115年嘉科實中教甄2-數學詳解

國立嘉科實驗高級中學11 5學年度高中部教師甄選

一、 填充題（每題 7 分，共 56 分）

解答:$$ \left( a+{2\over b} \right) \left( b+{2\over c} \right) \left( c+{2\over a} \right) =abc+{8\over abc} + 2 \left( a+{2\over b} \right) +2 \left( b+{2\over c} \right)+ 2 \left( c+{2\over a} \right) \\=abc+{8\over abc}+2\cdot 5+2\cdot 4+2\cdot {5\over 2}=abc+{8\over abc}+23=5\cdot 4\cdot {5\over 2}=50 \Rightarrow abc+{8\over abc}-27=0 \\ \Rightarrow (abc)^2-27(abc)+8=0 \Rightarrow abc= \bbox[red, 2pt]{27 \pm \sqrt{697} \over 2}$$

解答:$$假設\cases{B=\sqrt{m-184} \\ A=\sqrt{m+24}} \Rightarrow n=A+B \Rightarrow A^2-B^2=(A+B)(A-B)=208 \Rightarrow n(A-B)=208\\ \Rightarrow 兩數(A+B與A-B)相乘是偶數，而且兩數相加(2A)也是偶數 \Rightarrow 兩數皆為偶數\\ 由於208=104\times 2= 52\times 4= 26\times 8 \Rightarrow n=A+B要最大 \Rightarrow \cases{A+B=104 \\ A-B=2} \\ \Rightarrow A=53= \sqrt{m+24 } \Rightarrow m=53^2-24= \bbox[red, 2pt]{2785}$$

解答:$$t=x^2 \Rightarrow t^2+2(m-2)t+(m^2-5m+4)=0 有相異正實根\\ \Rightarrow \cases{\Delta=4(m-2)^2-4(m^2-5m+4)\gt 0\\ 兩根之和2(2-m)\gt 0\\ 兩根之積m^2-5m+4\gt 0} \Rightarrow \cases{m\gt 0\\m\lt 2\\ m\lt1或m\gt 4} \Rightarrow \bbox[red, 2pt]{0\lt m\lt 1}$$

解答:$$令  S_n = f(1) + f(2) + \dots + f(n) =n^2f(n) =S_{n-1}+f(n) \Rightarrow S_{n-1}=(n^2-1)f(n)\\ 又S_{n-1}=(n-1)^2f(n-1),因此(n-1)^2f(n-1)=(n^2-1)f(n) \Rightarrow (n-1)f(n-1)=(n+1)f(n) \\ \Rightarrow  f(n) = \frac{n-1}{n+1} f(n-1)   = \frac{n-1}{n+1} \cdot \frac{n-2}{n} \cdot \frac{n-3}{n-1} \dots \frac{2}{4} \cdot \frac{1}{3} \cdot f(1) ={2\over n(n+1)}f(1) \\ \Rightarrow f(2026) ={2\over 2026\cdot 2027}\cdot 2026= \bbox[red, 2pt]{2\over 2027}$$

解答:$$從 15 個相異正整數中任選  3  個,組合數為  C^{15}_3=455\\ 題意要求\cases{1 \le a\\ b - a \ge 3 \implies b \ge a + 3 \\c - b \ge 4 \implies c \ge b + 4 \\c \le 15}, 取\cases{x=a\\ y=b-2\\ z=c-5} \Rightarrow \cases{y\gt x\\ z\gt y\\z\le 10} \\ \Rightarrow 從 10 個相異正整數中任選  3  個,組合數為  C^{10}_3=120 \Rightarrow 機率為{120\over 466} =\bbox[red, 2pt]{24\over 91}$$

解答:$$取M=\overline{BC}中點\Rightarrow \overline{AM} \bot \overline{BC} \Rightarrow \overline{AB}^2=4=\overline{AM}^2 +\overline{BM}^2 \\ 對任意直角\triangle AP_kM: \overline{AP_k}^2= \overline{AM}^2+ \overline{MP_k}^2 \\ 因此\overline{BP_k} \cdot \overline{P_kC} = (\overline{BM} - \overline{MP_k})(\overline{BM} + \overline{MP_k}) = \overline{BM}^2 - \overline{MP_k}^2  \\ \Rightarrow  a_k = \overline{AP_k}^2 + \overline{BP_k} \cdot \overline{P_kC} =(\overline{AM}^2+ \overline{MP_k}^2)+ ( \overline{BM}^2 - \overline{MP_k}^2 ) =  \overline{AM}^2 + \overline{BM}^2 =4 \\ \Rightarrow a_k=4 \Rightarrow a_1+a_+ \cdots+ a_{100}=4\times 100= \bbox[red, 2pt]{400}$$

