107年新北市高中教甄聯招-數學詳解

 新北市立高級中等學校107學年度教師聯合甄選數學科 試題

一、 填充題： 70%，每題 10 分。

解：$$\cfrac{a}{x} +\cfrac{b}{x+\log 2} +\cfrac{c}{x+\log 5} =1 \Rightarrow \cfrac{a(x+\log 2)(x+\log 5)+ bx(x+ \log 5) +cx(x +\log 2)}{x(x+\log 2)(x+ \log 5)}=1\\ \Rightarrow a(x+\log 2)(x+\log 5)+ bx(x+ \log 5) +cx(x +\log 2)- x(x+\log 2)(x+ \log 5)=0\\ 令f(x)= a(x+\log 2)(x+\log 5)+ bx(x+ \log 5) +cx(x +\log 2)- x(x+\log 2)(x+ \log 5)\\\text{則題目可以改成 }f(x)=0的三根為3,7,11，求a+b+c:\\f(x)的x^2係數為-(a+b+c-1) \Rightarrow a+b+c-1= 3+7+11=21 \Rightarrow a+b+c=\bbox[red,2pt]{22}$$

解：$$圓O即為兩圖形\cases{x^2+y^2 +z^2=1 \\ y=z}的交集，即x^2+2y^2=1 \Rightarrow 圓O上的點Q(\cos \theta,{1\over \sqrt 2}\sin \theta,{1\over \sqrt 2}\sin \theta) \\ \Rightarrow \overline{PQ}^2 = (\cos \theta-4)^2 +({1\over \sqrt 2}\sin \theta-4)^2 +({1\over \sqrt 2}\sin \theta-12)^2 \\=\cos^2\theta -8\cos\theta+16 +{1\over 2}\sin^2\theta-{8\over \sqrt 2}\sin \theta+16 +{1\over 2}\sin^2\theta-{24\over \sqrt 2}\sin \theta +144\\ =-8\cos\theta-16\sqrt 2\sin\theta+177 =-24({1\over 3}\cos \theta+ {2\over 3}\sqrt 2\sin \theta)+177 \\ =-24 \sin(\theta+\eta)+177 \Rightarrow \overline{PQ}^2 的最大值為24+177=201 \Rightarrow \overline{PQ}的最大值為\bbox[red,2pt]{\sqrt{211}}$$

解：$$令h(x)=f(x)-g(x)-(x+1) \Rightarrow h(x)=0的3根為1,2,3\\ \Rightarrow f(x)-g(x)=2(x-1)(x-2)(x-3)+(x+1) \\令g(x)=ax^2+bx+c，則f(x)=2(x-1)(x-2)(x-3)+(x+1)+ax^2+bx+c\\ 由3f(1)-3f(2)+f(3)=5 \Rightarrow 3(2+a+b+c)-3(3+4a+2b+c)+4+9a+3b+c=5\\ \Rightarrow 1+c=5 \Rightarrow c=4 \Rightarrow f(0)= 2(-1)(-2)(-3)+1+c=-11+c=-11+4=\bbox[red,2pt]{-7}$$

解：$$\cfrac{10!}{4!2!2!2!}\times C^4_1 +\cfrac{10!}{3!3!2!2!}\times C^4_2 =75600+ 151200 = \bbox[red,2pt]{226800}$$

解：

$$作\overline{AB}上的高\overline{AG}，且\overline{AG}交\overline{DE}於F，並作\overline{DH}\bot \overline{AB}，見上圖；\\假設正\triangle ABC的邊長為a \Rightarrow \cases{\overline{AG}={\sqrt 3\over 2}a \\ \overline{AG}={1\over 2}a};\\ 直角\triangle CDF \Rightarrow \overline{CF} =\sqrt{7-({1\over 2})^2} ={3\over 2}\sqrt 3 \Rightarrow \overline{FG}= \overline{AG}-\overline{AF} ={\sqrt 3\over 2}a-{3\over 2}\sqrt 3 \\ 直角\triangle BDH \Rightarrow 2^2 =({a\over 2}-{1\over 2})^2 + ({\sqrt 3\over 2}a-{3\over 2}\sqrt 3)^2 \Rightarrow (a-1)^2 +(\sqrt 3a-3\sqrt 3)^2=16 \\ \Rightarrow a^2-5a+3=0 \Rightarrow a=\bbox[red,2pt]{5+\sqrt{13}\over 2}$$

解：

$$\pi\int_{-1}^1 (2-x^2)^2-(2-1)^2 \;dx =\pi \int_{-1}^1 x^4-4x^2+3\;dx =\pi \left. \left[ {1\over 5}x^5-{4\over 3}x^3+3x \right] \right|_{-1}^1 \\ =\pi({28\over 15}-(-{28\over 15})) = \bbox[red,2pt]{{56\over 15}\pi}$$

