115年金門高中教甄-數學詳解

 國立金門高中115 學年度第一次暨第二次教師甄選

一、 填充題 (每題 8 分)

解答:$$\prod_{k=4}^{255} {\log_k 5^{k^2-1} \over \log_{k+1} 5^{k^2-4}}  = \prod_{k=4}^{255} {{k^2-1} \over {k^2-4}} \cdot {\log (k+1) \over \log k} = \prod_{k=4}^{255} {(k-1)(k+1)\over (k-2)(k+2)} \cdot {\log (k+1) \over \log k} \\= \left( {3\cdot 5\over 2\cdot 6}\cdot {\log 5\over \log 4} \right) \cdot \left( {4\cdot 6 \over 3\cdot 7}\cdot {\log 6\over \log 5} \right)  \cdot \left( {5\cdot 7 \over 4\cdot 8}\cdot {\log 7 \over \log 6} \right) \cdots \left( {253\cdot 255 \over 252\cdot 256}\cdot {\log 255 \over \log 254} \right) \cdot \left( {254\cdot 256 \over 253\cdot 257}\cdot {\log 256 \over \log 255} \right) \\= {5\over 2} \cdot {254\over 257} \cdot {\log 256\over \log 4} = \bbox[red, 2pt]{2540\over 257}$$

解答:

$$\Gamma:{x^2\over 9}+{y^2\over 10} =1\Rightarrow \cases{a=\sqrt{10}\\b=3} \Rightarrow c=1 \Rightarrow \cases{F(0,1) \\A(3,0) \\O(0,0) \\P(3\cos \theta, \sqrt{10}\sin \theta)} \Rightarrow L=\overleftrightarrow{AF}: x+3y=3 \\ 由於\cases{OAPF面積=\triangle OAF+\triangle AFP\\ \triangle OAF面積固定} \Rightarrow OAPF面積最大值取決於 d(P,L)的最大值 \\ \Rightarrow d(P,L)={|3\cos \theta +3\sqrt{10}\sin \theta-3|\over \sqrt{10}} ={3\over \sqrt{10}}|\cos \theta+ \sqrt{10}\sin \theta-1|\\ 由於P在第一象限，也就是\cases{\cos \theta\ge0\\ \sin \theta\ge 0}，因此當\cases{\cos \theta=1/\sqrt{11}\\ \sin \theta= \sqrt{10}/ \sqrt{11}}時，d(P,L)有最大值\\ 此時P= \left( {3\over \sqrt{11}}, { {10}\over \sqrt{11}} \right) = \bbox[red, 2pt]{\left( {3\sqrt{11}\over 11},{10\sqrt{11}\over 11} \right)}$$

解答:$$餘弦定理: \overline{BC}^2= \overline{AB}^2+\overline{AC^2} -2\overline{AB}\cdot \overline{AC} \cos \angle BAC =3^2+5^2-30\cos 120^\circ=49 \Rightarrow \overline{BC}=7 \\ 外角平分線定理:{\overline{CE} \over \overline{BE}} ={\overline{AC} \over \overline{AB}} \Rightarrow {7+\overline{BE} \over \overline{BE}}={5\over 3} \Rightarrow \overline{BE}={21\over 2} \\餘弦定理: \cos \angle ABC={\overline{AB}^2+\overline{BC}^2 -\overline{AC}^2\over 2\cdot \overline{AB}\cdot \overline{BC}}={11\over 14} \Rightarrow \overline{BD} =\overline{AB}\cos \angle ABC={33\over 14} \\ \Rightarrow \overline{DE}= \overline{BD}+ \overline{BE}={33\over 14}+{21\over 2}= \bbox[red, 2pt]{90\over 7}$$

解答:$$\lim_{h\to 0}{\int_1^{2+h} f(t)\,dt-\int_1^2 f(t)\,dt \over h} =\lim_{h\to 0}{{d\over dh}(\int_1^{2+h} f(t)\,dt-\int_1^2 f(t)\,dt) \over {d\over dh}h}  =\lim_{h\to 0}{f(2+h)\over 1} \\=f(2)=(-3\cdot 16+5\cdot 8+2^2+5\cdot 2-2)^4=4^4= \bbox[red, 2pt]{256}$$

