國立新竹科學園區實驗高級中等學校
114 學年度第 1 次教師甄選
$$F=(c,0) \Rightarrow {c^2\over a^2}-{y^2\over b^2}=1 \Rightarrow y^2= {b^4\over a^2} \Rightarrow P(c,{b^2\over a}) \\ 又A=L\cap \{y={b\over a}x\} =(c,{bc\over a})\Rightarrow B(c,-{bc\over a}) \\ \overrightarrow{OP} =m \overrightarrow{OA} + n\overrightarrow{OB} \Rightarrow \left(c,{b^2\over a} \right) =m\left( c,{bc\over a}\right) + n\left(c,-{bc\over a} \right) =\left(c(m+n), {bc\over a}(m-n)\right) \\ \Rightarrow \cases{m+n=1\\ b=c(m-n)} \Rightarrow mn=m(1-m)={2\over 9} \Rightarrow (3m-2)(3m-1)=0 \Rightarrow \cases{m=2/3\\ m=1/3 不合, 違反m\gt n} \\\Rightarrow n=1-m={1\over 3} \Rightarrow b=c\left({2\over 3}-{1\over 3} \right) \Rightarrow c=3b \Rightarrow c^2=9b^2= b^2+a^2 \Rightarrow a^2=8b^2 \\\Rightarrow {\sqrt{a^2+b^2}\over a} ={c\over a}={3b\over 2\sqrt 2b} ={3\over 2\sqrt 2} = \bbox[red, 2pt]{3\sqrt 2\over 4}$$
解答:$$20張椅子5人先入坐，有5!排法,剩下15張椅子\\依規定頭尾各放一張,剩下13張; 兩人中間各擺一張，剩下9張 \\ 9張椅子可以在5人的六個間隔中擺放，有H^6_9擺法\\ 為符合「相鄰5張椅子至要有一人坐」，需扣除某個間隔塞入四張(含)以上的情形\\ 因此實際擺法有5!(H^6_9-C^6_1H^6_5+C^6_2H^6_1) = 120(2002-1512+90)= \bbox[red, 2pt]{69600}$$
解答:$$a_1=10 \Rightarrow a_2=10+d 為整數\Rightarrow 公差d為整數\\ 又S_n\le S_4 \Rightarrow d\lt 0 且\cases{a_5\lt 0 \\a_4 \gt 0} \Rightarrow \cases{10+4d\lt 0\\10+3d \gt 0} \Rightarrow d=-3 \\ \Rightarrow b_n={1\over a_na_{n+1}} =({1\over a_{n+1}-a_n})({1\over a_n}-{1\over a_{n+1}}) = {1\over d}({1\over a_n}-{1\over a_{n+1}}) \\ \Rightarrow \sum_{k=1}^{10} b_k = {1\over d}\sum_{k=1}^{10} ({1\over a_k}-{1\over a_{k+1}}) = {1\over d}({1\over a_1}-{1\over a_{11}}) ={1\over -3}({1\over 10}-{1\over 10-30}) = \bbox[red, 2pt]{-{1\over 20}}$$
解答:$$\cases{f(x,y)=3x^2+xy+y^2 \\ g(x,y)=4x^2-4xy+2y^2-1} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{6x+y=\lambda(8x-4y) \\x+2y=\lambda(-4x+4y) \\4x^2-4xy+2y^2=1} \\ \Rightarrow {6x+y\over x+2y} ={2x-y\over -x+y} \Rightarrow 8x^2-2xy-3y^2=0 \Rightarrow (4x-3y)(2x+y)=0\\ \Rightarrow \cases{x=3y/4 \Rightarrow y=\pm 2/\sqrt 5 \\ x=-y/2 \Rightarrow y=\pm 1/\sqrt 5} \Rightarrow \cases{f(\pm 3/2\sqrt 5, \pm 2/\sqrt 5) =11/4\\ f(\pm 1/2\sqrt 4, \mp1/\sqrt 5) =1/4} \Rightarrow {11\over 4}+{1\over 4}=\bbox[red, 2pt] 3$$
解答:
$$\cases{\sin \alpha=a\\ \cos \beta=b} \Rightarrow \cases{|\cos \alpha|=\sqrt{1-a^2} \\ |\sin \beta| = \sqrt{1-b^2}} \Rightarrow 原式:\sin \alpha\cdot \cos \beta+|\cos \alpha\cdot \sin \beta| = \sin \alpha\cdot |\cos \alpha| +|\sin \beta| \cdot \cos \beta \\ \Rightarrow \sin \alpha(\cos \beta-|\cos \alpha|) +|\sin \beta| (|\cos \alpha|-\cos \beta)=0 \Rightarrow (\sin \alpha-|\sin \beta|)(\cos \beta-|\cos \alpha|)=0 \\ \Rightarrow (a-\sqrt{1-b^2}) (b-\sqrt{1-a^2}) =0 \Rightarrow \cases{a=\sqrt{1-b^2} \\ b=\sqrt{1-a^2}} \Rightarrow a^2+b^2=1 \\ \Rightarrow k=(\tan \gamma-\sin \alpha)^2+(\cot \gamma-\cos \beta)^2= \overline{PQ}^2, 其中\cases{P\in \{(x,y)\mid xy=1\}\\ Q \in \{(x,y) \mid x^2+y^2=1\}} \\ \Rightarrow \min \{k\}=雙曲線與圓的最短距離的平方 =(1-{1\over \sqrt 2})^2+ (1-{1\over \sqrt 2})^2 = \bbox[red, 2pt]{3-2\sqrt 2}$$
解答:$$\left(\sqrt{m+\sqrt{m^2-n}} + \sqrt{m-\sqrt{m^2-n}}\right)^2=6^2 \Rightarrow 2m+2\sqrt n=36 \Rightarrow \sqrt n=18-m \\ \Rightarrow n為一完全平方數 \Rightarrow n=k^2 \Rightarrow k=18-m \Rightarrow m=18-k \Rightarrow 1\le k\le 17\\ 又根號內之值不得為負值 \Rightarrow \cases{m^2-n\ge 0\\ m-\sqrt{m^2-n} \ge 0} \Rightarrow (18-k)^2\ge k^2 \Rightarrow k\le 9 \\ \Rightarrow k=1,2,\dots,9 \Rightarrow n=1^2,2^2, \dots, 9^2 \Rightarrow \sum n=\sum_{k=1}^9 k^2 ={1\over 6}\cdot 9\cdot 10\cdot 19=\bbox[red, 2pt]{285}$$
解答:
$$(x+{1\over y})^2 +(y+{1\over 2x})^2 =\overline{PQ}^2, 其中\cases{P(-{1\over y},-y) \in\{(x,y) \mid xy=1, x\lt 0,y\lt 0\} \\Q(x,{1\over 2x}) \in \{(x,y) \mid xy=2,x\gt 0,y\gt 0\}} \\ \Rightarrow 最小值發生在\cases{P(-1,-1) \\ Q(1/\sqrt 2,1/\sqrt 2)} .此時\overline{PQ}^2= \bbox[red, 2pt]{3+2\sqrt 2}\\ \bbox[red, 2pt]{另解}: (x+{1\over y})^2+ (y+{1\over 2x})^2 =(x^2 +{1\over 4x^2}) +({1\over y^2}+y^2)+ ({2x\over y}+{ y\over x}) \\由於\cases{x^2+{1\over 4x^2} \ge 2\sqrt{1\over 4} =1\\ {1\over y^2}+y^2 \ge 2\sqrt{1}=2 \\ {2x\over y}+{y\over x} \ge 2\sqrt{2}} \Rightarrow 最小值=1+2+2\sqrt 2=3+2\sqrt 2$$
解答:$$最多鈍角\triangle 的情形:13個頂點緊密相鄰在圓周上，\\剩下兩個頂點(假設叫頂點1與頂點15)均分在剩下的周長上;\\ 頂點1至頂點14任選三點,有C^{14}_3 =364個鈍角三角形，\\頂點15與(頂點2-14)任選2點有C^{13}_2=78個鈍角三角形,因此共有364+78= \bbox[red, 2pt]{442}個$$
解答:$$令f(x)=(x-\sqrt 3)^{50} +(x+1)^{50},則所求即為f(i)的實部\\ f(i) =(-\sqrt 3+i)^{50}+(1+i)^{50} =\left[ 2e^{5\pi i/6}\right]^{50} + \left[ 2e^{\pi i/4}\right]^{50} =2^{50}e^{5\pi i/3} +2^{50}e^{\pi i/2} \\ \Rightarrow k=Re[f(i)] =2^{50} \cdot {1\over 2}+0=2^{49} \Rightarrow \log_4|k|=\log_4 (2\cdot 4^{24}) ={1\over 2} +24 = \bbox[red, 2pt]{49\over 2}$$
解答:
$$\cases{|z_1|=|z_1+z_2|=3 \\ |z_2-z_1|=3\sqrt 3} \Rightarrow \cases{z_1 = 3e^{i \theta} \\z_2=3e^{i(\theta+2\pi/3)}} \Rightarrow \cases{z_1\overline{z_2} =9e^{i(-2 \pi/3)} \\ \overline{z_1} z_2 = 9e^{i(2 \pi/3)}} \\ \Rightarrow \left( z_1 \overline{z_2} \right)^{2025} +\left( \overline{z_1} z_2 \right)^{2025} =9^{2025} \left( e^{i(-4050 \pi/3)} +e^{i(4050\pi/3)}\right) =3^{4050} \cdot 2\cos {4050\over 3}\pi =3^{4050} \cdot 2\cos 2\pi \\= \bbox[red, 2pt]{2\times 3^{4050} }$$
第二大題：計算與證明題，4 題，每題 10 分，共 40 分。
解答:$$\textbf{(1) }假設正方形邊長為2a,且A為原點，即\cases{A(0,0,0)\\ B(0,2a,0) \\D(2a,0,0)\\ C(2a,2a,0)} \Rightarrow E={B+ C\over 2} =(a,2a,0)\\ 又\cases{ \overline{BP}= \overline{BA}= 2a\\ \angle PBA=60^\circ} \Rightarrow P(0, 2a-2a\cos 60^\circ, 2a\sin 60^\circ)=(0,a,\sqrt 3 a) \\ \quad \Rightarrow \cases{ \overrightarrow{AE} =(a,2a,0) \\ \overrightarrow{PD}=(2a,-a,-\sqrt 3a)} \Rightarrow \overrightarrow{AE} \cdot \overrightarrow{PD}=0 \Rightarrow \overline{AE} \bot \overline{PD}\; \bbox[red, 2pt]{QED} \\\textbf{(2) }若\angle PBA=\alpha \Rightarrow P(0,2a(1-\cos \alpha), 2a \sin \alpha) \Rightarrow \vec n=\overrightarrow{AP} \times \overrightarrow{AD} =(0,-4a^2\sin \alpha, 4a^2(1-\cos \alpha)) \\ \quad \Rightarrow \cos(90^\circ-\theta) ={\vec n\cdot \overrightarrow{AE} \over |\vec n||\overrightarrow{AE}|} \Rightarrow \sin \theta= {-2\sin \alpha\over \sqrt 5\cdot \sqrt{2-2\cos \alpha}} =-{2\over \sqrt 5}\sin {\alpha\over 2} \Rightarrow \cos^2\theta=1-{4\over 5}\sin^2{\alpha\over 2} \\ \quad \Rightarrow \bbox[red, 2pt]{{1\over \sqrt 5}\lt \cos \theta \lt 1}$$
解答:$$\cases{\sqrt{S_n}= \lambda(a_n- 1)+1\\ 2a_2= a_1+a_3} \Rightarrow \cases{\sqrt{S_2} -\sqrt{S_1} =\lambda(a_2-a_1) \\ a_2-a_1=a_3-a_2 =d} \Rightarrow \sqrt{S_2}-\sqrt{S_1}=\lambda(a_3-a_2) =\sqrt{S_3} -\sqrt{S_2} \\ \Rightarrow 2\sqrt{S_2}= \sqrt{S_1}+ \sqrt{S_3} \Rightarrow 4S_2=S_1+S_3+2\sqrt{S_1S_3} \\\Rightarrow 4(a_1+a_2)= a_1+(a_1+ a_2+a_3) +2\sqrt{a_1(a_1+a_2+a_3)} \\ \Rightarrow 4(2a_1+d) =4a_1 +3d +2\sqrt{a_1(3a_1+3d)} \Rightarrow (4a_1+d)^2= 4a_1(3a_1+3d) \\ \Rightarrow 4a_1^2- 4a_1d+ d^2=0 \Rightarrow (2a_1-d)^2=0 \Rightarrow d= 2a_1 \Rightarrow \cases{a_2= 3a_1 \\ a_3=5a_1} \Rightarrow \cases{S_1=a_1\\ S_2=4a_1\\ S_3=9a_1}\\ \Rightarrow \lambda= {\sqrt{S_2}-1\over a_2-1} ={\sqrt{S_3}-1\over a_3-1} \Rightarrow {2\sqrt{a_1}-1\over 3a_1-1} ={3\sqrt{a_1}-1\over 