} \ge  \sqrt[4]{{2\over 27}m^3n}\\ \Rightarrow {81\over 256}\ge {{2\over 27}m^3n}  \Rightarrow m^3n\le \bbox[red, 2pt]{2187\over 512}$$

解答:$$A^{-1}不存在\Rightarrow \det(A)=ad-bc =0 \Rightarrow ad=bc=k:\cases{k=0: 49種 \\k=4:1種\\k=2:4種 \\ k=1:4種\\k=-1: 4種\\k=-2: 4種} \\ \Rightarrow \det(A)=0有49+1+4\times 4=66種\Rightarrow 機率為{66\over 4^4} = \bbox[red, 2pt]{33\over 128}$$

解答:$$A,D,E在xy平面上 \Rightarrow \triangle BCF \parallel \triangle ADE \Rightarrow B(0,y,z)到y軸的距離=|z|\\ 稜長為a的正八面體，其平行面之間的距離={\sqrt 6\over 3}a ={\sqrt 6\over 3}\cdot2= \bbox[red, 2pt]{2\sqrt 6\over 3}$$

解答:

$$\Gamma:{x^2\over 36}+{y^2\over 27} =1 \Rightarrow \cases{a=6\\b=3\sqrt 3} \Rightarrow c=3 \Rightarrow e={c\over a}={1\over 2} \Rightarrow 右準線:x={a^2\over c}=12\\ \Rightarrow \cases{右焦點F(3,0)\\ L:x=12為準線\\ A_0(6,0)為右頂點\\p=12-3=9} \Rightarrow {\overline{A_kF} \over d_k}=e  \Rightarrow {1\over d_k}= {e\over \overline{A_kF}} \\ 對任一焦弦\overline{PQ} \Rightarrow {1\over \overline{FP}}+{1\over  \overline{FQ}} ={2\over ep} ,依題意：A_k, F, A_{k+6} 在一直線上，即\overline{A_kA_{k+6}} 為焦弦\\ \Rightarrow {1\over d_k}+{1\over d_{k+6}} = e \left( {1\over \overline{A_kF}}+{1\over \overline{A_{k+6}F}} \right) =e\times {2\over ep}={2\over p} ={2\over 9} \\ 12個點剛好湊成6對，因此全部加總為6\times {2\over 9} = \bbox[red, 2pt]{4\over 3}$$

解答:

$$L:{x-5\over 3}={y-5\over 2} ={z-4\over -2} \Rightarrow \cases{L方向向量\vec u=(3,2,-2) \\ L通過P(5,5,4)} \\由題意可知:E_1及E_2相交於一直線L',L'經過原點O(0,0,0)且方向向量為\vec u \\ \Rightarrow d(L,L')= {\overrightarrow{OP} \times \vec u\over |\vec u|} ={\sqrt{833} \over \sqrt{17}} =7 \\ 假設E_1與E_2夾角為2\theta \Rightarrow \sin \theta={2\over 7} \Rightarrow \cos 2\theta=1-2\sin^2\theta ={41\over 49} \Rightarrow \sin 2\theta= \bbox[red, 2pt]{12\sqrt 5\over 49}$$

解答:$$f(x)+2f({x-3\over x+1})=x \Rightarrow f({x-3\over x+1}) +2f({{x-3\over x+1}-3\over {x-3\over x+1}+1})={x-3\over x+1} \Rightarrow f({x-3\over x+1})+2f({-x-3\over x-1})={x-3\over x+1} \\ 再作一遍: f({-x-3\over x-1}) +2f({{-x-3\over x-1}-3\over {-x-3\over x-1}+1}) ={-x-3\over x-1} \Rightarrow f({-x-3\over x-1}) +2f(x)={-x-3\over x-1} \\ 取\cases{A=f(x) \\B=f({x-3\over x+1}) \\C=f({-x-3\over x-1})} \Rightarrow \cases{A+2B=x\\ B+2C={x-3\over x+1} \\C+2A={-x-3\over x-1}} \Rightarrow A= \bbox[red, 2pt]{f(x)={x^3-6x^2-9x-18 \over 9(x^2-1)}}$$

