114年嘉義女中教甄-數學詳解

 國立嘉義女子高級中學 114 學年度第 1 次教師甄選

一、填充題（計 13 格，每格 6 分，共 78 分）：

解答:$$\cases{43^{100}-43^2 \Rightarrow 取f(x)=x^{100}-x^2 \\43^3-43^2+43 \Rightarrow 取g(x)=x(x^2-x+1) \Rightarrow (x+1)g(x)=x(x^3+1)} \\ \Rightarrow f(x)=g(x)p(x)+ ax^2+bx+c \Rightarrow f(0)=0=c \Rightarrow c=0 \\ \Rightarrow (x+1)(x^{100}-x^2)=x(x^3+1)p(x)+(ax^2+bx)(x+1) \\x^3=-1代入上式\Rightarrow (x+1)(-x-x^2) =(ax^2+bx)(x+1 )\Rightarrow \cases{a=-1\\ b=-1} \\ \Rightarrow f(x)=g(x)p(x)-x^2-x \Rightarrow f(43)=43^{100}-43^2 \equiv-43^2-43\pmod{ 43^3-43^2+43} \\ 餘數為-43^2-43相當於餘數為-43^2-43+(43^3-43^2+43)=43^3-2\cdot 43^2 \\=43^2\cdot 41= \bbox[red, 2pt]{75809}$$

解答:

$$\cases{假設B,B'對稱於\overleftrightarrow{AC} \\假設C,C'對稱於\overleftrightarrow{AB}} \Rightarrow \cases{D=\overline{AB} \cap \overline{B'C'} \\ E=\overline{AC} \cap \overline{B'C'}} \Rightarrow \cases{\overline{BE} = \overline{B'E} \\\overline{CD} =\overline{C'D}} \Rightarrow \overline{BE}+ \overline{DE}+ \overline{CD} =\overline{B'C'}為最小值\\ 由於\cases{B,B'對稱於\overleftrightarrow{AC} \\C,C'對稱於\overleftrightarrow{AB}} \Rightarrow \cases{ \cases{\angle B'AC= \angle CAB=30^\circ \\ \overline{AC'} =\overline{AC} =6}\\ \cases{\angle C'AB= \angle BAC =30^\circ \\ \overline{AB'}= \overline{AB} =10}} \Rightarrow \angle B'AC'=90^\circ\\假設\cases{\overline{AD}=a\\ \overline{AE} = b} \Rightarrow 直角AB'C'面積={1\over 2}\cdot 6\cdot 10=30 \Rightarrow \cases{30 =\triangle AC'D+ \triangle ADB' \\30=\triangle AC'E+ \triangle AEB'} \\ \Rightarrow \cases{3a\sin 30^\circ+5a\sin 60^\circ=30\\ 3b\sin 60^\circ+ 5b\sin 30^\circ =30} \Rightarrow (3+5\sqrt 3)a=(3\sqrt 3+5)b \Rightarrow {a\over b}={ 3\sqrt 3+5\over 3+5\sqrt 3} = \bbox[red, 2pt]{15+8\sqrt 3\over 33}$$

解答:$$假設\vec a與\vec c的夾角為\theta \Rightarrow \vec a在 \vec c的正射影長= |\vec a| \cos \theta= 3\cos \theta=\sqrt 5 \Rightarrow \cos \theta ={\sqrt 5\over 3} \\ 假設\vec b與\vec c的夾角為\rho \Rightarrow {\vec b\cdot \vec c\over \vec a\cdot \vec c} ={|\vec b||\vec c| \cos \rho\over |\vec a|| \vec c| \cos \theta} ={2\cos \rho \over \sqrt 5} ={1\over 2} \Rightarrow \cos \rho={\sqrt 5\over 4} \\ \vec a與\vec b的夾角為\cases{\rho+\theta\\ \rho-\theta} \Rightarrow \vec a在\vec b的正射影長為\cases{||\vec a| \cos(\rho+\theta)|={2\sqrt{11}-5\over 4} \\ ||\vec a|\cos(\rho -\theta)| ={2\sqrt{11}+5\over 4}} \\ \Rightarrow 正射影長為\bbox[red, 2pt]{2\sqrt{11}\pm 5\over 4}$$

