115年大直高中教甄-數學詳解

 臺北市立大直高級中學 115 學年度第一次專任教師甄選

一、填充題（每題 6 分，共 66 分）

解答:$$L:(3x - y) + a(x + y + 4) = 0 \Rightarrow  (a+3)x + (a-1)y + 4a = 0 \\\Rightarrow d(M,L)  = \frac{\vert{}(a+3)\cdot 1 + (a-1)\cdot 2 + 4a\vert{}}{\sqrt{(a+3)^2 + (a-1)^2}} =a \Rightarrow  \frac{\vert{}7a + 1\vert{}}{\sqrt{2a^2 + 4a + 10}} = a \\ \Rightarrow f(a)=  2a^4 + 4a^3 - 39a^2 - 14a - 1 = 0  \Rightarrow \cases{f(1)=-48\lt 0\\ f(2)=-121\lt 0\\ f(3)=-124\lt 0\\ f(4)=87 \gt 0} \\根據勘根定理 \Rightarrow 3\lt a\lt 4 \Rightarrow k= \bbox[red, 2pt]3$$

解答:

$$\cases{x - y \le 2\\x + 2y \ge 8 \\x \ge 2 \\y \le 5} \Rightarrow 區域\mathcal R的頂點\cases{A(2, 5) \\B(2, 3) \\ C(4, 2) \\D(7, 5)} \Rightarrow \cases{\triangle ABD=5\\ \triangle BCD= 9/2} \Rightarrow ABCD面積=5+{9\over 2}=9.5\\ \Rightarrow 面積A=  \text{Area} = \frac{1}{2} \vert{}(2\times 3 + 2\times 2 + 4\times 5 + 7\times 5) - (5\times 2 + 3\times 4 + 2\times 7 + 5\times 2)\vert{}=9.5 \\ 又(2k+4)x + (k-1)y = 7k+5 \Rightarrow k(2x+y-7)+(4x-y-5) =0 \\ \Rightarrow 該直線恆過兩直線 \cases{2x + y - 7 = 0 \\4x - y - 5 = 0}\; 的交點(2,3) =B\\ 由於\triangle ABD\gt \triangle BCD \Rightarrow P\in \overline{AD}: y=5,滿足\triangle ABP ={1\over 2}ABCD=9.5/2=4.75 \\ \triangle ABP ={1\over 2}\cdot \overline{AB}\cdot \overline{AP} \Rightarrow 4.75={1\over 2}\cdot 2\cdot \overline{AP} \Rightarrow \overline{AP} =4.75 \Rightarrow P(4.75+2,5)=P({27\over 4},5) \\ 將P代入 k(2x + y - 7) + (4x - y - 5) = 0 \Rightarrow  k\left(2 \cdot \frac{27}{4} + 5 - 7\right) + \left(4 \cdot \frac{27}{4} - 5 - 5\right) = 0 \\ \Rightarrow  k = \bbox[red, 2pt]{-\frac{34}{23}}$$

解答:$$複數平面上四點  0, 1, i, 1+i ，這四點剛好構成單位正方形的四個頂點\\令 z_{n-1} = x + yi 於此正方形區域內\\\Rightarrow 最靠近的頂點座標即為將 x 與 y 分別四捨五入至最接近的整數(0或1)\\ 因此z_n = 2z_{n-1} - \omega_n代表 \cases{\text{Re}(z_n) = \{2\text{Re}(z_{n-1})\} \\ \text{Im}(z_n) = \{2\text{Im}(z_{n-1})\}}, 其中 \{k\} 代表 k 的小數部分 \\ \Rightarrow \cases{ x_0 = \frac{1}{7} \\x_1 = \frac{2}{7} \\x_2 = \frac{4}{7} \\x_3 = \frac{8}{7} \pmod 1 = \frac{1}{7} = x_0} \Rightarrow 實部x_n的循環數為3 \\\Rightarrow \cases{y_0 = \frac{1}{5} \\y_1 = \frac{2}{5} \\ y_2 = \frac{4}{5} \\y_3 = \frac{8}{5} \pmod 1 = \frac{3}{5}\\y_4 = \frac{6}{5} \pmod 1 = \frac{1}{5} = y_0} \Rightarrow 虛部y_n的循環數為4 \\ 因此z_n的循環數為3\times 4=12 \Rightarrow  2026 = 12 \times 168 + 10 \Rightarrow z_{2026} =z_{10} \\ \Rightarrow \cases{x_{10}: 因為  10 \equiv 1 \pmod 3$，故  x_{10} = x_1 = \frac{2}{7} \\ y_{10} : 因為 10 \equiv 2 \pmod 4，故 y_{10} = y_2 = \frac{4}{5}} \Rightarrow z_{2026}= {2\over 7}+{4\over 5}i \Rightarrow 70z_{2026} = \bbox[red, 2pt]{20+56i}$$

