111年中壢高中教甄-數學詳解

國立中央大學附屬中壢高級中學111學年度第1次教師甄選

一、填充題：(每題7分，共計70分。)

解答：$${1\cdot 3\cdot 5\over 3^6-64} +{3\cdot 5\cdot 7\over 5^6-64} +{5\cdot 7\cdot 9\over 7^6-64} +\cdots +{19\cdot 21\cdot 23\over 21^6-64} =\sum_{k=1}^{10}{(2k-1)(2k+1)(2k+3)\over ((2k+1)^3)^2-8^2} \\ =\sum_{k=1}^{10}{(2k-1)(2k+1)(2k+3)\over ((2k+1)^3)-8)((2k+1)^3)+8)} \\ =\sum_{k=1}^{10}{(2k-1)(2k+1)(2k+3)\over (2k-1)((2k+1)^2+2(2k+1)+4)(2k+3)((2k+1)^2-2(2k+1)+4)} \\ =\sum_{k=1}^{10}{ (2k+1) \over  ((2k+1)^2+2(2k+1)+4) ((2k+1)^2-2(2k+1)+4)} \\=\sum_{k=1}^{10}{ (2k+1) \over  (4k^2+3)(4k^2+8k+7) } =\sum_{k=1}^{10}{ (2k+1) \over  (4k^2+3)(4(k+1)^2+3) } \\={1\over 4}\sum_{k=1}^{10} \left({ 1\over  (4k^2+3)}-{1\over (4(k+1)^2+3) }\right) ={1\over 4}\left({1\over 7}-{1\over 487} \right) ={1\over 4}\cdot {480\over 3409} =\bbox[red,2pt]{120\over 3409}$$

解答：

$$f(x)= 3x^4+4x^3-12x^2-1 \Rightarrow f'(x)=12x^3+12x^2-24x=0 \Rightarrow f''(x)=36x^2+24x-24\\ 若f'(x)=0 \Rightarrow x(x^2+x-2)= x(x+2)(x-1)=0 \Rightarrow 0,1,-2為f'(x)=0的根\\且\cases{f''(-2)=18 \gt 0\\ f''(0)= -24\lt 0\\ f''(1)=36\gt 0} \Rightarrow \cases{f(-2)= -33為極小值\\f(0)=-1 為極大值\\ f(1)= -6為極小值};\\由於f(x)為四次式且首項係收為正值，因此圖形為凹向上，而三極值皆為負值；\\因此f(x)有兩相異實根x_1,x_2，且\cases{x_1\lt -2\\ x_2\gt 1}，再考慮\cases{f(-3)=26\gt 0\\ f(2)=31 \gt 0}\Rightarrow \cases{-3\lt x_1\lt -2\\ 1\lt x\lt 2}；\\g(x)=f(f(x))=0 \Rightarrow \cases{-3\lt f(x)\lt -2 \Rightarrow 4相異實根\\ 1\lt f(x) \lt 2\Rightarrow 2相異實根} \Rightarrow 共\bbox[red, 2pt]{6}相異實根$$

解答：$$\cases{L_1:x+y=0\\ L_2:x+3y=0} \Rightarrow \cases{A\in L_1\\ B\in L_2\\ 重心G=L_1\cap L_2} \Rightarrow \cases{A(a,-a)\\ B(-3b,b)\\ G(0,0)},a,b\in \mathbb{R}\\ G=(A+B+C)\div 3 \Rightarrow C(-a+3b,a-b) \Rightarrow \cases{\overrightarrow{CA}= (2a-3b, -2a+b)\\ \overrightarrow{CB} =(a-6b,-a+2b)}\\ 因此\cases{\overline{AB}=18\\ \overrightarrow{CA}\cdot \overrightarrow{CB}=0} \Rightarrow \cases{(a+3b)^2+(-a-b)^2=18^2 \\ (2a-3b)(a-6b)+(-2a+b)(-a+2b)=0} \Rightarrow ab=18\\ \Rightarrow \triangle ABC =3\triangle GAB = {3\over 2} \begin{Vmatrix}a & -a\\ -3b & b \end{Vmatrix} =3ab =3\times 18= \bbox[red, 2pt]{54}$$

