113年北一女教甄-數學詳解

臺北市立第一女子高級中學 113 學年度第一次正式教師甄選

填充題

解答:$$已知44^2=1936,45^2=2025,46^2=2116\\ 假設a_1=\sqrt{2024} \Rightarrow a_2=\sqrt{2024+a_1} \Rightarrow a_3 =\sqrt{2024+a_2} \Rightarrow x=a_{2024}\\ \sqrt{1936} \lt \sqrt{2024}\lt \sqrt{2025} \Rightarrow 44\lt a_1\lt 45 \\ \Rightarrow \sqrt{2024+44} \lt \sqrt{2024+a_1} \lt \sqrt{2024+45} \Rightarrow 45\lt a_2\lt 46 \\ \Rightarrow \sqrt{2024+45} \lt \sqrt{2024+a_2} \lt \sqrt{2024+46} \Rightarrow 45 \lt a_3\lt 46\\ \Rightarrow \sqrt{2024+45} \lt \sqrt{2024+a_3} \lt \sqrt{2024+46} \Rightarrow 45 \lt a_5\lt 46 \\ \dots \dots \\\\ \Rightarrow \sqrt{2024+45} \lt \sqrt{2024+a_{2023}} \lt \sqrt{2024+46} \Rightarrow 45 \lt a_{2024}\lt 46 \\ \Rightarrow [x]=\bbox[red, 2pt]{45}$$

解答:$$\log(x^2)+ \sqrt{(\log x)-1} \lt 5 \Rightarrow (5-2\log x)^2 \gt (\log x)-1 \ge 0 \\ \Rightarrow \cases{\log x\ge 1\\4(\log x)^2-21\log x+26\gt 0}  \Rightarrow \cases{x\ge 10 \cdots(1)\\ (4\log x-13)(\log x-2)\gt 0 \Rightarrow x\lt 100 或x\gt 10^{13/4}  \cdots(2)} \\ (1)\cap (2) \Rightarrow \bbox[red, 2pt]{10\le x\lt 100}$$

解答:$$\cases{8反0正:1\\7反1正:C^8_1=8\\ 6反2正:C^7_2=21\\ 5反3正:C^6_3=20\\ 4反4正:C^5_4=5} \Rightarrow 合計1+8+21+20+5=55 \Rightarrow 機率={2^8-55\over 2^8} =\bbox[red,2pt]{201\over 256}$$

解答:

$$令\cases{A(0,3) \\B(0,0) \\C(4,0)},由\cases{\overline{BD}: \overline{DC}=1:3\\ \overline{CF}: \overline{FG}: \overline{GA}=1:1:2 \\ \overline{AH}: \overline{HB}= 1:2} \Rightarrow \cases{D(1,0)\\ F(3,3/4)\\ G(2,3/2)\\ H(0,2)} \Rightarrow \cases{\overleftrightarrow{DG}=L_1:y=3x/2-3/2\\ \overleftrightarrow{FH}=L_2: y=-5x/12+2} \\ \Rightarrow P=L_1\cap L_2 =({42\over 23},{57\over 46});假設 \Rightarrow \overline{FP} : \overline{PH} =a:b \Rightarrow P={aH+bF\over a+b} \Rightarrow \cases{{3b\over a+b}={42\over 23} \\ {2a+3b/4\over a+b}={57\over 46}} \\\Rightarrow 92a+{345b\over 2}=141(a+b) \Rightarrow 49a={63\over 2}b \Rightarrow a:b=\bbox[red, 2pt]{9:14}$$

