108年松山工農教甄-數學詳解

臺北市立松山高級工農職業學校108學年度正式教師甄選

一、填充題

解答：$$30=2\times 3\times 5 =(1+1)\times (2+1)\times (4+1)，取N=2^4\cdot 3^2\cdot 5=\bbox[red,2pt]{720}$$

解答：$$f(x)在x=2有極大值5 \Rightarrow f'(2)=0；\\在切點(5,f(5))的切線斜率為-7 \Rightarrow f'(5)=-7；\\因此\int_2^5 f''(x)\;dx = \left. f'(x) \right|_2^5= f'(5)-f'(2)=-7-0= \bbox[red,2pt]{-7}$$

解答：$$\cos \angle DAF ={\overrightarrow{AD}\cdot \overrightarrow{AF} \over \overline{AD}\cdot \overline{AF}} ={11\over 25}；\\又\cos \angle CBE= \cos \angle DAF ={11\over 25}={ \overline{BC}^2+ \overline{BE}^2-\overline{CE}^2 \over 2\cdot \overline{BE}\cdot \overline{BE}} ={50-\overline{CE}^2 \over 50} \\ \Rightarrow \overline{CE}^2 =28 \Rightarrow \cos \angle CAE ={\overline{AC}^2+\overline{AE}^2 -\overline{CE}^2 \over 2\cdot \overline{AC}\cdot \overline{AE}} ={50+50-28\over 2\cdot 5\sqrt 2\cdot 5\sqrt 2} ={72\over 100} \\\Rightarrow \overrightarrow{AC}\cdot \overrightarrow{AE} =\overline{AC} \cdot \overline{AE} \cdot \cos \angle CAE =5\sqrt 2\cdot 5\sqrt 2\cdot {72\over 100} =\bbox[red,2pt]{36}$$

解答：

$$與x+y=1、x軸、y軸同時相切的圓，共有四個，假設圓心分別為A、B、C、D\\，其中\cases{A,D在x=y上且圓A較小\\ B,C在x=-y上且兩圓有相同半徑}\\ 令P\in x=y \Rightarrow P(t,t),t\in \mathbb{R} \Rightarrow d(P,x+y=1)=t \Rightarrow {|2t-1|\over \sqrt 2}=t \Rightarrow 2t^2-4t+1=0\\ \Rightarrow t=2\pm \sqrt 2 \Rightarrow \cases{A((2-\sqrt 2)/2,(2-\sqrt 2)/2) \\ D((2+\sqrt 2)/2,(2+\sqrt 2)/2)}；\\同理，令Q\in x=-y \Rightarrow Q(t,-t) \Rightarrow d(Q,x+y=1)=|t|\\ \Rightarrow {1\over \sqrt 2}=|t| \Rightarrow t=\pm {1\over \sqrt 2} \Rightarrow \cases{B(-1/\sqrt 2,1/\sqrt 2)\\ C(1/\sqrt 2,-1/\sqrt 2)}\Rightarrow 圓半徑分別為\bbox[red,2pt]{{2\pm \sqrt 2\over 2},{\sqrt 2\over 2}}$$

解答：$$令z=x+iy, x,y\in \mathbb{R} \Rightarrow {z^2-2\over z^2+2} =1-{4\over z^2+2} =i \\\Rightarrow {4\over (x+iy)^2+2}= {4\over x^2-y^2+2 +2xyi}=1-i\\ \Rightarrow ((x^2-y^2+2)+2xyi)(1-i)=4\\ \Rightarrow (x^2-y^2+2+2xy)-(x^2-y^2+2-2xy)i=4 \Rightarrow \cases{x^2-y^2+2+2xy=4\\ x^2-y^2+2-2xy=0} \\ \Rightarrow \cases{xy=1\\ x^2-y^2=0} \Rightarrow (x,y)=(1,1),(-1,-1) \Rightarrow z=\bbox[red,2pt]{1+i或-1-i}$$

