113年嘉義中學教甄-數學詳解

 國立嘉義高級中學113學年度第1次教師甄選

一、 填充題：共 18 題，每題 5 分，合計 90 分。

解答: $$假設\log_a b=t \Rightarrow t+ {6\over t}=5  \Rightarrow t^2-5t+6=0 \Rightarrow (t-3)(t-2)=0 \\ \Rightarrow \cases{\log_a b=3 \Rightarrow b=a^3 \Rightarrow (a,b) =(2,8), (3,27),\dots,(12,1728),共11組\\ \log_a b=2 \Rightarrow b=a^2 \Rightarrow (a,b)=(2,4), (3,9),\dots, (44,1936),共43組} \Rightarrow ,共11+43= \bbox[red, 2pt]{54}組$$

解答:

$$\cases{Q(a_2,b_2),Q'(a_2+3,b_2+4) \in L_3 \Rightarrow m_3=4/3\\ R(a_3,b_3) ,R'(a_3+5,b_3-12) \in L_2 \Rightarrow m_2=-12/5}\\ m_1=\overline{QR}斜率= \overline{Q'R'}斜率 \Rightarrow \cases{{b_3-b_2\over a_3-a_2} =m_1 \cdots(1)\\ {b_3-b_2-16\over a_3-a_2+2} =m_1 \cdots(2)} \\ 由(1) \Rightarrow b_3-b_2=m_1 (a_3-a_2)代入(2) \Rightarrow {m_1(a_3-a_2)-16\over a_3-a_2 +2}=m_1 \Rightarrow -16=2m_1 \Rightarrow m_1=-8\\ \Rightarrow (m_1,m_2,m_3) =\bbox[red, 2pt]{ \left(-8,-{12\over 5},{4\over3} \right)}$$

解答: $$\left|{5-12i\over 13} -z^3\right| = \left|{5-12i\over 13} -z^5\right|  \Rightarrow z^4= {5-12i\over 13}\\ 假設z^2=e^{i\theta} \Rightarrow z^4=e^{i2\theta} ={5-12i\over 13} \Rightarrow \tan 2\theta={-12/13\over 5/13} =-{12\over 5} ={2\tan \theta\over 1-\tan^2 \theta} \\ \Rightarrow 6\tan^2 \theta-5\tan \theta-6=0 \Rightarrow (3\tan \theta+2) (2\tan \theta-3) =0 \Rightarrow {b\over a} =\tan \theta= \bbox[red, 2pt]{-{2\over 3}} \\ z^2在第二象限\Rightarrow \tan \theta\lt 0 \Rightarrow \tan \theta \ne {3\over 2}$$

解答: $$轉換矩陣A=\begin{bmatrix} 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0 \\1 & 0 & 1 & 0 & 0 & 1 & 0 & 0 \\0 & 1 & 0 & 1 & 0 & 0 & 1 & 0 \\1 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\1 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\0 & 1 & 0 & 0 & 1 & 0 & 1 & 0 \\0 & 0 & 1 & 0 & 0 & 1 & 0 & 1 \\0 & 0 & 0 & 1 & 1 & 0 & 1 & 0\end{bmatrix}\\ \Rightarrow A^9= \begin{bmatrix}0 & 4921 & 0 & 4921 & 4921 & 0 & 4920 & 0 \\4921 & 0 & 4921 & 0 & 0 & 4921 & 0 & 4920 \\0 & 4921 & 0 & 4921 & 4920 & 0 & 4921 & 0 \\4921 & 0 & 4921 & 0 & 0 & 4920 & 0 & 4921 \\4921 & 0 & 4920 & 0 & 0 & 4921 & 0 & 4921 \\0 & 4921 & 0 & 4920 & 4921 & 0 & 4921 & 0 \\4920 & 0 & 4921 & 0 & 0 & 4921 & 0 & 4921 \\0 & 4920 & 0 & 4921 & 4921 & 0 & 4921 & 0 \end{bmatrix}\\ \Rightarrow A^9\begin{bmatrix}1 \\0 \\0 \\0 \\0 \\0 \\0 \\0\end{bmatrix} =\begin{bmatrix}0 \\4921 \\0 \\4921 \\4921 \\0 \\4920 \\0 \end{bmatrix} \Rightarrow A\to G有\bbox[red, 2pt]{4920}種路線$$

