114年嘉義高中教甄-數學詳解

 國立嘉義高級中學 114 學年度第一學期第 1 次教師甄選

一、 填充題： (每題 6 分，共 90 分)

解答:

$$\overline{AC}有最大值\Rightarrow A,O,C三點在一直線上\\ \angle AOB=117^\circ-57^\circ=60^\circ \Rightarrow \cos \angle AOB={1\over 2}={4^2+3^2-\overline{AB}^2\over 2\cdot 3\cdot 4} \Rightarrow \overline{AB} =\sqrt{13} \\ \angle BOC=180^\circ-\angle AOB=120^\circ \Rightarrow \cos \angle BOC =-{1\over 2}={3^2+5^2-\overline{BC}^2 \over 2\cdot 3\cdot 5} \Rightarrow \overline{BC}=7 \\ \Rightarrow \cos A={13+9^2-7^2\over 18\sqrt{13}} ={5\over 2\sqrt{13}} \Rightarrow \sin A={3\sqrt 3\over 2\sqrt{13}} \Rightarrow {7\over 3\sqrt 3/2\sqrt{13}} =2R \\ \Rightarrow R={7\sqrt{13} \over 3\sqrt 3} = \bbox[red, 2pt]{7\sqrt{39}\over 9}$$

解答:$$令\cases{x=1-p\\ y=1-q\\ z=1-r} \Rightarrow \cases{xyz={1\over 4}  \cdots(1)\\ (1-x)yz+x(1-y)z+ xy(1-z)={11\over 24} \cdots(2)\\ (1-x)(1-y)z+ x(1-y)(1-z)+ (1-x)y(1-z)= {1\over 4} \cdots(3)} \\(2) \Rightarrow yz+xz+xy-3xyz=yz+xz+xy-3\cdot {1\over 4}={11\over 24} \Rightarrow xy+yz+zx= {29\over 24} \\ (3) \Rightarrow (x+y+z)-2(xy+yz+zx) +3xyz =x+y+z-{29\over 12}+{3\over 4} ={1\over 4} \Rightarrow x+y+z={23\over 12} \\ \Rightarrow \cases{xyz=1/4\\ xy+yz+zx= 29/24\\ x+y+z=23/12} \Rightarrow x,y,z 為f(\alpha)=\alpha^3-{23\over 12}\alpha^2 +{29\over 24}\alpha-{1\over 4}=0的三根 \\ \Rightarrow f(\alpha)=24(2\alpha-1)(2\alpha-3)(3\alpha-4)=0 \Rightarrow \alpha={1\over 2},{2\over 3},{3\over 4} \Rightarrow (p,q,r) = \bbox[red, 2pt]{({1\over 2},{1\over 3}, {1\over 4})}$$

解答:$$除了[2,4],[4,8]沒有包含正奇數平方外，其餘\bbox[red, 2pt]{12}個皆符合要求$$

解答:$$f({4\over 3}) =f({2\over 3} +{2\over 3}) = {1\over 2}(f({2\over 3}))^2 \Rightarrow 1=f(2)=f({4\over 3}+{2\over 3}) ={1\over 2}f({4\over 3})f({2\over 3}) ={1\over 4}f({3\over 3})^3 \\ \Rightarrow f({2\over 3})^3= 4 \Rightarrow f({2\over 3}) = \bbox[red, 2pt]{\sqrt[3]4}$$

解答:$$f(x)={1\over 1-x} =\sum_{k=0}^\infty x^k \Rightarrow f'(x)={1\over (1-x)^2}=\sum_{k=0}^\infty kx^{k-1} \Rightarrow g(x)=xf'(x)={x\over (1-x)^2} =\sum_{k=0}^\infty kx^k \\ \Rightarrow g'(x)={1\over (1-x)^2}+{2x\over (1-x)^3} =\sum_{k=0}^\infty k^2x^{k-1} \Rightarrow g'({2\over 3}) =45 =\sum_{k=0}^\infty k^2({2\over 3})^{k-1} \\ \Rightarrow \sum_{k=1}^\infty k^2({2\over 3})^{k-1}\cdot {1\over 3} = \bbox[red, 2pt]{15}$$

解答:$${x/2+x/2+2(1+y)\over 3} \ge \sqrt[3]{x^2(1+y)\over 2} \Rightarrow {5\over 3}\ge \sqrt[3]{x^2(1+y)\over 2} \Rightarrow {125\over 27}\ge {x^2(1+y)\over 2} \\ \Rightarrow x^2(1+y)\le {250\over 27} \Rightarrow x\cdot \sqrt{1+y} \le {5\sqrt{10} \over 3\sqrt 3} =\bbox[red, 2pt]{5\sqrt{30} \over 9}\\ \bbox[cyan,2pt]{另解: }\cases{f(x,y)= x\sqrt{1+y} \\ g(x,y)=x+2y-3} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{\sqrt{1+y} =\lambda\\ {x\over 2\sqrt{1+y} } =2\lambda\\ x+2y=3} \Rightarrow {2(1+y) \over x} ={1\over 2} \Rightarrow x=4(1+y) \\ \Rightarrow 4(1+)y+2y=3 \Rightarrow y=-{1\over 6} \Rightarrow x={10\over 3} \Rightarrow f(10/3,-1/6) = \bbox[red, 2pt]{{5\over 9}\sqrt{30}}$$