解答：$$\begin{array}{}k & \sqrt[3]k &小計 \\\hline 1-7& 1 & 7\\ 8-26& 2& 2\times 19=38\\ 27-63& 3 & 3\times 37=111\\ 64-124& 4 & 4\times 61=244\\ 125& 5 & 5\\\hline\end{array} \\ \Rightarrow 7+38+11+ 244+5= \bbox[red,2pt]{405}$$

解答：$$本題\bbox[blue,2pt]{送分}$$

解答：

$$\cases{\tan  A={3/4-1/4\over 1+3/16}={8\over 19} \\ \tan B={1/4-(-3/4)\over 1-3/16} ={16\over 13}}；令\cases{\overline{BD} \bot \overline{AC}且\overline{BD}=h\\ \overline{AD}=a}，則\cases{{h\over a}={8\over 19}\\ {h\over 51-a}={16\over 13}} \\\Rightarrow h= {8\over 19}a={16\over 13}(51-a) \Rightarrow a=38 \Rightarrow h=16 \Rightarrow \triangle ABC面積={1\over 2}\cdot 51\cdot 16 =\bbox[red,2pt]{408}$$

解答：$$\begin{array}{} 樣式 & 數量 & 累計\\\hline 3個6& 1 & 1\\ 2個6 & 3\times 5=15& 16\\ 1個6& 3\times 5^2=75 & 91\\\hdashline 231 & 3!=6 & 97\\ 232& 3& 100\\ 233 & 3& 103\\ 234& 3!=6& 109\\ 235& 3!=6& 115\\\hdashline 341 & 3!=6 & 121\\ 343& 3 & 124\\ 344 & 3 & 127 \\ 345 & 3!=6& 133\\\hline \end{array}\\ \Rightarrow 機率為{133\over 6^3} =\bbox[red,2pt]{133\over 216}$$

解答：$$\cases{y\le -{1\over n}x+n \\x,y \ge 0},n\in \mathbb{N},x,y \in \mathbb{Z};\\ \begin{array}{}y & x & 數量\\\hline 0 & 0-n^2 & n\cdot n+1\\ 1 & 0-n(n-1) & n(n-1)+1\\ 2 & 0-n(n-2) &  n(n-2)+1\\ \cdots & \cdots & \cdots\\ n-1& 0-n & n\cdot 1+1\\ n & 0 & n\cdot 0+1\\\hline\end{array}\\ 合計=n(n+(n-1)+(n-2)+\cdots + 1+0)+ n+1=n\cdot {n(n+1)\over 2}+n+1\\ ={n^3+ n^2\over 2}+n + 1=\bbox[red,2pt]{n^3+n^2 +2n +2\over 2}$$

解答：

$$y=x^{3/2}+1 \Rightarrow x=(y-1)^{2/3} \Rightarrow 繞y軸旋轉體積=\int_1^9 x^2\pi\;dy =\int_1^9 x^2\pi\;dy \\=\int_1^9 (y-1)^{4/3}\pi\;dy = \left. \left[{3\over 7}(y-1)^{7/3} \right] \right|_1^9 = \bbox[red,2pt]{{384\over 7}\pi}$$

解答：$$令g(x)=\sqrt[3]{ax+b}-2 \Rightarrow \displaystyle \lim_{x\to 1} {g(x)\over x-1}=1 \Rightarrow \cases{g(1)=0\\ g'(1)=1} \Rightarrow \cases{\sqrt[3]{a+b}=2\Rightarrow a+b=8\cdots(1)\\ {a\over 3}\cdot {1\over \sqrt[3]{(a+b)^2}}=1\cdots(2)} \\ 將(1)代入(2)\Rightarrow a=12 \Rightarrow b=-4 \Rightarrow (a,b)=\bbox[red,2pt]{(12,-4)}$$