解答:$$取\cases{x_k= \sqrt{a_k} \\ y_k= \sqrt{a_k^3}}\quad代入柯西不等式:  \left(\sum_{k=1}^n x_k^2\right) \left(\sum_{k=1}^n y_k^2\right) \ge \left(\sum_{k=1}^n x_k y_k\right)^2 \\ \Rightarrow  \left(\sum_{k =1}^n a_k\right) \left(\sum_{k =1}^n a_k^3\right) \ge \left(\sum_{k= 1}^n a_k^2\right)^2 \Rightarrow 24\times 54\ge 36^2 \Rightarrow 1296\ge 1296\\ \Rightarrow 等號成立代表 {\sqrt{a_k^3} \over \sqrt{a_k}} =c \Rightarrow a_k\sqrt{a_k}=c\sqrt{a_k} \Rightarrow c=a_k \Rightarrow a_1=a_2=\cdots =a_n=c \\ \Rightarrow \cases{ \sum_{k=1}^n a_k = n \cdot c = 24 \\  \sum_{k=1}^n (a_k)^2 = n \cdot c^2 = 36 } \Rightarrow c={36\over 24} ={3\over 2} \Rightarrow 首項a_1=c={3\over 2} \Rightarrow n\cdot {3\over 2}=24 \Rightarrow n=16 \\ \Rightarrow (a_1, n)= \bbox[red, 2pt] {\left( {3\over 2},16 \right)}$$

解答:$$ f(x) = x^3 + ax^2 + bx + 115  為遞增函數 \Rightarrow f'(x)=3x^2+2ax+b \ge 0 \\\Rightarrow 判別式 \Delta=4a^2-12b\le 0 \Rightarrow b\ge {1\over 3}a^2, 另一條件為2a-b+24 \ge 0 \\ 欲求 b^2={1\over a^2}與2a-b+24=0的交點 :  \frac{1}{3}a^2 = 2a + 24 \Rightarrow a^2-6a-72=0 \\ \Rightarrow (a-12)(a+6)=0 \Rightarrow a=-6,12 \Rightarrow 交集\text{面積} = \int_{-6}^{12} \left( (2a + 24) - \frac{1}{3}a^2 \right) da \\ = \left[ a^2 + 24a - \frac{1}{9}a^3 \right]_{-6}^{12} =240-(-84)= \bbox[red, 2pt]{324}$$

解答:$$函數  g(b) ，在  b = 0  附近的泰勒展開式為:g(b) = g(0) + g'(0)b + \frac{g''(0)}{2!}b^2 + \dots \\ 取g(b)=f(1+2a+3b) \Rightarrow g(0)=f(1+2a) \Rightarrow g'(0)=3f'(1+2a ) \Rightarrow g''(0)=9f''(1+2a) \\ 因此 f(1+2a+3b) = f(1+2a) + 3b \cdot f'(1+2a) +  O(b^2) \\ \Rightarrow f(1+3b)=f(1)+3b \cdot f'(1)+ O(b^2)\\ \Rightarrow f(1+2a+3b)-f(1+3b)=f(1+2a)-f(1)+3b \left[ f'(1+2a)-f'(1) \right]+O(b^2) \\ \Rightarrow  L = \lim_{a \to 0} \left( \lim_{b \to 0} \frac{f(1+2a+3b) - f(1+2a) - f(1+3b) + f(1)}{ab} \right)  \\  = \lim_{a \to 0} \left( \lim_{b \to 0} \frac{3b \left[ f'(1+2a)-f'(1)\right]  +O(b^2)}{ab} \right)  = \lim_{a \to 0}   \frac{3 \left[ f'(1+2a)-f'(1) \right]}{a}  =3\cdot 2f''(1)  =6f''(1) \\令g(x) = f(2x+3) = 1 + \sum_{k=1}^{115} \frac{x^k}{k}  \Rightarrow g'(x)=2f'(2x+3) \Rightarrow g''(x)=4f''(2x+3) \\ \Rightarrow g''(-1)=4f''(1) \Rightarrow f''(1)={1\over 4}g''(-1) \\ 由於g(x) =1 + \sum_{k=1}^{115} \frac{x^k}{k}  = 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \dots + \frac{x^{115}}{115} \Rightarrow  g'(x) = 1 + x + x^2 + x^3 + \dots + x^{114} \\ \Rightarrow g''(x)   = 1 + 2x + 3x^2 + 4x^3 + \dots + 114x^{113} \Rightarrow g''(-1)= (1-2)+(3-4)+\cdots +(113-114) \\=(-1) \times 57 =-57 \Rightarrow f''(1)= {1\over 4}\cdot (-57)=-{57\over 4} \Rightarrow L= 6f''(1)= \bbox[red, 2pt]{-{171\over 2}}$$

