106年松山工農教甄-數學詳解

 臺北市立松山高級工農職業學校106學年度第 1 次教師甄選

一、填充題

解答： $${1\over 2}(x^2+6x+10)^3-{1\over 2} \Rightarrow 最高次方項為{1\over 2}x^6 \Rightarrow f(x)最高次方項為\left({1\over 2}x^6 \right)' =3x^5 \Rightarrow a=3\\ \int_b^b f(t)\;dt=0 \Rightarrow {1\over 2}(b^2+6b+10)^3-{1\over 2} =0 \Rightarrow b^2+6b+10=1 \Rightarrow (b+3)^2=0 \Rightarrow b=-3\\ \Rightarrow (a,b)= \bbox[red,2pt]{(3,-3)}$$

解答： $$2017^{2017} =(183\times 11+4)^{2017}  \equiv 4^{2017} \mod 11\\ 4^{2017}=2^{4034} = (2^{10})^{403}\cdot 2^4 =(93\times 11+1)^{403} \cdot 2^4 \equiv 2^4 \mod 11 =\bbox[red,2pt]{5}$$

解答： $$\begin{array}{} 甲& 乙 & 機率\\\hline 紅& 紅 & {4\over 15}\cdot {3\over 14}\\ 白紅& 紅 & {1\over 15}\cdot {4\over 14} \cdot {3\over 13} \\ 紅& 白紅 & {4\over 15}\cdot {1\over 14} \cdot {3\over 13} \\\hline\end{array} \\ \Rightarrow 機率={4\times 3\times 13 +12+12 \over 15\times 14\times 13} =\bbox[red, 2pt]{6\over 91}$$

解答： $$\cases{x=a+b+2c \\ y=a+2b+c \\ z=a+b+c} \Rightarrow \cases{a=3z-x-y \\b=y-z\\ c=x-z}\\ \Rightarrow  {2b-2c\over a+b+2c} +{2a+4c \over a+ 2b+c } +{b\over a+b+c} ={2y-2x \over x} +{2x-2y+2z \over y} +{y-z\over z} \\= \left({2y\over x}+{2x\over y}\right)+ \left({2z\over y} +{y\over z}\right)-5\\ \ge 2 \sqrt {{2y\over x}\cdot {2x\over y}} +2\sqrt{  {2z\over y} \cdot {y\over z}} -5 = 4+2\sqrt 2  -5 =\bbox[red,2pt]{2\sqrt 2-1}$$

解答：

$$假設立方體邊長為1，各頂點坐標如上圖，則\cases{A'=A+t\overrightarrow{AB}=(1, t,0)\\ C'=C+ t\overrightarrow{CD}=(0,1,t)\\ E'=E+ t\overrightarrow{EF} =(t,0,1)} \\\Rightarrow \cases{\vec u = \overrightarrow{A'C'}=(-1,1-t,t)\\ \vec v= \overrightarrow{A'E'}=(t-1,-t,1)} \\ \Rightarrow \triangle A'C'E' = {1\over 2}\sqrt{|\vec u|^2|\vec v|^2 -(\vec u\cdot \vec v)^2} ={1\over 2}\sqrt{4(t^2-t+1)^2-(t^2-t+1)^2} \\ ={\sqrt 3\over 2}(t^2-t+1) = {\sqrt 3\over 2}((t-{1\over 2})^2+{3\over 4})\Rightarrow 最小值為{\sqrt 3\over 2}\times {3\over 4} =\bbox[red,2pt]{3\sqrt 3\over 8}$$

解答：

$$\triangle EDA為等腰直角\Rightarrow \cases{\overline{AD}=2\sqrt 2\\ \overline{DAE}=45^\circ} \Rightarrow \angle DAB=105^\circ -45^\circ =60^\circ \\ \Rightarrow \cos \angle DAB={1\over 2} ={(2\sqrt 2)^2+(\sqrt 6+\sqrt 2)^2 -a^2 \over 4\sqrt 2(\sqrt 6+\sqrt 2)} \Rightarrow a=2\sqrt 3 \\\Rightarrow \overline{BC}^2+\overline{BD}^2 = 2^2+a^2=16 = \overline{CD}^2\\ \Rightarrow \angle CBD=90^\circ \Rightarrow ABCDE面積= \triangle AED+\triangle ABD +\triangle BCD = 2 +(3+\sqrt 3)+ 2\sqrt 3 \\= \bbox[red,2pt]{5+3\sqrt 3}$$

