115年台中女中教甄-數學詳解

 臺中市立臺中女子高級中等學校 115 學年度第一次教師甄選

一、 填充題（每題 5 分，共 80 分）

解答:$$假設在 7 個位置中放入了 k 顆紅球。為了保證紅球不相鄰，最多只能放入 4 顆紅球，因此0\le k\le 4\\ 現在我們有k個紅球及7-k個非紅球，每個非紅球的位置不是白球就是黑球，有2種可能，\\共有2^{7-k}種排法; 剩下8-k個間隙選出k個插入紅球,有{8-k\choose k}種選法 \\ \cases{k=0 \Rightarrow 2^7\cdot {8-0\choose 0}=128\\ k=1 \Rightarrow 2^6\cdot {8-1\choose 1} =448\\ k=2\Rightarrow 2^5\cdot {8-2\choose 2} =480\\ k=3 \Rightarrow 2^4\cdot {8-3\choose 3} =160\\k=4\Rightarrow 2^3\cdot {8-4\choose 4}=8} \Rightarrow 合計= \bbox[red, 2pt]{1224}$$

解答:$$假設\cases{M(m,0)\\ N(0,n)} \Rightarrow \cases{P={1\over 11}(7M+4N)= (7m/11,4n/11) \\ m^2+n^2=11^2=121} \\ \Rightarrow 四邊形OAPB面積=\triangle OAP+\triangle OBP ={1\over 2} \begin{vmatrix} 0&0& 1\\ 7&0&1\\ 7m/11& 4n/11& 1 \end{vmatrix} +{1\over 2} \begin{vmatrix} 0&0& 1\\ 0& 4& 1\\ 7m/11& 4n/11& 1 \end{vmatrix} \\=f(m,n)= {14\over 11}(m+n) \\在已知m^2+n^2=121的條件下求f(m,n)的最大值,\\當m=n時,即2m^2=121 \Rightarrow 當m=n={11\over \sqrt 2}時, f有最大值={14\over 11}({11\over \sqrt 2}+{11\over \sqrt 2}) = \bbox[red, 2pt]{14\sqrt 2}$$

解答:

$$假設\cases{左圓圓心O_1\\ 右圓圓心O_2\\ 兩圓半徑皆為r} 及\cases{左圓與\overline{AC}相切於D \\ 右圓與\overline{AC}相切於E} \Rightarrow 公切線\overline{DE} =\overline{O_1O_2} =2r \\ \angle B=90^\circ \Rightarrow \overline{AC}= \sqrt{5^2+ 12^2}=13 \Rightarrow \cases{\sin A=12/13\\ \cos A=5/13\\ \sin C=5/13\\ \cos C=12/13} \Rightarrow \cases{\tan (A/2)= \sin A/(1+\cos A)=2/3\\ \tan (C/2)= \sin C/(1+\cos C)=1/5}\\ \cases{\overline{AO_1}為\angle A的角平分線\Rightarrow \overline{AD} = r\cot (A/2) ={3\over 2}r \\ \overline{CO_2}為\angle C的角平分線\Rightarrow \overline{CE}=r\cot (C/2)=5r} \Rightarrow \overline{AC}={3r\over 2}+2r+5r=13 \Rightarrow r={26\over 17} \\ \Rightarrow 2r=\bbox[red, 2pt]{52\over 17}$$

解答:$$假設\cases{A(0,0,0)\\ \overline{AB}在x軸上} ,由於\overline{AB}=1 \Rightarrow B(1,0,0); \\假設C在xy-平面上 \Rightarrow C(2\cos 60^\circ,2\sin 60^\circ, 0)  \Rightarrow C(1,\sqrt 3, 0) \\ 假設D(x,y,z), 由於\overline{AD}=3 \Rightarrow x^2+y^2+z^2=9;\\ 又\angle DAB=60^\circ \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AD} =\overline{AB}\cdot \overline{AD} \cos 60^\circ \Rightarrow (1,0,0) \cdot (x,y,z)=1\times 3\times {1\over 2} \Rightarrow x={3\over 2} \\ \angle CAD =60^\circ  \Rightarrow \overrightarrow{AC} \cdot \overrightarrow{AD} =\overline{AC}\cdot \overline{AD} \cos 60^\circ \Rightarrow (1,\sqrt 3,0)\cdot (x,y,z)=2\times 3\times {1\over 2} \\ \qquad \Rightarrow x+\sqrt 3y=3 \Rightarrow y={\sqrt 3\over 2} \Rightarrow x^2+y^2+z^2={9\over 4}+{3\over 4}+z^2=9 \Rightarrow z=\sqrt 6 \Rightarrow D({3\over 2},{\sqrt 3\over 2}, \sqrt 6)\\ \Rightarrow E=平面BCD: 2\sqrt 6x-z-2\sqrt 6=0 \Rightarrow d(A,E)= \bbox[red, 2pt]{2\sqrt 6\over 5}$$

