107年政大附中教甄-數學詳解

國立政治大學附屬高級中學 107 年度第 1 次代理教師甄選

解答： $$原等比數列:a,ar,ar^2 \Rightarrow \cases{a,ar+4,ar^2 成等差\\ a,ar+4,ar^2+32成等比} \Rightarrow \cases{a+ar^2=2(ar+4)\\ a(ar^2+32) =(ar+4)^2} \\ \Rightarrow \cases{ar^2+a=2ar+8 \cdots(1)\\ r=4-2/a \cdots(2)}，將(2)代入(1) \Rightarrow a(4-{2\over a})^2+a =2a(4-{2\over a})+8 \\ \Rightarrow 9a^2-20a+4=0 \Rightarrow (9a-2)(a-2)=0 \Rightarrow \cases{a=2 \Rightarrow r=3\\ a=2/9 \Rightarrow r=-5} \\ \Rightarrow 原數列:\bbox[red,2pt]{2,6,18或{2\over 9},-{10\over 9},{50\over 9}}$$

解答： $$\cases{a=\sqrt[3]{\sqrt 3+2}\\ b=\sqrt[3]{\sqrt 3-2}} \Rightarrow \cases{ab=-1 \\ a^3-b^3=4}\\ 又(a-b)(a^2+ab+b^2)=a^3-b^3  \Rightarrow (a-b)((a-b)^2+3ab)=4 \Rightarrow \alpha(\alpha^2-3)=4\\ \Rightarrow \alpha^3-3\alpha-4=0，即\bbox[red, 2pt]{x^3-3x-4=0}$$

解答： $$本題\bbox[red,2pt]{送分}$$

解答：

$$令\cases{a=\overline{AF} =\overline{AE} \\ 1=\overline{BF} =\overline{BD} =\overline{CD} =\overline{CE} } \Rightarrow \cases{\triangle AEF={a^2\over (a+1)^2}\triangle ABC\\ \triangle BDF =\triangle CDE={1\over 2(a+1)}\triangle ABC}\\ 由於\triangle AEF=\triangle BDF+\triangle CDE \Rightarrow {a^2\over (a+1)^2} ={1\over a+1} \Rightarrow a^2=a+1 \Rightarrow a=\bbox[red, 2pt]{1+\sqrt 5\over 2}$$

解答：

(1) $$f(x)= (23x^3+75x^2 +61x+8)^4 =((23 x + 6)(x^2+3x+2)-3x-4)^4\\ \Rightarrow (-3x-4)^4= (9x^2+24x+ 16)^2 =(9(x^2+3x+2)-3x-2)^2 \\ \Rightarrow (-3x-2)^2=9x^2+12x+4 =9(x^2+3x+2)-15x-14 \\ \Rightarrow 餘式為\bbox[red,2pt]{-15x-14}$$ (2) $$令N=101\times 102 \Rightarrow (23756108)^4= (2305\times N+9998)^4 =(2306\times N-304)^4 \\ =(-304)^4 \mod N = (92416)^2 \mod N= (8N+10000)^2 \mod N = (9N-302)^2 \mod N\\ =(-302)^2 \mod N =91204 \mod N = \bbox[red,2pt]{8788}\mod N$$

解答： $$x^4+y^4=(x^2+y^2)^2-2x^2y^2 =((x+y)^2-2xy)^2-2x^2y^2 = (x+y)^4-4xy(x+y)^2+2x^2y^2\\ \Rightarrow 97=625-100xy +2x^2y^2 \Rightarrow x^2y^2-50xy+264=0 \Rightarrow (xy-6)(xy-44)=0\\ \Rightarrow \cases{xy=6 \Rightarrow x+{6\over x}=5 \Rightarrow x^2-5x+6=0 \Rightarrow (x,y)=(2,3),(3,2)\\ xy=44 \Rightarrow x+{44\over x}=5 \Rightarrow x^2-5x+44=0 \Rightarrow (x,y)=({5\pm \sqrt{151}i\over 2}, {5\mp \sqrt{151}i\over 2})}\\ \Rightarrow (x,y)=\bbox[red, 2pt]{(2,3),(3,2),({5+ \sqrt{151}i\over 2}, {5- \sqrt{151}i\over 2}),({5- \sqrt{151}i\over 2}, {5+ \sqrt{151}i\over 2}) }$$

解答： $$假設拋物線\Gamma: x^2=4cy \Rightarrow y'={x\over 2c}及兩切點\cases{A(a,{a^2\over 4c})\\ B(b,{b^2\over 4c})} 及該切點斜率\cases{m_a={a\over 2c}\\ m_b={b\over 2c}}\\ 由於兩切線互垂，因此m_am_b=-1 \Rightarrow {ab\over 4c^2}=-1 \Rightarrow ab=-4c^2 \cdots(1)\\ 兩切線方程式:\cases{L_a:y={a\over 2c}(x-a)+{a^2\over 4c} \\ L_b:y={b\over 2c}(x-b)+{b^2\over 4c}} ，求交點 \Rightarrow  {a\over 2c}(x-a)+{a^2\over 4c}={b\over 2c}(x-b)+{b^2\over 4c}\\ \Rightarrow {a-b\over 2c}x= {a^2-b^2\over 2c}-{a^2-b^2\over 4c} \Rightarrow x=a+b-{a+b\over 2} ={a+b\over 2}\\ \Rightarrow y={a\over 2c}({a+b\over 2}-a) +{a^2\over 4c} ={ab\over 4c} \cdots(2)\\ 將(1)代入(2) \Rightarrow y=-{4c^2\over 4c}=-c \Rightarrow 兩垂直切線的交點為P({a+b\over 2},-c)落在準線y=-c上\\，\bbox[red,2pt]{故得證}$$

解答： $$\bbox[blue,2pt]{略}$$

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