解答:$$g(x)=(x-2)^3+20(x-2)+121 \Rightarrow g'(x)=3(x-2)^2+20 \Rightarrow g''(x)=6(x-2) \\ g''(x)=0 \Rightarrow x=2 \Rightarrow g(2)=121 \Rightarrow 反曲點(2,121) \Rightarrow g'(2)=20 \Rightarrow \cases{f(2)=121\\ f'(2)=20}\\ 又(1,115)為f(x)的反曲點\Rightarrow \cases{f(1)=115\\ f'(1)=f''(1)=0} \Rightarrow f'(x)=A(x-1)^3+B(x-1)^2 \\ \Rightarrow f(x)= \int f'(x)\,dx ={A\over 4}(x-1)^4+{B\over 3} (x-1)^3+C \Rightarrow f(1)=C=115 \\ \Rightarrow \cases{f(2)=A/4+B/3+115=121\\ f'(2)=A+B=20} \Rightarrow \cases{A=8\\ B=12} \Rightarrow f(x)= 2(x-1)^4+4(x-1)^3+115 \\ \Rightarrow f(-2)= 2\cdot(-3)^4+4\cdot (-2)^3+115= \bbox[red, 2pt]{169}$$

解答:$$\begin{pmatrix}1+{\sqrt 3\over 2}&-{1\over 2} \\ {1\over 2}&1+{\sqrt 3\over 2} \end{pmatrix}^6 = \left[ \begin{pmatrix}1&0\\0&1 \end{pmatrix} + \begin{pmatrix}\cos 30^\circ& -\sin 30^\circ\\ \sin30^\circ& \cos 30^\circ \end{pmatrix}\right]^6 =(I+A)^6 \\\begin{pmatrix}1-{\sqrt 3\over 2}&{1\over 2} \\ -{1\over 2}&1-{\sqrt 3\over 2} \end{pmatrix}^6 = \left[ \begin{pmatrix}1&0\\0&1 \end{pmatrix} - \begin{pmatrix}\cos 30^\circ& -\sin 30^\circ\\ \sin30^\circ& \cos 30^\circ \end{pmatrix}\right]^6 =(I-A)^6 \\ (I+A)^6+(I-A)^6= 2I^6+ 30I^4A^2 +30I^2A^4 +2A^6 =2I+30A^2+30A^4+2A^6 \\=2  \begin{pmatrix}1&0\\0&1 \end{pmatrix}+30 \begin{pmatrix}{1\over 2}&-{\sqrt 3\over 2}\\{\sqrt 3\over 2}&{1\over 2} \end{pmatrix} +30 \begin{pmatrix}-{1\over 2}&-{\sqrt 3\over 2}\\{\sqrt 3\over 2}& -{1\over 2} \end{pmatrix}+2 \begin{pmatrix}-1& 0\\0 &-1 \end{pmatrix} = 30 \begin{pmatrix} 0& -\sqrt 3\\ \sqrt 3 &0\end{pmatrix} \\= \bbox[red, 2pt]{\begin{pmatrix} 0& -30\sqrt 3\\ 30\sqrt 3 &0\end{pmatrix} }$$

解答:$$假設\cases{f(x)=x^4+3x^2+x-2 \\g(x)=x^4+3x^3 +x^2 +a+2},由於a是f(x)=0的一根\Rightarrow f(a)= a^4+3a^3+a-2=0 \\ \Rightarrow a^4+3a^3+  a= 2  \Rightarrow g(a)=a^2+4, 同理可得\cases{g(b)=b^2+4\\ g(c)=c^2+4 \\g(d)=d^2+4} \\ \Rightarrow 欲求 g(a)g(b)g(c)g(d) = (a^2+4)(b^2+4)(c^2+4) (d^2+4)\\ a,b,c,d為f(x)=0的四根\Rightarrow f(x)=(x-a)(x-b)(x-c)(x-d) \\\Rightarrow \cases{f(2i)=(2i-a) (2i-b)(2i-c)(2i-d) \\f(-2i)=(-2i-a)(-2i-b)(-2i-c)(-2i-d)} \\\Rightarrow f(2i)\cdot f(-2i)=(a^2+4)(b^2+4) (c^2+4) (d^2+4), 其中\cases{f(2i)=14-22i\\ f(-2i)=14+22i} \\ \Rightarrow (14-22i)(14+22i)=14^2+22^2= \bbox[red, 2pt]{680}$$

