國立新竹高商106學年度第一次數學科教師甄選
一、填充題
解：$$x^4+ax^2 +bx+c = (x-1)^3(x-c) =(x^3-3x^2+3x-1)(x-c) \\ \Rightarrow x^3係數=0=-c-3 \Rightarrow c=-3 \Rightarrow \cases{a=3c +3=-6 \\ b=-3c-1=8} \Rightarrow (a,b,c)=\bbox[red,2pt]{(-6,8,-3)}$$
解：$$此題相當於求兩圖形\cases{y=f(x)=x/3\\ y=g(x)=\cos(\pi x)}有幾個交點\\ \cases{g(x)=1 \Rightarrow x=2k,k\in Z \\ g(x)=-1 \Rightarrow x= 2k+1 ,k\in Z} 及\cases{f(x)\ge 1 \Rightarrow x\ge 3\\ f(x)\le -1 \Rightarrow x\le -3} \Rightarrow x在區間[-3, 3)有交點\\ 且在x >0 有3個交點，x < 0有4個交點，共\bbox[red,2pt]{7}個交點$$
解：$$\cases{\overline{AB} =10\sin \theta\\ \overline{BO}= 10\cos \theta } \Rightarrow \overline{BH} =\overline{AB}\times \overline{BO} \div \overline{OA} = 100\sin \theta \cos \theta \div 10 = 10\sin \theta \cos \theta \\ \Rightarrow \overline{AB} +\overline{BO} +\overline{BH} =10(\sin \theta +\cos \theta) +10\sin \theta \cos \theta =10\sqrt 2 \sin(\theta+45^\circ)+5\sin (2\theta)\\ 顯然\theta=45^\circ 時，有最大值 \bbox[red,2pt]{10\sqrt 2+5}$$
解：$$\cases{n=1\\ m=2} \Rightarrow 7n^2-m^2=7-4=\bbox[red,2pt]{3}為最小值$$此題並無特定解法!
解：$$\cases{21x +22y+27z =50 \cdots(1)\\ 22x+23y+28 z=51\cdots(2)\\ 23x +24y+25z=52 \cdots(3)} \xrightarrow{(2)-(1),(3)-(2)} \cases{x+y+z =1\\ x+y-3z=1} \Rightarrow \cases{z=0\\ x+y=1} 代回(1) \\ \Rightarrow 21x+ 22(1-x)=50 \Rightarrow x=-28 \Rightarrow y=1-x=29 \Rightarrow (x,y,z)= \bbox[red,2pt]{(-28,29,0)}$$
解：$${1\over x}+{1\over y}={1\over 6^6} \Rightarrow {x+y\over xy}={1\over 6^6} \Rightarrow xy-6^6(x+y)=0 \Rightarrow (x-6^6)(y-6^6) =6^{12} \\而6^{12}=2^{12}\times 3^{12} 的因數有(1+12)(1+12)=169個，因此有\bbox[red,2pt]{169}組解$$
解：$$取1,4,7,...,3k+1,...,2017，共有\bbox[red,2pt]{673}個。$$
解：$$\lim_{n\to \infty}{1\over n^2} \sum_{k=1}^n\sqrt{16n^2-(4k)^2} =\lim_{n\to \infty}{1\over n} \sum_{k=1}^n\sqrt{16-16({k\over n})^2} = \int_0^1 \sqrt{16-16x^2}\;dx\\ =\int_0^1 4\sqrt{1-x^2}\;dx = \int_0^{\pi/2}4\cos^2\theta \;d\theta (x=\sin \theta \Rightarrow dx = \cos\theta \;d\theta) = \int_0^{\pi/2} 2\cos(2\theta)+2 \;d\theta\\ =\left. \left[ \sin (2\theta) +2\theta\right] \right|_0^{\pi/2} =\bbox[red,2pt]{\pi}$$
