國立新竹女中 107 學年度代理教師甄選
一、填充題(每格 5 分，共 70 分)
解答：$$令g(x)=-x-k，則f(x)+x+k=0 有二實根，相當 f(x)=g(x)有兩交點；\\y=g(x)=-x-k為一直線，且斜率為-1；\\因此兩圖形\cases{y=g(x)\\ y=e^x,x\le 0}僅在k\ge -1時有一交點，\\ 而兩圖形\cases{y=g(x)\\ y=\ln x,x\gt 0}一定有一交點，因此有二實根的條件，即\bbox[red, 2pt]{k\ge -1}$$
解答：$$x^4+ 2\sqrt 3(\log_2 k)x^2+4-(\log_2 k)^2=0 \Rightarrow x^2={-2\sqrt 3(\log_2 k) \pm \sqrt{12(\log_2 k)^2-16+4(\log_2k)^2}\over 2}\\ =-\sqrt 3(\log_2 k)\pm 2\sqrt{(\log_2 k)^2-1} \Rightarrow \cases{(\log_2 k)^2 -1 \gt 0 \\ -\sqrt 3(\log_2 k)- 2\sqrt{(\log_2 k)^2-1}\gt 0} \\ \Rightarrow \cases{(\log_2 k)^2 \gt 1\\ -\sqrt 3(\log_2 k)\gt 2\sqrt{(\log_2 k)^2-1} \\ \log_2 k \lt 0} \Rightarrow \cases{\log_2 k\gt 1或\log_2 k\lt -1 \\ 3(\log_2 k)^2 \gt 4(\log_2 k)^2-4 \Rightarrow 4\gt (\log_2 k)^2 \\k\lt 1} \\ \Rightarrow \cases{\log_2 k\gt 1或\log_2 k \lt -1\\ 2\gt \log_2 k\gt -2\\ k\lt 1} \Rightarrow \cases{1\lt \log_2 k \lt 2 \\-2\lt \log_2 k\lt -1 \\ k\lt 1} \Rightarrow \bbox[red, 2pt]{ 1/4\lt k\lt 1/2}$$
解答：$$令a_n =\overline{A_nA_{n+1}} =2^{4-n},n\in \mathbb{N} \\x_\infty =8+{a_1\over \sqrt 2}-a_2 -{a_3\over \sqrt 2}+ a_4 +{a_5\over \sqrt 2}-a_6-{a_7 \over \sqrt 2}+a_8 +\cdots\\ =8+(a_4+a_8+ a_{12}+\cdots )+{1\over \sqrt 2}(a_1+a_5 +a_9+\cdots )-{1\over \sqrt 2}(a_3+a_7+ a_{11}+\cdots )-(a_2+a_6+a_{10}+ \cdots)\\ =8+({a_4\over 1-1/16}) +{1\over \sqrt 2}({a_1\over 1-1/16}) -{1\over \sqrt 2}({a_3\over 1-1/16})-({a_2\over 1-1/16}) \\ =8 +{16\over 15} +{1\over \sqrt 2} \cdot 8\cdot {16\over 15}-{1\over \sqrt 2}\cdot 2\cdot {16\over 15}-4\cdot {16\over 15} =8+{16\over 5}(\sqrt 2-1)\\ y_\infty = {a_1\over \sqrt 2}-{a_3\over \sqrt 2} +{a_5\over \sqrt 2}-{a_7\over \sqrt 2}+{a_9\over \sqrt 2}-\cdots \\= {1\over \sqrt 2}(a_1+a_5+a_9+ \cdots) -{1\over \sqrt 2}(a_3+a_7+a_{11} +\cdots)\\= {1\over \sqrt 2}({a_1\over 1-1/16}) -{1\over \sqrt 2}({a_3\over 1-1/16}) = {1\over \sqrt 2}\cdot 8\cdot {16\over 15}-{1\over \sqrt 2}\cdot 2\cdot {16\over 15}={16\sqrt 2\over 5}\\ 因此x_\infty+y_\infty =8+{16\over 5}(\sqrt 2-1)+{16\sqrt 2\over 5} =\bbox[red,2pt]{24+32\sqrt 2\over 5}$$
解答：$$令\cases{午餐做飯的人是A、洗碗的是B \\晚餐做飯的人是C、洗碗的是D}，則\cases{A\ne C\\ B\ne D\\ A\ne B\\ C\ne D}\\ 因此A有7位人選\xrightarrow{A\ne B} B有6位人員\cases{\xrightarrow{B=C}C只有1種選擇 \xrightarrow{D\ne C}D有6種選擇\\ \xrightarrow{B\ne C}C有5種選擇 \xrightarrow{D\ne C,D\ne B}D有5種選擇} \\ 因此共有7\times 6\times 6+7\times 6\times 5\times 5=252+1050 =\bbox[red, 2pt]{1302}種安排$$