解答:$$z=x+yi \Rightarrow Arg(z+3) =Arg((x+3)+yi)={3\pi\over 4} \Rightarrow {y\over x+3}=\tan {3\pi\over 4}=-1 \Rightarrow y=-x-3\\ \Rightarrow d=|z+6|+ |z-3i|= \sqrt{(x+6)^2+y^2}+\sqrt{x^2+ (y-3)^2} \\=\sqrt{(x+6)^2+(-x-3)^2}+\sqrt{x^2+ (-x-6)^2} = \sqrt{2x^2+18x+45}+\sqrt{2x^2+12x+36} \\ \Rightarrow d'={4x+18\over 2\sqrt{2x^2+18x+45}} +{4x+12\over 2\sqrt{2x^2+12x+36}} =0 \Rightarrow {(2x+9)^2\over 2x^2+18x+45} ={(2x+6)^2\over 2x^2+12x+36} \\ \Rightarrow x^2+10x+24=0 \Rightarrow(x+4)(x+6)=0 \Rightarrow x=-4 \Rightarrow y=-(-4)-3=1 \Rightarrow z= \bbox[red, 2pt]{-4+i}$$

二、 綜合題 (每題 12 分)

解答:$$  2\cos^2\theta + \frac{2}{\sqrt{3}}\cos\theta - \frac{2}{3} = 0 解出的兩根是\cos \theta的兩種可能值，並不是\cos \theta 與\sin \theta的值\\ 正確作法: \cases{兩根和:\sin \theta+ \cos \theta=-p/2\\ 兩根積:\sin \theta \cos \theta=q/2} \Rightarrow (\sin \theta-\cos \theta)^2= \left( {1\over \sqrt 3} \right)^2 \\\Rightarrow 1-2\sin \theta \cos \theta={1\over 3} \Rightarrow \sin \theta\cos \theta={1\over 3} ={q\over 2} \Rightarrow q={2\over 3}\\ 又(\sin \theta+\cos \theta)^2+ (\sin \theta-\cos \theta)^2=2( \sin^2\theta+ \cos^2\theta)=2 \Rightarrow {p^2\over 4}+{1\over 3}=2 \\ \Rightarrow p^2={20\over 3} \Rightarrow p^2-8q={20\over 3}-{16\over 3}= \bbox[red, 2pt]{4\over 3}$$

解答:

$$取\cases{平面OCD:z=0\\ M=\overline{CD}中點=(0,0,0)} \Rightarrow \cases{C(1,0,0) \\ D(-1,0,0)} \Rightarrow \cases{\overline{OC}=\sqrt 6/2\\ \overline{CM}=1} \Rightarrow \overline{OM}=\sqrt{6/4-1} ={\sqrt 2\over 2} \\ \Rightarrow O(0,{\sqrt 2\over 2},0) ,並假設折疊後的A(B)=P(x,y,z)\\ \Rightarrow \cases{\overline{PC}=\overline{BC}=\sqrt 2\\ \overline{PD}=\overline{AD}=\sqrt 2\\ \overline{PO}=\overline{OA}=\sqrt 6/2} \Rightarrow \overline{PC}=\overline{PD} \Rightarrow P(0,y,z) \Rightarrow \cases{\overline{PC}^2=1+y^2+z^2=2 \\ \overline{PO}^2= (y-\sqrt 2/2)^2+z^2=3/2} \\ \Rightarrow y=0 \Rightarrow z=1 \Rightarrow P(0,0,1) \Rightarrow \cases{平面OCD:z=0 \Rightarrow 法向量\vec n_1=(0,0,1) \\ 平面PCD:y=0 \Rightarrow 法向量\vec n_2=(0,1,0)} \\ \Rightarrow \vec n_1\cdot \vec n_2=0\Rightarrow 平面 OCD 與平面 ACD 垂直 \quad \bbox[red, 2pt]{故得證}$$

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