解答:$$91x^2-24xy+84y^2= [x,y]\begin{bmatrix}91 & -12 \\-12 & 94 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} = [x,y] \begin{bmatrix}3/5 & -4/5 \\4/5 & 3/5 \end{bmatrix} \begin{bmatrix}75 & 0 \\0 & 100 \end{bmatrix} \begin{bmatrix}3/5 & 4/5 \\-4/5 & 3/5 \end{bmatrix} \begin{bmatrix}x\\ y \end{bmatrix} \\= \begin{bmatrix} (3x+4y)/5, (-4x+3y)/5\end{bmatrix} \begin{bmatrix}75 & 0 \\0 & 100 \end{bmatrix} \begin{bmatrix} (3x+4y)/5\\ (-4x+3y)/5 \end{bmatrix} =[x',y'] \begin{bmatrix}75 & 0 \\0 & 100 \end{bmatrix} \begin{bmatrix} x'\\ y' \end{bmatrix} \\  P\in \Gamma':91x^2-24xy+84y^2-300=0 \cap \{x=0\}\Rightarrow P=(0,{5\over \sqrt7})=(x,y)\\ (x',y')=({3x+4y \over 5},{-4x+3y\over 5}) = ({20\over 5\sqrt 7},{15\over 5\sqrt 7}) =\bbox[red, 2pt]{\left({4 \sqrt 7\over 7},{3\sqrt 7\over 7} \right)}$$

解答:$$f(x)=x^3+ax^2+bx+c \Rightarrow \cases{f(-1)=a-b+c-1\\ f(0)=c\\ f(1)= a+b+c+1\\ f(2)=4a+2b+c+8\\ f(3) =9a+3b+c+27} \\f(-1),f(0),f(1)成等差 \Rightarrow f(-1)+f(1) =2f(0) \Rightarrow 2(a+c)=2c \Rightarrow a=0\\ f(0),f(1),f(2),f(3)成等比 \Rightarrow {f(1)\over f(0)} = {f(2)\over f(1)} = {f(3)\over f(2)} \Rightarrow {b+c+1\over c}={2b+c+8 \over b+c+1} ={3b+ c+27\over 2b+c+8} \\ \Rightarrow \cases{(b+c+1)^2=c(2b+c+8) \\ (2b+c+8)^2=(b+c+1) (3b+c+27)} \Rightarrow \cases{(b+1)^2=6c \cdots(1)\\ b^2+2b-12c+37=0 \cdots(2)} \\ 將(1)代入(2) \Rightarrow (b+1)^2-12c+36=6c-12c+36=0 \Rightarrow c=6 \Rightarrow b=5\\ \Rightarrow f(x)= \bbox[red, 2pt]{x^2+5x+6}$$

解答:$$\int_0^{10} \lfloor x \lfloor x\rfloor \rfloor\,dx =\int_0^1 \lfloor x \lfloor x\rfloor \rfloor\,dx + \int_1^2 \lfloor x \lfloor x\rfloor \rfloor\,dx + \cdots + \int_9^{10} \lfloor x \lfloor x\rfloor \rfloor\,dx  \\= 0+ \int_1^2 \lfloor x\rfloor\,dx + \int_2^3 \lfloor 2x\rfloor \,dx +\cdots +\int_9^{10} \lfloor 9x\rfloor \,dx \\=1+ \left( \int_2^{5/2} \lfloor 2x\rfloor \,dx+\int_{5/2}^3 \lfloor 2x\rfloor \,dx\right)+\left( \int_3^{10/3} \lfloor 3x\rfloor \,dx+ \int_{10/3}^{11/3} \lfloor 3x\rfloor \,dx+ \int_{11/3}^{4} \lfloor 3x\rfloor \,dx\right) +\cdots \\\qquad +\left( \int_9^{9+1/9}  \lfloor 9x\rfloor \,dx + \int_{9+1/9}^{9+2/9}  \lfloor 9x\rfloor \,dx+ \cdots +\int_{9+8/9}^{9+9/9}  \lfloor 9x\rfloor \,dx\right) \\=1+{1\over 2}(4+5)+{1\over 3} (9+10 +11)+\cdots + {1\over 9}(81+82+ \cdots +89) \\=\sum_{k=1}^9 {1\over k}\sum_{m=0}^{k-1}(k^2+m) =\sum_{k=1}^9 \left(k^2+{k-1\over 2} \right)=285+18 =\bbox[red, 2pt]{303}$$

解答:$$明天待續.....$$