105年台中文華高中教甄-數學詳解

國立臺中文華高級中學 105 學年度第二次教師甄選

一、 填充題(每題 5 分， 共 80 分，全對才給分。 )

解答：

$$A=\int_3^8 \log_2 32x\;dx =\int_3^8 \left(5+{1\over \ln 2} \ln x\right)dx  \\ B=\int_1^6 \log_2(x+2)dx =\int_1^6{1\over \ln 2}\ln(x+2)dx = \int_3^8{1\over \ln 2}\ln u\;du,其中u=x+2\\ \Rightarrow A-B=\int_3^8 5\;dx=\bbox[red, 2pt]{25}$$

解答：

$$令\cases{A(z+2)\\ P(z)\\ B(z-2)}，依題意可得\cases{\overline{AP}=2\\ \overline{BP}=2\\ \angle A=\pi/3\\ \angle B=\pi/6} \Rightarrow \triangle OAB為直角\triangle ，其中\cases{\angle A=60^\circ\\ \angle B=30^\circ \\\angle AOB=90^\circ }\\ \Rightarrow  \overline{OA} =\overline{AB}\sin 30^\circ=2 \Rightarrow \cases{\overline{AA'}= 2\sin 60^\circ =\sqrt 3 \\\overline{OA'}=2\sin 30^\circ =1} \Rightarrow z+2=1+\sqrt 3i \Rightarrow z= \bbox[red, 2pt]{-1+\sqrt 3i}$$

解答：

$$對同弧的圓周角相等，即\angle BCE=\angle BAE=30^\circ \Rightarrow \triangle AEC\sim \triangle CED(AAA) \\ \Rightarrow {\overline{AE} \over \overline{CE}} ={\overline{CE} \over \overline{DE}} \Rightarrow \overline{CE}=\sqrt{8\times 2}=4 \Rightarrow \cos \angle EAC={\overline{AE}^2 +\overline{AC}^2-\overline{EC} \over 2\times \overline{AE}\times \overline{AC}} \\\Rightarrow {\sqrt 3\over 2}={64+\overline{AC}^2-16\over 16\overline{AC}}  \Rightarrow \overline{AC}=4\sqrt 3；同理，\overline{AB}=4\sqrt 3；\\因此\triangle ABC面積={1\over 2}\times \overline{AB}\times \overline{AC} \sin \angle A= {1\over 2}\times (4\sqrt 3)^2 \times {\sqrt 3\over 2} =\bbox[red,2pt]{12\sqrt 3}$$

解答： $$\overrightarrow{AG} = \overrightarrow{AE}+\overrightarrow{EG}=\overrightarrow{AE}+\overrightarrow{EF} +\overrightarrow{FG} \Rightarrow |\overrightarrow{AG}|^2 =(\overrightarrow{AE}+\overrightarrow{EF} +\overrightarrow{FG}) \cdot (\overrightarrow{AE}+\overrightarrow{EF} +\overrightarrow{FG})\\ =|\overrightarrow{AE}|^2 +|\overrightarrow{EF}|^2 +|\overrightarrow{FG}|^2 +2(\overrightarrow{AE}\cdot \overrightarrow{EF} +\overrightarrow{EF}\cdot \overrightarrow{FG} +\overrightarrow{FG}\cdot \overrightarrow{AE} ) \\ =5^2+7^2+6^2 +2(|\overrightarrow{AE}||\overrightarrow{EF} | \cos \angle AEF\\ \qquad\qquad + |\overrightarrow{EF}|| \overrightarrow{FG}| \cos(180^\circ-\angle HEF) +|\overrightarrow{FG}| | \overrightarrow{AE}|\cos (180^\circ- \angle AEH )\\ =110+2(5\cdot 7\cdot {3\over 5} +7\cdot 6\cdot (-{2\over 3})+6 \cdot 5 \cdot (-{2\over 5}) =110-10=100\\ \Rightarrow \overline{AG}=\sqrt{100}= \bbox[red,2pt]{10}$$