解答: $$假設t=\cos x+ \sin x \Rightarrow t^2=1+2\sin x\cos x \Rightarrow \sin x\cos x={t^2-1\over 2} \\ \Rightarrow \cases{\tan x+\cot x=\displaystyle {\sin x\over \cos x}+{\cos x\over \sin x} ={1\over \cos x\sin x} ={2\over t^2-1} \\  \sec x+\csc x=\displaystyle {1\over \cos x}+{1\over \sin x} ={\sin x+\cos x\over \cos x\sin x} ={t\over (t^2-1)/2} ={2t\over t^2-1}} \\ \Rightarrow 原式f(x)=g(t)=\left| t+{2\over t^2-1} + {2t\over t^2-1}\right|=\left| t+{2(t+1)\over t^2-1}  \right|=\left| t+{2 \over t-1}  \right|  \\=\left| t-1+{2 \over t-1} +1 \right| \ge \left|-2\sqrt 2+1 \right| =\bbox[red, 2pt]{2\sqrt 2 -1}$$

解答: $$f(x)=-x^3-2x \Rightarrow f'(x)=-3x^2-2 \lt 0 \Rightarrow f(x)為嚴格遞減\\ 因此f(ax^2-3ax) \gt f(a+26) \Rightarrow ax^2-3ax \lt a+26 \Rightarrow g(x)=ax^2-3ax-a-26\lt 0 \\ \Rightarrow \cases{判別式\lt 0 \Rightarrow 9a^2+4a(a+26) \lt 0 \Rightarrow 13a(a+8)\lt 0 \Rightarrow -8\lt a\lt 0 \\ a=0 \Rightarrow -26\lt 0符合要求} \\ \Rightarrow a=-7,-6,\dots,-1,0，共\bbox[red, 2pt]8個整數$$

解答: $$點數和=10的情形:(a,b,c) =(1,3,6),(1,4,5),(2,2,6),(2,3,5), (2,4,4),(3,3,4)\\ 其排列數總和=6+6+3+6+3+3=27 \Rightarrow 機率p={27\over 6^3} ={1\over 8} \\ \Rightarrow 期望值= 50\cdot p+50\cdot p^2+\cdots =50(p+p^2+ \cdots) =50\cdot {p\over 1-p} = \bbox[red, 2pt]{50\over 7}$$

解答: $$取g(x)=xf(x)-pqr \Rightarrow g(p)= g(q) =g(r)=0 \Rightarrow g(x)=xf(x)-pqr = a(x-p)(x-q)(x-r) \\ \Rightarrow f(x)={1\over x}(ax^3-a(p+q+r)x^2+ a(pq+qr+rp)x-apqr +pqr) \\ 由於f(x)為二次式，因此-apqr +pqr=0 \Rightarrow a=1 \Rightarrow f(x)=x^2-(p+q+r)x+pq+qr+ rp \\ \Rightarrow f(p+q+r) =(p+q+r)^2-(p+q+r)^2+pq+qr+rp = \bbox[red, 2pt]{pq+qr+rp}$$

解答: $${1\over n} \sum_{k=1}^{10n} {400n^2\over 400n^2+(2k-1)^2} ={1\over n} \sum_{k=1}^{20n} {400n^2\over 400n^2+k^2} -{1\over n} \sum_{k=1}^{10n} {400n^2\over 400n^2+(2k)^2} \\={1\over n} \sum_{k=1}^{20n} {400\over 400+(k/n)^2} -{1\over n} \sum_{k=1}^{10n} { 400\over  400+4(k/n)^2} =\int_0^{20} {400\over 400+x^2}\,dx -\int_0^{10} {400\over 400+4x^2}\,dx \\ =\int_0^{20} {1\over 1+(x/20)^2}\,dx -\int_0^{10} {1 \over 1+(x/10)^2}\,dx =\left. \left[ 20\tan^{-1} {x\over 20}\right] \right|_0^{20} -\left. \left[ 10 \tan^{-1}{x\over 10}\right] \right|_0^{10}  \\=20\cdot {\pi\over 4}-10\cdot {\pi\over 4} = \bbox[red, 2pt]{{5\over 2}\pi}$$

