113年桃園市高中聯合教甄-數學詳解

桃園市立高級中等學校113學年度教師聯合甄選筆試試題

科目： 數學科

第壹部份：填充題（ 共5題，占40分）

解答: $$f(x)=x^6+ax^5+bx^4+cx^3+dx^2+ex+f \Rightarrow f(x)=0的六根中,任四根乘積之和=d\\ 因此取f(x)= (x-\sin{2\pi \over 7})(x-\sin{4\pi \over 7}) (x-\sin{6\pi \over 7}) (x-\sin{8\pi \over 7})(x-\sin{10\pi \over 7})(x-\sin{12\pi \over 7})\\ \sin 7\theta=-64\sin^7\theta+112\sin^5\theta -56\sin^3\theta +7\sin \theta \\\qquad =-64\sin \theta(\sin^6 \theta-{7\over 4}\sin^4\theta+{7\over 8}\sin^2 \theta-{7\over 64})\\當 \theta={2\pi\over 7},{4\pi\over 7},\dots,{12\pi\over 7}時, \sin 7\theta=0,但\sin \theta\ne 0, 也就是說\\ \sin{2\pi\over 7}, \sin {4\pi\over 7},\dots, \sin{12\pi\over 7}是x^6-{7\over 4}x^4+{7\over 8}x^2-{7\over 64}=0的六個根\\ 因此任四根相乘之和=\bbox[red, 2pt]{7\over 8}\\ 註:可用\cos 7\theta+ i\sin 7\theta= (\cos \theta+i\sin \theta)^7 =\sum_{k=0}^7C^7_k \cos^k \theta(i\sin \theta)^{7-k},\\ 左右兩式虛部相等可求出\sin 7\theta$$

解答: $$\cases{P(7,1,\sqrt{15}) \\ Q(x,0,0)\\ R(0,y,0)} \Rightarrow 周長=\overline{PQ}+ \overline{PR}+ \overline{QR}= \sqrt{(x-7)^2+16} +\sqrt{(y-1)^2 +64} +\sqrt{x^2+y^2} \\ =\overline{Q'A}+\overline{R'B} +\overline{Q'R'}, 其中\cases{Q'(x,0)\\ R'(0,y)\\ A(7,4)\\ B(8,1)}, 因此取\cases{A'(7,-4)=A對稱x軸的對稱點 \\B'(-8,1) =B對稱y軸的對稱點} \\ \Rightarrow 最小值=\overline{A'B'}=\sqrt{15^2+5^2} =\bbox[red, 2pt]{5\sqrt{10}}\\ 這答案應該是錯誤,但學校沒有更改答案,僅供參考$$

解答: $$原式f(x)=(C^2_2+ C^3_2x+C^4_2+\cdots+C^7_2x^5)^4 = \left(\sum_{n=0}^5 C^{n+2}_2 x^n \right)^4\\ 假設g(x)= {1\over 1-x}= \sum_{n=0}^\infty x^n \Rightarrow x^2g(x)= \sum_{n=0}^\infty x^{n+2} \Rightarrow \frac{\text{d} }{\text{d}x}\left( x^2g(x)\right) = \sum_{n=0}^\infty (n+2)x^{n+1} \\ \Rightarrow \frac{\text{d}^2 }{\text{d}x^2 }\left( x^2g(x)\right) = \sum_{n=0}^\infty (n+2)(n+1)x^{n} =2\sum_{n=0}^\infty C^{n+2}_2 x^n \Rightarrow {1\over (1-x)^3}=\sum_{n=0}^\infty C^{n+2}_2 x^n\\ 因此原式x^5係數=\left({1\over (1-x)^3} \right)^4 ={1\over (1-x)^{12}} =(1+x+x^2+ \cdots)^{12}的x^5係數\\ 也就是X_1+X_2+\cdots+ X_{12}=5非負整數解個數= H^{12}_5=C^{16}_5= \bbox[red, 2pt]{4368}$$

解答: $${8\cdot C^{12}_4\over C^{16}_8} =8\cdot {1\over 26}= \bbox[red, 2pt]{4\over 13}$$

解答: $$f(n)=n+\lfloor {n\over 4}\rfloor +\lfloor {n\over 4^2}\rfloor+\lfloor {n\over 4^3}\rfloor+\cdots \ge 100 \\ \Rightarrow \cases{f(60)=60+15+0=75\lt 100\\ f(70)=70+17+4+1=91\lt 100\\ f(80)=80+20+5+1=106\gt 100} \Rightarrow \cases{f(75)=75+18+4+1=98\\ f(76)=76+19+4+1=100} \\ \Rightarrow 買\bbox[red, 2pt]{76}瓶,可額外換24瓶,可以喝到100瓶$$