解答：

$$\log_{1/2}(-x^2-y^2+2y) \lt \log_{1/2}(2y-1) \Rightarrow \cases{-x^2-y^2+2y \gt 2y-1\\ -x^2-y^2+2y \gt 0 \\2y-1 \gt 0} \Rightarrow \cases{x^2+y^2 \lt 1\\ x^2+(y-1)^2 \lt 1\\ y\gt 1/2}\\ \Rightarrow 所圍區域就是上圖著色區域 ；\\先求兩圓\cases{x^2+y^2=1\\ x^2+(y-1)^2=1}的交點\cases{P(\sqrt 3/2,1/2)\\ Q(-\sqrt 3/2,1/2)} \Rightarrow \angle POQ=120^\circ \\\Rightarrow \cases{扇形OPQ面積 ={1\over 3}\pi \\\triangle OPQ={\sqrt 3\over 4}} \Rightarrow 著色面積=扇形OPQ-\triangle OPQ=\bbox[red, 2pt]{ {1\over 3}\pi -{\sqrt 3\over 4} }$$

解答： $$\sqrt{x^2-10\sqrt 3x+95}+\sqrt{x^2+10\sqrt 3x+95}=20 \\\Rightarrow \sqrt{(x-5\sqrt 3)^2+(0-2\sqrt 5)^2} + \sqrt{(x+5\sqrt 3)^2+(0-2\sqrt 5)^2}=20\\ 相當於求橢圓\Gamma:{x^2\over 100}+ {(y-2\sqrt 5)^2\over 25}=1 \equiv \cases{a=10,b=5,c=5\sqrt 3\\ F_1(-5\sqrt 3,2\sqrt 5)\\ F_2(5\sqrt 3,2\sqrt 5)}與x軸的交點\\因此，將y=0代入\Gamma \Rightarrow {x^2\over 100}+ {(0-2\sqrt 5)^2\over 25}=1 \Rightarrow {x^2\over 100}= {1\over 5} \Rightarrow x=\bbox[red,2pt]{\pm 2\sqrt 5}$$

解答： $${4大人取2\times 6小孩取3\qquad\over 10人取5}  ={C^4_2C^6_3\over C^{10}_5} ={120\over 252}=\bbox[red,2pt]{10\over 21}$$

解答： $$假設f(x)為n次多項式，f(x^3)與x^6f(x)次數相同，即3n=6+n \Rightarrow n=3\\ \Rightarrow f(x)=ax^3+bx^2+cx+d\\ \Rightarrow \cases{8f(x^3)=8ax^9+8bx^6 +8cx^3+8d\\ x^6f(2x)=8ax^9+ 4bx^8 +2cx^7+dx^6\\ 2f(x^2)=2ax^6 +2bx^4+2cx^2 +2d}  \Rightarrow  8f(x^3)-x^6f(2x)-2f(x^2)+12 \\=-4bx^8-2cx^7+(8b-d-2a)x^6-2bx^4+8cx^3-2cx^2+6d+12=0 \\ \Rightarrow \cases{b=c=0\\ 8b-d-2a=0\\ 6d+12=0} \Rightarrow \cases{b=c=0\\ d=-2\\a=1} \Rightarrow f(x)=\bbox[red,2pt]{x^3-2}$$

解答： $$令f(x)=x^3-18x^2+107x-210，a,b,c為f(x)=0的三根\Rightarrow  a+b+c=18 \\ 令s=(a+b+c)\div 2=9 \Rightarrow \triangle ABC面積=\sqrt{s(s-a)(s-b)(s-c)} =\sqrt{sf(s)} \\ =\sqrt{9\cdot 24} =\bbox[red,2pt]{6\sqrt 6}$$

解答： $$E:x-2y+3z-16=0 \Rightarrow \vec n=(1,-2,3)\\令A(1,2,-3)在E的投影點為A'，則\overrightarrow{AA'}=t\vec n \Rightarrow A'(1+t,2-2t, -3+3t),t\in \mathbb{R}\\ A'在E上\Rightarrow (1+t)-2(2-2t)+3(-3+3t)-16=0 \Rightarrow t=2 \Rightarrow A'(3,-2,3)\\ B(5,-4,1)在E上，因此P=(A'+B)/2=\bbox[red,2pt]{(4,-3,2)}$$

解答： $$x^2-2xy+105y^2=2016 \Rightarrow (x-y)^2+104y^2=2016\\ y=2代入上式 \Rightarrow (x-2)^2=416=2016 \Rightarrow (x-2)^2=1600 \Rightarrow x=42\\ \Rightarrow (x,y)= \bbox[red, 2pt]{(42,2)}$$