解答:$$圓C: x^2 + y^2 - 4x - 1 = 0 \implies (x - 2)^2 + y^2 = 5 \Rightarrow \cases{圓心K(2,0)\\ 半徑R=\sqrt 5} \\ \overline{MN}為圓心直徑 \Rightarrow K=\overline{MN}中點\Rightarrow \overrightarrow{KN} =-\overrightarrow{KM} \\ \cases{\overrightarrow{PM}= \overrightarrow{PK}+ \overrightarrow{KM} \\ \overrightarrow{PN}= \overrightarrow{PK}+ \overrightarrow{KN}= \overrightarrow{PK}- \overrightarrow{KM}} \Rightarrow  \overrightarrow{PM} \cdot  \overrightarrow{PN} =(\overrightarrow{PK}+ \overrightarrow{KM}) \cdot (\overrightarrow{PK}- \overrightarrow{KM}) \\  = |\overrightarrow{PK}|^2 - |\overrightarrow{KM}|^2 = |\overrightarrow{PK}|^2 - R^2 = |\overrightarrow{PK}|^2 - 5 \\ 由於P(x,y)\in \Gamma:{x^2\over 25}+{y^2\over 41}=1 \Rightarrow y^2 = 41 \left(1 - \frac{x^2}{25}\right) = 41 - \frac{41}{25}x^2  \\ \Rightarrow |\overrightarrow{PK}|^2 =(x-2)^2+y^2 = (x^2 - 4x + 4) + \left(41 - \frac{41}{25}x^2\right)  = -\frac{16}{25}x^2 - 4x + 45  \\ 令g(x) = -\frac{16}{25}x^2 - 4x + 45 \Rightarrow 極值發生在x=-{25\over 8} \Rightarrow \cases{g(-25/8)= {205\over 4} \\g(5) =9\\ g(-5) =49} \\\Rightarrow \cases{當 x=-25/8, g(x)有最大值205/4\\ 當x=5, g(x)有最小值9}\Rightarrow  \overrightarrow{PM} \cdot  \overrightarrow{PN}\cases{最大值a=205/4-5= 185/4 \\ 最小值b=9-5=4}\\ \Rightarrow (a,b)= \bbox[red, 2pt]{\left( {185\over 4},4 \right)}$$

解答:$$假設黑板上整數 i 的個數為 c_i\Rightarrow  初始時,  c_0 = c_1 = c_2 = c_3 = c_4 = c_5 = c_6 = 1 \\每擦去一個  n , n-1 的數量 c_{n-1}$ 就會增加  c_{n-1}  個， 即  c_{n-1}  變為原來的 2 倍 ,即c_{n-1} \to 2c_{n-1} \\若黑板上有 c_n 個 n，將所有的 n 全部擦完後，n-1 的數量會被連續翻倍 c_n 次,即c_{n-1} \to 2^{c_n} c_{n-1}\\ 處理n=6(此時c_6=1) \Rightarrow c_5\to 1\cdot 2^1=2 \Rightarrow 處理n=5(此時c_5= 2) \Rightarrow c_4\to 1\cdot 2^2=4  \\ \Rightarrow 處理n=4(此時c_4= 4) \Rightarrow c_3\to 1\cdot 2^4= 16 \Rightarrow  處理n=3(此時c_3= 16) \Rightarrow c_2\to 1\cdot 2^16= 2^{16}  \\ \Rightarrow  處理n=2(此時c_2= 2^{16}) \Rightarrow c_1\to 1\cdot 2^{2^{16}}= 2^{2^{16}} \Rightarrow 處理n=1(此時c_1 = 2^{2^{16}})\Rightarrow c_0\to 1\cdot 2^{2^{2^{16}}} \\ \Rightarrow A = 2^{2^{2^{16}}} \Rightarrow \log_{10}A = 2^{2^{16}} \log _{10}2 \Rightarrow  \log_{10}\left(\log_{10} A\right) = \log_{10}\left(2^{2^{16}} \cdot \log_{10} 2\right) \\= 2^{16} \cdot \log_{10} 2 + \log_{10}(\log_{10} 2)   \approx 65536 \cdot 0.301+\log_{10}(0.301) \approx 19728.3-0.521 \approx 19727.8 \\ 由於 4 < \log_{10}(19727.8) < 5 \Rightarrow \log_{10}\log_{10}\log_{10} A$ 的整數部分為 \bbox[red, 2pt]4$$