第貳部份： 計算題（共3題，占30分）

解答: $$\textbf{(1)}\; a_n=\left({3n+1\over 3n-1} \right)^n \Rightarrow \ln a_n=n \ln{3n+1\over 3n-1} ={\ln(3n+1/(3n-1)\over 1/n} \\ \Rightarrow \lim_{n\to \infty} \ln a_n=\lim_{n\to \infty} {\ln(3n+1/(3n-1)\over 1/n} =\lim_{n\to \infty} {(\ln({1+{2\over 3n-1}}))'\over (1/n)'} \\=\lim_{n\to \infty} {{3n-1\over 3n+1} \cdot (-{6\over (3n-1)^2})\over -{1\over n^2}} ={18\over 27} ={2\over 3}  \Rightarrow \lim_{n\to \infty} a_n= \bbox[red, 2pt]{e^{2/3}}\\ \textbf{(2)} \; 假設\lim_{n\to \infty} a_n= x \Rightarrow x=\sqrt{6+x} \Rightarrow x^2-x-6=0 \Rightarrow (x-3)(x+2)=0 \\\qquad \Rightarrow x= \bbox[red, 2pt]3 (x\gt 0,負值不合)$$

解答: $$u=x^2 \Rightarrow du=2xdx \Rightarrow I=\int_0^1 {x\over \sqrt{1+x^4}}\,dx = \int_0^1 {1/2 \over \sqrt{1+u^2}}\,du\\ 再取u=\tan \theta \Rightarrow du =\sec^2 \theta \,d\theta \Rightarrow I= {1\over 2}\int_0^{\pi/4}{\sec^2 \theta \over \sec \theta}\,d\theta = {1\over 2}\int_0^{\pi/4} \sec \theta \,d\theta \\={1\over 2}\left .\left[ \ln|\tan \theta+\sec \theta|\right] \right|_0^{\pi/4} = \bbox[red, 2pt]{{1\over 2}(\sqrt 2+1)}$$

解答: $$f(x,y)=2x^2-4x+y^2-4y+1 \Rightarrow \cases{f_x=4x-4\\ f_y=2y-4\\ f_{xx}=4 \gt 0\\ f_{xy}=0 \\ f_{yy}=2} \Rightarrow d(x,y)=f_{xx}f_{yy}-f_{xy}^2 =8 \gt 0\\ 又\cases{f_x=0\\ f_y=0} \Rightarrow (x,y)=(1,2) \Rightarrow f(1,2)=-5為極小值,邊界點\cases{f(0,0)= 1\\ f(0,2)=-3} \\ \Rightarrow \bbox[red, 2pt]{\cases{極大值1\\ 極小值-5}}$$

第參部份： 證明題（共 3 題，占 30 分）

解答: $$\textbf{(1)}\; \sqrt n \gt \ln n \Rightarrow {1\over \sqrt n \cdot \ln n} \gt {1\over \sqrt n \cdot \sqrt n} ={1\over n} \Rightarrow \sum_{n=2}^\infty {1\over \sqrt n\cdot \ln n} \gt \sum_{n=2}^\infty{1\over n} :調合級數發散\\ \qquad \Rightarrow  \sum_{n=2}^\infty {1\over \sqrt n\cdot \ln n}\bbox[red, 2pt]{發散} \\ \textbf{(2)}\; \sum_{n=1}^\infty {7\over (2n+5)^n} \lt \sum_{n=1}^\infty {7\over 7^n}:公比為{1\over 7}的等比級數收斂 \Rightarrow \sum_{n=1}^\infty {7\over (2n+5)^n}\bbox[red, 2pt]{收斂}$$

解答: $$\bbox[red, 2pt]{反證}:若\cases{v_1=(1,0,0,0) \\v_2=(0,1,0,0) \\v_3=(0,0,1,0)\\ v_4=(0,0,0,1)} \Rightarrow \cases{|v_1 v_2 v_3 v_4|= \begin{vmatrix}1& 0& 0 & 0 \\0 & 1& 0 & 0\\ 0 & 0 & 1& 0\\ 0 & 0 & 0 & 1 \end{vmatrix} =1 \\[1ex] |v_4 v_1 v_2 v_3|= \begin{vmatrix}0 & 0 & 0 & 1\\1& 0& 0 & 0 \\0 & 1& 0 & 0\\ 0 & 0 & 1& 0  \end{vmatrix}=-1} \\ \Rightarrow |v_1 v_2 v_3 v_4| \ne |v_4 v_1 v_2 v_3|$$

解答: $$令S(n)=前n天吃的餅乾數總和,則1\le S(1)\lt S(2) \lt \cdots \lt S(7)\le 13\\ 若存在S(k)=7, 1\le k\le 7,則得證\\ 若S(k)\ne 7, \forall  k, 由於S(1)至S(7)一定有有兩個S(i), S(j)除以7的餘數相同, \\若S(i)\lt S(j), 則(S(j)-S(i))除以7餘0,即第i天至第j天吃了7片餅干,故得證$$