114年內壢高中教甄-數學詳解

 桃園市立內壢高級中等學校114學年度教師甄選

第一部份： 填充題（共 8 題，占 40 分）
說明： 作答時請將答案依照順序寫在答案本上。 本部分1到7題只考慮實數系。

解答:$$(x-\sqrt{x^2-2011})(y+\sqrt{y^2-2011})+2011=0\\ \Rightarrow 2011(y+ \sqrt{y^2-2011})+2011(x+\sqrt{x^2-2011})=0 \\ \Rightarrow 2011(x+y+\sqrt{x^2-2011} +\sqrt{y^2-2011})=0 \Rightarrow x+y= -(\sqrt{x^2-2011} +\sqrt{y^2-2011}) \cdots(1)\\ 同理, (x-\sqrt{x^2-2011})2011+ 2011(y-\sqrt{y^2-2011})=0 \\ \Rightarrow 2011(x+y-(\sqrt{x^2-2011}+ \sqrt{y^2-2011})) =0   \Rightarrow x+y=\sqrt{x^2-2011}+ \sqrt{y^2-2011} \cdots(2)\\ 由(1)及(2)可得\sqrt{x^2-2011}+ \sqrt{y^2-2011}=0 \Rightarrow x^2=2011 \Rightarrow x=\pm \sqrt{2011} \\ \Rightarrow 2x+y=x+y+x=0+x= \bbox[red, 2pt]{\pm \sqrt{2011}}$$

解答:$$(1-x)^{1001} =C^{1001}_0-C^{1001}_1x+C^{1001}_2x^2- \cdots -C^{1001}_{1001} x^{1001} \\ \Rightarrow \int_0^1 (1-x)^{1001}\,dx = \left. \left[C^{1001}_0x-{1\over 2}C^{1001}_1x^2 +{1\over 3}C^{1001}_2x^3- \cdots -{1\over 1002} C^{1001}_{1001} x^{1002}\right] \right|_0^1 \\ \Rightarrow \left.\left[ -{1\over 1002} (1-x)^{1002}\right] \right|_0^1 =C^{1001}_0-{1\over 2}C^{1001}_1  +{1\over 3}C^{1001}_2 - \cdots -{1\over 1002}C^{1001}_{1001} \\ \Rightarrow C^{1001}_0-{1\over 2}C^{1001}_1  +{1\over 3}C^{1001}_2 - \cdots -{1\over 1002}C^{1001}_{1001} = \bbox[red, 2pt]{1\over 1002}$$

解答:$$f(x)為一對一 \Leftrightarrow f(x)為遞增或遞減函數\\ f(x)=x^3+2x^2+kx-1 \Rightarrow f'(x)=3x^2+4x +k =3(x+{2\over 3})^2+k-{4\over 3}\ge 0 \\ \Rightarrow \bbox[red, 2pt]{k\ge {4\over 3}}$$

解答:$$\cases{u=\ln(x+1)\\ dv=dx/x^2} \Rightarrow \cases{du=dx/(x+1) \\ v=-1/x} \Rightarrow \int{\ln(x+1)\over x^2}\,dx =-{1\over x}\ln(x+1)+ \int {1\over x^2+x}\,dx \\=-{1\over x}\ln(x+1)+ \int ({1\over x}-{1\over x+1})\,dx =\bbox[red, 2pt]{-{1\over x}\ln(x+1)+ \ln |x|-\ln|x+1|+ C}$$

解答:$${x^3-x+360\over (x-1)(x+1)} ={x^3-x+360\over x^2-1} =x+{360\over x^2-1} \Rightarrow x^2-1=360 \Rightarrow x=\bbox[red, 2pt]{19}$$

解答:$$\cases{O(0,0,0) \\A(0,1,1)\\ P(x,y,0)} \Rightarrow \cases{\overrightarrow{AO} =(0,-1,-1) \\ \overrightarrow{AP} =(x,y-1,-1)} \Rightarrow \cos \angle OAP ={(0,-1,-1) \cdot (x,y-1,-1) \over |(0,-1,-1)||(x, y-1,-1)|} \\ \Rightarrow {\sqrt 3\over 2} ={2-y\over \sqrt 2\cdot \sqrt{x^2+(y-1)^2+1}} \Rightarrow {3\over 4}={(y-2)^2 \over 2\cdot (x^2+(y-1)^2+1)} \Rightarrow 3x^2+y^2+2y=2\\用\text{ Lagrange's }算子求極值, 令\cases{f(x,y)=x(y+1)\\g(x,y)= 3x^2+y^2+2y-2} \Rightarrow \cases{f_x =\lambda g_x\\ f_y= \lambda g_y\\ g=0} \\ \Rightarrow \cases{y+1 =\lambda(6x) \\x=\lambda(2y+2)} \Rightarrow {y+1\over x}={3x\over y+1} \Rightarrow (y+1)^2=3x^2 代入g(x,y)=0 \\ \Rightarrow 3x^2+(y+1)^2=3 \Rightarrow \cases{3x^2+ 3x^2=3 \\(y+1)^2+ (y+1)^2=3} \Rightarrow \cases{x=\pm 1/\sqrt 2\\ y+1 =\pm \sqrt{3/2}} \\ \Rightarrow \cases{M={1\over \sqrt 2}\cdot \sqrt{3\over 2} =\sqrt 3/2 \\ m=-{1\over \sqrt 2} \cdot \sqrt{3\over 2}=-\sqrt 3/2} \Rightarrow (M,m) =\bbox[red, 2pt]{\left({\sqrt 3\over 2},-{\sqrt 3\over 2} \right)}$$