解答:$$S_n=  \sum_{k=1}^{n} \frac{k^2}{2^k} =  \frac{1^2}{2} + \frac{2^2}{2^2} + \frac{3^2}{2^3} + \dots + \frac{n^2}{2^n}   \Rightarrow  \frac{1}{2}S_n = \frac{1^2}{2^2} + \frac{2^2}{2^3} + \dots + \frac{(n-1)^2}{2^n} + \frac{n^2}{2^{n+1}} \\ \Rightarrow S_n-{1\over 2}S_n=  \frac{1}{2} + \frac{2^2 - 1^2}{2^2} + \frac{3^2 - 2^2}{2^3} + \dots + \frac{n^2 - (n-1)^2}{2^n} - \frac{n^2}{2^{n+1}} \\ \Rightarrow   \frac{1}{2}S_n = \frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + \dots + \frac{2n-1}{2^n} - \frac{n^2}{2^{n+1}} \cdots(1) \\取 T_n = \frac{1}{2} + \frac{3}{2^2} + \frac{5}{2^3} + \dots + \frac{2n-1}{2^n} \Rightarrow {1\over 2}T_n  = \frac{1}{2^2} + \frac{3}{2^3} + \dots + \frac{2n-3}{2^n} + \frac{2n-1}{2^{n+1}} \\ \Rightarrow T_n-{1\over 2}T_n={1\over 2}T_n  = \frac{1}{2} + \left( \frac{2}{2^2} + \frac{2}{2^3} + \dots + \frac{2}{2^n} \right) - \frac{2n-1}{2^{n+1}}  \\ = \frac{1}{2} + 1 - \frac{1}{2^{n-1}} - \frac{2n-1}{2^{n+1}} =  \frac{3}{2} - \frac{4}{2^{n+1}} - \frac{2n-1}{2^{n+1}} \Rightarrow T_n = 3 - \frac{2n+3}{2^n} 代回(1) \\ \Rightarrow  \frac{1}{2}S_n = T_n - \frac{n^2}{2^{n+1}} =  3 - \frac{2n+3}{2^n} - \frac{n^2}{2^{n+1}}  \Rightarrow  S_n = \bbox[red, 2pt]{6 - \frac{n^2 + 4n + 6}{2^n} }$$

解答:$$ \cos^2 80^\circ + \cos^2 160^\circ + \cos 80^\circ \cos 160^\circ   = (\cos 80^\circ + \cos 160^\circ)^2 - \cos 80^\circ \cos 160^\circ \\ = \left[  2 \cos\left(\frac{80^\circ + 160^\circ}{2}\right) \cos\left(\frac{80^\circ - 160^\circ}{2}\right)  \right]^2-{1\over 2} \left( \cos(80^\circ+160^\circ)+ \cos(160^\circ-80^\circ) \right)   \\= \left[   2 \cos 120^\circ \cos(-40^\circ)  \right]^2-{1\over 2}(\cos 240^\circ +\cos 80^\circ) = \left[  - \cos 40^\circ \right]^2-{1\over 2}(-{1\over 2}+ \cos 80^\circ) \\ =  {1\over 4}-{1\over 2} \cos 80^\circ+\cos^2 40^\circ =  {1\over 4}-{1\over 2} \cos 80^\circ+{1\over 2}(\cos 80^\circ +1) ={1\over 4}+0+{1\over 2} =\bbox[red, 2pt]{3\over 4}$$

二、 計算證明題（每題 10 分，共 20 分）

解答:$$ \triangle ABC  的外接圓面積為  25\pi \Rightarrow 外接圓半徑R=\sqrt{25}=5\\  a^2 + b^2 = c^2 \Rightarrow \angle C=90^\circ \Rightarrow \cases{c=2R =10 \\ \angle A+ \angle B=90^\circ \Rightarrow \sin B=\cos A}\\  \sin A 與 \sin B 為方程式 (m + 5)x^2 - (2m - 5)x + 12 = 0 的兩根 \Rightarrow   \cases{\sin A + \sin B = \frac{2m - 5}{m + 5} \\ \sin A \sin B = \frac{12}{m + 5}} \\ \Rightarrow (\sin A + \sin B)^2 = \sin^2 A + \sin^2 B + 2\sin A \sin B = \sin^2 A + \cos^2 A + 2\sin A \sin B \\=1+2\sin A\sin B \Rightarrow  \left( \frac{2m - 5}{m + 5} \right)^2 = 1 + 2 \left( \frac{12}{m + 5} \right) \Rightarrow m^2 - 18m - 40 = 0 \\ \Rightarrow (m-20)(m+2) =0 \Rightarrow m=20 \Rightarrow \cases{\sin A+\sin B=7/5\\ \sin A\sin B=12/25} \\ \Rightarrow \sin A,\sin B 是t^2-{7\over 5}t+{12\over 25}=0的兩根\Rightarrow 25t^2-35t+12=0 \Rightarrow (5t-3)(5t-4)=0 \\ \Rightarrow \cases{(\sin A, \sin B)=({3\over 5},{4\over 5})  \Rightarrow \cases{a=6\\b =8}\\ (\sin A,\sin B)=({4\over 5},{3\over 5}) \Rightarrow \cases{a=8\\ b=6}} \Rightarrow \bbox[red, 2pt]{\cases{m=20,a=6,b=8,c=10\\ m=20,a=8,b=6,c=10}}$$