解答：

$$每一內角為{(6-2) \cdot 180\over 6}=120度，因此假設\cases{A(0,0)\\ B(2,0)} \Rightarrow C(2+ 4\cos 60^\circ, 4\sin 60^\circ)=(4,2\sqrt 3) \\ \Rightarrow D(4-6\cos 60^\circ, 2\sqrt 3+6\sin 60^\circ) =(1,5\sqrt 3) \Rightarrow E(1-4,5\sqrt 3)=(-3, 5\sqrt 3)\\ \Rightarrow \cases{\overleftrightarrow{EF} 斜率=\sqrt 3\\ \overleftrightarrow{AF} 斜率=-\sqrt 3} \Rightarrow \cases{\overleftrightarrow{EF} :y=\sqrt 3(x+3)+5\sqrt 3\\ \overleftrightarrow{AF} :y=-\sqrt 3 x} \Rightarrow F= \overleftrightarrow{EF} \cap \overleftrightarrow{AF} =(-4,4\sqrt 3)\\ \Rightarrow \cases{\overline{EF}=\sqrt{1+3}=2 \\ \overline{AF} =\sqrt{16+48}=8} \Rightarrow \overline{EF}+\overline{FA} =\bbox[red, 2pt]{10}$$

解答：$$\cases{|b-c|\cos(A/2) = 8\sqrt 3 \cdots(1)\\ (b+c)\sin (A/2)=13 \cdots(2)},(1)^2+(2)^2 =(b-c)^2\cos^2(A/2) +(b+c)^2\sin^2(A/2)= 361\\ \Rightarrow b^2+c^2-2bc(\cos^2{(A/2)}- \sin^2(A/2)) =361 \Rightarrow b^2+c^2-361=2bc \cos A\\ \Rightarrow \cos A={b^2+c^2-19^2\over 2bc} \Rightarrow a= \bbox[red, 2pt]{19}$$

解答：$$兩平面\cases{x+y+z+4=0\\ x+y+z-2=0} 的距離={6\over \sqrt 3}=2\sqrt 3 \\ \Rightarrow 稜長=2\sqrt 3\times \sqrt{3\over 2} =3\sqrt 2 \Rightarrow 體積={\sqrt 2\over 3}\times (3\sqrt 2)^3 = \bbox[red, 2pt]{36}$$

解答：$$A在直線L:{x-4\over -1} ={y+4\over 2} ={z+14\over 6}上\Rightarrow A(-t+4,2t-4,6t-14),t\in \mathbb{R}\\ \Rightarrow \cases{\vec p=\overrightarrow{AP} =(t-1,-2t+4,-6t+19)\\ \vec q=\overrightarrow{AQ} =(t+7,-2t-5,-6t-4)\\ L的方向向量\vec h= \overrightarrow{AH}=(-1,2,6)} \Rightarrow \cases{\cos \angle PAH = \vec p\cdot \vec h/(|\vec p||\vec h|) \\ \cos \angle QAH =\vec q\cdot \vec h/(|\vec q||\vec h||} \\ 由於\angle PAH=\angle QAH\Rightarrow {\vec p \cdot \vec h \over |\vec p||\vec h|} ={\vec q\cdot \vec h\over |\vec q||\vec h|}\\ \Rightarrow {(-41t+123)^2 \over (t-1)^2 +(2t-4)^2 +(6t-19)^2} ={(-41t-41)^2 \over (t+7)^2+(2t+5)^2+ (6t+4)^2}\\ \Rightarrow  {41^2(t-3)^2\over 41(t-3)^2+9} ={41^2(t+1)^2 \over 41(t+1)^2+49}\\ \Rightarrow 41(t+1)^2(t-3)^2+ 49(t-3)^2 = 41(t+1)^2(t-3)^2 +9(t+1)^2\\ \Rightarrow 49(t-3)^2 = 9(t+1)^2 \Rightarrow 7(t-3) =\pm 3(t+1) \Rightarrow \cases{t=6\\ t=9/5 \Rightarrow \cases{\vec p\cdot \vec h\gt 0\\ \vec q\cdot \vec h\lt 0}，不合} \\ \Rightarrow A=(-6+4,12-4,36-14)= \bbox[red, 2pt]{(-2,8,22)}$$