解答：

$$\overline{EF}為對摺線\Rightarrow \cases{\overline{EA} =\overline{ED}=a\\ \overline{FA} =\overline{FD}=b \\ \angle EDA= \angle A=60^\circ} \Rightarrow \cases{\overline{EB}=3-a\\ \overline{FC}=3-b} \\ 餘弦定理\Rightarrow \cases{\cos \angle B= {1\over 2}={(3-a)^2+1-a^2 \over 2(3-a)} \\ \cos \angle C={1\over 2}= {(3-b)^2+4-b^2 \over 4(3-b)}} \Rightarrow \cases{a=7/5\\ b=7/4}\\ 同理,\cos \angle EDF ={1\over 2} ={a^2+b^2-\overline{EF}^2 \over 2ab} \Rightarrow {49\over 25}+{49 \over 16}-\overline{EF}^2={49\over 20} \Rightarrow \overline{EF} = \bbox[red,2pt]{7\sqrt{21}\over 20}$$

解答：

$${(x-2)^2\over 25} +{y^2\over b^2}=1 \Rightarrow a=5 \Rightarrow \overline{BF_1}=\overline{BF_2}=5\\ 令\overline{AF_2}=m，由於\overline{AF_1}+\overline{AF_2}=2a=10，可得\overline{AF_1}=10-m\\ \angle F_1=90^\circ \Rightarrow \overline{AB}^2 = \overline{AF_1}^2 +\overline{BF_1}^2 \Rightarrow (m+5)^2=(10-m)^2+25 \Rightarrow m=10/3\\ 餘弦定理\Rightarrow \cos \angle B ={(m+5)^2+5^2-(10-m)^2 \over 10(m+5)} ={5^2+5^2-\overline{F_1F_2}^2 \over 50} \Rightarrow {3\over 5}={50-\overline{F_1F_2}^2 \over 50} \\ \Rightarrow \overline{F_1F_2}=\sqrt{20} = \bbox[red,2pt]{2\sqrt 5}$$

解答： $$X\sim B(n,p) \Rightarrow \cases{EX=np\\ Var(X)=  E(X^2)-(EX)^2 =np(1-p)} \Rightarrow EX^2=np(1-p)+(np)^2 \\ 現在X\sim B(8,1/5) \Rightarrow \sum_{k=1}^8 k^2C^8_k({1\over 5})^k({4\over 5})^{8-k}=EX^2 =8\cdot {1\over 5}\cdot {4\over 5}+ (8\cdot {1\over 5})^2 =\bbox[red,2pt]{96\over 25}$$

解答： $$L:\cases{x+y+z=3 \\ x+y+12z=3} \Rightarrow \cases{x+y=3\\ z=0} \Rightarrow (t,3-t,0)\in L,t\in \mathbb{R} \Rightarrow 方向向量\vec u=(1,-1,0)\\ E:3x-2y+z=6 \Rightarrow 法向量\vec n=(3,-2,1)\\ L':x={y-p\over m} ={z-q\over n} \Rightarrow 方向向量\vec v=(1,m,n)\\ 由於L'在E上\Rightarrow \vec n\bot \vec v \Rightarrow (3,-2,1)\cdot (1,m,n)=0 \Rightarrow 3-2m+n=0\cdots(1) \\ 令\vec w=\vec u \times \vec v =(-n,-n,m+1) \Rightarrow \vec w\bot \vec n \Rightarrow (-n,-n,m+1)\cdot (3,-2,1)=0\\ \Rightarrow m-n+1=0 \Rightarrow n=m+1代入 (2) \Rightarrow   (m,n)= \bbox[red, 2pt]{(4,5)}$$

解答：

$$直角\triangle ABD\Rightarrow  \overline{BC} =\sqrt{\overline{BC}^2-\overline{AB}^2} =\sqrt{17^2-15^2}=8\\ 令\cases{\overline{BD}=a \\ \overline{CD}=b} \Rightarrow \triangle BCD面積={1\over 2}(a+b+17)\cdot {10\over 3}={1\over 2}\cdot 15 \cdot b \Rightarrow b=(2a+34)/7 \\ 又直角\triangle BAD\Rightarrow a^2=15^2+(8+b)^2 =289+16b+b^2 = 289+16\cdot {2a+34\over 7} +({2a+34\over 7})^2\\ ={19125\over 49} +{360\over 49}a+{4\over 49}a^2 \Rightarrow a^2-8a-425=0 \Rightarrow (a-25)(a+17)=0 \Rightarrow a=\bbox[red,2pt]{25}$$

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