113年高師大附中教甄-數學詳解

國立高雄師範大學附屬中學113學年度教師甄選

解答:

$$假設\cases{圓半徑=r\\  P=\overline{BC} \cap \overline{AO} \\ \overline{AO}=t} ,由於\angle BAC=120^\circ  \Rightarrow t\le {r\over 2}\\ 因此\overrightarrow{AO} ={r\over t} \overrightarrow{AP} ={r\over t} (m\overrightarrow{AB}+ n\overrightarrow{AC}),其中m+n=1 \\ \Rightarrow x+y={r\over t}(m+n)={r\over t} \ge {r\over r/2}=2 \Rightarrow x+y的最小值=\bbox[red, 2pt]2$$

解答: $$假設t=x+{1\over x} \Rightarrow x^2-tx+1=0有實根\Rightarrow t^2-4\ge 0 \Rightarrow t^2\ge 4\\ 又 t^2=x^2+{1\over x^2}+2 \Rightarrow f(x)=0 \Rightarrow t^2+at+b-2=0  \Rightarrow at+b = 2-t^2  \\ 柯西不等式:(a^2+ b^2)(t^2+1^2) \ge (at+b)^2 =(2-t^2)^2 \\ \Rightarrow a^2+b^2 \ge {(t^2+1-3)^2\over t^2+1} =(t^2+1)+{9\over t^2+1}-6 \ge (4+1)+{9\over 4+1}-6= \bbox[red, 2pt]{4\over 5}$$

解答:

$$f(x)有相異極值點 \Rightarrow f'(x)=e^x(x-ae^x)+ e^x(1-ae^x) =e^x(1+x-2ae^x) =0有相異二實根 \\ \Rightarrow 兩圖形\cases{\Gamma_1:y=1+x\\ \Gamma_2: y=2ae^x} 有相異二交點,而\cases{\Gamma_1與y軸交於A(0,1)\\ \Gamma_2與y軸交於B(0,2a)} \\ 因此0\lt \overline{AB}\lt 1 \Rightarrow 0\lt 1-2a\lt 1 \Rightarrow \bbox[red, 2pt]{0\lt a\lt {1\over 2}}$$

解答:

$$\cases{D(0,0,0)\xrightarrow{折起後} D'\\ A(0,2\sqrt 3,0)\\ B(6,2\sqrt 3,0)\\ C(6,0,0)} \Rightarrow \cases{\angle DAC=60^\circ \\ \angle DCA=30^\circ} \Rightarrow \overline{DE}=3,其中\overline{DE}\bot \overline{AC} \\ \overline{D'E}= \overline{DE} =3 \Rightarrow \cases{\overline{D'E}在x-y平面的投影長=\overline{EF}= 3\cos 60^\circ=3/2\\ D'的高度= \overline{D'F} =3\sin 60^\circ=3\sqrt 3/2} \\ \Rightarrow D'=(\overline{DF} \cos \angle EDC, \overline{DF}\sin \angle EDC,\overline{D'F})=((3+{3\over 2})\cos 60^\circ, (3+{3\over 2})\sin 60^\circ, {3\sqrt 3\over 2})\\=({9\over 4},{9\over 4}\sqrt 3,{3\over 2}\sqrt 3)  \Rightarrow \overline{BD'}=\sqrt{({15\over 4})^2+ ({\sqrt 3\over 4})^2+({3 \sqrt 3\over 2})^2} = \bbox[red, 2pt]{\sqrt{21}}$$