解答:$$\cases{a_{n+1} =a_n+2b_n\\ b_{n+1} =2a_n+b_n} \Rightarrow  \cases{兩式相加\Rightarrow a_{n+1}+b_{n+1} =3(a_n+b_n) \Rightarrow \langle a_n+b_n \rangle 為公比3的等比級數 \\ 兩式相減\Rightarrow a_{n+1}-b_{n+1}=-(a_n-b_n) \Rightarrow \langle a_n-b_n \rangle為公比-1的等比級數}\\ \Rightarrow \cases{a_1 +b_1=5+1=6 \Rightarrow a_n+b_n=6\cdot 3^{n-1} =2\cdot 3^n \\a_1-b_1=5-1=4 \Rightarrow a_n-b_n=4\cdot (-1)^{n-1}} \Rightarrow 兩式相加:2a_n=2\cdot 3^n+4\cdot(-1)^{n-1} \\ \Rightarrow a_n=3^n+2\cdot(-1)^{n-1} = \bbox[red, 2pt]{3-2\cdot (-1)^n}$$

解答:$$f(x)=(1+x)^n = \sum_{k=0}^n C^n_k x^k \Rightarrow f'(x)=n(1+x)^{n-1} = \sum_{k=1}^n kC^n_k x^{k-1} \\ \Rightarrow g(x)=x^2f'(x)=nx^2(1+x)^{n-1} = \sum_{k=1}^n kC^n_k x^{k+1} \Rightarrow g(-3)=9n(-2)^{n-1}= \sum_{k=1}^n kC^n_k (-3)^{k+1} \\ \Rightarrow |9n(-2)^{n-1}|\lt 10^5 \Rightarrow 9n\cdot 2^{n-1}\lt 100000 \Rightarrow S(n)=n\cdot 2^{n-1}\lt{100000\over 9} \approx 11111.1 \\ \Rightarrow \cases{S(10)=5120 \lt 11111.1\\ S(11)=11264 \gt 11111.1} \Rightarrow n=\bbox[red, 2pt]{10}$$

解答:$$\angle B+\angle D=180^\circ \Rightarrow \cos \angle B=-\cos \angle D \Rightarrow \cases{\triangle ABC: \overline{AC}^2=2^2+3^2-2\cdot 2\cdot 3 \cos \angle B \\ \triangle ACD: \overline{AC}^2=5^2+4^2-2\cdot 5\cdot 4 \cos \angle D} \\ \Rightarrow 13-12\cos \angle B=41+40\cos \angle B \Rightarrow \cos \angle B=-{7\over 13} \Rightarrow \overline{AC}^2={253\over 13} \\ 同理, \cases{\triangle ABD: \overline{BD}^2=5^2+2^2-2\cdot 5\cdot 2 \cos \angle A\\ \triangle CBD:\overline{BD}^2 = 4^2+3^2-2\cdot 4\cdot 3\cos \angle C} \Rightarrow \cos \angle A={1\over 11} \Rightarrow \overrightarrow{AB}\cdot \overrightarrow{AD}={10\over 11} \\ \triangle ABC:  \overrightarrow{AB}\cdot  \overrightarrow{AC} = \overline{AB}\cdot \overline{AC} \cos \angle BAC =   {\overline{AB}^2+ \overline{AC}^2 -\overline{BC}^2 \over 2 } ={94\over 13} \\ 同理,\triangle ACD: \overrightarrow{AC} \cdot \overrightarrow{AD} ={\overline{AC^2}+ \overline{AD}^2-\overline{CD}^2 \over 2} ={185\over 13}\\ \overrightarrow{AC} =m\overrightarrow{AB}+ n\overrightarrow{AD} \Rightarrow \overrightarrow{AC} \cdot \overrightarrow{AB} =m\overline{AB}^2+ n\overrightarrow{AD} \cdot \overrightarrow{AB} \Rightarrow {94\over 13}=4m+{10\over 11}n \cdots(1) \\ \overrightarrow{AC} \cdot \overrightarrow{AD} =m \overrightarrow{AB} \cdot \overrightarrow{AD} +n\overline{AD}^2 \Rightarrow {185\over 13}={10\over 11}m  +25n \cdots(2) \\ 由式(1)及(2) \Rightarrow (m,n)= \bbox[red, 2pt]{ \left( {22\over 13},{33\over 65} \right)}$$