解：$$\cases{a^2+b^2=4 \\ (c-5)^2+(d-12)^2=36} \Rightarrow \cases{(a^2+b^2)(12^2+(-5)^2) \ge (12a-5b)^2 \\(a^2+(-b)^2)((d-12)^2+(c-5)^2) \ge (a(d-12)-b(c-5))^2}\\ \Rightarrow \cases{4\times 169 \ge (12a-5b)^2 \\4\times 36 \ge (ad-bc-12a +5b)^2} \Rightarrow \cases{26 \ge 12a-5b \ge -26 \\ 12+12a-5b \ge ac-bd \ge -12+12a-b} \\ \Rightarrow 12+26 \ge ac-bd \ge -12-26 \Rightarrow 38 \ge ac-bd \ge -38 \\ \Rightarrow ac-bd的最大值為\bbox[red,2pt]{38}$$
解：$$\cases{A+B+C+D =16\\ A,B,C,D \in N\\ A > B} \Rightarrow \cases{A+B+C+D= 12\\ A,B,C,D \in \{0,1,2,\cdots,12\} \\ A >B} \\ \Rightarrow \begin{array}{} A & B & C+D & 個數\\\hline 1 & 0 & 11 & H^2_{11}=C^{12}_{11} \\ 2 & 0 & 10 & H^2_{10} =C^{11}_{10} \\ \cdots\\ 12& 0 & 0 & H^2_0=C^1_0\\\hdashline 2 & 1& 9 & H^2_9=C^{10}_9\\ \cdots\\ 11& 1 & 0 & H^2_0 =C^1_0 \\\hdashline \cdots \\\hdashline 6 & 5 & 1 & H^2_1=C^2_1\\7 & 5 & 0 & H^2_0=C^1_0 \\\hline\end{array}\\ \Rightarrow 總和= \sum_{n=1}^{6}\sum_{k=1}^{2n}k =\sum_{n=1}^{6}n(2n+1) =\sum_{n=1}^{6}2n^2 +\sum_{n=1}^{6}n =282+21 =\bbox[red,2pt]{303} $$
解：$$1,\sin\theta,\sqrt 3\cos \theta 為x^3+ax+b=0的三根\Rightarrow \cases{1+\sin\theta+\sqrt 3\cos \theta=0 \cdots(1)\\ \sin\theta+ \sqrt 3\sin\theta \cos\theta +\sqrt 3\cos \theta = a\cdots(2)\\ \sqrt 3\sin\theta \cos \theta= -b \cdots(3)}\\ (1) \Rightarrow \sin\theta+\sqrt 3\cos \theta=-1 \Rightarrow {1\over 2}\sin\theta+ {\sqrt 3 \over 2}\cos \theta= -{1\over 2} \Rightarrow \sin(\theta +30^\circ)=-{1\over 2}\\ \Rightarrow \theta = 150^\circ (\because 0 < \theta < \pi)代入(2)及(3)，可得\cases{a={1\over 2}-{\sqrt 3\over 2}\cdot {\sqrt 3\over 2}-\sqrt 3\cdot {\sqrt 3\over 2} =-{7\over 4} \\b= {\sqrt 3\over 2}\cdot {\sqrt 3\over 2}={3\over 4}} \\ \Rightarrow (a,b)=\bbox[red,2pt]{(-{7\over 4} ,{3\over 4})}$$
解：$$圓與坐標軸分別交於\cases{A(0,2)\\ B(-2,0)\\ C(0,-2) \\ D(2,0)}，兩直線斜率均為1，因此兩直線即為\overleftrightarrow{AB}及\overleftrightarrow{CD}\\ \Rightarrow m^2+n^2=2^2+(-2)^2 =\bbox[red,2pt]{8}$$
