臺北市立大安高級工業職業學校115學年度第1次教師甄選
一、 填充題(每題 3 分)
解答:$$N=[\sqrt[3]k] \Rightarrow N= \begin{cases} 1, &1\le k\lt 2^3 \\ 2,& 8\le k\lt 3^3 \\3,& 27\le k\lt 4^3 \\ 4, & 64\le k\lt 5^3 \\ 5, & k=5^3\end{cases} \Rightarrow \sum_{k=1}^{125} \left[\sqrt[3] k \right] =1\times 7+2\times 19+ 3\times 37+ 4\times 61+5= \bbox[red, 2pt]{405}$$
解答:$$y\le -{1\over n}x+n \Rightarrow x\le n^2-ny \Rightarrow \cases{y=0,1,2,\dots, n \\x=0,1,2,\dots,n^2-ny} \\ \Rightarrow a_n= \sum_{y=0}^n (n^2-ny+1) =\sum_{y=0}^n(n^2+1)-\sum_{y=0}^nny =(n^2+1)(n+1)-{n^2(n+1) \over 2} \\= \bbox[red, 2pt]{(n+1)(n^2+2) \over 2}$$
解答:$$\Gamma:y^2=x^3+tx^2+1 \Rightarrow 取f(x,y)=x^3+tx^2+1-y^2 \\ \Rightarrow \bullet類:\cases{f(A)=t-1\\ f(C)=4t\\ f(E)= t+2\\ f(G)=9t+3} \Rightarrow \blacktriangle類:\cases{f(B)=9t-27\\f(D)=t-4\\f(F)=t-7} \\ \textbf{Case I: }f(\bullet)\gt 0且f(\blacktriangle)\lt 0 \Rightarrow \cases{t-1\gt0\\ 4t\gt0\\t+2\gt0\\9t+3\gt 0} \Rightarrow t\gt 1 且\cases{9t-27\lt 0\\t-4\lt 0\\ t-7\lt 0} \Rightarrow t\lt 3\\ \qquad \Rightarrow 1\lt t\lt 3 \\ \textbf{Case II: }f(\bullet)\lt 0且f(\blacktriangle)\gt 0 \Rightarrow \cases{t-1\lt0\\ 4t\lt0\\t+2\lt0\\9t+3\lt 0} \Rightarrow t\lt -2 且\cases{9t-27\gt 0\\t-4\gt 0\\ t-7\gt 0} \Rightarrow t\gt 7 \\ \qquad t\lt -2且t\gt 7, 矛盾 \\ \Rightarrow \text{Case I 成立} \Rightarrow \bbox[red, 2pt]{1\lt t\lt 3}$$
解答:$$假設\cases{\overline{BC}=a \\ \overline{AC}=b\\ \overline{AB}=c \\ \triangle ABC外心O \\ 外接圓半徑R=8} \Rightarrow \cases{d(O,\overline{BC}) =7 =R\cos A\\ d(O,\overline{AB}) =2 =R\cos C} \Rightarrow \cases{\cos A=7/8\\ \cos C=1/4} \Rightarrow \cases{\sin A=\sqrt{15}/8\\ \sin C=\sqrt{15}/4} \\ \cos B=\cos(\pi-(A+C))= \sin A\sin C-\cos A\cos C={1\over 4} \Rightarrow \sin B={\sqrt{15} \over 4} \\ \Rightarrow \cases{a= 2R\sin A=2\sqrt{15} \\b=2R\sin B=4\sqrt{15} \\ c=2R\sin C=4\sqrt{15}} \Rightarrow s={1\over 2}(a+b+c) =5\sqrt{15} \\\Rightarrow \triangle ABC面積= \sqrt{s(s-a)(s-b)(s-c)} =15\sqrt{15} =rs=r\times 5\sqrt{15}\Rightarrow r= \bbox[red, 2pt]3$$