解答:$$\cos A={6^2+4^2-5^2\over 2\cdot 6\cdot 4} ={9\over 16} \Rightarrow \sin A=\sqrt{1-({9\over 16})^2} = {5\sqrt 7\over 16} \\ \Rightarrow 正弦定理: {5\over {5\sqrt 7\over 16}} =2R \Rightarrow 2R= {16\over \sqrt 7} \Rightarrow \overline{AH}=2R\cos A={16\over \sqrt 7}\times {9\over 16}= \bbox[red, 2pt]{9\sqrt 7\over 7}$$

解答:$$\alpha^2+3\beta^2+4\gamma^2-2\alpha\gamma -6\beta \gamma =(\alpha^2-2\alpha\gamma+\gamma^2)+(3\beta^2-6\beta \gamma+3\gamma^2) =(\alpha-\gamma)^2+ 3(\beta-\gamma)^2=0 \\ \Rightarrow \alpha-\gamma=\pm \sqrt 3 i(\beta-\gamma) \Rightarrow {\alpha-\gamma\over \beta-\gamma}=\pm \sqrt 3 i \Rightarrow  \left| {\alpha-\gamma\over \beta-\gamma} \right|=\sqrt 3\\ \Rightarrow \cases{\angle C=90^\circ\\ \overline{AC}=\sqrt3\cdot \overline{BC}} \Rightarrow \triangle ABC 為30^\circ-60^\circ-90^\circ的直角三角形 \\ \Rightarrow \overline{AB}^2= \overline{AC}^2+ \overline{BC}^2 \Rightarrow 10^2=(\sqrt 3a)^2+a^2 \Rightarrow a=5 \Rightarrow b=5\sqrt 3 \Rightarrow 面積={1\over 2}ab= \bbox[red, 2pt]{25\sqrt 3\over 2}$$

解答:$$\href{https://brilliant.org/wiki/titus-lemma/}{\text{Titu's Lemma:}} {a_1^2\over b_1} +{a_2^2\over b_2} +\cdots+{a_n^2\over b_n} \ge {(a_1+a_2+\cdots +a_n)^2 \over b_1+b_2+ \cdots+ b_n} \\ \Rightarrow {y^2\over x-1}+{x^2\over y-4}\ge {(x+y)^2 \over x+y-5}\\ 取u=x+y-5 \Rightarrow {(x+y)^2 \over x+y-5}={(u+5)^2\over u} = 10+u+{25\over u} \ge 10+2\sqrt{u\cdot {25\over u}}=20\\ \Rightarrow {(x+y)^2 \over x+y-5} 最小值為20 \Rightarrow {y^2\over x-1}+{x^2\over y-4}最小值為\bbox[red, 2pt]{20}$$

解答:

$$\overline{AD}為\angle A的角平分線 \Rightarrow \angle BAD=\angle DAC=\theta\\  \overline{PE} 為\overline{AD}的中垂線 \Rightarrow \angle ADP=\angle DAP \Rightarrow \theta+\angle C=\theta+\angle BAP \Rightarrow \angle BAP=\angle C \\ \Rightarrow \triangle CAP \sim \triangle ABP (AAA) \Rightarrow {\overline{PB} \over \overline{PA}} ={\overline{AB} \over \overline{AC}} ={4\over 6} \Rightarrow \overline{PA} ={3\over 2} \overline{PB} \Rightarrow\cases{\overline{PB}=a\\ \overline{PA}=3a/2} \\ \Rightarrow {\overline{PB} \over \overline{PA}} ={\overline{PA} \over \overline{PC}} \Rightarrow \overline{PA}^2=\overline{PB} \cdot \overline{PC} \Rightarrow ({3a\over 2})^2 =a\cdot (7+a) \Rightarrow a= \bbox[red, 2pt]{28\over 5}$$

