臺中市立臺中第二高級中等學校114學年度第1次教師甄選
一、 填充題：每格 5 分， 共 10 格， 合計 50 分
解答:$$f(x)= \sqrt{x^4-x^2-6x+10} -\sqrt{x^4-3x^2+4} =\sqrt{(x^2-1)^2+(x-3)^2} -\sqrt{(x^2-2)^2 +(x-0)^2} \\=\overline{PA} -\overline{PB} \le \overline{AB} = \bbox[red, 2pt]{\sqrt{10}},其中\cases{P(x^2,x) \in \Gamma:x=y^2 \\ A(1,3) \\B(2,0)}$$
解答:$$\cases{\sigma_X=10\\ \sigma_Y=3\\ \sigma_{X+Y} =12} \Rightarrow \cases{Var(X) =100\\ Var(Y)=9\\ Var(X+Y) =144} \Rightarrow Var(X+Y) =Var(X)+Var(Y)+ 2Cov(X,Y) \\ \Rightarrow 144=100+9+2Cov(X,Y) \Rightarrow Cov(X,Y) ={35\over 2} \Rightarrow 相關係數\rho={Cov(X,Y) \over \sigma_X \sigma_Y} ={35/2\over 10\cdot3 } =\bbox[red, 2pt]{7\over 12}$$
解答:$$\cases{抽到兩張5 \Rightarrow 點數和=10 \Rightarrow 機率=1/10 \Rightarrow 期望值=1\\抽到兩張4 \Rightarrow 點數和=8 \Rightarrow 機率=3/10 \Rightarrow 期望值=12/5\\ 抽到一張5一張4 \Rightarrow 點數和=9 \Rightarrow 機率=6/10 \Rightarrow 期望值=27/5}\\ \Rightarrow 期望值={1+(12/5)+(27/5)\over 6/10} = \bbox[red, 2pt]{44\over 3}$$
解答:
$$假設\cases{B在\overline{AC}的投影點B' \\D在\overline{AC}的投影點D' \\ P=\overline{AC}\cap \overline{BD}} \Rightarrow \triangle PB'B \sim \triangle PD'D (AAA) \Rightarrow {\overline{PB} \over \overline{PD}} ={\overline{BB'}\over \overline{DD'}} ={16\over 7} ={\triangle OAB\over \triangle OAD} \\ \Rightarrow \overrightarrow{AP} ={7\over 23}\overrightarrow{AB} +{16\over 23} \overrightarrow{AD} \Rightarrow \overrightarrow{AC} =t\overrightarrow{AP}\\ 由於 \overrightarrow{AC} \cdot \overrightarrow{AB} =\cases{ (\overrightarrow{AB} +\overrightarrow{BC}) \cdot \overrightarrow{AB} =(4\sqrt 5)^2+0=80 \\ t\overrightarrow{AP} \cdot \overrightarrow{AB} =t({7\over 23} \overrightarrow{AB} +{16\over 23} \overrightarrow{AD}) \cdot \overrightarrow{AB} =t({7\over 23}\cdot 80+{16\over 23}\cdot 11)} \\\Rightarrow t\cdot {736\over 23}=80 \Rightarrow t={80\over 32} ={5\over 2} \Rightarrow \overrightarrow{AC} ={5\over 2} \left({7\over 23}\overrightarrow{AB} +{16\over 23}\overrightarrow{AD} \right) ={35\over 46}\overrightarrow{AB} +{40\over 23}\overrightarrow{AD} \\ \Rightarrow (r,s) =\bbox[red, 2pt]{\left({35\over 46},{40\over 23} \right)}$$
解答:
$$假設正方體頂點坐標\cases{O(0,0,0) \\A(3,0,0)\\ B(0,3,0)\\ C(3,3,0) \\ D(3,0,3) \\E(0,3,3) \\ F(0,0,3)} \Rightarrow 平面E=\triangle ABF:x+y+z=3 \\ \Rightarrow \cases{D在平面E的投影D'=(2,-1,2) \\ F在平面E的投影F'=F} \Rightarrow 正六邊形邊長\overline{D'F'}=\sqrt 6 \\ \Rightarrow 正六邊形面積=6個正三角形(邊長\sqrt 6)面積=6\times {\sqrt 3\over 4} (\sqrt 6)^2= \bbox[red, 2pt]{9\sqrt 3}$$
解答:
$$\cases{f(x)+f(y) = x^2+6x+1+ y^2+6y+1 \le 0\\ f(x)-f(y)=x^2+6x+1-y^2-6y-1\le 0} \Rightarrow \cases{(x+3)^2+ (y+3)^2 \le 4^2\\ (x^2-y^2)+6(x-y)\le 0 \Rightarrow (x-y)(x+y+6)\le 0}\\ 兩直線\cases{x=y\\ x+y+6=0}互垂\Rightarrow 所圍面積=半圓= \bbox[red, 2pt]{8\pi}$$