解：$$C,F,D為切點 \Rightarrow \cases{\overline{AC}= \overline{AD}=a+2\\ \overline{BC}= \overline{BF}=a-2\\ \overline{ED}= \overline{EF}=1 } \Rightarrow \cases{\overline{AB}=4 \\ \overline{AE}=a+1 \\ \overline{BE}=a-1} \\ 直角\triangle ABE \Rightarrow \overline{AB}^2 = \overline{AE}^2 +\overline{BE}^2 \Rightarrow 4^2=(a+1)^2 +(a-1)^2 \Rightarrow a^2=7 \Rightarrow a= \bbox[red,2pt]{\sqrt{7}}$$
解：$$a+b+c \ge 3\sqrt[3]{abc} \Rightarrow {1\over 3} \ge \sqrt[3]{abc} \Rightarrow \sqrt[3]{abc}的最大值為{1\over 3}\\\sqrt{a^2+b^2} +\sqrt{b^2+c^2} +\sqrt{c^2+a^2} \ge 3\sqrt[3]{\sqrt{a^2+b^2}\sqrt{b^2+c^2}\sqrt{c^2+a^2}} \ge 3\sqrt[3]{\sqrt{2ab} \cdot \sqrt{2bc} \cdot \sqrt{2ca}}\\ = 3\sqrt 2\sqrt[3]{abc} \Rightarrow \sqrt{a^2+b^2} +\sqrt{b^2+c^2} +\sqrt{c^2+a^2} \ge 3\sqrt 2\sqrt[3]{abc} \ge \bbox[red,2pt]{\sqrt 2}$$
解：$$\cases{原數1000a+100b+10c+d \\ 新數1000d+100c+10b+a} \Rightarrow 1000d+100c+10b+a-(1000a+100b+10c+d)=4725\\ \Rightarrow 999d+90c-90b-999a=4725 \Rightarrow 111(d-a)-10(b-c)=525 \Rightarrow \cases{d-a=5 \\b-c=3} \\ \Rightarrow \cases{(d,a)=(6,1),(7,2),(8,4),(9,5)\\ (b,c)=(3,0),(4,1),(5,2), (6,3),(7,4),(8,5),(9,6)} \Rightarrow 共有4\times 7= \bbox[red,2pt]{28}個四位數$$
解：$$f(93)=90^2-f(90) = 90^2-(87^2-f(87))=90^2-87^2 +f(87)= 90^2-87^2 +84^2-f(81) \\ =90^2-87^2 +84^2-\cdots+30^2-f(30)=93 \Rightarrow f(30)=90^2-87^2 +84^2-\cdots+30^2-93 \\ =\sum_{k=0}^{10} (30+6k)^2-\sum_{k=0}^{10} (33+6k)^2+93^2-93=\sum_{k=0}^{10}(-36k-189)+93^2-93 \\= -36\times 55 -189\times 11+93^2-93= \bbox[red,2pt]{4497}$$
解：$$由長除法可知: f(x)=(x+1)(x^2+(a-1)x+(b-a+1))+c-b+a-1\\ 又\lim_{x\to -1}{f(x)\over x+1}=3 \Rightarrow \cases{f(x)=(x+1)g(x)\\ g(-1)=3} \Rightarrow g(x)= x^2+(a-1)x+(b-a+1) \\\Rightarrow g(-1)= 1+1-a+b-a+1=3 \Rightarrow b=2a\cdots(1)\\ y=f(x)無極值 \Rightarrow f'(x)=3x^2+2ax +b=0 無相異實根\Rightarrow 判別式\le 0 \Rightarrow 4a^2-12b \le 0 \\ \stackrel{式(1)}{\Rightarrow} 4a^2-12\cdot (2a) \le 0 \Rightarrow 4a(a-6) \le 0 \Rightarrow \bbox[red,2pt]{0\le a\le 6}$$
解：$$令\log_2{y\over x^2} =M為最大值 \Rightarrow y=2^Mx^2，依題意意相當於求\cases{y=2^Mx^2\\ 2x+y+2=0} 的交點；\\由於M為最大值，兩圖形僅交於一點，即相切，也就是2x+2^Mx^2+2=0僅有一解，\\判別式:2^2-8\cdot 2^M=0 \\ \Rightarrow 2^M={1\over 2} \Rightarrow M= \bbox[red,2pt]{-1}$$