解答： $$\cases{A(57,23)\\ B(7,-2)\\ C(5,12) \\ D(x,y)} \Rightarrow \cases{\overrightarrow{AB} =(-50,-25)\\ \overrightarrow{AC} =(-52,-11) \\ \overrightarrow{AD} =(x-57,y-23)} \Rightarrow \cases{7\overrightarrow{AB} =(-350,-175) \\ 3\overrightarrow{AC} =(-156,-33)\\ 4\overrightarrow{AD} = (4x-228, 4y-92)} ；\\ 4\overrightarrow{AD} = 7\overrightarrow{AB}- 3\overrightarrow{AC} \Rightarrow \cases{4x-228= -350+156=-194\\ 4y-92 = -175+33=-142} \Rightarrow \cases{x=17/2\\ y=-25/2} \\ \Rightarrow D(x,y)= \bbox[red,2pt]{\left({17\over 2},-{25\over 2} \right)}$$

解答： $$\sqrt{105\sqrt{104\sqrt{103\sqrt{102\sqrt{101\sqrt{100\times 98+1}+1}+1}+1}+1}+1} \\ =\sqrt{105\sqrt{104\sqrt{103\sqrt{102\sqrt{101\sqrt{(99+1)(99-1)+1}+1}+1}+1}+1}+1} \\= \sqrt{105\sqrt{104\sqrt{103\sqrt{102\sqrt{101\sqrt{99^2 -1 +1}+1}+1}+1}+1}+1} \\ = \sqrt{105\sqrt{104\sqrt{103\sqrt{102\sqrt{101\times 99+1}+1}+1}+1}+1} \\ = \sqrt{105\sqrt{104\sqrt{103\sqrt{102\times 100+1}+1}+1}+1} \\ = \sqrt{105\sqrt{104\sqrt{103 \times 101+1}+1}+1}  = \sqrt{105\sqrt{104 \times 102 +1}+1}\\  = \sqrt{105 \times 103+1}= \bbox[red,2pt]{104}$$

解答： $$\cases{直線L:y=x\\ 曲線\Gamma: y=x^3-3x^2+ax}兩圖形相切，令切點為P；P在L上，也在\Gamma上\\ \Rightarrow y=k=k^3-3k^2+ak \Rightarrow k^3-3k^2+(a-1)k=0 \Rightarrow k(k^2-3k+a-1)=0 \cdots(1)\\ 又切點斜率為1 \Rightarrow y'(k)=1 \Rightarrow 3k^2-6k+a=1 \Rightarrow 3k^2-6k+a-1=0 \cdots(2)\\ (1)\Rightarrow \cases{k=0\\ k^2-3k=1-a} 代入(2) \Rightarrow \cases{a-1=0 \\ 3k^2-6k=k^2-3k} \Rightarrow \cases{a=1\\ k(2k-3)=0 \Rightarrow k=0,3/2}\\ 將k=3/2代入(2) \Rightarrow {27\over 4}-9+a-1=0 \Rightarrow a=13/4\\ 因此a=\bbox[red,2pt]{1或13/4}$$

解答： $$z^{12}=64 的根為z_k=\sqrt 2e^{ik\pi/6},k=0-11 \Rightarrow 正實部的根為z_0,z_1,z_2,z_{11},z_{10} \\ \Rightarrow z_0+(z_1+z_{11} ) +(z_2+z_{10}) = \sqrt 2+({\sqrt 6\over 2}+{\sqrt 6\over 2})+({\sqrt 2\over 2}+{\sqrt 2\over 2}) = \bbox[red, 2pt]{2\sqrt 2+\sqrt 6}$$