解答：$$P(香蕉先分完)= P(香蕉比芭樂先分完) +P(香蕉比鳳梨先分完)-P(香蕉比其中之一先分完) \\ ={3\over 5+3} +{4\over 5+4} -{3+4\over 5+3+4} = \bbox[red,2pt]{7\over 72}$$
$$延長\overrightarrow{PC}，使得\overrightarrow{PD}=2\overrightarrow{PC}；並令\overrightarrow{PQ} =\overrightarrow{PA} +\overrightarrow{PD}，則B為\overline{PQ}中點，即符合題意要求，如上圖；\\因此\cases{\cos \angle APQ ={3^2+6^2-(3\sqrt 6)^2\over 2\times 3\times 6}=-{1\over 4} \Rightarrow \sin \angle APQ={\sqrt{15}\over 4} \\ \cos \angle DPQ ={6^2+ (3\sqrt 6)^2-3^2 \over 2\times 6\times 3\sqrt 6} ={3\over 8}\sqrt 6 \Rightarrow \sin \angle DPQ ={\sqrt{10}\over 8}} \\ \Rightarrow PABC面積=\triangle PAB+\triangle PBC ={1\over 2}(3\times 3\times {\sqrt{15}\over 4} +3\times \sqrt 6\times {\sqrt{10} \over 8}) = \bbox[red, 2pt]{3\sqrt{15}\over 2}$$
解答：$$正弦定理: {3\over \sin C}=2\times 6=12 \Rightarrow \sin C={1\over 4} \Rightarrow \cos C={\sqrt{15}\over 4}\\ 令\triangle =\begin{vmatrix} 1 & \cos A & \cos B\\ \cos A & -1 & \cos C\\ \cos B & \cos C & -1\end{vmatrix} =1+ 2\cos A\cos B\cos C+\cos^2 A +\cos^2 B-\cos^2 C\\ 三角形三內角和=A+B+C =180^\circ \Rightarrow \cos C=-\cos(A+B)\\ \Rightarrow \cos^2 C=\cos^2(A+B) =(\cos A\cos B-\sin A\sin B)^2\\ =\cos^2 A\cos^2 B-2\cos A\cos B\sin A\sin B+(1-\cos^2A)(1-\cos^2 B)\\ =1+2\cos^2 A\cos^2B-2\cos A\cos B\sin A\sin B-\cos^2 A-\cos^2 B\\ =1+2\cos A\cos B(\cos A\cos B-\sin A\sin B)-\cos^2 A-\cos^2B\\ =1+2\cos A\cos B\cos(A+B)-\cos^2 A-\cos^2 B \\ \Rightarrow \cos^2 A+\cos^2 B=1+2\cos A\cos B\cos(A+B)-\cos^2 C \\\Rightarrow \triangle = 2+2\cos A\cos B(\cos C+\cos(A+B))-2\cos^2 C= 2+0-2\cos^2 C\\ =2-2\times {15\over 16} =\bbox[red, 2pt]{1\over 8}$$
解答：$$令\cases{\vec u=(1,2,3)\\ \vec v=(3,-1,2)}及L與M夾角為\theta \Rightarrow \cos \theta={\vec u\cdot \vec v\over |\vec u||\vec v|} ={1\over 2} \Rightarrow L與M夾角為60^\circ 或120^\circ \\ \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{CD} = |\overrightarrow{AB}|| \overrightarrow{CD}|\cos\theta =6\cdot 5\cdot (\pm {1\over 2})= \pm 15\\又A、C為垂足\Rightarrow \cases{\overline{AC}\bot \overline{AB} \\ \overline{AC} \bot \overline{CD}} \Rightarrow \cases{\overrightarrow{AC} \cdot \overrightarrow{AB} =0\\ \overrightarrow{AC} \cdot \overrightarrow{CD} =0} \\ \overline{BD}^2 = |\overrightarrow{BD}|^2 =(\overrightarrow{BA} +\overrightarrow{AC} +\overrightarrow{CD} )\cdot (\overrightarrow{BA} +\overrightarrow{AC} +\overrightarrow{CD} ) \\= |\overrightarrow{BA}|^2+ |\overrightarrow{AC}|^2 + |\overrightarrow{CD}|^2 +2( \overrightarrow{BA} \cdot \overrightarrow{AC}+ \overrightarrow{AC} \cdot \overrightarrow{CD} +\overrightarrow{CD} \cdot \overrightarrow{BA}) =6^2 +4^2+5^2 +2(0+0 \pm 15)\\ =77\pm 30 = 107或47 \Rightarrow \overline{BD}= \bbox[red,2pt]{\sqrt{107}或\sqrt{47}}$$
解答：
$$\cases{\overline{AB}為直徑\Rightarrow \angle APB=90^\circ \\\overline{AD}為直徑\Rightarrow \angle APD=90^\circ} \Rightarrow B、P、D在一直線上；又\cases{\angle A=90^\circ \\ \overline{AD} =\sqrt 3\times \overline{AB}} \Rightarrow \cases{\angle ABD=60^\circ \\ \angle ADB=30^\circ}\\ 令\cases{E為\overline{AB}中點\\ F為\overline{AD}中點}，則\cases{\angle AEP=120^\circ \\ \angle AFP=60^\circ} \Rightarrow \cases{扇形EAP面積=({\sqrt 6\over 2})^2\pi \times {120^\circ \over 360^\circ}={1 \over 2}\pi \\扇形FAP面積=({3\sqrt 2\over 2})^2\pi \times {60^\circ \over 360^\circ}={3 \over 4}\pi}\\ 又\cases{\triangle EAP面積={1\over 2}\times \overline{AP}\times \overline{BP}= {1\over 4}\times {3\over 2}\sqrt 2 \times {1\over 2}\sqrt 6={3\over 8}\sqrt 3\\ \triangle APD面積={1\over 4}\times \overline{AP}\times \overline{PD}= {1\over 2}\times {3\over 2}\sqrt 2 \times {3\over 2}\sqrt 6={9\over 8}\sqrt 3}\\ \Rightarrow 欲求之舖色面積= \left( {1 \over 2}\pi-{3\over 4}\sqrt 3\right) +\left( {3 \over 4}\pi-{9\over 4}\sqrt 3\right) = \bbox[red, 2pt]{{5\over 4}\pi-{3\over 2}\sqrt 3}$$
解答：
$$\triangle ABD \Rightarrow \cases{\angle DAB = \angle CAB-\angle CAD= \gamma-\alpha\\ \angle ABD=90^\circ +\angle CBD=90^\circ +\beta} \\\Rightarrow \angle ADB =180^\circ -\angle DAB-\angle ABD= 90^\circ+\alpha-\beta-\gamma \\ 再利用正弦定理 \Rightarrow {\overline{AB} \over \overline{AD}} ={\sin(90^\circ+\alpha-\beta-\gamma) \over \sin (90^\circ +\beta)} ={\cos (\alpha-\beta-\gamma) \over \cos \beta} ={\cos (\alpha-\beta-\gamma) \over 8/17} \\ 由於\overline{AD}= \overline{AC}，所以{\overline{AB} \over \overline{AD}} ={\overline{AB} \over \overline{AC}} =\cos \gamma ={\cos (\alpha-\beta-\gamma) \over 8/17}\\ \Rightarrow {8\over 17} \cos \gamma= \cos (\alpha-\beta-\gamma) =\cos(\alpha-\beta) \cos \gamma+ \sin(\alpha-\beta) \sin\gamma\\ =(\cos \alpha \cos \beta + \sin \alpha \sin \beta)\cos \gamma + (\sin \alpha \cos \beta - \sin \beta\cos \alpha)\sin \gamma \\ =({4\over 5}\cdot {8\over 17}+ {3\over 5}\cdot {15\over 17})\cos \gamma + ({3\over 5} \cdot {8\over 17}-{15\over 17} \cdot{4\over 5}) \sin \gamma ={77\over 85}\cos \gamma -{36\over 85}\sin \gamma \\ \Rightarrow {36\over 85}\sin \gamma ={37\over 85} \cos \gamma \Rightarrow \tan \gamma ={\sin \gamma \over \cos \gamma} ={37/85 \over 36/85} = \bbox[red, 2pt]{37\over 36}$$
解答：$$令z=(x+y-|x-y|)\div 2，則\\\begin{array}{ccc|ccc} x & y & z &x & y & z \\\hline 1 & 1 & 1 & 4 & 1& 1\\ 1& 2& 1 & 4 & 2& 2 \\ 1& 3& 1& 4& 3& 3 \\ 1& 4& 1& 4& 4& 4\\ 1& 5 & 1 & 4 & 5& 4\\ 1& 6 & 1 & 4 & 6 & 4 \\\hdashline 2 & 1& 1 & 5 & 1& 1 \\ 2& 2& 2 & 5 & 2& 2\\ 2& 3& 2& 5 & 3& 3\\ 2& 4& 2 & 5 & 4& 4\\ 2 & 5 & 2& 5& 5& 5\\ 2& 6 & 2 & 5& 6& 5 \\\hdashline 3 & 1& 1 & 6 & 1& 1\\ 3 & 2& 2 & 6 & 2& 2\\ 3 & 3& 3 & 6 & 3& 3\\ 3 & 4& 3 & 6 & 4& 4\\ 3 & 5& 3 & 6 & 5& 5\\ 3 & 6& 3 & 6 & 6& 6\\\hline & \sum & 32& &\sum &59\end{array} \\ \Rightarrow 期望值={32+59\over 6^2} =\bbox[red, 2pt]{61\over 36}$$
解答：$$\cases{\tan \alpha+\tan \beta=-(m+1) \\ \tan \alpha\tan \beta=m+4 \\ 有實根:(m+1)^2-4(m+4)\ge 0} \Rightarrow \cases{ \tan(\alpha+\beta) ={-(m+1)\over 1-(m+4)} ={m+1\over m+3} \\(m-5)(m+3)\ge 0 \Rightarrow m\ge 5或m\le -3 \cdots(1)} \\ \sum_{n=1}^\infty \tan^{n-1}(\alpha+\beta) 收斂\Rightarrow |\tan(\alpha+\beta)| \lt 1 \Rightarrow|{m+1\over m+3}|\lt 1 \Rightarrow (m+1)^2 \lt (m+3)^2\\ \Rightarrow m\gt -2 \cdots(2)；\\ 同時符合條件(1)及(2) \Rightarrow \bbox[red,2pt]{m\ge 5}$$
解答：$$令g(x)=\sqrt{x+b}-a，則\cases{g(2)=0 \\ \lim_{x\to 2}{g(x)\over x-2}={1\over 6}} \Rightarrow \cases{\sqrt{b+2}=a\\ g'(2)=1/6 \Rightarrow b=7} \Rightarrow a=3\\ \Rightarrow (a,b)= \bbox[red, 2pt]{(3,7)}$$
解答：$$區間[0,4]切成n段，每段長度為4/n \Rightarrow \cases{U_n={4\over n}(({4\over n})^3+({8\over n})^3+ \cdots +({4n\over n})^3) \\ L_n={4\over n}(0^3+({4\over n})^3+ \cdots +({4(n-1)\over n})^3) } \\ \Rightarrow U_n-L_n= {4\over n}\cdot 4^3= {256\over n} \lt 0.01 \Rightarrow n\gt 25600 \Rightarrow n=\bbox[red, 2pt]{25601}$$
二、計算或證明題(每題 10 分，共 30 分，須有計算或證明過程)