解答:

$$\begin{bmatrix}{7\over 5} \cos \theta &  -{7\over 5} \sin \theta\\ {7\over 5} \sin \theta & {7\over 5} \cos \theta\end{bmatrix} ={7\over 5}\begin{bmatrix} \cos \theta &    \sin \theta\\   \sin \theta &  \cos \theta\end{bmatrix} 即旋轉\theta後再將與原點距離變為原來的{7\over 5}倍\\ A(10,10) \Rightarrow \overline{OA} =10\sqrt 2 \Rightarrow \overline{OA'} =10\sqrt 2\times {7\over 5}=14\sqrt 2 \Rightarrow 變換後的邊長變為14\sqrt 2\cdot \sqrt 2=28\\ \Rightarrow 變換後的面積變為28\cdot 28={1\over 2}\cdot 14\sqrt 2\cdot 10\sqrt 2\sin \theta\cdot 4 +{1\over 2}\cdot 14\sqrt 2\cdot 10\sqrt 2\cos \theta\cdot 4 \\ \Rightarrow 784=560(\sin \theta+ \cos \theta) \Rightarrow \sin \theta+ \cos \theta={7\over 5} \Rightarrow 1+2\sin \theta\cos \theta={49\over 25} \\ \Rightarrow \sin \theta\cos \theta={12\over 25} \Rightarrow \cos \theta={12\over 25\sin \theta} \Rightarrow \sin \theta+{12\over 25\sin \theta}={7\over 5} \Rightarrow 25\sin^2\theta-35\sin \theta+12=0 \\ \Rightarrow (5\sin \theta-4) (5\sin \theta-3)=0 \Rightarrow \sin \theta= \bbox[red, 2pt]{4\over 5}\\  {\pi \over 4}\lt \theta\lt {\pi\over 2} \Rightarrow {\sqrt 2\over 2}\lt \sin \theta\lt 1 ,而{3\over 5}\lt {\sqrt 2\over 2}, 因此\sin \theta={3\over 5}不合$$

解答: $$P\in L:{x-1\over 2} ={y\over 1} ={z-3\over -2} \Rightarrow P(2t+1,t,-2t+3) \\\Rightarrow \overline{PA} +\overline{PB} =\sqrt{(2t-6)^2+ (t-6)^2 +(-2t)^2} +\sqrt{(2t-4)^2 +(t+1)^2 +(-2t+1)^2} \\=\sqrt{9t^2-36t+72} +\sqrt{9t^2-18t+18} =3\left(\sqrt{t^2-4t+8} +\sqrt{t^2-2t+2} \right) \\=3 \left(\sqrt{(t-2)^2+2^2} +\sqrt{(t-1)^2+1^2}\right) =3(\overline{QC} +\overline{QD}),其中\cases{Q(t,0) \in x軸\\ C(2,2) \\ D(1,1)} \\ D對稱x軸的對稱點D'(1,-1) \Rightarrow Q=\overline{CD'}\cap x軸 =({4\over 3},0) \Rightarrow t={4\over 3} \Rightarrow P=\bbox[red, 2pt]{({11\over 3},{4\over 3},{1\over 3})}$$

解答:

$$假設圓半徑r \Rightarrow \sin \theta={2\over 2+\sqrt{4r^2-1}} ={1\over 2r} \Rightarrow 12r^2-16r+5=0 \Rightarrow (6r-5)(2r-1)=0\\ \Rightarrow r={5\over 6} (r={1\over 2}不合,違反斜邊最長) \Rightarrow 直徑=\bbox[red, 2pt]{5\over 3}$$

解答: $$P\in \Gamma:{(x-1)^2\over 4} +{(y+2)^2\over 9}=1 \Rightarrow P(2\cos \theta+1, 3\sin \theta-2) \\ \Rightarrow d(P,L) ={|4\cos \theta-3\sin \theta+10 |\over \sqrt 5} \Rightarrow 當\cases{\cos \theta=4/5 \\ \sin \theta=-3/5}時，d(P,L)有最大值\\,此時P\bbox[red, 2pt]{({13\over 5}, -{19\over 5})}$$

解答: $$\cases{A(1,2,3) \\ B(2,3,1)\\ C(2t,1,-1+t)} \Rightarrow \cases{\overrightarrow{AB} =(1,1,-2) \\ \overrightarrow{AC} =(2t-1,-1,-4+t)} \Rightarrow \vec n=\overrightarrow{AB} \times \overrightarrow{AC} = (-6+t,6-5t,-2t) \\ \Rightarrow S_{\triangle ABC} ={1\over 2}||\vec n|| \le 3\sqrt 5 \Rightarrow (t-6)^2+(5t-6)^2+4t^2 \le 36\cdot 5 \Rightarrow 5t^2-12t-18 \le 0 \\ \Rightarrow {6-3\sqrt{14}\over 5}(\approx-1.04)  \le t\le {6+3\sqrt{14}\over 5} (\approx 3.44) \Rightarrow t=-1,0,1,2,3，共\bbox[red, 2pt]5個$$