解答:

$$2x+2^x=5 \Rightarrow 2(x-1)+2\cdot 2^{(x-1)} =3 \Rightarrow (x-1)+2^{x-1} ={3\over 2} \Rightarrow f(x)=2^{x-1}={5\over 2}-x \\ 2x+2\log_2(x-1)=5 \Rightarrow 2(x-1)+2\log_2(x-1)=3 \Rightarrow g(x)=\log_2(x-1)={5\over 2}-x\\ f(x)與g(x)互為反函數\Rightarrow 兩圖形對稱於y=x-1 \\\Rightarrow 兩點\cases{A(\alpha,f(\alpha)) \\ B(\beta, g(\beta))} 的中心點P在直線y=x-1上\Rightarrow P({\alpha+\beta\over 2},{\alpha+\beta\over 2}-1) \\ 同時P也在y={5\over 2}-x上\Rightarrow {\alpha+\beta\over 2}-1={5\over 2}-{\alpha+\beta\over 2} \Rightarrow \alpha+\beta= \bbox[red, 2pt]{7\over 2}$$

解答:

$$假設\cases{\overline{AD}=a \\\overline{BD}=b \\ 內切圓半徑r} \Rightarrow \cases{\triangle ACD面積={1\over 2} 9a\sin A={1\over 2}r(9+6+a) \\ \triangle BCD面積={1\over 2}8b\sin B={1\over 2}r(8+6+b)} \\ \Rightarrow {\triangle ACD\over \triangle BCD}= {9a\sin A\over 8b\sin B}={9a\cdot {8\over a+b} \over 8b\cdot {9\over a+b}}={a\over b} ={r(15+a)\over r(14+b)} ={15+a\over 14+b} \Rightarrow {a\over b}=\bbox[red, 2pt]{15\over 14}$$

解答:

$$B(t^2+4,t) \Rightarrow B至直線x-y=0的距離={t^2-t+4\over \sqrt 2} ={(t-1/2)^2+15/4\over \sqrt 2} \ge {15\over 4\sqrt 2} \\ \Rightarrow \overline{AB}的最小值={15\over 4\sqrt 2} =\bbox[red, 2pt]{15\sqrt 2\over 8}$$

解答:$$f_n(x) =\cos x \cos 2x \cos 3x\cdots \cos nx \Rightarrow f_k(x)= f_{k-1}(x) \cos kx \\\Rightarrow f'_k(x) =f'_{k-1}(x)\cos kx-kf_{k-1}\sin kx \Rightarrow f''_k(x)=f''_{k-1}(x)\cos kx-kf'_{k-1}(x)\sin kx-k^2f_{k-1}\cos kx \\ \Rightarrow f_k''(0)=f''_{k-1}(0)-k^2 \Rightarrow \cases{f_n''(0)=f_{n-1}''(0)-n^2\\ f_{n-1}''(0)=f_{n-2}''(0)-(n-1)^2 \\ \cdots\\ f_2''(0)=f_{1}''(0)-2^2} \\ \Rightarrow 合部相加可得: f''_{n}(0)=f_1''(0)-(2^2+3^2+ \cdots+n^2)\quad (f_1=\cos x \Rightarrow f_1''(0)=-\cos(0)=-1) \\ \Rightarrow |f_n''(0)|= \sum_{k=1}^n k^2 ={n(n+1)(2n+1)\over 6} \ge 2025 \Rightarrow n\ge \bbox[red, 2pt]{18}$$

解答:

$$假設\cases{\angle APD= \alpha \\\angle OPD=\beta }  \Rightarrow \overrightarrow{PA} \cdot \overrightarrow{PD} = |\overrightarrow{PA} || \overrightarrow{PD}| \cos  \alpha =4\cdot 5\cos \beta \cos \alpha \\=10(\cos (\alpha+\beta)+ \cos(\alpha-\beta)) =10({4\over 5}+ \cos(\alpha-\beta)) \le 10({4\over 5}+1) = \bbox[red, 2pt]{18}$$