解答：

$$三角錐底面為\triangle ABC，並令B為原點，則\cases{B(0,0,0)\\ C(24,0,0)\\ A(12,16,0)}；由於頂點D在平面x=12y上，\\因此D(12,y,z)；再由\cases{\overline{BD}=25 \\ \overline{AD}=25} \Rightarrow \cases{144+y^2+z^2 =625\\ (y-16)^2+z^2 = 625} \Rightarrow \cases{y=7/2\\ z=25\sqrt 3/2} \\ \Rightarrow 三角錐體積={1\over 3}\times \triangle ABC\times h = {1\over 3}\times {1\over 2}\times {24}\times 16\times {25\sqrt 3 \over 2}= \bbox[red,2pt]{800\sqrt 3}$$

解答：

$${(x-8)^2\over 121}+ {(y-15)^2\over 100} =1 \Rightarrow \cases{中心點A(8,15)\\ a=11\\ b=10} \Rightarrow \overline{OA}=\sqrt{8^2 +15^2} =17;\\ 又\overleftrightarrow{OA} 與橢圓的交點為P與Q，如上圖；因此\cases{6\lt \overline{OP} \lt 7 \\ 27\lt \overline{OQ} \lt 28} \\\Rightarrow 橢圓與原點的整數距離為7,8,\dots,27，共有21種；由於左右對稱，共有\bbox[red,2pt]{42}個點與原點距離為整數$$

解答： $$\sqrt{2014}x^3- 4029x^2+2=0 \Rightarrow \cases{x_1x_2 +x_2x_3 +x_1x_3= 0\\(\sqrt{2014}x-1)(x^2-2\sqrt{2014}x-2)=0}\\ \Rightarrow x={1\over \sqrt{2014}}, {\sqrt{2014} \pm \sqrt{2016}  } \Rightarrow {\sqrt{2014} - \sqrt{2016}  }\lt {1\over \sqrt{2014}} \lt {\sqrt{2014} + \sqrt{2016}  } \\ \Rightarrow \cases{x_1= {\sqrt{2014} - \sqrt{2016}  }\\ x_2={1\over \sqrt{2014}} \\ x_3={\sqrt{2014} + \sqrt{2016}  } } \Rightarrow x_2(x_1+x_3) = {1\over \sqrt{2014}} \cdot 2\sqrt{2014}= \bbox[red,2pt]{2}$$

解答：

$$||x|-1| +||y|-1|=1 \Rightarrow \begin{cases} |x-1|+|y-1|=1 & x,y\ge 0\\ |x-1|+|-y-1|=1 & x\ge 0,y\le 0\\ |-x-1|+ |y-1| = 1 & x \le 0,y\ge 0\\ |-x-1|+ |-y-1|=1  & x,y\le 0\end{cases} \\ \Rightarrow \cases{以(1,1)為中心的菱形\\ 以(1,-1)為中心的菱形\\ 以(-1,1)為中心的菱形\\ 以(-1,-1)為中心的菱形}，每個菱形都是邊長為\sqrt 2的正方形，見上圖；\\因此該圖形與圓:x^2+y^2=2有\bbox[red,2pt]{8}個交點$$

