114年高科實中教甄-數學詳解

 國立高科實驗高級中等學校籌備處 114學年度第1次教師甄選

第壹部份:單選題 (每題5分,共4題,共20分)

解答:$$S_n =3n^2-2n \Rightarrow S_{25}=1825 \Rightarrow {1825-a_k\over 24}=74 \Rightarrow a_k=1825-74\times 24=49 \\ \Rightarrow a_k= S_k-S_{k-1} =3k^2-2k-(3(k-1)^2-2(k-1)) =6k-5=49 \Rightarrow k=9,故選\bbox[red, 2pt]{(B)}$$

解答:

$$對同弧的圓周角相等\Rightarrow \cases{\angle ADC= \angle B= 60^\circ \\ \angle ADB =\angle C=60^\circ} \Rightarrow \overline{AD}為\angle BDC的角平分線\Rightarrow  {\overline{DB}\over \overline{DC}} ={\overline{BE} \over \overline{EC}} ={{3\over 5}} \\ \Rightarrow \cases{\overline{BE}=3k\\ \overline{EC}=5k} \Rightarrow \overline{AB}=\overline{AC} =5k+3k=8k\\又\triangle ABE \sim \triangle CDE(AAA) \Rightarrow {\overline{AB} \over \overline{CD}} ={\overline{BE}\over \overline{DE}} \Rightarrow {8k\over 5}={3k\over \overline{DE}} \Rightarrow \overline{DE} ={15\over 8}  \\\text{托勒密定理(Ptolemy's theorem)}:\overline{AD} \times \overline{BC}= \overline{AB}\times \overline{CD}+ \overline{BD}\times \overline{AC} \Rightarrow \overline{AD}\times 8k=40k+24k \\ \Rightarrow \overline{AD}=8\quad \Rightarrow \overline{AE}=8-{15\over 8} ={49\over 8}, 故選\bbox[red, 2pt]{(B)}$$

解答:

$$假設ABCD為正方形,且邊長為4 \Rightarrow \cases{A(0,4) \\B(0,0)\\ C(4,0) \\D(4,4)} \Rightarrow \cases{E=(B+C)/2=(2,0) \\F=(3D+C)/4=(4,3)} \\ \Rightarrow \cases{L_1= \overleftrightarrow{AE}: 2x+y=4\\ L_2=\overleftrightarrow{BF}: 3x-4y=0} \Rightarrow G=L_1\cap L_2 =({16\over 11}, {12\over 11}) \\ \Rightarrow GECF面積=\triangle GEC+\triangle GCF ={1\over 2}\cdot 2\cdot {12\over 11}+{1\over 2}\cdot 3\cdot (4-{16\over 11}) ={54\over 11} \\ \Rightarrow GECF面積:ABCD面積= {54\over 11}: 16=27:88 \Rightarrow n+m=115, 故選\bbox[red, 2pt]{(D)}$$

解答:$$E: kx+(k+1)y+(k+2)z=1 \Rightarrow \cases{A_k({1\over k},0,0) \\B_k(0,{1\over k+1},0) \\C_k(0,0,{1\over k+2})} \Rightarrow V_k={1\over 6}\cdot {1\over k} \cdot{1\over k+1}\cdot {1\over k+3} \\ \Rightarrow S=\sum_{k=1}^{18}V_k = {1\over 6}\sum_{k=1}^{18} {1\over k(k+1)(k+2)} ={1\over 12}\sum_{k=1}^{18}   \left[ \left( {1\over k}-{1\over k+1} \right) -\left( {1\over k+1}-{1\over k+2} \right) \right] \\={1\over 12}\sum_{k=1}^{18} (b_k-b_{k+1})={1\over 12} \left( b_1-b_{19} \right) ,b_k={1\over k+1}-{1\over k+2} \Rightarrow \cases{b_1=1-1/2=1/2\\ b_{19}=1/19-1/20} \\ \Rightarrow S={1\over 12} \left( {1\over 2}-{1\over 19}+{1\over 20} \right) ={63\over 1520} \Rightarrow m+n=1583, 故選\bbox[red, 2pt]{(A)}$$

第貳部份:填充題(每題6分,共10題,共60分)

解答:$$2025-19ab=5(1+9+a+b) \Rightarrow 2025-(1900-10a-b) =5(10+a+b) \\ \Rightarrow 15a+6b=75 \Rightarrow \cases{a=1 \Rightarrow b=10 (\times), b必須小於 10 \\ a=2 \Rightarrow b不是整數(\times) \\a=3\Rightarrow b=5\\ a=4 \Rightarrow b不是整數(\times) \\ a=5 \Rightarrow b=0} \Rightarrow 小高出生年份為\bbox[red, 2pt]{1935或1950}$$