解答：$$\cases{5P+ 2Q= A\\ P+Q=I_2} \Rightarrow \cases{P=(A-2I_2)\div 3\\ Q=(5I_2-A)\div 3} \Rightarrow \cases{P= \left[\begin{matrix}3 & 3\\-2 & -2\end{matrix}\right] \\[1ex] Q=\left[\begin{matrix}-2 & -3\\2 & 3\end{matrix}\right]}\\ 又A=\left[\begin{matrix}11 & 9\\-6 & -4\end{matrix}\right] \Rightarrow A^2= \left[\begin{matrix}67 & 63\\-42 & -38\end{matrix}\right] \Rightarrow A^4=\left[\begin{matrix}1843 & 1827\\-1218 & -1202\end{matrix}\right] =aP+bQ \\ =\left[\begin{matrix}3a-2b & 3a-3b\\-2a+2b & -2a+3b\end{matrix}\right] \Rightarrow \cases{a=625 \\ b=16} \Rightarrow \log_{10} ab= \log_{10} 10000= \bbox[red, 2pt]{4}$$

解答：$$ z={32\over x-yi} ={32\over x^2+y^2}(x+yi) 代入圓(x+2)^2+(y-2)^2=8 \\\Rightarrow \left({32x\over x^2+y^2}+2 \right)^2+\left({32y\over x^2+y^2}-2 \right)^2=8 \\ \Rightarrow (2x^2+2y^2+32x)^2 +(2x^2+2y^2-32y)^2 =8(x^2+y^2)^2 \\ \Rightarrow (x^2+y^2+ 16x)^2 +(x^2+y^2- 16y)^2 = 2(x^2+y^2)^2 \\ \Rightarrow (x^2+y^2)^2 +32x(x^2+y^2)+256x^2 +(x^2+y^2)^2 -32y(x^2+y^2)+ 256y^2 = 2(x^2+y^2)^2\\ \Rightarrow 32x(x^2+y^2)-32y(x^2+y^2)+256x^2+256y^2 =0 \\ \Rightarrow (x^2+y^2)(32x-32y+256)=0 \Rightarrow 32x-32y+256=0 \Rightarrow \bbox[red, 2pt]{x-y+8=0}$$

解答：$$\cases{P(A\to B)= P(A\to C)= 1/2 \\ P(B\to C)= P(B\to A)= 1/2\\ P(C\to A)= P(C\to B)=1/2\\ P(A\to A)= P(B\to B) = P(C\to C)=0} \Rightarrow 轉移矩陣A=\begin{bmatrix} 0 &1/2 & 1/2\\ 1/2 & 0 & 1/2 \\ 1/2 & 1/2 & 0\end{bmatrix} \\\Rightarrow A^5= \begin{bmatrix} 5/16 &11/32 & 11/32\\ 11/32 & 5/16 & 11/32 \\ 11/32 & 11/32 & 5/16\end{bmatrix} \Rightarrow \cases{A^5 \begin{bmatrix} 1\\ 0\\ 0\end{bmatrix} = \begin{bmatrix} 5/16\\ 11/32\\ 11/32\end{bmatrix} \\[1ex]A^5 \begin{bmatrix} 0\\ 1\\ 0\end{bmatrix} = \begin{bmatrix}11/32\\ 5/16\\  11/32\end{bmatrix} \\[1ex]A^5 \begin{bmatrix} 0\\ 0\\ 1\end{bmatrix} = \begin{bmatrix}  11/32\\ 11/32 \\5/16 \end{bmatrix} } \\ \Rightarrow  同時在A的機率= 同時在B的機率= 同時在C的機率={5\over 16}\cdot ({11\over 32} )^2 ={605\over 16384} \\ \Rightarrow 三人在相同位置的機率=3\times {605\over 16384} =\bbox[red, 2pt]{1815\over 16384}$$