解答:$$\begin{array} {} x& y& z & 機率p& 排列數k &xyz & k\cdot p\cdot xyz \\\hline 3& 1& 1& (3/6)^3(2/6)(1/6)=1/144& 5!/3!=20 & 3& 5/12 \\ 2& 2& 1 & (3/6)^2 (2/6)^2(1/6) = 1/216& 5!/2!2!=30 & 4 & 5/9 \\ 2& 1& 2& (3/6)^2(2/6)(1/6)^2 = 1/432& 30&4 & 5/18\\ 1& 2& 2 & (3/6)(2/6)^2(1/6)^2= 1/648& 30& 4& 5/27 \\1& 1& 3 & (3/6)(2/6)(1/6)^3= 1/1296 & 20&3& 5/108 \\1& 3& 1& (3/6)(2/6)^3 (1/6)=1/324 & 20& 3& 5/27\\\hline\end{array} \\ \Rightarrow 期望值={5 \over 12}+{5\over 9}+{5\over 18}+ {5\over 27} +{5\over 108} +{5\over 27}= \bbox[red, 2pt]{5\over 3}$$

解答:$$g(x)=ax^3+ bx^2+cx+d \Rightarrow f(x)=(x^2-2x+2)(x^2+1)p(x)+ ax^3+bx^2+cx+d \\ \Rightarrow \cases{f(1+i) =a(1+i)^3+b(1+i)^2+c(1+i)+d=5\\ f(i) =ai^3+ bi^2+ ci+d= 10} \Rightarrow \cases{(-2a+c+d)+ (2a+2b+c)i=5\\ (d-b)+(c-a)i=10} \\ \Rightarrow \cases{-2a+c+d=5\\ 2a+2b+c=0\\ d-b=10\\ c-a=0} \Rightarrow \cases{-a+d=5\\ 3a+2b=0\\ d-b=10} \Rightarrow \cases{a=2\\ b=-3\\ c=2\\d=7} \Rightarrow g(x)=2x^3-3x^2+2x+7 \\\Rightarrow g'(x)=6x^2-6x+2 \Rightarrow g'(-1/2)={13\over 2} \Rightarrow \lim_{x\to -1/2} {g(x)-5\over 2x+1} =\lim_{x\to -1/2} {g'(x)\over 2} =\bbox[red, 2pt]{13\over 4}$$

第二部份： 計算證明題（ 共 6 題，占 60 分）
說明： 每大題10分(需有計算過程只寫答案不給分)
作答時請將答案依照順序寫在答案本上。 本部分1到5題只考慮實數系。

解答:$$\textbf{(a) }  f(x)=(a_1+b_1x)^2 +(a_2+b_2x)^2 +(a_3+b_3x)^2  \\=(b_1^2+b_2^2 +b_3^2)x^2 +2(a_1b_1+ a_2b_2+ a_3b_3)x+ a_1^2+a_2^2+a_3^2 \\ 由於f(x)\ge 0 \Rightarrow 判別式4(a_1b_1+ a_2b_2+ a_3b_3)^2-4(a_1^2+a_2^2+a_3^2) (b_1^2+b_2^2 +b_3^2) \le 0 \\ \Rightarrow (a_1^2+a_2^2+a_3^2) (b_1^2+b_2^2 +b_3^2) \ge (a_1b_1+ a_2b_2+ a_3b_3)^2\\ 將上述內容a_i換成x_i, b_i換成y_i即為所求, \bbox[red, 2pt]{故得證} \\\textbf{(b) }  令\cases{a=\sqrt[3]{x_1}\\ b=\sqrt[3]{x_2} \\ c=\sqrt[3]{x_3}} \Rightarrow \cases{x_1 =a^3\\ x_2 =b^3\\ x_3 =c^3} \Rightarrow {x_1+ x_2 +x_3\over 3}-\sqrt[3]{x_1x_2x_3} ={1\over 3}(x_1+ x_2 +x_3-3\sqrt[3]{x_1x_2x_3}) \\={1\over 3}(a^3+b^3+c^3-3abc) ={1\over 3}(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \\={1\over 6}(a+b+c)(2a^2+2b^2+2c^2-2ab-2bc-2ca) \\={1\over 6}(a+b+c)((a-b)^2+(b-c)^2)+(c-a^2) \ge 0 \\ \Rightarrow  {x_1+ x_2 +x_3\over 3}-\sqrt[3]{x_1x_2x_3} \ge 0 \Rightarrow   {x_1+ x_2 +x_3\over 3}\ge\sqrt[3]{x_1x_2x_3} \quad \bbox[red, 2pt]{故得證}$$