解答:$$取\cases{u=14x+13y-14\\ v=13x+14y+13} \Rightarrow \cases{|u|+|v|\le 27 \Rightarrow 面積=2\cdot 27^2\\ \displaystyle \left|{\partial (u,v)\over \partial(x,y)} \right| = \begin{Vmatrix}14& 13\\ 13& 14 \end{Vmatrix} =27} \\ \Rightarrow (x,y)所圍面積={2\cdot 27^2\over 27} = \bbox[red, 2pt]{54}$$

解答:

$$假設\cases{\triangle AEF=a\\ \triangle ADF=b} \Rightarrow \cases{\displaystyle  {\overline{AE} \over \overline{EB}} ={\triangle AFE\over \triangle BFE} ={\triangle CAE \over \triangle CEB} \\ \displaystyle {\overline{AD} \over \overline{DC}} ={  \triangle DFA\over \triangle DFC} = {\triangle BDA\over \triangle BDC}} \Rightarrow \cases{\displaystyle {a\over 10}={a+b+15\over 10+20}={a+b+15\over 30} \\ \displaystyle {b\over 15} ={a+b+10\over 15+20} ={a+b+10\over 35}} \\ \Rightarrow \cases{2a-b-15=0\\3a-4b+30=0} \Rightarrow \cases{a=18\\ b=21} \Rightarrow AEFD面積=a+b= \bbox[red, 2pt]{39}$$

解答:

$$正\triangle ABC 邊長\overline{BC}= \overline{BP}+ \overline{CP}=12+8=20 \Rightarrow \cases{A(10,10\sqrt 3)\\ B(0,0)\\ C(20,0)\\P(12,0) \\Q(4, 4\sqrt 3) \\\ \overline{CR}=a} \Rightarrow R(20-{a\over 2},{\sqrt 3a\over 2}) \\ \Rightarrow \cases{\overrightarrow{PQ}=(-8,4\sqrt 3) \\ \overrightarrow{PR}=(8-{a\over 2},{\sqrt 3a\over 2})} \Rightarrow \overrightarrow{PQ} \cdot \overrightarrow{PR} =-64+4a+6a=0 \Rightarrow a= \bbox[red, 2pt]{32\over 5}$$

解答:$$只有1組兩數之和為10: \cases{(0,10,-,-), 其中(-,-)不得為(1,9),(2,8), (3,7),(4,6) \Rightarrow 共有C^9_2-4 \\(1,9,-,-), 其中(-,-)不得為(0,10),(2,8), (3,7),(4,6) \Rightarrow 共有C^9_2-4 \\ \cdots\\(4,6,-,-), 其中(-,-)不得為(0,10),(1,9),(2,8), (3,7)  \Rightarrow 共有C^9_2-4} \\\qquad \Rightarrow 合計5(C^9_2-4) =5C^9_2-20 \\ 有2組數字和為10: 從(0,10), (1,9),(2,8), (3,7),(4,6)挑2組, 有C^5_2=10種 \\ 因此任兩數字和不為10的機率=1-{ 5C^9_2-20+10\over C^{11}_4} =1-{170\over 330} = \bbox[red, 2pt]{16\over 33}$$

解答:$$ x= \sqrt[3]{45-3\sqrt{222}} +\sqrt[3]{45+3\sqrt{222}}  \Rightarrow x^3=90+ 3\sqrt[3]{27} \left(  \sqrt[3]{45- 3\sqrt{222}} +\sqrt[3]{45+3 \sqrt{222}} \right) =90 +9x \\ \Rightarrow x^3=90 +9x =9(x+ 10)\Rightarrow x^6= 81(x+10)^2 \\ \Rightarrow  2x^6-36x^4 +162x^2+114=162(x+10)^2-36x^4+162x^2+114 \\=162x^2+ 3240x+16200 -36x^4+162x^2+114=-36x^4+324x^2 +3240x+ 16314 \\=-36(90x+9x^2) +324x^2 +3240x+16314= \bbox[red, 2pt]{16314}$$