解答:$$假設x+y+z=k \Rightarrow {1\over x}+{1\over y}+{1\over z}=1 \Rightarrow yz+xz+xy=xyz \\因此 (x+y+z)^2=x^2+y^2+z^2 +2(xy+yz+ zx) \Rightarrow k^2={3\over 2}+2xyz \Rightarrow xyz=(k^2-3/2)/2\\ 利用公式 (x+y+z)^3= x^3+y^3+z^3+ 3(x+y) (y+z)(z+x) \\ \Rightarrow k^3=1+3(k-z)(k-x)(k-y) =1+3 [k^3-k^2(x+y+z)+k(xy+yz+zk)-xyz] \\\qquad =1+3[k^3-k^3+kxyz-xyz] =1+3xyz(k-1)=1+3\cdot {k^2-3/2\over 2}\cdot (k-1) \\ \Rightarrow 2k^3-6k^2-9k+13=0 \Rightarrow (k-1)(2k^2-4k-13)=0 \Rightarrow k=\bbox[red, 2pt]1$$
解答:$$依題意，四面體在z軸投影的長度為6。假設此四面體的邊長為a，且內接於一個邊長為k的立方體\\ 將立方體不相鄰的4個頂點連線構成一個正四面體。因此a=\sqrt 2k \Rightarrow k={a\over \sqrt 2} \le 6\\ \Rightarrow a\le \bbox[red, 2pt]{6\sqrt 2}$$
解答:$$\cases{鋼琴相鄰排列數:5!\times 2=240 \\小提琴相鄰排列數:5!\times 2=240 \\歌唱相鄰排列數:5!\times 2=240 } \Rightarrow \cases{鋼琴相鄰且小提琴相鄰排列數:4!\times 2\times 2=96 \\歌唱相鄰且小提琴相鄰排列數:4!\times 2\times 2=96 \\鋼琴相鄰且歌唱相鄰排列數:4!\times 2\times 2=96} \\ \Rightarrow 三種表演同時相鄰:3!\times 2\times 2\times 2=48 \\ \Rightarrow 至少有一種表演相鄰:240\times 3-96\times 3+48=480\\ \Rightarrow 表演完全不相鄰=6!-480= \bbox[red, 2pt]{240}$$
解答:$$假設\cases{A(z)\\ B(0) \\C(z-1-\sqrt 2-i)} \Rightarrow \cases{\overrightarrow{AB} =-z\\ \overrightarrow{BC}=z-1-\sqrt 2-i} \\ 正八邊形的內角={(8-2)\times 180^\circ\over 8}=135^\circ \Rightarrow 外角=45^\circ \Rightarrow \overrightarrow{BC}= \overrightarrow{AB}旋轉45^\circ \\ \Rightarrow z-1-\sqrt 2-i= -z(\cos 45^\circ+i\sin 45^\circ)=-z({\sqrt 2\over 2}+{\sqrt 2i\over 2}) \\ \Rightarrow z \left( 1+{\sqrt 2\over 2}+{\sqrt 2i\over 2} \right)=1+\sqrt 2+i \Rightarrow z=\bbox[red, 2pt]{\sqrt 2}$$
解答:
$$此題相當於以P(x,y)為圓心, 半徑r=|x-y|, 限制條件就是P到正方形四個邊\ge r \\ \Rightarrow \cases{x\ge |x-y|\\ 2-x\ge |x-y|\\ y\ge |x-y|\\ 2-y\ge |x-y|} \Rightarrow \cases{y\ge x\Rightarrow y\le 2x且y\le {1\over 2}x+1 \\x\le y \Rightarrow x\le 2y且x\le {1\over 2}y+1}\Rightarrow 區域R頂點\cases{O(0,0) \\P(4/3,2/3)\\ B(2,2)\\Q(2/3,4/3)} \\ \Rightarrow \triangle OPB面積= {1\over 2}\begin{vmatrix} 0&0 & 1\\ 4/3& 2/3& 1\\ 2& 2& 1\end{vmatrix} ={2\over 3} \Rightarrow R面積=2\times {2\over 3}= \bbox[red, 2pt]{4 \over 3}$$