解答:

解答:

$$16-a^2 =27-(5-a)^2 \Rightarrow a={7\over 5} \Rightarrow h= \sqrt{16-a^2} = {3\over 5} \sqrt{39} \\ \Rightarrow 四角錐體積={1\over 3}\cdot 6\cdot 5\cdot {3\over 5} \sqrt{39}= \bbox[red, 2pt]{6\sqrt{39}}$$

解答:

$$此題相當於求兩圖形\cases{y=f(x)=\begin{cases} x^2-2x &x\ge 2\\ -x^2+2x & x\le 2\end{cases} \\y=g(x) =4x+k}只有一個交點時，k的範圍?\\ 直線y=g(x)的斜率為4 \Rightarrow \cases{x\ge 2\Rightarrow f'(x)=2x-2=4 \Rightarrow x=3 \Rightarrow f(3)=3\\ x\le 2 \Rightarrow f'(x) =-2x+2=4\Rightarrow x=-1 \Rightarrow f(-1)=-3} \\ \Rightarrow \cases{g(3)=12+k\lt 3 \Rightarrow k\lt -9\\ g(-1) =-4+k\gt -3 \Rightarrow k\gt 1} \Rightarrow \bbox[red, 2pt]{k\lt -9 或k\gt 1}$$

解答: $$借貸兩年本利和=10^6\times 1.02^{24}元，若每月底償還a元，則{a(1.02^{24}-1)\over 1.02-1} \ge 10^6\times 1.02^{24} \\ \Rightarrow a\ge {10^6\times 1.02^{24}\times 0.02\over  1.02^{24}-1} \approx 52871.5 \Rightarrow a=\bbox[red, 2pt]{52872}$$

解答:

$$假設圓柱體\cases{高=h\\ 底面圓半徑=r} \Rightarrow {r\over 3} ={12-h\over 12} \Rightarrow r=3-{h\over 4} \\ \Rightarrow 圓柱體體積=f(h)=\left(3-{h\over 4} \right)^2\pi\cdot h \Rightarrow f'(h)=-{1 \over 2}\left(3-{h\over 4} \right) \pi\cdot h +\left(3-{h\over 4} \right)^2\pi =0 \\ \Rightarrow \pi(3-{h\over 4})(3-{3\over 4}h)=0 \Rightarrow h=4 \;(h=12不合) \Rightarrow f(h)=\bbox[red, 2pt]{16\pi}$$

二、 計算證明題： 兩小題， 配分寫在各小題後， 共 10 分

解答: $$假設\cases{A(x_a,y_a)\\ B(x_b,y_b)\\ C(x_c,y_c)}經變換後\cases{A'(ax_a+ by_a, cx_a+dy_a) \\B'(ax_b+ by_b, cx_b+dy_b) \\C'(ax_c+ by_c, cx_c+dy_c) } \\ \Rightarrow \cases{\Delta={1\over 2} \begin{Vmatrix}x_a&y_a &1 \\x_b&y_b &1 \\x_c&y_c &1 \end{Vmatrix} = {1\over 2} \begin{Vmatrix}0&0 &1 \\x_b-x_a&y_b-y_a &1 \\x_c-x_a &y_c-y_a &1 \end{Vmatrix}\\ \Delta'={1\over 2} \begin{Vmatrix}ax_a+by_a &cx_a+ dy_a &1 \\ax_b+ by_b& cx_b+ dy_b &1 \\ax_c+ by_c& cx_c+ dy_c &1 \end{Vmatrix} ={1\over 2} \begin{Vmatrix}0 &0 &1 \\a(x_b-x_a)+ b(y_b-y_a) & c(x_b-x_a)+ d(y_b-y_a) &1 \\a(x_c-x_a)+ b(y_c-y_a)& c(x_c-x_a) + d(y_c-y_a) &1 \end{Vmatrix} } \\ \Rightarrow \Delta'=(ad-bc)\Delta (用力算),\bbox[red, 2pt]{故得證}$$

解答: $$令\cases{u=2x+y-113\\ v=x+3y-2024} \Rightarrow \left|{\partial(u,v) \over \partial (x,y)} \right| =\begin{Vmatrix} 2& 1 \\1 & 3\end{Vmatrix} =5\\ |u|+|v|=5所圍面積=50 \Rightarrow |2x+y-113|+ |x+3y-2024|所圍面積={50\over 5}=\bbox[red, 2pt]{10}$$

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解題僅供參考，其他教甄試題及詳解