解答:$$\cases{x^2+xy+y^2=19 \cdots(1)\\ y^2+yz+ z^2=28 \cdots(2)\\ z^2+zx+x^2=37 \cdots(3)} \Rightarrow (2)-(1)=x^2-z^2+xy-yz=-9 \Rightarrow (x+y+z)(z-x)=9 \\ \Rightarrow z-x={9\over x+y+z}; 同理, (3)-(2) \Rightarrow (x+y+z)(x-y)=9 \Rightarrow x-y={9\over x+y+z} \\ \Rightarrow z-x=x-y \Rightarrow y+z=2x \Rightarrow S=x+y+z=3x \Rightarrow x={S\over 3} \Rightarrow {S\over 3}-y={9\over S} \\ \Rightarrow y={S\over 3}-{9\over S}, 同理z-{S\over 3}={9\over S} \Rightarrow z={S\over 3}+{9\over S} \Rightarrow \cases{x=S/3\\ y=S/3-9/S\\ z=S/3+9/S}\\ \Rightarrow x^2+xy+y^2= {S^2\over 9}+{S\over 3} \left( {S\over 3}-{9\over S} \right)+ \left(  {S\over 3}-{9\over S} \right)^2=19 \Rightarrow {S^2\over 3}+{81\over S^2}=28 \\ \Rightarrow S^4-84S^2+243=0 \Rightarrow (S^2-81)(S^2-3)=0 \Rightarrow S= \bbox[red, 2pt]9 \\ 若S=\sqrt 3 \Rightarrow y={\sqrt 3\over 3}-{9\over \sqrt 3} \lt 0, 違反y\gt 0$$

解答:$$\cases{f(a,b,c)=abc-3ab\\ g(a,b,c)=a+b+c-114} \Rightarrow \cases{f_a= \lambda g_a\\ f_b=\lambda g_b\\ f_c= \lambda g_c\\ g=0} \Rightarrow \cases{b(c-3)=\lambda\\ a(c-3)=\lambda\\ ab= \lambda\\ a+b+c=114} \Rightarrow a=b=c-3 \\ \Rightarrow c-3+c-3+c=114 \Rightarrow c=40\Rightarrow a=b=37 \Rightarrow f(37,37,40) =37^2(40-3)= \bbox[red, 2pt]{37^3}$$

解答:

$$\angle A, \angle B, \angle 成等差 \Rightarrow 2\angle B=\angle A+\angle C \Rightarrow \angle A+\angle B +\angle C=3\angle B=180^\circ \Rightarrow \angle B=60^\circ\\\Rightarrow 假設\cases{\angle A=60^\circ-\theta\\ \angle C=60^\circ+\theta}\;及\cases{\overline{BC}=a\\ \overline{AC}=b \\ \overline{AB}=c} \Rightarrow 正弦定理:{b\over \sin 60^\circ} =2\cdot {14\sqrt 3\over 3} \Rightarrow b= 14 \\ \Rightarrow 周長=a+ b+c= a+14+c=36 \Rightarrow c=22-a \Rightarrow {a\over \sin(60^\circ+\theta)} ={22-a\over \sin(60^\circ-\theta)} ={28\sqrt 3\over 3} \\ \Rightarrow a={28\sqrt 3\over 3} \sin(60^\circ+\theta) =22-{28\sqrt 3\over 3}\sin(60^\circ- \theta) \Rightarrow {28\sqrt 3\over 3} (\sin(60^\circ+\theta)+ \sin (60^\circ-\theta)) =22 \\ \Rightarrow {28\sqrt 3\over 3} \cdot 2 \sin60^\circ \cos\theta =22 \Rightarrow \cos \theta= {11\over 14} \Rightarrow \sin \theta={5\sqrt 3\over 14} \Rightarrow a={28\sqrt 3\over 3} \sin(60^\circ+\theta) =16 \\ \Rightarrow c=22-a=6 \Rightarrow \triangle ABC面積={1\over 2} ac\sin \angle B={1\over 2} \cdot 16\cdot 6\cdot {\sqrt 3\over 2} = \bbox[red, 2pt]{24\sqrt 3}$$

解答:$$x={\sqrt 3+i\over 2} =e^{\pi i/6} \Rightarrow x^6=-1 \Rightarrow x^2=e^{\pi i/3} ={1\over 2}+{\sqrt 3\over 2}i \Rightarrow 114x^2=57+57\sqrt 3 i \\ \Rightarrow f(e^{\pi i/6}) =(-1+1-1+\cdots -1)+57+57\sqrt 3i=-1+57+57\sqrt 3i= \bbox[red, 2pt]{56+57\sqrt 3i}$$

解答:$$\bbox[cyan,2pt]{略}$$

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