解答:

$$此題相當於求最大的r,滿足\cases{y=x^4\\ x^2+(y-r)^2=r^2}有三相異實根\\ 而x^2+(y-r)^2=r^2 \Rightarrow x^2+y^2-2ry= 0 \Rightarrow r={x^2+y^2 \over 2y} \\ 也就是在y=x^4條件下，欲求f(x,y)={x^2+y^2 \over 2y}的最大值 \\ \Rightarrow f(x,y)=g(x)={x^2+x^8\over 2x^4} ={1\over 2x^2}+{x^4\over 2} \Rightarrow g'(x)=-{1\over x^3}+2x^3=0 \Rightarrow x=({1\over 2})^{1/6} \\ \Rightarrow y=({1\over 2})^{2/3} \Rightarrow f(({1\over 2})^{1/6},({1\over 2})^{2/3})={({1\over 2})^{1/3}+({1\over 2})^{4/3} \over 2\cdot ({1\over 2})^{2/3}}=\bbox[red, 2pt]{ {3 \over 4} \sqrt[3]2}$$

解答: $$f(x)=\left|{x^2+24\over 8} \right| -\sqrt{{x^4\over 64}+{x^2\over 2}-{9\over 2}x+{145\over 16}} \\\qquad = \left|{x^2\over 8} -(-3)\right| -\sqrt{\left({x^2\over 8}-{1\over 4} \right)^2+\left( {3\over 4}x-3\right)^2} \\=|x'-(-3)|- \sqrt{(x'-{1\over 4})^2 +(y'-3)^2}\\ 其最大值就是({1\over 4},3)至直線x=-3的距離,即{1\over 4}+3= \bbox[red, 2pt]{13\over 4}\\ 當然也可以直接算:f'(x)=0\Rightarrow {9\over 4}x^3-{81\over 16}x^2-36x+81=0 \\\Rightarrow x=\pm 4,{9\over 4} \Rightarrow f(4)=5-\sqrt{49\over 16} ={13\over 4}$$

解答: $$1-9有5個奇數、4個偶數\Rightarrow 條件A的個數:C^5_3+C^4_3=14\\ 條件B的樣本\cases{公差1:123,234,...,789，共7個\\ 公差2:135,246,357,468,579，共5個\\ 公差3:147,258,369，共3個\\ 公差4:159，只有1個} \Rightarrow 共16個\\ A\cap B=135,246,357,468,579,159，共6個\\ 因此期望值={1\over C^9_3}(400\times 6+100\times (14-6)+ 100\times(16-6)) ={ 4200\over 84} =\bbox[red, 2pt]{50}$$

解答:

$$所圍面積I=4\int_0^{1/2} \sqrt{-4(x-1/2)}\,dx =8 \int_0^{1/2} \sqrt{{1\over 2}-x} \,dx \\取u={1\over 2}-x \Rightarrow du=-dx \Rightarrow I=8\int_0^{1/2} \sqrt u\,du =8\left. \left[ {2\over 3} u^{3/2} \right]\right|_0^{1/2} =\bbox[red, 2pt]{4\sqrt 2\over 3}$$

解答:

$$\lim_{n\to \infty}C_n =\overline{AB} ={5\over 6}\pi-{\pi\over 6}= \bbox[red, 2pt]{{2\over 3}\pi}$$

解答: $$內接正方形四頂點顯然為\cases{A(-1,1)\\ B(1,1)\\ C(1,-1)\\ D(-1,-1)} \Rightarrow 欲求之橢圓:\cases{A,C為焦點,B,D為頂點\\ B,D為焦點,A,C為頂點}\\ \textbf{Case I } A,C為焦點,B,D為頂點:2a=\overline{BA}+\overline{BC} =4 \Rightarrow a=2 \\\qquad \Rightarrow 橢圓方程式 \sqrt{(x+1)^2+(y-1)^2} +\sqrt{(x-1)^2+(y+1)^2} =4 \Rightarrow 3x^2+3y^2+2xy=8\\ \textbf{Case II } B,D為焦點,A,C為頂點: 2a=\overline{AB} +\overline{AD}=4 \Rightarrow a=2 \\\qquad \Rightarrow 橢圓方程式 \sqrt{(x-1)^2+(y-1)^2} +\sqrt{(x+1)^2+(y+1)^2} =4 \Rightarrow 3x^2+3y^2-2xy=8 \\ \Rightarrow 橢圓方程式\;\bbox[red, 2pt]{3x^2\pm 2xy+3y^2=8}$$

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