解答：

$$令\cases{\overline{BC}= a=4\\ \overline{AB}=c=6\\ \overline{CA}=b =5}及s=(a+b+c) \div 2 ={15\over 2}\\ 由於\cases{\overline{AP} =\overline{AF}\\ \overline{BQ} =\overline{BF}}\Rightarrow \overline{CP} +\overline{CQ} = (\overline{AC}+ \overline{AF}) +(\overline{BC}+ \overline{BF}) = a+b+c =2s，\\又 \overline{CP} =\overline{CQ}\Rightarrow \overline{CP} =\overline{CQ}=s \Rightarrow \cases{\overline{AF}=\overline{CP}-\overline{AC} =s-b={15\over 2}-5 ={5\over 2}\\ \overline{BF}=\overline{CQ}-\overline{BC} =s-a ={15\over 2}-4= {7\over 2}}\\ 同理可得\cases{\overline{AE} =s-c ={3\over 2}\\ \overline{CE}=s-a ={7\over 2}\\ \overline{CD}=s-b ={5\over 2} \\ \overline{BD}=s-c ={3\over 2}} \Rightarrow {\triangle DEF \over \triangle ABC} ={\triangle ABC-(\triangle AEF +\triangle BDF +\triangle CDE) \over \triangle ABC} \\ =1-\left({\overline{AE}\cdot \overline{AF}\over \overline{AC}\cdot \overline{AB}} +{\overline{BD}\cdot \overline{BF}\over \overline{BC}\cdot \overline{AB}} +{\overline{CD}\cdot \overline{CE}\over \overline{AC}\cdot \overline{BC}} \right) =1-\left({15/4\over 30} +{21/4 \over 24} +{35/4 \over 20} \right) 1-{25\over 32} = \bbox[red,2pt]{7\over 32}$$ 二、問答題:以下是本校學生解題時常犯的錯誤，(1)請寫出錯誤之處及正確觀念為何？(2 分)(2)寫出正確答案(3 分，不需要寫算式)。

解答： $$x=-1只是其中一種特例，並非所有的答案\\2^{(x^2-1)} =5^{(x+1)} \Rightarrow (x^2-1)\log 2= (x+1)\log 5\Rightarrow  (x+1)\left( (x-1)\log 2-\log 5\right)=0 \\ \Rightarrow \cases{x+1=0 \\ (x-1)\log 2=\log 5 =1-\log 2} \Rightarrow \cases{x=-1\\ x={1-\log 2\over \log 2} +1 ={1\over \log 2}}\\ \Rightarrow  x= \bbox[red,2pt]{-1或{1\over \log 2}}$$

解答： $$迴歸直線不一定會經過樣本，但一定會經過(\bar x,\bar y);\\ \bar x=(1+2+4+5) \div 4=3,迴歸直線: y={1\over 2}x+{3\over 2} = {1\over 2}(x-3)+3 \Rightarrow \bar y=3 \\ \Rightarrow \bar y=(2+a+b+5) \div 4=3 \Rightarrow a+b=5\cdots(1);\\迴歸直線斜率{1\over 2}= {\sum(x-\bar x)(y-\bar y)\over \sum (x-\bar x)^2} ={(-2)(-1)+(-1)(a-3)+ 1\cdot(b-3)+2\cdot 2\over 4+1+1+4}\\ ={6-a+b \over 10}  \Rightarrow a-b=1 \cdots(2);\\ 由(1)及(2)可知:(a,b)=\bbox[red,2pt]{(3,2)}$$

解答： $$(AB)^{-1} =B^{-1}A^{-1} =\begin{bmatrix} 0& -1\\ 3& -1\end{bmatrix}\begin{bmatrix} 1& 2\\ -1 & 1\end{bmatrix} = \bbox[red,2pt]{\begin{bmatrix} 1& -1\\ 4& 5\end{bmatrix}}$$

解答： $$\cases{甲=9，乙+丙+丁=3 \Rightarrow H^3_3= C^5_3=10\\ 甲=10，乙+丙+丁=2 \Rightarrow H^3_2= C^4_2=6\\ 甲=11，乙+丙+丁=1 \Rightarrow H^3_1= C^3_1=3\\ 甲=12，乙+丙+丁=0 \Rightarrow H^3_0= 1\\ } \\\Rightarrow 甲拿9個以上物品，有10+6+3+1=20種分法\\ 全部的分法為H^4_{12} =C^{15}_{12}=455，扣除甲拿9個以上，即455-20=\bbox[red,2pt]{435}種分法$$

解答： $$\lim_{x\to \pi/2}{\sin 2x\over \cos x} =\lim_{x\to \pi/2}{2\sin x \cos x\over \cos x} = \lim_{x\to \pi/2}2\sin x  = \bbox[red,2pt]{2}$$