解答:$$\theta ={2\pi\over 7},{4\pi\over 7},{6\pi\over 7} 都滿足7\theta =2k\pi \Rightarrow 4\theta=2k\pi-3\theta \Rightarrow \cos 4\theta=\cos 3\theta\\ 假設\cos \theta=x \Rightarrow \cases{\cos 3\theta=4x^3-3x \\ \cos 2\theta =2x^2-1 } \Rightarrow \cos 4\theta= 2(2x^2-1)^2-1=8x^4-8x^2 +1\\ \Rightarrow \cos 4\theta = \cos 3\theta \Rightarrow 8x^4-8x^2+1=4x^3-3x \Rightarrow 8x^4-4x^3-8x^2+3x+1=0 \\ \Rightarrow (x-1)(8x^3+4x^2-4x-1)=0 \Rightarrow f(x)=8x^3+4x^2-4x-1 \\ \Rightarrow \cases{a=8\\ b=4\\ c=4\\ d=1} \Rightarrow 20\log_2 a-2\log_2 b-6\log_2 c-12\log_2 d=20\cdot 3-2\cdot 2-6\cdot 2-12\cdot 0= \bbox[red, 2pt]{44}$$

解答:$$\cases{\overline{BC}=a \\ \overline{AC}=b\\ \overline{AB}=c} \Rightarrow \cases{ \overrightarrow{AB} \cdot  \overrightarrow{AC} =cb\cos A\\  \overrightarrow{BA} \cdot  \overrightarrow{BC} =ca\cos B} \Rightarrow 2cb\cos A=ca \cos B \Rightarrow 2b\cos A=a\cos B \\ \Rightarrow 2(2R\sin B) \cos A= 2R\sin A\cos B \Rightarrow 2\sin B\cos A=\sin A\cos B \Rightarrow  2{\sin B\cos A\over \cos A\cos B}={\sin A\cos B \over \cos A\cos B} \\ \Rightarrow 2\tan B=\tan A \\已知\cos C={2\over \sqrt{13}} \Rightarrow \sin C={3\over \sqrt{13}} \Rightarrow \tan C={3\over 2} \Rightarrow \tan(A+B) =-\tan C=-{3\over 2} \\ \Rightarrow \tan(A+B) ={\tan A+\tan B\over 1-\tan A\tan B} \Rightarrow -{3\over 2}={\tan A+{1\over 2}\tan A\over 1-\tan A\cdot ({1\over 2}\tan A)} \Rightarrow \tan^2A-2\tan A-2=0 \\ \Rightarrow \tan A=\bbox[red, 2pt]{1+\sqrt 3} \; (\tan A\lt 0 \Rightarrow \tan  B\lt 0 不可能)$$

解答:$$假設\cases{x=a+b+c\\ y=4a+2b+c\\ z=9a+3b+c} \Rightarrow \cases{0\le x\le 4\\ -2\le y\le 2\\ -4\le z\le 0} \Rightarrow a-2b+4c= f(x,y,z) ={35\over 2}x-21y+{15\over 2}z \\ \Rightarrow \cases{M=f(4,-2,0) =112\\ m=f(0,2,-4)=-72} \Rightarrow (M,m)= \bbox[red, 2pt]{(112,-72)}$$

解答:$$x^2+2xy+2y^2=1為一橢圓, 而x^2+y^2就是距原點距離的平方，最大值發生在橢圓長軸上的頂點\\ x^2+2xy+2y^2 = [x\; y] \begin{bmatrix}1& 1\\1& 2 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} \Rightarrow \begin{bmatrix}1& 1\\1& 2 \end{bmatrix} =P \begin{bmatrix}(3-\sqrt 5)/2& 0\\0& (3+\sqrt 5)/2 \end{bmatrix} P^{-1} \\ 將橢圓旋轉，讓長軸在x軸上，旋轉後的橢圓方程式:{3-\sqrt 5\over 2}x^2+{3+\sqrt 5\over 2}y^2=1 \\ \Rightarrow a^2={2\over 3-\sqrt 5}={2(3+\sqrt 5)\over 4} = \bbox[red, 2pt]{3+\sqrt 5\over 2}$$