解答:$$\cases{L_1:y=0\\ L_2:3x-4y=0 \Rightarrow y=3x/4} \Rightarrow \cases{L_1與x軸角度為0^\circ\\ L_2與x軸角度為\theta,\tan \theta=3/4} \\ \tan \theta={3\over 4} \Rightarrow \cases{\cos (\theta/2) =3/\sqrt{10} \\ \sin (\theta/2)=1/\sqrt{10}} \\ 旋轉矩陣A= \begin{bmatrix}\cos \alpha& -\sin \alpha\\ \sin \alpha& \cos \alpha \end{bmatrix} \Rightarrow \cases{\alpha=-\theta/2 \Rightarrow A= \begin{bmatrix}3/ \sqrt{10} & 1/\sqrt{10} \\-1/\sqrt{10} & 3/\sqrt{10}\end{bmatrix} \\ \alpha= \pi/2-\theta/2 \Rightarrow A= \begin{bmatrix}1/\sqrt{10} & -3/\sqrt{10} \\ 3/\sqrt{10}& 1/\sqrt{10} \end{bmatrix} \\ \alpha=\pi-\theta/2 \Rightarrow A= \begin{bmatrix}-3/ \sqrt{10}&-1/\sqrt{10} \\1/\sqrt{10} & -3/\sqrt{10} \end{bmatrix} \\ \alpha=3\pi/2-\theta/2 \Rightarrow A= \begin{bmatrix}-1/\sqrt{10}& 3/\sqrt{10} \\ -3/\sqrt{10} & -1/\sqrt{10}\end{bmatrix}} \\ \Rightarrow A= \bbox[red, 2pt]{\pm {1\over \sqrt{10}} \begin{bmatrix}3& 1\\-1& 3 \end{bmatrix}或 \pm {1\over \sqrt{10}} \begin{bmatrix}1&-3\\3& 1 \end{bmatrix}}$$
解答:$$A+B+C=180^\circ \Rightarrow A+B=180^\circ-C \Rightarrow \tan (A+B)=\tan(180^\circ-C) \\ \Rightarrow {\tan A+\tan B\over 1-\tan A\cdot \tan B}=-\tan C \Rightarrow \tan A+\tan B+\tan C=\tan A\cdot \tan B\cdot \tan C \\ \Rightarrow 1+2+3=1\cdot 2\cdot 3 =\bbox[red, 2pt]6$$
解答:$$\textbf{(1) }m=2 \Rightarrow x^3-6x^2+10x=2x \Rightarrow x^3-6x^2+8x=0 \Rightarrow x(x-2)(x-4)=0 \Rightarrow x=\bbox[red, 2pt]{0,2,4} \\ \textbf{(2) }面積=\int_0^4 |x^3-6x^2+8x|\,dx = \int_0^2(x^3-6x^2+8x)\,dx + \int_2^4 -(x^3-6x^2+8x)\,dx \\=4+4= \bbox[red, 2pt]8 \\\textbf{(3) }x^3-6x^2+10x=mx \Rightarrow x(x^2-6x+(10-m)) =0 \Rightarrow x^2-6x+10-m=0有兩相異正根\\ \Rightarrow \cases{判別式\gt 0 \Rightarrow m\gt 1\\ 兩根之積大於0 \Rightarrow 10-m\gt 0\Rightarrow m\lt 10 } \Rightarrow 1\lt m\lt 10 \Rightarrow \bbox[red, 2pt] {\cases{a=1\\ b=10}}$$
解答:$$\textbf{(1) }1\cdot f(1)=3-2+2+0\Rightarrow f(1)= \bbox[red, 2pt]3 \\\textbf{(2) }{d\over dx} \left( xf(x) \right) = {d\over dx} \left( 3x^4-2x^3+2x^2+ \int_1^x f(t)\,dt \right) \Rightarrow f(x)+xf'(x)=12x^3-6x^2+ 4x+f(x)\\ \Rightarrow f'(x)=12x^2-6x+4 \Rightarrow f(x)= \int (12x^2-6x+4)\,dx =4x^3-3x^2+4x+C \\ \Rightarrow f(1)=4-3+4+C=3\Rightarrow C=-2 \Rightarrow \bbox[red, 2pt]{f(x) =4x^3-3x^2+4x-2} \\ \textbf{(3) } \int_0^a (4x^3-3x^2+4x-2)\,dx =a^4-a^3+2a^2-2a \\ \Rightarrow 欲證a^4-a^3+2a^2-2a=1在a\gt 1時恰有一實根,因此取g(a)=a^4-a^3+2a^2-2a-1 \\ \Rightarrow \cases{g(1)=-1\lt 0\\ g(2)=11\gt 0} \Rightarrow 在區間(1,2)內至少有一實根c,使得g(c)=0\\ 接著g'(a)=4a^3-3a^2+4a-2 \Rightarrow g''(a) =12a^2-6a+4=12(a-{1\over 4}^2)+{52\over 16}\gt 0 \\ \Rightarrow g'(a)為嚴格遞增,且g'(1)=3\gt 0 \Rightarrow g(a)在區間[1,\infty)是嚴格遞增\Rightarrow 恰有一實根\bbox[red, 2pt]{故得證}$$
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解題僅供參考，其他教甄試題及詳解
解答:$$\cases{L_1:y=0\\ L_2:3x-4y=0 \Rightarrow y=3x/4} \Rightarrow \cases{L_1與x軸角度為0^\circ\\ L_2與x軸角度為\theta,\tan \theta=3/4} \\ \tan \theta={3\over 4} \Rightarrow \cases{\cos (\theta/2) =3/\sqrt{10} \\ \sin (\theta/2)=1/\sqrt{10}} \\ 旋轉矩陣A= \begin{bmatrix}\cos \alpha& -\sin \alpha\\ \sin \alpha& \cos \alpha \end{bmatrix} \Rightarrow \cases{\alpha=-\theta/2 \Rightarrow A= \begin{bmatrix}3/ \sqrt{10} & 1/\sqrt{10} \\-1/\sqrt{10} & 3/\sqrt{10}\end{bmatrix} \\ \alpha= \pi/2-\theta/2 \Rightarrow A= \begin{bmatrix}1/\sqrt{10} & -3/\sqrt{10} \\ 3/\sqrt{10}& 1/\sqrt{10} \end{bmatrix} \\ \alpha=\pi-\theta/2 \Rightarrow A= \begin{bmatrix}-3/ \sqrt{10}&-1/\sqrt{10} \\1/\sqrt{10} & -3/\sqrt{10} \end{bmatrix} \\ \alpha=3\pi/2-\theta/2 \Rightarrow A= \begin{bmatrix}-1/\sqrt{10}& 3/\sqrt{10} \\ -3/\sqrt{10} & -1/\sqrt{10}\end{bmatrix}} \\ \Rightarrow A= \bbox[red, 2pt]{\pm {1\over \sqrt{10}} \begin{bmatrix}3& 1\\-1& 3 \end{bmatrix}或 \pm {1\over \sqrt{10}} \begin{bmatrix}1&-3\\3& 1 \end{bmatrix}}$$
解答:$$假設\overline{PA}=a \Rightarrow {\overrightarrow{PA} \over |\overrightarrow{PA}|} +{\overrightarrow{PB} \over |\overrightarrow{PB}|} +{\overrightarrow{PC} \over |\overrightarrow{PC}|} = \vec 0 \Rightarrow \angle APB = \angle BPC= \angle CPA =120^\circ \\ \Rightarrow \cases{\overline{AB}^2= a^2+132^2-2a\cdot 132\cos 120^\circ= a^2+132a+132^2\\ \overline{AC}^2= a^2+14^2-28a\cos120^\circ = a^2+14a+14^2\\ \overline{BC}^2=132^2+14^2-28\cdot 132 \cos 120^\circ= 132^2 +14\cdot 132+ 14^2} \\ \angle A =90^\circ \Rightarrow \overline{AB}^2+ \overline{AC}^2 = \overline{BC}^2 \Rightarrow 2a^2+146a+ 132^2+ 14^2=132^2+14\cdot 132+14^2\\ \Rightarrow a^2+73a-924=0 \Rightarrow (a-11)(a+84)=0 \Rightarrow a=\bbox[red, 2pt]{11}$$