解：$$內接拋物線\Gamma :y^2=4cx正\triangle 三頂點\cases{A({\alpha^2\over 4c},\alpha) \\ B({\beta^2 \over 4c},\beta) \\ C({\gamma^2\over 4c},\gamma) } \Rightarrow \cases{重心G({\alpha^2+\beta^2 +\gamma^2\over 12c},{\alpha +\beta +\gamma \over 3})\\ \overleftrightarrow{AB}斜率m_1={\beta-\alpha\over (\beta^2-\alpha^2)/4c}={4c\over \beta+\alpha} \\\overleftrightarrow{BC}斜率m_2= {\gamma-\beta\over (\gamma^2-\beta^2)/4c}={4c\over \gamma+\beta} \\ \overleftrightarrow{CA}斜率m_3={\alpha-\gamma\over (\alpha^2-\gamma^2)/4c}={4c\over \alpha+\gamma} }\\ \Rightarrow \tan 60^\circ = \sqrt 3= {m_2-m_1\over 1+m_1m_2 }= {m_3-m2\over 1+m_2m_3 }= {m_1-m_3\over 1+m_1m_3 } \\ \Rightarrow {{4c\over \gamma+\beta}-{4c\over \beta+\alpha} \over 1+{16c^2\over (\alpha+\beta)(\beta +\gamma)}} ={{4c\over \alpha+\gamma}-{4c\over \beta+\gamma} \over 1+{16c^2\over (\alpha+\gamma)(\beta +\gamma)}} = {{4c\over \alpha+\beta}-{4c\over \alpha+\gamma} \over 1+{16c^2\over (\alpha+\beta)(\alpha +\gamma)}} =\sqrt 3\\ \Rightarrow {4c(\alpha-\gamma) \over 16c^2+(\alpha+\beta)(\beta+\gamma)} ={4c(\beta-\alpha) \over 16c^2+(\beta+\gamma)(\gamma+\alpha)} = {4c(\gamma-\beta) \over 16c^2 +(\alpha+\beta)(\alpha+\gamma)} =\sqrt 3\\ \Rightarrow {4c(\alpha-\gamma)+4c(\beta-\alpha) + 4c(\gamma-\beta) \over 48c^2+(\alpha+\beta) (\beta+\gamma)+ (\beta+\gamma)(\gamma+\alpha) +(\gamma+\alpha) (\alpha+\beta)} =\sqrt 3\\ 上式分子為0，因此分母亦為0，即48c^2+(\alpha+\beta) (\beta+\gamma)+ (\beta+\gamma)(\gamma+\alpha) +(\gamma+\alpha) (\alpha+\beta)=0 \\ \Rightarrow 48c^2+ \alpha^2+\beta^2+\gamma^2 +3(\alpha\beta +\beta\gamma +\gamma\alpha)=0 \Rightarrow \alpha\beta +\beta\gamma +\gamma\alpha=-16c^2-(\alpha^2+\beta^2+\gamma^2)/3\\ 重心G({\alpha^2+\beta^2 +\gamma^2\over 12c}=x,{\alpha +\beta +\gamma \over 3}=y) \Rightarrow (\alpha+\beta +\gamma)^2 =9y^2 =\alpha^2+\beta^2+\gamma^2 +2(\alpha\beta +\beta\gamma +\gamma\alpha)\\ =12cx+2(\alpha\beta +\beta\gamma +\gamma\alpha) =12cx+2(-16c^2-(\alpha^2+\beta^2+\gamma^2)/3)=12cx+2(-16c^2-4cx) \\ =4cx-32c^2 \Rightarrow \bbox[red,2pt]{9y^2=4cx-32c^2}$$
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