解答:$$\cases{H_0: p=0.1\\ H_1: p\lt 0.1\\ \alpha=0.1} \quad;  X\sim Geo(p=0.1) \Rightarrow \cases{P(X=k) =(1-p)^{k-1}p\\ P(X\le k)=1-(1-p)^k} \\ \Rightarrow P(X\ge k)=1-P(X\le k)=(1-p)^k \le \alpha \Rightarrow 0.9^{k-1} \le 0.1 \Rightarrow (k-1)\log 0.9\le \log 0.1 \\ \Rightarrow (k-1)(2\log 3-1) \le -1 \Rightarrow (k-1)(2\cdot 0.4771-1) \le -1 \Rightarrow k-1\ge {1\over 0.0458} \approx 21.8 \\ \Rightarrow k\ge 22.8 \Rightarrow k\ge 23 \Rightarrow 拒絕域: \bbox[red, 2pt]{\{ X\in \mathbb N \mid X\ge 23\}}$$

解答:$$\cos^2 x+\cos^2 y={1\over 2}(1+\cos 2x) +{1\over 2}(\cos 2y+1) =1+{1\over 2}(\cos 2x+\cos 2y) \\=1+\cos(x+y) \cos(x-y) =1+\cos(x+y)\cos 60^\circ =1+{1\over 2}\cos(x+y) \Rightarrow \cases{最大值1+1/2=3/2\\ 最小值1-1/2=1/2} \\ \Rightarrow (M,m)= \bbox[red, 2pt]{\left( {3\over 2},{1\over 2} \right)}$$

解答:$$x^2-(3a+1)x-2a+2=0有虛根z \Rightarrow z^2-(3a+1)z-2a+2 \Rightarrow z^2=(3a+1)z+2a-2 \\ \Rightarrow z^3 =(3a+1)z^2+(2a-2)z =(3a+1)[(3a+1)z+ 2a-2]+(2a-2)z \\ \Rightarrow z^3=[(3a+1)^2+2a-2]z+ (3a+1)(2a-2)為實數 \Rightarrow (3a+1)^2+2a-2=0 \\ \Rightarrow 9a^2+8a-1=0 \Rightarrow (9a-1)(a+1)=0 \Rightarrow a=\bbox[red, 2pt]{{1\over 9}或-1}$$

解答:$$\cases{A(2,4) \\B(8,13) \\C(1,1) \\P(x,y)} \Rightarrow \cases{ \overrightarrow{AP} =(x-2,y-4) \\\overrightarrow{BP}=(x-8,y-13) \\\overrightarrow{CP} =(x-1,y-1) }\\ \Rightarrow \overrightarrow{AP} \cdot \overrightarrow{BP} = |\overrightarrow{CP}|^2 \Rightarrow (x-2)(x-8)+ (y-4)(y-13) =(x-1)^2+ (y-1)^2 \\ \Rightarrow 8x+15y=66 \Rightarrow 取直線L:8x+15y-66=0 \\ |2\overrightarrow{AP} +\overrightarrow{BP}| =|2(P-A)+P-B|=|3P-2A-B|=3 \left|P-{2A+B\over 3} \right|=3 \left|P-{(4,8)+(8,13)\over 3} \right| \\=3|P-(4,7)|最小值=3\times d((4,7),L) =3\times {71\over 17} = \bbox[red, 2pt]{213\over 17}$$

解答:$$\vec u在\vec v上的正射影向量為\vec w, 則(\vec u-\vec w) \bot \vec w \Rightarrow (\vec u-\vec w) \cdot \vec w=0 \Rightarrow \vec u\cdot \vec w=|\vec w|^2\\ 假設\vec a=(x,y,z) \Rightarrow \cases{(x,y,z)\cdot (1,2,2)= |(1,2,2)|^2 \\ (x,y,z)\cdot (2,3,6)= |(2,3,6)|^2 \\ (x,y,z)\cdot (3,1,4)= |(3,1,4)|^2  } \Rightarrow \cases{x+2y+2z=9\\ 2x+3y+6z=49\\ 3x+y+4z=26} \\ \Rightarrow \cases{x=-7\\ y=-5\\z=13} \Rightarrow \vec a= \bbox[red, 2pt]{(-7,-5,13)}$$

解答:$$f(x)=x^3-{3\over 2}x^2-{1\over 4}x+1 \Rightarrow f(1-x)=(1-x)^3-{3\over 2}(1-x)^2-{1\over 4}(1-x)+1 \\= -x^3+{3\over 2}x^2+{1 \over 4}x+{1\over 4} \Rightarrow f(x)+f(1-x)={5\over 4} \\ \Rightarrow \sum_{k=1}^{2026}f \left( {k\over 2026} \right) = \left[ f({1\over 2026})+f({2025\over 2026}) \right] + \left[ f({2\over 2026})+f({2024\over 2026}) \right] +\cdots+\left[ f({1012\over 2026})+f({1014\over 2026}) \right] \\\qquad +f({1013\over 2026}) +f({2026\over 2026}) =1012\times {5\over 4}+ f({1\over 2})+f(1) =1265+{5\over 8}+{2\over 8} = \bbox[red, 2pt]{10127\over 8}$$