解答:$$A= \begin{bmatrix}1 & \frac{2}{21} \\0 & \frac{22}{21} \end{bmatrix}= \begin{bmatrix} 1 & 2 \\0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\0 & \frac{22}{21} \end{bmatrix} \begin{bmatrix} 1 & -2 \\0 & 1 \end{bmatrix} \Rightarrow A^n= \begin{bmatrix} 1 & 2 \\0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\0 & (\frac{22}{21})^n \end{bmatrix} \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}  \\ = \begin{bmatrix} 1 & \frac{-2\cdot 3^n\cdot 7^n+2\cdot 2^n\cdot 11^n}{3^n\cdot 7^n} \\ 0 & \frac{2^n\cdot 11^n}{3^n\cdot 7^n} \end{bmatrix} =\begin{bmatrix} a_n& c_n\\ b_n& d_n\end{bmatrix} \Rightarrow c_n \gt d_n \Rightarrow -2\cdot 3^n\cdot 7^n+2\cdot 2^n\gt  2^n\cdot 11^n \\ \Rightarrow 2^n\cdot 11^n \gt 2\cdot 3^n\cdot 7^n \Rightarrow n(\log2+ \log 11)\gt \log 2+n(\log 3+ \log 7) \\ \Rightarrow n\gt {\log 2\over \log 2+\log 11-\log 3-\log 7} \approx 14.9 \Rightarrow n= \bbox[red, 2pt]{15}$$

解答:

$$假設A在\overline{BC}的垂足為O，又\cos C={1\over \sqrt{10}}, 因此假設\overline{BO}=a\\則可假設\cases{O(0,0) \\C(1,0) \\A(0,3) \\B(-a,0)} \Rightarrow \overrightarrow{AB} \cdot (\overrightarrow{AC}+ 2\overrightarrow{BC}) =(-a,-3)\cdot ((1,-3)+2(1+a,0)) \\= -2a^2-3a+9=0 \Rightarrow (-2a+3)(a+3)=0 \Rightarrow a={3\over 2} \Rightarrow \cases{\overline{AB}={3\sqrt 5/2} \\ \overline{BC}=5/2} \\ \Rightarrow \cos A={1\over \sqrt 2} \Rightarrow \tan A:\tan B:\tan C= \bbox[red, 2pt]{1:2:3}$$

解答:

$$2(\log_a c+ \log_b c) =9\log_{ab}c \Rightarrow {2\log c\over \log a}+ {2\log c\over \log b} ={9\log c\over \log a+ \log b}\\  \Rightarrow {2\over \log a}+ {2 \over \log b} ={9 \over \log a+ \log b} \Rightarrow 2(\log a+\log b)^2 = 9\log a \log b \\ \Rightarrow 2(\log a)^2-5(\log a)(\log b)+ 2(\log b)^2=0 \Rightarrow (2\log a-\log b)(\log a-2\log b)=0 \\ \Rightarrow a^2=b \; (b^2=a 不合, 因b\gt a\gt 1) \\ \sqrt{16a^2+(4b-1)^2} +\sqrt{(4a-6)^2+(4b-11)^2} = 4\left(\sqrt{a^2+(b-{1\over 4})^2} +\sqrt{(a-{3 \over 2})^2 +(b-{11\over 4})^2}\right) \\= 4(\overline{PA}+ \overline{PB}),其中\cases{P(a,b) \in \Gamma:a^2=b 為一拋物線,準線L:y=-1/4\\A(0,1/4)剛好是\Gamma的焦點\\ B(3/2,11/4)} \\ \Rightarrow \overline{PA}+ \overline{PB}的最小值= d(B,L)=3 \Rightarrow 4(\overline{PA}+ \overline{PB}) = \bbox[red, 2pt]{12}$$