解答：$$(2n-1)(2n+1)=4n^2-1 \lt 4n^2 \Rightarrow (2n-1)(2n+1) \lt (2n)^2\\ 令S=1\times 3\times 5\times \cdots (2n-1) \\\Rightarrow (2n+1)S^2 = 1^2\times 3^2 \times 5^2\times \cdots \times (2n-1)^2(2n+1)\\ =(1\times 3)(3\times 5)(5\times 7)\cdots ((2n-3)(2n-1))((2n-1)(2n+1)) \\ \lt 2^2\times 4^2 \times 6^2\times \cdots \times (2n)^2 \\ \Rightarrow (2n+1)S^2 \lt 2^2\times 4^2 \times 6^2\times \cdots \times (2n)^2 \Rightarrow \sqrt{2n+1}S \lt 2\times 4\times 6\times \cdots \times (2n) \\ \Rightarrow 0\lt {S\over 2\times 4\times 6\times \cdots \times (2n)} \lt {1\over \sqrt{2n+1}} \equiv 0\lt a_n \lt {1\over \sqrt{2n+1}} \\ \Rightarrow 0\lt \lim_{n\to \infty} a_n \lt \lim_{n\to \infty}{1\over \sqrt{2n+1}}=0 \Rightarrow \lim_{n\to \infty} a_n =\bbox[red, 2pt]{0}$$
解答：$$y=x^3+ax^2+x+1 \Rightarrow y'=3x^2+2ax+1\\ 令切點P(t,t^3+at^2+t+1)，則斜率為3t^2+2at+1\\，過P之切線L: y=(3t^2+2at+1)(x-t)+t^3+at^2+t+1\\又L過(0,0) \Rightarrow -3t^3-2at^2-t+t^3+at^2+t+1=0 \Rightarrow f(t)=2t^3+at^2-1=0 有三相異根\\ \Rightarrow f'(t)=6t^2+2at=0 \Rightarrow 2t(3t+a)=0 \Rightarrow t=0,-a/3有極值 \Rightarrow f(0)f(-a/3)\lt 0\\ \Rightarrow -1(-2a^3/27+a^3/9-1)\lt 0 \Rightarrow {a^3\over 27}-1 \gt 0 \Rightarrow {a\over 3} \gt 1 \Rightarrow \bbox[red,2pt]{a\gt 3}$$
解答： (1)
$$\angle D'BP+ \angle CB'M=90^\circ = \angle CB'M+\angle B'MC \Rightarrow \angle D'BP = \angle B'MC\\ \Rightarrow \triangle CMB' \sim \triangle DB'P (AAA) ，因此令\cases{\overline{CM}=a \\ \overline{CB'}=b \\ \overline{B'M}=c=1-a} \Rightarrow \cases{\overline{B'D}=ka=1-b \Rightarrow k={1-b\over a}\\ \overline{DP}=kb\\ \overline{B'P}=kc} \\ \Rightarrow \triangle PB'D周長= k(a+b+c) = {1-b\over a}(a+b+1-a) = {1-b^2\over a}= {1-(c^2-a^2)\over a}\\= {1-((1-a)^2-a^2)\over a}={2a\over a} =2，\bbox[red, 2pt]{故得證}$$(2)$$\triangle MB'C三邊長: \cases{\overline{MC} =a\\ \overline{MB'}=c=1-a\\ \overline{B'C}=b}，由於\angle C=90^\circ \Rightarrow b^2= (1-a)^2-a^2 = 1-2a\\ \Rightarrow b=\sqrt{1-2a} \Rightarrow \triangle MB'C面積=f(a)={1\over 2}\times a\times \sqrt{1-2a}\\ 因此f'(a)=0 \Rightarrow \sqrt{1-2a} -{a\over \sqrt{1-2a}} =0 \Rightarrow {1-3a\over \sqrt{1-2a}}=0 \Rightarrow a={1\over 3}\\ \Rightarrow f({1\over 3})= {1\over 2}\times {1\over 3}\times \sqrt{1\over 3} =\bbox[red, 2pt]{\sqrt 3\over 18}$$
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