解答:$$假設7人分別為甲、乙、丙、丁、戊、己、庚，3個景點為A,B,C\\依題意可假設7人的選擇結果:\cases{甲\to A\\ 乙\to B\\ 丙\to C\\丁\to AB \\ 戊 \to BC\\ 己\to AC\\ 庚\to 不選} \\\Rightarrow 三人選擇的結果恰包含三個景點的情形\cases{甲乙(丙,戊,己):3種\\ 甲丙(丁戊):2種\\ 甲丁(戊,己):2種\\ 甲戊(己,庚):2種 }及\cases{乙丙(丁己):2種\\ 乙丁(戊,己):2種\\ 乙戊已:1種\\ 乙己庚:1種\\ 丙丁(戊,己,庚):3種\\ 丙戊己:1種\\ 丁戊(己,庚):2種\\ 丁己庚:1 種\\ 戊己庚:1種}\\ \Rightarrow 合計23種\Rightarrow 機率={23\over C^7_3} = \bbox[red, 2pt]{23\over 35}$$

解答:

$$三個集合依交集區域分成7個獨立區塊,如圖(a), 依題意需滿足\cases{d+g= e+g=f+g=2\\ a+d+f+g=2\\ b+d+e+g=2\\ c+e+f+g=2} \\ \cases{\textbf{Case I }圖(b):有8!=40320種分配方法\\ \textbf{Case II }圖(c):有C^8_2C^6_2C^4_2= 2520種分配方法\\ \textbf{Case III }圖(d):也是有2520種分配方法} \Rightarrow 合計\bbox[red, 2pt]{45360}$$

解答:$$X\sim B(n=101,p={1\over 6}) \Rightarrow \cases{P(X=k) \ge P(X=k+1) \\ P(X=k) \ge P(X=k-1)} \Rightarrow \cases{C^n_kp^k(1-p)^{n-k} \ge C^n_{k+1}p^{k+1}(1-p)^{n-k-1} \\C^n_kp^k(1-p)^{n-k} \ge C^n_{k-1}p^{k-1}(1-p)^{n-k+1}} \\ \Rightarrow \cases{\displaystyle {1-p\over n-k} \ge {p\over k+1} \\ \displaystyle {p\over k} \ge {1-p\over n-k+1}} \Rightarrow \cases{(1-p)(k+1)\ge p(n-k) \\ p(n-k+1)\ge k(1-p)} \Rightarrow \cases{k\ge (n+1)p-1 =16\\ k\le (n+1)p=17} \\ \Rightarrow 16\le k\le 17 \Rightarrow k= \bbox[red, 2pt]{16或17}$$

解答:$$x^4+5x^3+mx^2+nx+4 =0 \Rightarrow \cases{四根之和=-5\\ 兩相異實根和為-5} \Rightarrow 四根為\alpha,\beta,\gamma,-\gamma 且\alpha+\beta=-5\\ \Rightarrow \cases{m=\alpha\beta-\gamma^2\\ n= (\alpha+\beta) \gamma^2=-5r^2 \in \mathbb N \Rightarrow \gamma^2 \lt 0 \Rightarrow \cases{-\gamma^2=1,2,3,\dots\\\gamma為虛根}\\ 4=-\alpha\beta \gamma^2} \\ \Rightarrow 當\cases{\alpha+\beta=-5\\  \alpha\beta=1} \Rightarrow x^2+5x+1=0 \Rightarrow x={-5\pm \sqrt{21} \over 2} \Rightarrow 最小實根 \bbox[red, 2pt]{-5-\sqrt{21} \over 2}$$

二、 計算證明題： (共 10 分) 需有計算過程。

解答:

$$\textbf{(1) }\begin{bmatrix} {9\over 10} \cos \theta& -{9\over 10} \sin \theta\\ {9\over 10} \sin \theta& {9\over 10} \cos \theta\end{bmatrix} ={9\over 10}\begin{bmatrix}   \cos \theta& -  \sin \theta\\ \sin \theta&   \cos \theta\end{bmatrix} \Rightarrow 旋轉\theta, 再將與原點距離變為原來的{9\over 10} \\ \Rightarrow \cases{\overline{OC} ={9\over 10}\overline{OA} =36\\ \overline{OD} ={9\over 10}\overline{OB} =36} \Rightarrow \cases{\overline{OA} =\overline{OB}=40\\ \overline{OC}=\overline{OD} =36 \\ \angle AOC= \angle BOD= \theta} \Rightarrow \triangle OAC\cong \triangle OBD (SAS) \\\textbf{(2) } \triangle OAC\cong \triangle OBD \Rightarrow \angle OAC=\angle OBD \Rightarrow \angle OBC+ \angle OAC= \angle CBD=90^\circ\\ 又\angle COD= \angle AOB \Rightarrow  \cos \angle COD =\cos \angle AOB \\\quad \Rightarrow {36^2+36^2-\overline{CD}^2\over 2\cdot 36\cdot 36}={40^2+40^2-20^2\over 2\cdot 40\cdot 40}={7\over 8} \Rightarrow \overline{CD}= 18 \\ \Rightarrow \overline{BC}^2+ \overline{AC}^2=\overline{BC}^2+ \overline{BD}^2= \overline{CD}^2=18^2= \bbox[red, 2pt]{324}$$

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