解答:$$f(x)=\int_1^x f(t)\,dt-x^4-4x^3-x^2+9x+116 \Rightarrow   \Rightarrow f'(x)=f(x)-4x^3-12x^2-2x+9\\ 令f(x)=ax^3+bx^2+cx+d \Rightarrow f'(x)=3ax^2+2bx+c \\\Rightarrow f'(x)-f(x) =-ax^3+(3a-b)x^2+(2b-c)x+(c-d)= -4x^3-12x^2-2x+9 \\ \Rightarrow \cases{a=4\\ b=24\\ c=50\\ d=41} \Rightarrow f(x)=4x^3+24x^2+50x+41 \Rightarrow f'(x)=12x^2+48x+50\\ \Rightarrow f''(x)=24x+48=0 \Rightarrow x=-2 \Rightarrow 對稱中心(-2,f(-2))=(-2, 5) \\ 由於{8+2\over 2}=5 \Rightarrow (s,f(s))與(k,f(k))剛好對稱(-2,f(-2)) \Rightarrow {s+k\over 2}=-2 \Rightarrow s+k= \bbox[red, 2pt]{-4}$$

解答:

$$圖形有24個頂點，任兩個頂點可決定一個平行四邊形，但需扣除在一直線的任二頂點\\ C^{24}_2-4C^3_2(三個頂點的直線有四條, 上圖藍線)-6C^4_2(四個頂點的直線有6條,上圖紅線)\\ \qquad -6C^5_2(五個頂點的直線有6條,上圖綠線)-C^6_2(六個頂點的直線有1條,上圖棕線) =\bbox[red, 2pt]{153}$$

解答:$$符合條件的點數和為10或19\\ (1)點數和為10的樣本:8個1及1個2的排列數=9\\ (2)點數和為19的樣本:\cases{5個3,4個1:排列數C^9_4=126 \\ 4個3,2個2,3個1:排列數9!/(4!2!3!)=1260 \\ 3個3,4個2,2個1:排列數9!/(4!2!3!)=1260 \\ 2個3,6個2,1個1:排列數9!/(2!6!)= 252 \\1個3,8個2:排列數9!/8!= 9 } \\ 因此機率為{9+126+1260+1260+252 +9\over 3^9} ={2916\over 3^9} = \bbox[red, 2pt]{4\over 27}$$

解答:$$|z|=1 \Rightarrow z=\cos \theta+i\sin \theta \Rightarrow z^2+z-6= \cos 2\theta+\cos \theta-6+i(\sin 2\theta+\sin \theta) \\ \Rightarrow |z^2+z-6|=\sqrt{(\cos 2\theta+ \cos \theta-6)^2+(\sin 2\theta+\sin \theta)^2} \\=\sqrt{38+2(\cos 2\theta \cos \theta+ \sin 2\theta \sin \theta)-12(\cos 2\theta+\cos \theta)} =\sqrt{38+2 \cos \theta -12(\cos 2\theta+\cos \theta)} \\= \sqrt{38-12\cos 2\theta-10\cos \theta} =\sqrt{38-12(2\cos^2\theta-1)-10\cos \theta} \\=\sqrt{-24\cos^2\theta-10\cos \theta+50} =\sqrt{-24(\cos\theta+{5\over 24})^2+50+{25\over 24}} \Rightarrow 最大值=\sqrt{50+{25\over 24}} \\= \bbox[red, 2pt]{35\sqrt 6\over 12}$$

解答:$$a+b=k \Rightarrow a=k-b \Rightarrow (k-b)^3-6(k-b)^2+15(k-b)+2025=0 \\ \Rightarrow b^3+(6-3k)b^2-(-3k^2+12k-15)b-(k^3-6k^2+15k+2025)=0 \\=b^3-15b^2+78b-2197 \Rightarrow 6-3k=-15 \Rightarrow k= \bbox[red, 2pt]7$$