解答:$$取\cases{X= \cos 2\pi/7+ \cos 4\pi/7+ \cos 8\pi/7\\ Y=\sin 2\pi/7+ \sin 4\pi/7 + \sin 8\pi/7\gt 0} \Rightarrow 欲求S=z+z^2+z^4= X+iY \\ 由於\cos {8\pi\over 7} =\cos (2\pi-{6\pi\over 7}) = \cos {6\pi\over 7}\Rightarrow X =\cos {2\pi\over 7} +\cos {4\pi\over 7} +\cos {6\pi\over 7} \\ \Rightarrow 2\sin{\pi\over 7}\cdot X= \left( \sin{3\pi\over 7}-\sin{\pi\over 7} \right)+ \left( \sin{5\pi\over 7}-\sin{3\pi\over 7} \right)+ \left( \sin{7\pi\over 7} -\sin {5\pi\over 7}\right)=-\sin{\pi\over 7} \\ \Rightarrow X=-{1\over 2} \\ |S|^2 = S\cdot\bar S=(z+z^2+z^4)(z^6+z^5+z^3)= 3+(z+z^2+z^3+z^4+z^5+z^6) =3+(-1)=2 \\S=X+iY \Rightarrow |S|^2=X^2+Y^2=(-{1\over 2})^2+Y^2=2 \Rightarrow Y^2={7\over 4} \Rightarrow Y={\sqrt{7} \over 2} \\ \Rightarrow S=X+iY=-{1\over 2}+{\sqrt 7i\over 2} =\bbox[red, 2pt]{-1+\sqrt 7i\over 2}$$
解答:$$A+B+C=180^\circ \Rightarrow A+B=180^\circ-C \Rightarrow \tan (A+B)=\tan(180^\circ-C) \\ \Rightarrow {\tan A+\tan B\over 1-\tan A\cdot \tan B}=-\tan C \Rightarrow \tan A+\tan B+\tan C=\tan A\cdot \tan B\cdot \tan C \\ \Rightarrow 1+2+3=1\cdot 2\cdot 3 =\bbox[red, 2pt]6$$
解答:$$\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx \Rightarrow I=\int_0^{\pi/2} {\sin^3 x\over \sin^3 x+ \cos^3 x} dx =\int_0^{\pi/2} {\sin^3 ({\pi\over 2}-x)\over \sin^3 ({\pi\over 2}-x)+ \cos^3 ({\pi\over 2}-x)} dx \\= \int_0^{\pi/2} {\cos^3 x\over \cos^3 x+\sin^3 x} \,dx \Rightarrow 2I =\int_0^{\pi/2} {\sin^3 x+ \cos^3 x\over \sin^3 x+ \cos^3 x} dx =\int_0^{\pi/2} 1\,dx ={\pi\over 2} \Rightarrow I= \bbox[red, 2pt]{\pi\over 4}$$
解答:$$取\cases{\vec u=(a,b) \\ \vec v=(-d,c)} \Rightarrow \cases{|\vec u|=\sqrt{a^2+b^2} =3\\ |\vec v|=\sqrt{c^2+d^2}=4 \\ \vec u\cdot \vec v=-ad+bc=12} \Rightarrow \cos \theta={ \vec u\cdot \vec v\over |\vec u||\vec v|} ={12\over 3\cdot 4}=1 \Rightarrow \theta=0^\circ \\ \Rightarrow \vec v={4\over 3} \vec u \Rightarrow (-d,c)={4\over 3}(a,b) \Rightarrow \cases{c=4b/3\\d=-4a/3} \Rightarrow bd=b\cdot (-{4a\over 3}) =-{4ab\over 3} \\ a^2+b^2=9 \Rightarrow \cases{a=3\cos \alpha\\ b=3\sin \alpha} \Rightarrow bd=-{4\over 3}\cdot 3\cos \alpha \cdot 3\sin \alpha =-6\sin 2\alpha \Rightarrow 最大值\bbox[red, 2pt]6$$