解答:$$2a\cos(B+C)+c\cos B+b\cos C=2a\cos(B+C)+a= -2a\cos A+a=0 \\ \Rightarrow \cos A={1\over 2} \Rightarrow A=60^\circ \Rightarrow {a\over \sin A} ={a\over \sqrt 3/2} =2\cdot 2 \Rightarrow a=2\sqrt 3\\ 又b\ge a \Rightarrow 4\sin B\ge 2\sqrt 3 \Rightarrow \sin B\ge {\sqrt 3\over 2} \\\Rightarrow 2b-c=2\cdot 4\sin B-4\sin C =8\sin B-4\sin(120^\circ-B) =8\sin B-4({\sqrt 3\over 2}\cos B+ {1\over 2}\sin B) \\=6\sin B-2\sqrt 3\cos B =4\sqrt 3\sin (B-30^\circ) \Rightarrow 2b-c \in \bbox[red, 2pt]{[2\sqrt 3, 4\sqrt 3)} \; (\because 60^\circ\le B\lt 120^\circ)$$
解答:
$$\cases{x=x'\\ y=\sqrt 2y'} \;代入橢圓\Gamma \Rightarrow {x'^2 \over 16}+{y'^2\over 16}=1為一圓\Gamma', 其中\cases{圓心(0,0)\\ 圓半徑=4} \\又\cases{x=x'\\ y=\sqrt 2y'} \Rightarrow \cases{x'=x\\ y'=y/\sqrt 2} \Rightarrow A(\sqrt 6+\sqrt 2,2\sqrt 3-2) \in \Gamma\\ \Rightarrow A'(\sqrt 6+\sqrt 2,(2\sqrt 3-2)/\sqrt 2) =(\sqrt 6+\sqrt 2,\sqrt 6-\sqrt 2) \Rightarrow \overline{OA'} =4 \\ \Rightarrow \overline{A'B'}=4\cdot {3\over 2}\cdot {2\over \sqrt 3} =4\sqrt 3 \Rightarrow 正\triangle A'B'C'={\sqrt 3\over 4}(4\sqrt 3)^2 =12\sqrt 3\\ J=\left|{\partial (x,y)\over \partial(x',y')} \right|= \begin{vmatrix} 1& 0\\0& \sqrt 2\end{vmatrix} =\sqrt 2 \Rightarrow \triangle ABC=J\cdot\triangle A'B'C'= \bbox[red, 2pt]{12\sqrt 6}$$
解答:
$$y=f(x)=x^3+ax \Rightarrow f''(x)=6x=0 \Rightarrow 對稱中心為原點(0,0)\\ \Rightarrow 內接正方形ABCD對角線\cases{L_1= \overleftrightarrow{AC}: y=mx\\ L_2= \overleftrightarrow{BD}: y=-x/m} \Rightarrow \cases{A=L_1\cap y=f(x) \\ B=L_2\cap y=f(x)} \Rightarrow \cases{x^3+ax=mx\\ x^3+ax=-x/m} \\ \Rightarrow \cases{A(\sqrt{m-a}, m\sqrt{m-a}) \\B \left(\sqrt{-{1\over m}-a},-{1\over m }\sqrt{-{1\over m}-a} \right)} \Rightarrow B=A逆時針旋轉90^\circ \Rightarrow -m\sqrt{m-a}= \sqrt{-{1\over m}-a} \\ \Rightarrow a={m^3\over m^2-1}+{1\over m(m^2-1)} ={m^2\over m-1/m}+{1/m^2\over m-1/m} ={(m-1/m)^2+2 \over (m-1/m)} =m-{1\over m}+{2\over m-1/m} \\ \Rightarrow |a|\ge 2\sqrt 2 \Rightarrow a= \bbox[red, 2pt]{-2\sqrt 2}$$
解答:$$\textbf{(1) }S為\triangle ABC面積\Rightarrow \cases{r=\displaystyle {2S\over a+b+c} \\ R=\displaystyle {abc\over 4S}} \Rightarrow S={1\over 2}(a+b+c)r={abc\over 4R} \\ \Rightarrow r= {abc \over 2R(a+b+c)} = {2R\sin A\cdot 2R\sin B\cdot 2R\sin C\over 2R(2R\sin A+2R\sin B+ 2R\sin C)}= 2R\cdot {\sin A\sin B\sin C\over \sin A+\sin B+ \sin C} \\=2R\cdot {2\sin{A\over 2} \cos {A\over 2} \cdot 2\sin{B\over 2} \cos{B\over 2} \cdot 2\sin{C\over 