解答:$$a_1=S_1=2a_1-3\times 2^1+4 \Rightarrow a_1=2 \\ a_n=S_n-S_{n-1} =2a_n-3\times 2^n+4-\left(2a_{n-1} -3\times 2^{n-1}+4 \right) =2a_n-2a_{n-1}-3\times 2^{n-1} \\ \Rightarrow a_n=2a_{n-1} +3 \times 2^{n-1} =2^2 a_{n-2}+ 2(3\times 2^{n-1}) =\cdots=2^{n-1}a_1+ (n-1)(3\times 2^{n-1}) \\=2^n+(n-1)(3\times 2^{n-1}) =2^{n-1}\left( 2+3n-3\right) = \bbox[red, 2pt]{(3n-1)2^{n-1}}$$

解答:$$\cases{f(x)=\sqrt{4x-2} \\g(x)= \displaystyle{x^2+2\over 4}} \Rightarrow f(x)與g(x)互為反函數\Rightarrow f(x)=g(x)的解在直線y=x上 \\ \Rightarrow x={x^2+2\over 4} \Rightarrow x^2-4x+2=0 \Rightarrow x={4\pm \sqrt{8}\over 2} = \bbox[red, 2pt]{2\pm \sqrt 2}$$

二、非選擇題與證明題（計 3 題，共 22 分）：

解答:$${a'^2+c'^2-b'^2 \over b'^2+c'^2-a'^2} ={2a'c'\cos B'\over 2b'c'\cos A'} ={a' \cos \beta\over b'\cos(\alpha -\beta)} ={\sin(\alpha-\beta) \cos \beta\over \sin \beta\cos(\alpha -\beta)} \\={{1\over 2}(\sin \alpha +\sin(\alpha-2\beta)) \over {1\over 2}(\sin \alpha+ \sin(2\beta-\alpha))} ={ \sin \alpha +\sin(\alpha-2\beta) \over  \sin \alpha+ \sin(2\beta-\alpha)} ={a+b\over a-b}, \bbox[red, 2pt]{故得證}$$

解答:$${1\over x}+{1\over y}+{1\over z}=2 \Rightarrow 3-\left( {1\over x}+{1\over y}+{1\over z}\right)=3-2 \Rightarrow {x-1\over x}+{y-1\over y}+{z-1\over z}=1 \\ \Rightarrow \left( {x-1\over x}+{y-1\over y}+{z-1\over z}\right)(x+y+z) \ge (\sqrt{x-1}+ \sqrt{y-1}+\sqrt{z-1})^2 \\ \Rightarrow \sqrt{x+y+z} \ge \sqrt{x-1}+ \sqrt{y-1}+\sqrt{z-1}, \bbox[red, 2pt]{故得證}$$

解答:$$E_4法向量\vec n_4=(1,a,b), 正四面體相鄰平面夾角的餘弦值相等 \\ \Rightarrow \cases{a+b+1=\pm(1-a+b) \Rightarrow \cases{a=0\\ b=-1}\\ a+b+1=\pm(1-a-b) \Rightarrow a+b=0} \Rightarrow \cases{a=1\\ b=-1} \Rightarrow E_4:x+y-z=c\\ 又正四面體的高h ={\sqrt 6\over 3}\times 稜長={\sqrt 6\over 3}\times 6\sqrt 2=4\sqrt 3 \Rightarrow d(P,E_4)=h\\ \Rightarrow {|1+2-4-c| \over \sqrt 3} =4\sqrt 3 \Rightarrow |-1-c|=12 \Rightarrow c=11,-13 \\\Rightarrow E_4: \bbox[red, 2pt]{x+y-z=11,x+y-z=-13}$$

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