解答:$$E_1,E_2,E_3的法向量皆為\vec n_1=(1,1,1) \Rightarrow E_1\parallel E_2\parallel E_3 \Rightarrow \cases{A\in L_1=E_1\cap E_4\\ B\in L_2=E_2\cap E_4\\ C\in L_3=E_3\cap E_4} \\ \Rightarrow \cases{d(E_1,E_2)=2\sqrt 3\\ d(E_2,E_3)=4\sqrt 3} \Rightarrow E_4的法向量\vec n_4=(1,-1,1) \Rightarrow E_4與其他平面的夾角均為\theta \\ \Rightarrow \cos \theta= {\vec n_1\cdot \vec n_2\over |\vec n_1||\vec n_2|} ={1\over 3} \Rightarrow \sin \theta={2\sqrt 2\over 3} \Rightarrow \cases{d(L_1,L_2) =\displaystyle {d(E_1,E_2)\over \sin \theta} = 3\sqrt 6/2\\ d(L_2,L_3) =\displaystyle {d(E_2,E_3) \over \sin \theta} =3\sqrt 6}\\ 假設\cases{\overline{AB}=a\\ d(A,L_2)=p\\ d(C,L_2)=q} \Rightarrow 3a^2=4(p^2+pq+q^2) = 4\left( ({3\sqrt 6 \over 2})^2+{3\sqrt 6\over 2}\cdot 3\sqrt 6+(3\sqrt 6)^2 \right) =378\\ \Rightarrow a^2=126 \Rightarrow \triangle ABC面積={\sqrt 3\over 4}a^2= \bbox[red, 2pt]{63\sqrt 3\over 2}\\\bbox[cyan,2pt]{公式說明}:假設B(0,0)且L_2為x軸, 則\cases{A(x_A,-p) \\C(x_C,q)} \Rightarrow \cases{\overline{AB}^2=x_A^2+p^2 =a^2\cdots(1) \\ \overline{BC}^2=x_C^2+q^2=a^2 \cdots(2)\\ \overline{CA}^2=(x_A-x_C)^2+(p+q)^2 =a^2 \cdots(3)} \\ 將(1),(2)代入(3) \Rightarrow 3a^2=4(p^2+pq+q^2)$$

二、 計算證明題（ 須詳細寫出計算過程， 共 20 分）

解答:$$\textbf{(1) }f(x)-g(x)=k, k為常數\Rightarrow \cases{f(x)=px^2+qx+r_1 \\ g(x)=px^2+qx+r_2}\Rightarrow 兩圖形\cases{y=f(x)\\y=g(x)}頂點x坐標均為-{q\over 2p} \\ \cases{f(a)=f(b) \Rightarrow y=f(x)的頂點x坐標={a+b\over 2} \\ g(c)=g(d) \Rightarrow y=g(x)的頂點x坐標={c+d\over 2}} \Rightarrow {a+b\over 2}=-{q\over 2p} ={c+d\over 2} \\ \Rightarrow {a+b\over 2}  ={c+d\over 2} \Rightarrow a+b=c+d\quad \bbox[red, 2pt]{故得證} \\\textbf{(2) } 令m={a+b\over 2} ={c+d\over 2} \Rightarrow \cases{f(x)=p(x-m)^2+k_1\\ g(x)=p(x-m)^2+k_2} \\ \Rightarrow \cases{f(a)=p(a-m)^2+k_1\\ f({a+b\over 2}) =f(m)=k_1}\Rightarrow f(a)-f({a+b\over 2})=p(a-m)^2 =p(a-{a+b\over 2})^2 =p\cdot {(a-b)^2\over 4}\\ 同理, \cases{g(c)=p(c-m)^2+k_2\\ g({c+d\over 2}) =k_2} \Rightarrow g(c)-g({c+d\over 2}) =p(c-m)^2 =p(c-{c+d\over 2})^2 =p\cdot {(c-d)^2\over 4} \\ \Rightarrow {f(a)-f({a+b\over 2}) \over g(c)-g({c+d\over 2})} ={p\cdot {(a-b)^2\over 4}\over p\cdot {(c-d)^2\over 4}} ={(a-b)^2\over (c-d)^2} = \left( {a-b\over c-d} \right)^2\quad \bbox[red, 2pt]{故得證}$$

解答:$$\log_4(x^2)={\log_2 x^2\over \log_2 4} ={2\log_2|x|\over 2\log_2 2} =\log_2 |x| \\\Rightarrow 此題相當於欲求兩圖形\cases{y=f(x) = \sqrt{12+8\log_2|x|-4(\log_2|x|)^2} \\ y=g(x)=1-x}的交點數 \\ f(x)=\sqrt{16-4(\log_2|x|-1)^2} \Rightarrow \cases{f(x)\ge 0\\最大值 f(\pm 2)=4\\ x\in[-8,-1/2] \cup [1/2,8]} \\ y=g(x)圖形為一左上右下的直線且\cases{g(0.5)\gt f(0.5) 且g(1)\lt f(0) \Rightarrow 在y軸右側有一交點\\ g(-8)\gt f(-8), g(-2)\lt f(-2), g(-1/2)\gt f(-1/2) \\\qquad \Rightarrow 在y軸左側有2個交點} \\ \Rightarrow 共有3個交點 \Rightarrow \bbox[red, 2pt]3個不同的實根$$

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解題僅供參考，其他教甄試題及詳解