解答:$$(1+x+x^2 +x^3)^{10} = \left( {1-x^4\over 1-x} \right)^{10} =(1-x^4)^{10}(1-x)^{-10} \\ \cases{(1-x^4)^{10}=1-10x^4+\cdots \\ (1-x)^{-10} = \sum_{k= 0}^\infty H^{10}_k x^k = \sum_{k=0}^\infty {k+9\choose k}x^k} \Rightarrow (1-x^4)^{10}(1-x)^{-10} \\= \left( 1-10x^4+\cdots \right) \left( \sum_{k=0}^\infty {k+9\choose k}x^k\right) \Rightarrow x^6係數=1\times {6+9\choose 6}-10\times {2+9\choose 2} \\={15\choose 6}-10{11\choose 2} =5005-10\times 55= \bbox[red, 2pt]{4455}$$
解答:$$假設經過n次傳球後\cases{a_n代表球最後在甲手上的傳球方法數\\ b_n代表球最後不在甲手上的傳球方法數} \\ 每一次傳球，持球者都可以自由傳給剩下的 (k-1) 個人。\\因此，傳球 n 次的總方法數=a_n+b_n=(k-1)^n \\要讓第 n 次球回到甲的手上，第 n-1 次時，球必定在不在甲手上，而這個人要把球傳給甲 \\ \Rightarrow a_n=b_{n-1}\times 1 \Rightarrow a_{n-1}+b_{n-1} =(k-1)^{n-1} \Rightarrow a_{n-1}+ a_n=(k-1)^{n-1} \\ \Rightarrow a_n-{1\over k}(k-1)^n=- \left( a_{n-1}-{1\over k}(k-1)^{n-1} \right) \\ \Rightarrow a_n-{1\over k}(k-1)^n= \left( a_1-{1\over k}(k-1) \right)\times (-1)^{n-1} ={k-1\over k}\times (-1)^n \;(\because a_1=0) \\ \Rightarrow \bbox[red, 2pt]{a_n= {1\over k} \left( (k-1)^n +(k-1)(-1)^n \right)}$$
解答:$$取\cases{A(0,0,0) \\B(1,0,1)\\ C(1,1,0) \\ D(0,1,1)} \Rightarrow \cases{E=(3A+B)/4= (1/4,0,1/4) \\ F=(3C+D)/4 =(3/4,1,1/4)} \Rightarrow \cases{\overrightarrow{DE}=(1/4,-1,-3/4) \\ \overrightarrow{BF} =(-1/4,1,-3/4)} \\ \Rightarrow \cos \theta = {\overrightarrow{DE} \cdot \overrightarrow{BF} \over |\overrightarrow{DE}| |\overrightarrow{BF}|} ={-1/2\over 13/8} =-{4\over 13} \Rightarrow \sin \theta= {3\sqrt{17} \over 13} = \bbox[red, 2pt]{\sqrt{153}\over 13}$$
解答:$$\lim_{x\to \infty} \left( {4x+a\over 4x-a} \right)^{3x+1} =\lim_{x\to \infty} \left( {1+a/4x\over 1-a/3x} \right)^{3x+1} = {\lim_{x\to \infty} \left( 1+{a/4\over x} \right)^{3x+1} \over \lim_{x\to \infty} \left( 1-{a/4\over x} \right)^{3x+1}} ={e^{3a/4} \over e^{-3a/4}} \\=e^{3a/2}=3 \Rightarrow {3a\over 2} =\ln 3 \Rightarrow a= \bbox[red, 2pt]{{2\over 3}\ln 3}$$
二、簡答題(每題 10 分)
解題僅供參考，其他教甄試題及詳解
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