解答:$$ \lim_{n\to\infty} \sum_{k=1}^{n} \frac{n}{(n+k)\sqrt{nk+k^2}} = \lim_{n\to\infty} \sum_{k=1}^{n} \frac{n}{n^2 \left(1+\frac{k}{n}\right) \sqrt{\frac{k}{n} +\left(\frac{k}{n}\right)^2}} = \\\lim_{n\to\infty} \sum_{k=1}^{n} \frac{1}{n} \cdot \frac{1}{\left(1+\frac{k}{n}\right) \sqrt{\frac{k}{n}+\left(\frac{k}{n}\right)^2}} = I = \int_{0}^{1} \frac{1}{(1+x)\sqrt{x+x^2}} \, dx \\取u=\sqrt x \Rightarrow x=u^2 \Rightarrow dx=2udu \Rightarrow I=  \int_{0}^{1} \frac{1}{(1+u^2)\sqrt{u^2(1+u^2)}} \cdot 2u \, du \\= \int_{0}^{1} \frac{2u}{(1+u^2) \cdot u\sqrt{1+u^2}} \, du = \int_{0}^{1} \frac{2}{(1+u^2)^{3/2}} \, du \\ 取u=\tan \theta \Rightarrow du=\sec^2 \theta\,d\theta \Rightarrow I=  \int_{0}^{\pi/4} \frac{2 \sec^2 \theta}{(\sec^2 \theta)^{3/2}} \, d\theta   = \int_{0}^{\pi/4} 2 \cos \theta \, d\theta \\= \left. \left[ 2\sin \theta \right] \right|_0^{\pi/4}= \bbox[red, 2pt]{\sqrt 2}$$

三、申論題（每題 12 分，共 24 分）

解答:$$\textbf{方法一: }\cases{a^2+c^2 \ge 2ac\\ b^2+d^2\ge 2bd} \Rightarrow a^2+b^2+c^2+d^2 \ge 2(ac+bd) \Rightarrow 1+1\ge 2(ac+bd) \Rightarrow ac+ bd\le 1 \\ \textbf{方法二: }\cases{a^2+b^2=1\\ c^2+d^2=1} \Rightarrow \cases{a =\cos \theta, b=\sin \theta\\ c=\cos \phi, d=\sin \phi} \Rightarrow  ac + bd = \cos \theta \cos \phi + \sin \theta \sin \phi  =\cos(\theta-\phi) \le 1\\ \qquad \Rightarrow ac+bd\le 1$$

解答:$$\textbf{(1) }z= \overrightarrow{BO} \cdot \overrightarrow{OA} = -\overrightarrow{OB} \cdot \overrightarrow{OA} \Rightarrow  \overrightarrow{OA} \cdot \overrightarrow{OB}=-z \\ x=\overrightarrow{OA} \cdot \overrightarrow{AB} =\overrightarrow{OA} \cdot (\overrightarrow{OB} - \overrightarrow{OA}) = \overrightarrow{OA} \cdot \overrightarrow{OB} - \vert{} \overrightarrow{OA} \vert{}^2  =-z- \vert{}\overrightarrow{OA}\vert{}^2 \Rightarrow \bbox[red, 2pt]{\vert{}\overrightarrow{OA}\vert{}^2=-(x+z)} \\ y= \overrightarrow{AB} \cdot \overrightarrow{BO}  =(\overrightarrow{OB} - \overrightarrow{OA}) \cdot (-\overrightarrow{OB}) = -\vert{} \overrightarrow{OB} \vert{}^2 + \overrightarrow{OA} \cdot \overrightarrow{OB}  = -\vert{}\overrightarrow{OB} \vert{}^2-z \Rightarrow \bbox[red, 2pt]{\vert{}\overrightarrow{OB} \vert{}^2 = -(y+z)} \\ \textbf{(2) } \triangle OAB  \text{面積} = \frac{1}{2} \sqrt{\vert{}\overrightarrow{OA}\vert{}^2 \vert{}\overrightarrow{OB}\vert{}^2 - (\overrightarrow{OA} \cdot \overrightarrow{OB})^2} = \frac{1}{2} \sqrt{  [-(x+z)][-(y+z)] - (-z)^2 } \\= \bbox[red, 2pt]{\frac{1}{2}  \sqrt{xy+yz+ zx}}$$

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解題僅供參考，其他教甄試題及詳解