2}\cos {C\over 2} \over 2\sin{A+B\over 2} \cos{A-B\over 2} +2\sin{C\over 2}\cos {C\over 2}} \\=8R\cdot {\sin{A\over 2} \sin{B\over 2} \sin {C\over 2} \cos{A\over 2}\cos {B\over 2} \cos {C\over 2} \over \cos{C\over 2}\cos{A-B\over 2}+\cos{A+B\over 2} \cos{C\over 2}} =8R\cdot {\sin{A\over 2} \sin{B\over 2} \sin {C\over 2} \cos{A\over 2}\cos {B\over 2} \over \cos{A-B\over 2}+\cos{A+B\over 2} } \\= 8R\cdot {\sin{A\over 2} \sin{B\over 2} \sin {C\over 2} \cdot {1\over 2}(\cos{A-B\over 2}+\cos{A+B\over 2}) \over \cos{A-B\over 2}+\cos{A+B\over 2} } =4R\sin{A\over 2} \sin{B\over 2} \sin {C\over 2} \\ \Rightarrow r=4R\sin{A\over 2} \sin{B\over 2} \sin {C\over 2} \quad \bbox[red, 2pt]{QED} \\\textbf{(2) }(b-c)^2 \ge 0 \Rightarrow b^2+c^2\ge 2bc \Rightarrow \cos A={b^2+c^2-a^2 \over 2bc} \ge {b^2+c^2-a^2 \over b^2+c^2} =1-{a^2\over b^2+c^2} \\ \Rightarrow \cos A=1-2\sin^2{A\over 2} \ge 1-{a^2\over b^2+c^2} \Rightarrow {a^2\over b^2+ c^2}\ge 2\sin^2{A\over 2} \;,同理可得 \cases{{b^2\over a^2+c^2} \ge 2\sin^2{B\over 2} \\ {c^2\over a^2+b^2}\ge 2\sin^2{C\over 2}} \\ \Rightarrow {a^2\over b^2+c^2} \cdot {b^2\over c^2+a^2} \cdot {c^2\over a^2+b^2} \ge 8 \sin^2{A\over 2} \sin^2{B\over 2} \sin^2{C\over 2}\\ \Rightarrow {abc\over \sqrt{(a^2+b^2)(b^2+ c^2)( c^2+a^2)}} \ge 2\sqrt 2 \sin{A\over 2} \sin{B\over 2}\sin {C\over 2}\\將\textbf{(1)}的結果: \sin{A\over 2} \sin{B\over 2} \sin{C\over 2}={r\over 4R}代入上式，可得\\{abc\over \sqrt{(a^2+b^2)(b^2+ c^2)( c^2+a^2)}} \ge 2\sqrt 2\cdot {r\over 4R} \Rightarrow {abc\over \sqrt{2(a^2+b^2)(b^2+ c^2)( c^2+a^2)}} \ge {r\over 2R}\; \bbox[red, 2pt]{QED}$$
解答:$$\textbf{(1) } |S|=112 \Rightarrow |T| =C^{112}_2 +C^{112}_4+ \cdots+C^{112}_{112} = \bbox[red, 2pt]{2^{111}-1 } \\\cases{(1+1)^{112} =1+C^{112}_1+C^{112}_2+C^{112}_3+\cdots +C^{112}_{112} \\ (1-1)^{112} =1-C^{112}_1 +C^{112}_2-C^{112}_3+ \cdots+C^{112}_{112}} \\兩式相加 2^{112} =2(1+C^{112}_2+C^{112}_4+\cdots+C^{112}_{112}) \Rightarrow |T|=2^{111}-1 \\\textbf{(2) }f(x)=(x+{1\over 2})(x+{1\over 3}) \cdots(x+{1\over 113}) \Rightarrow \cases{x^{110} 係數為符合|A_i|=2的元素相乘積\\ x^{108} 係數為符合|A_i|=4的元素相乘積 \\ \cdots\\ x^2係數為符合|A_i|=110的元素相乘積 \\常數項即為|A_i|=112的元素相乘積} \\\Rightarrow M={1\over 2}(f(1)+f(-1))-1={1\over 2} \left({3\over 2}\cdot {4\over 3}\cdot {5\over 4} \cdots{114\over 113}+{1\over 2}\cdot {2\over 3} \cdots {112\over 113} \right) -1 \\\qquad ={1\over 2} \left(57+{1\over 113} \right)-1= \bbox[red, 2pt]{3108\over 113}$$
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解題僅供參考，其它教甄試題及詳解
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