二、 計算證明題，共 28 分

解答:$$\alpha, \beta 是a\cos x+ b\sin x-c=0的兩根\Rightarrow \cases{a\cos \alpha+ b\sin \alpha=c\\ a\cos \beta+ b\sin \beta =c} \\ \Rightarrow \cases{兩式相加\Rightarrow a(\cos \alpha+ \cos \beta)+b(\sin \alpha+\sin \beta)=2c \\ 兩式相減\Rightarrow a(\cos \alpha-\cos \beta)+b(\sin \alpha-\sin \beta)=0} \\\Rightarrow \cases{a \displaystyle \left( 2\cos{\alpha+\beta \over 2} \cos {\alpha-\beta\over 2} \right)+ b\left( 2\sin {\alpha+\beta\over 2} \cos{\alpha-\beta\over 2}  \right)=2c \\ a \displaystyle \left( -2\sin{\alpha+\beta \over 2} \sin{\alpha-\beta\over 2} \right) +b \left( 2\cos{\alpha+\beta\over 2} \sin {\alpha-\beta\over 2} \right) =0}  \\ \Rightarrow \cases{\displaystyle a\cos{\alpha+\beta\over 2}+b\sin {\alpha+\beta\over 2} ={c\over \cos{\alpha-\beta\over 2}} \cdots(1)\\ \displaystyle -a\sin {\alpha+\beta\over 2} + b\cos{\alpha+\beta\over 2}=0 \cdots(2)}  \quad 式(1)^2+式(2)^2 \Rightarrow a^2+b^2={c^2\over \cos^2{\alpha-\beta\over 2}} \\ \Rightarrow \cos^2 {\alpha-\beta\over 2} ={c^2\over a^2+b^2}\quad \bbox[red, 2pt]{故得證}$$

解答:$$\textbf{方法一: 利用拋物線的標準式}\\ \qquad (x-h)^2=4c(y-k) \Rightarrow 將頂點(h=2,k=-1-c)代入可得(x-2)^2 =4c(y+1+c) \\\qquad 又通過(-1,3) \Rightarrow (-1-2)^2=4c(3+1+c) \Rightarrow 4c^2+16c-9=0 \Rightarrow (2c-1)(2c+9)=0 \\\qquad \Rightarrow \cases{c=1/2 \Rightarrow (x-2)^2=2(y+3/2) \\ c=-9/2 \Rightarrow (x-2)^2=-18(y-7/2)} \\\textbf{方法二: 利用拋物線的幾何定義} \\\qquad 假設準線y=L \Rightarrow \sqrt{(x-2)^2+(y-(-1))^2}=|y-L|, 將(-1,3)代入可得5=|3-L| \\ \qquad \Rightarrow \cases{L=-2 \Rightarrow \sqrt{(x-2)^2+(y+1)^2} =|y+2| \Rightarrow (x-2)^2=2(y+3/2) \\L=8 \Rightarrow \sqrt{(x-2)^2+(y+1)^2} =|y-8| \Rightarrow (x-2)^2=-18(y-7/2)} \\ 傾向使用\textbf{方法二}，讓用學生理解拋物線的定義，且計算較為簡單$$

解答:$$\textbf{(1) }b,c為實數\Rightarrow x^2+(a-2)x+(a^2-a-2)=0有實根\Rightarrow \Delta=(a-2)^2-4(a^2-a-2)\ge0\\ \qquad \Rightarrow a^2-4\le 0 \Rightarrow \bbox[red, 2pt]{-2\le a\le 2} \\ \textbf{(2) }\cases{兩根之和=b+c=2-a\\ 兩根之積=bc=a^2-a-2} \Rightarrow b^3+c^3=(b+c)^3-3bc(b+c)=(2-a)^3-3(a^2-a-2)(2-a) \\ \qquad \Rightarrow b^3+c^3=2a^3-3a^2 -12a+20 \Rightarrow (a+1)^3+b^3 +c^3=f(a)= (a+1)^3+ 2a^3-3a^2 -12a+20 \\\qquad \Rightarrow f(a)=3a^3-9a+21 \Rightarrow f'(a)=9a^2-9=0 \Rightarrow a=\pm 1 \Rightarrow \cases{f(1)=15\\f(-1)=27} \Rightarrow 最小值\bbox[red, 2pt]{15}$$

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解題僅供參考，其他教甄試題及詳解