解：$$\sqrt{1-{1729\over p^n}} = {b\over a} ,a,b\in {Z}且(a,b)=1 \Rightarrow 1-{1729\over p^n}={b^2\over a^2} \Rightarrow a^2p^n-1729 a^2=b^2 p^n \\ \Rightarrow p^n(a^2-b^2) =1729a^2 \Rightarrow p^n(a+b)(a-b) = 7\cdot 13\cdot 19\cdot a^2\\ (a,b) =1 \Rightarrow \cases{(a,a+b)=1\\ (a,a-b)=1}又n是正偶數 \Rightarrow \cases{p^n=a^2 \\ (a+b)(a-b)= 7\cdot 13\cdot 19} \\ \begin{array}{c|c|l} a+b & a-b & a \\\hline 7 & 13\cdot 19 & 127 \\ 13\cdot 19 & 7 & \\\hdashline 13 & 7\cdot 19 & 73 \\ 7\cdot 19 & 13 & \\\hdashline 19 & 7\cdot 13 & 55(非質數)\\ 7\cdot 13 & 19 \\\hdashline 1 & 1729 & 865(非質數) \\ 1729 & 1 & \\\hline\end{array} \Rightarrow p^n=\cases{127^2\\ 73^2} \Rightarrow p=\bbox[red,2pt]{73,127}$$

解：

(a)$$\begin{vmatrix}x & x^3+ax^2 & b\\ y & y^3+ay^2 & b\\ z & z^3+az^2 & b \end{vmatrix} =b\begin{vmatrix}x & x^3+ax^2 & 1\\ y & y^3+ay^2 & 1\\ z & z^3+az^2 & 1 \end{vmatrix} =b\begin{vmatrix}x-y & x^3-y^3+a(x^2-y^2) & 0\\ y & y^3+ay^2 & 1\\ z & z^3+az^2 & 1 \end{vmatrix}\\ =b\begin{vmatrix}x-y & x^3-y^3+a(x^2-y^2) & 0\\ y & y^3+ay^2 & 1\\ z & z^3+az^2 & 1 \end{vmatrix}  = b\begin{vmatrix}x-y & x^3-y^3+a(x^2-y^2) & 0\\ y-z & y^3-z^3+a(y^2-z^2) & 0\\ z & z^3+az^2 & 1 \end{vmatrix} \\= b((x-y)(y^3-z^3+a(y^2-z^2) -(y-z)(x^3-y^3+a(x^2-y^2)) \\ =b[(x-y)((y-z)(y^2+yz+z^2)+a(y-z)(y+z)-(y-z)((x-y)(x^2+xy+y^2) +a(x-y)(x+y)] \\ =b(x-y)(y-z)[y^2+yz+z^2 +ay+az-(x^2+xy +y^2+ax+ay)] \\ =b(x-y)(y-z)(yz+z^2+az-x^2-xy-ax) \\ =b(x-y)(y-z)((z+x)(z-x)+ y(z-x)z+a(z-x)) \\ =\bbox[red,2pt]{b(x-y)(y-z)(z-x)(x+y+z+a)}$$(b)$$三點\cases{A_i(x_i,y_i)\\ A_j(x_j,y_j) \\ A_k(x_k,y_k)} 共線的充要條件為\begin{vmatrix} x_i & y_i & 1 \\ x_j & y_j & 1 \\ x_k & y_k & 1 \end{vmatrix}=0;\\ 由(a)知\begin{vmatrix} x & x^3+ax^2 & b \\ y & y^3+ay^2 & b \\ z & z^3+az^2 & b \end{vmatrix}= (x-y)(y-z)(z-x)(x+y+z+a);\\ 因此我們取\cases{b=1\\ a=-110\\ A_i(i,i^3-110i^2)\\ A_j(j,j^3-110j^2)\\ A_k(k,k^3-110k^2)} \Rightarrow \begin{vmatrix} i & i^3-110i^2 & 1 \\ j & j^3-110j^2 & 1 \\ k & k^3-110k^2 & 1 \end{vmatrix}= (i-j)(j-k)(k-i)(i+j+k-110)\\ 因此任相異三點(即i,j,k互異)共線的充要條件為i+j+k-110=0，即i+j+k=110；\\也就是我們已經找到107個相異點A_n(n,n^3-110n^2),1\le n \le 107滿足其要求，\bbox[red,2pt]{故得證}。$$

解：

$$\cases{\triangle APB: \cos \angle APB={\overline{AP}^2 +\overline{BP}^2-\overline{AB}^2 \over 2\overline{AP}\times \overline{BP}} \\ \triangle APC: \cos \angle APC={\overline{AP}^2 +\overline{CP}^2-\overline{AC}^2 \over 2\overline{AP}\times \overline{CP}}};\\由於\angle APB+\angle APC=180^\circ \Rightarrow \cos \angle APB= -\cos \angle APC \\\Rightarrow \overline{CP}(\overline{AP}^2 +\overline{BP}^2-\overline{AB}^2) =-\overline{BP}(\overline{AP}^2 +\overline{CP}^2-\overline{AC}^2) \\ \Rightarrow \overline{AP}^2(\overline{CP}+\overline{BP}) +\overline{CP} \cdot \overline{BP}(\overline{BP} +\overline{CP}) =\overline{AB}^2\cdot \overline{CP} +\overline{AC}^2\cdot \overline{BP} \\ \Rightarrow \overline{AP}^2 \cdot \overline{BC} +\overline{CP} \cdot \overline{BP}\cdot \overline{BC} = \overline{AB}^2\cdot \overline{CP} +\overline{AC}^2\cdot \overline{BP} \\ \Rightarrow (\overline{AP}^2 + \overline{BP} \cdot \overline{CP})\overline{BC} =\overline{AB}^2\cdot \overline{CP} +\overline{AC}^2\cdot \overline{BP} ，\bbox[red,2pt]{故得證}。$$