解答:$$\Leftarrow\\利用積分審斂法\text{(integral test)}: p\gt 1 \Rightarrow \int_1^\infty{1\over x^p}\,dx = \left. \left[ {1\over -p+1}x^{-p+1}\right] \right|_1^\infty ={1\over p-1} 存在\\\qquad \Rightarrow \sum_{n=1}^\infty {1\over n^p}收斂\\ \Rightarrow \\ 利用\text{ Cauchy’s condensation test} :\sum_{n= 1}^\infty {1\over n^p}收斂 \text{ iff }\sum_{n=1}^\infty 2^n\cdot {1\over (2^n)^p} =\sum_{n=1}^\infty \left({1\over 2^{p-1}} \right)^n收斂 \\\qquad \Leftrightarrow {1\over 2^{p-1}}\lt 1 \Leftrightarrow p\gt 1\\ \bbox[red, 2pt]{\text{QED}}$$

解答:$$15! =(15\cdot 14\cdot 13)\cdot (12\cdot 11\cdot 10)\cdot (9\cdot 8\cdot 7)\cdot (6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1)\\\equiv[(146 \text{ mod }323) \cdot (28 \text{ mod }323)] \cdot [(181 \text{ mod }323) \cdot (74 \text{ mod }323)] \\\equiv (212 \text{ mod }323) \cdot (151 \text{ mod 323 }) \equiv \bbox[red, 2pt]{35} \text{ mod }323$$

解答:$$a_i\in \mathbb Z \Rightarrow -1\le {1\over a_i}\le 1 \Rightarrow -4\le n\le 4 \\ \Rightarrow\cases{ n=4:(a_1,a_2,a_3,a_4) = (1,1,1,1)\\n=2:(a_1,a_2,a_3,a_4) =(1,1,1,-1)\\ n=0:(a_1,a_2,a_3,a_4)= (1,1,-1,-1)\\ n=-2:(a_1,a_2,a_3,a_4) =(-1,-1,-1,1)\\ n=-4:(a_1,a_2,a_3,a_4) =(-1,-1,1,1)} \Rightarrow n= \bbox[red, 2pt]{\pm 4,\pm 2,0}$$

解答:$$a^{\log_b (a^{-32})} =b^{\log_a(ba^{-6})} \Rightarrow e^{\log_b (a^{-32})\cdot \ln a} = e^{\log_a(ba^{-6})\cdot \ln b} \Rightarrow \log_b (a^{-32})\cdot \ln a= \log_a(ba^{-6})\cdot \ln b \\ \Rightarrow -32\log_b a=(\log_a b-6)\cdot {\ln b\over \ln a} \Rightarrow {-32\over \log_a b} =(\log_a b-6)\cdot \log_a b \\ \Rightarrow (\log_a b)^3-6(\log_a b)^2+32=0 \Rightarrow (\log_a b+2)(\log_a b-4)^2=0 \Rightarrow \log_a b=4 \Rightarrow a^4=b \\ \Rightarrow (a,b)=(2,16), (3,81),(4,256),(5,625),(6,1296) \Rightarrow 共\bbox[red, 2pt]5組$$

解答:

$$假設\cases{O(0,0) \\ A(\omega) \\ B(\omega^2) \\C(\lambda \omega)\\ 圓\Gamma:x^2+y^2=25} 並 依題意\triangle ABC為正三角形 \Rightarrow \cases{\overline{AB} =\overline{AC}=5(\lambda-1) \\\angle BAC=60^\circ }\\ \Rightarrow \angle OAB =180^\circ-60^\circ=120^\circ \Rightarrow \triangle \cos \angle OAB ={\overline{OA}^2+\overline{AB}^2-\overline{OB}^2 \over 2\cdot \overline{OA}\cdot \overline{AB}} \\ \Rightarrow  -{1\over 2} ={5^2+(5(\lambda-1))^2-25^2\over 2\cdot 5\cdot 5(\lambda-1)} \Rightarrow \lambda^2 -\lambda-24=0 \Rightarrow \lambda= \bbox[red, 2pt]{1+\sqrt{97}\over 2}$$

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解題僅供參考，其他教甄試題及詳解