解答：$$\cases{{1\over x}+ {1\over y}+{1\over z}=1 \Rightarrow yz+xz+ xy= xyz\cdots(1)\\ x^2+y^2+z^2 =3/2\cdots(2) \\ x^3+y^3 +z^3=1 \cdots(3)} \\ 因此 \cases{(x+y+z)^2 =x^2 +y^2 +z^2 +2(xy+yz +zx)\\ x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^z+z^2 -(xy+ yz+zx)}\\ \Rightarrow \cases{(x+y+z)^2 =3/2+ 2xyz \\ 1-3xyz=(x+y +z) (3/2-xyz)} \Rightarrow \left({1-3xyz\over 3/2-xyz} \right)^2 ={3\over 2}+2xyz \\\Rightarrow ({1-3a\over 3/2-a})^2={3\over 2}+2a,其中a=xyz \Rightarrow 16a^3-108a^2+48a+19=0 \Rightarrow (4a+1)(4a^2-28a+19)=0\\ xyz=a=-{1\over 4}(\because xyz\in \mathbb{Z}) \Rightarrow x+y+z ={1-3xyz\over 3/2-xyz} ={7/4\over 7/4} =\bbox[red,2pt]{1}$$

解答：

$$A=\begin{bmatrix} 4 & -3\\ 3 & 4\end{bmatrix} =5\begin{bmatrix} 4/5 & -3/5\\ 3/5 & 4/5\end{bmatrix} =5\begin{bmatrix} \cos \theta & -\sin \theta\\ \sin \theta & \cos \theta\end{bmatrix} ，其中\cases{\cos\theta =4/5\\ \sin \theta = 3/5} \\\Rightarrow \cases{\angle P_2OP_1= \angle P_3OP_2 = \theta\\ \overline{OP_1}=a\\ \overline{OP_2}=5\cdot \overline{OP_1}= 5a\\ \overline{OP_3}= 5\cdot \overline{OP_2}=25a} \Rightarrow \triangle P_1P_2P_3 =\triangle OP_1P_2 +\triangle OP_2P_3- \triangle OP_1P_3 \\ ={1\over 2}\left(\overline{OP_1}\cdot \overline{OP_2}\sin \theta+ \overline{OP_2}\cdot \overline{OP_3}\sin \theta -\overline{OP_1}\cdot \overline{OP_3}\sin 2\theta \right)\\ ={1\over 2}\left(5a^2\cdot{3\over 5}+125a^2\cdot {3\over 5}-25a^2\cdot 2\cdot {3\over 5}\cdot {4\over 5} \right)={1\over 2}\cdot 54a^2= \bbox[red,2pt]{27a^2}$$

解答：$${1\over 1!}+{2\over 3!}+{3\over 5!}+{4\over 7!}+\cdots =\sum_{k=0}^\infty {k+1\over (2k+1)!} ={1\over 2}\sum_{k=0}^\infty {(2k+1)+1\over (2k+1)!}\\ ={1\over 2}\sum_{k=0}^\infty \left({1\over (2k)!}+{ 1\over (2k+1)!} \right)=\bbox[red,2pt]{ {1\over 2}e}$$

解答：

$$\cases{A(0,0,6)\\ B(0,0,20)\\ P(x,y,0)} \Rightarrow \cases{\overline{AB}=14\\ \overline{AP}=\sqrt{x^2+y^2+36} =\sqrt{a+ 36}\\ \overline{BP}= \sqrt{x^2+y^2+400}=\sqrt{a+400}}，其中a=x^2+y^2\\ \Rightarrow \cos \angle APB= {a+36+a+400-14^2\over 2\sqrt{a+36} \cdot \sqrt{a+400}} \le \cos 30^\circ ={\sqrt 3\over 2} \Rightarrow a^2-348a+14400 \le 0\\ \Rightarrow (a-48)(a-300)\le 0 \Rightarrow 48\le a\le 300\\ 本題相當於求聯立方程式\cases{48\le x^2+y^2\le 300\\ 0\le x\le 15\\ 0\le y\le 15}所圍面積=2\left({75\over 2}\sqrt 3-4\pi\right)+{1\over 12}(300-48)\pi\\ =\bbox[red,2pt]{75\sqrt 3+13\pi}$$

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