5a_1-1} \Rightarrow a_1=1 \Rightarrow \lambda= {1\over 2} \\ \Rightarrow \sqrt{S_n}={1\over 2}(a_n-1)+1 \Rightarrow a_n=2\sqrt{S_n}-1 \Rightarrow S_n-S_{n-1}=2\sqrt{S_n}-1 \\ \Rightarrow S_n-2\sqrt{S_n} +1=S_{n-1} \Rightarrow (\sqrt{S_n}-1)^2=S_{n-1} \Rightarrow \sqrt{S_n} =\sqrt{S_{n-1}}+1 \\ \Rightarrow {a_n+1\over 2} ={a_{n-1}+1\over 2}+1 \Rightarrow a_n=a_{n-1}+2 =a_{n-2}+4= \cdots =a_1+2(n-1) \\ \Rightarrow a_n=1+2(n-1) \Rightarrow \bbox[red, 2pt]{a_n=2n-1}$$
解答:$$t=\sin^2 x \Rightarrow \cos^2 x=1-t \Rightarrow f(x)=\sin^{12}x+ \cos^{12}x=g(t)=t^6+(1-t)^6 \\ \Rightarrow g'(t)=6t^5-6(1-t)^5 \Rightarrow g''(t)=30t^4+30(1-t)^4\\若g'(t)=0 \Rightarrow t={1\over 2} \Rightarrow g''(1/2)\gt 0 \Rightarrow m=g(1/2) ={1\over 64}+{1\over 64}={1\over 32} \\ 邊界值\cases{g(0)=1\\ g(1)=1} \Rightarrow M=1 \Rightarrow \bbox[red, 2pt]{\cases{M=1\\ m=1/32}}$$
解答:$$\overline{BL}: \overline{LC}=c:b \Rightarrow \overrightarrow{AL}={b\over b+c} \overrightarrow{AB}+{c\over b+c} \overrightarrow {AC} \\ \overline{CM}: \overline{MA}= a:c \Rightarrow \overrightarrow{BM} = -\overrightarrow{AB}+ \overrightarrow{AM} =-\overrightarrow{AB}+ {c\over a+c}\overrightarrow{AC}\\ \overline{AN}: \overline{NB}=b: a \Rightarrow \overrightarrow{CN} = \overrightarrow{CB} + \overrightarrow{BN} =\overrightarrow{AB}-\overrightarrow{AC}-{a\over a+b} \overrightarrow{AB} ={b\over a+b} \overrightarrow{AB}-\overrightarrow{AC} \\ \Rightarrow b\overrightarrow{BM}+ c\overrightarrow{CN}+ a\overrightarrow{AL} =b \left( -\overrightarrow{AB} +{c\over a +c}\overrightarrow{AC} \right) +c\left( {b\over a+b}\overrightarrow{AB}-\overrightarrow{AC}\right) +a\left( {b\over b+c}\overrightarrow{AB}+{c\over b+c}\overrightarrow{AC} \right) \\=\left(-b+{bc \over a+b}+ {ab\over b+c}\right) \overrightarrow{AB} + \left({bc \over a+c}-c+ {ac\over b+c}\right) \overrightarrow{AC} =0 \\ \Rightarrow -b+{bc \over a+b}+ {ab\over b+c}=0 \Rightarrow {c\over a+b}+{a\over b+c}=1 \Rightarrow c(b+c)+a(a+b)=(a+b)(b+c) \\ \Rightarrow a^2+c^2-b^2=ac \Rightarrow \cos \angle B={1\over 2} \Rightarrow \angle B=60^\circ \\ 同理,{bc \over a+c}-c+ {ac\over b+c}=0 \Rightarrow a^2+b^2-c^2=ab \Rightarrow \cos \angle C={1\over 2} \Rightarrow \angle C=60^\circ \\ \Rightarrow \triangle ABC為一\bbox[red, 2pt]{正三角形}$$
解題僅供參考，其他教甄試題及詳解
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