二、 計算證明題：（ 每題 10 分， 共計 30 分。 ）

解答：$$f(x)={x\over x^2-2x+3} +{x^2+3\over 4x} \Rightarrow f'(x)={ (x^2-3)((x^2-2x+3)^2-4x^2) \over 4x^2(x^2-2x+3)^2} \\ 因此f'(x) = 0\Rightarrow \cases{x=\sqrt 3\\ x^2-2x+3=\pm 2x \Rightarrow x=1,3} \Rightarrow \cases{f(\sqrt 3)=(1+3\sqrt 3)/4\\ f(1)=f(3) = 3/2}\\ \Rightarrow \cases{(1)\bbox[red, 2pt]{3/2}\\ (2)\bbox[red, 2pt]{1或3}}$$

解答：$$a_n:長度為n的不同字串數目\Rightarrow \cases{a_1= 7(1,2,3,4,5,6,a)\\ a_2=13 (\#A,A\#,AA\Rightarrow  6+6+1=13,\#代表數字)}\\ 由於\cases{在長度為n-1的字串的最左邊加上A就是長度為n的字串\\ 在長度為n-2的字串的最左邊加上\#A就是長度為n的字串}\\ \Rightarrow a_n = a_{n-1}+6a_{n-2} \Rightarrow a_n-a_{n-1}-6a_{n-2}=0 \Rightarrow \lambda^2-\lambda-6=0\\ \Rightarrow \lambda =3,-2 \Rightarrow a_n =C_13^n +C_2(-2)^n \Rightarrow \cases{a_1=3C_1-2C_2= 7\\ a_2= 9C_1+4C_2 = 13} \Rightarrow \cases{C_1=9/5\\ C_2=-4/5} \\ \Rightarrow a_n={9\over 5}\cdot 3^n-{4\over 5}\cdot (-2)^n \Rightarrow a_5={9\over 5}\cdot 3^5-{4\over 5}\cdot (-2)^5 ={2187+ 128\over 5}=463 \\\Rightarrow \cases{(1)\bbox[red, 2pt]{463}\\ (2) \bbox[red, 2pt]{a_n ={9\over 5}\cdot 3^n-{4\over 5}\cdot (-2)^n}}$$

解答：

(1)$$y=f(x)=|||x|-1|-2|-3的圖形如上，繞x軸旋轉圖形如下$$

$$y=\begin{cases}x-2 & 0\le x\le 1\\ -x& 1\le x\le 3\\ x-6 & 3\le x\le 6 \end{cases} \Rightarrow 右半部繞x軸旋轉體積=\\ \int_0^1 (x-2)^2\pi\,dx +\int_1^3 (-x)^2 \,dx +\int_3^6 (x-6)^2\,dx =\pi \left(\left. \left[{1\over 3}(x-2)^3 \right] \right|_0^1 + \left. \left[ {1\over 3}x^3\right] \right|_1^3 + \left. \left[{1\over 3}(x-6)^3 \right] \right|_3^6 \right)\\ =\pi \left( {7\over 3} +{26\over 3} +9\right) =20\pi \Rightarrow 左右半部合計=20\pi \times 2= \bbox[red, 2pt]{40\pi}$$

(2)$$x=\begin{cases}y+2 & -2\le y\le -1\\ -y & -3\le y\le -1\\ y+6 & -3\le y\le 0 \end{cases} \Rightarrow 繞y軸旋轉體積=\int_{-3}^0 (y+6)^2\pi\,dy -\int_{-3}^{-1} (-y)^2 \,dy +\int_{-2}^{-1} (y+2)^2\,dy\\ =\pi \left(\left. \left[{1\over 3}(y+6)^3 \right] \right|_{-3}^0 - \left. \left[ {1\over 3}y^3\right] \right|_{-3}^{-1} + \left. \left[{1\over 3}(y+2)^3 \right] \right|_{-2}^{-1} \right) ={\pi \over 3}\left( 189-26+1\right) =  \bbox[red, 2pt]{164\pi \over 3}$$ 

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解題僅供參考，其他教甄試題及詳解