解答:$$\cases{f(x,y)=x^2+ 2y^2 \\g(x,y)=x^2-2xy+3y^2-2} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{2x= \lambda(2x-2y) \\ 4y=\lambda(-2x+6y) \\ x^2-2xy+3y^2=2} \Rightarrow {2x\over 4y} ={2x-2y\over -2x+6y} \\ \Rightarrow x^2-xy-2y^2=0 \Rightarrow (x-2y)(x+y)=0 \Rightarrow \cases{x=2y \Rightarrow 4y^2-4y^2+3y^2=2 \Rightarrow y^2=2/3\\ x=-y \Rightarrow y^2+2y^2+3y^2=2 \Rightarrow y^2=1/3} \\ \Rightarrow \cases{y=\pm \sqrt{2/3} \Rightarrow x=\pm 2\sqrt{2/3} \Rightarrow x^2+ 2y^2=4\\ y=\pm \sqrt{1/3} \Rightarrow x= \mp \sqrt{1/3} \Rightarrow x^2+2y^2=1} \Rightarrow (M,m)= \bbox[red, 2pt]{(4,1)}$$

解答:

$$ x^2 + y^2 - 6y + 1 = 0 \Rightarrow   x^2 + (y-3)^2 = 8 \Rightarrow \cases{圓心O(0,3)\\ 半徑R=2\sqrt 2} \\ 假設第一象限的切點P(x_0,y_0) \Rightarrow \overline{OP}斜率={y_0-3 \over x_0} \\ y=kx^2 \Rightarrow y'=2kx \Rightarrow 在P點的切線斜率=2kx_0 \Rightarrow 法線斜率=-{1\over 2kx_0}={y_0-3 \over x_0} \\ \Rightarrow y_0-3=-{1\over 2k}, 又P在y=kx^2上\Rightarrow y_0=kx_0^2 \Rightarrow k={y_0\over x_0^2} \Rightarrow y_0-3=-{x_0^2\over 2y_0} \\ \Rightarrow x_0^2=6y_0-2y_0^2代入圓\Rightarrow  (6y_0 - 2y_0^2) + y_0^2 - 6y_0 + 1 = 0 \Rightarrow  1 - y_0^2 = 0 \Rightarrow y_0=1 \\ \Rightarrow x_0=2 \Rightarrow k={1\over 2^2}={1\over 4} \Rightarrow 拋物線:y={1\over 4}x^2 \\ \Rightarrow 旋轉體積V= \pi \int_{-2}^2 \left( \left[ 3-\sqrt{8-x^2} \right]^2- \left[ {1\over 4}x^2 \right]^2 \right)\,dx = \pi \int_{-2}^2 \left( 17-x^2-{1\over 16}x^4-6\sqrt{8-x^2} \right) \,dx \\=  2\pi \left( \frac{284}{15} - 6\pi \right) =\bbox[red, 2pt]{ \frac{568\pi }{15}-12\pi^2}$$

解答:$$x_1滿足2x+2^x=15 \Rightarrow 2x_1+2^{x_1}=15\cdots(1) \\x_2滿足2x+2\log_2(x+3)=7 \Rightarrow 2x_2+2\log_2(x_2+3) =7 \Rightarrow x_2+\log_2(x_2 +3)={7\over 2} \\ 取y=x_2+3 \Rightarrow y-3+\log_2  y={7\over 2 } \Rightarrow y+\log_2 y={13\over 2} \\ 取t=\log_2y \Rightarrow 2^t+t={13 \over 2} \Rightarrow 2t+2^{t+1}=13\\ 取z=t+1 \Rightarrow 2(z-1)+2^z=13 \Rightarrow 2z+2^z=15 \cdots(2) \\ 式(1)與式(2)有相同的解 \Rightarrow z=x_1 \Rightarrow z=t+1= x_1 \Rightarrow t=x_1-1 \Rightarrow y=2^t=2^{x_1-1} ={1\over 2}2^{x_1} \\={1\over 2}(15-2x_1) ={15\over 2}-x_1 \Rightarrow x_2+3= {15\over 2}-x_1 \Rightarrow x_1+x_2= \bbox[red, 2pt]{9\over 2}$$

二、非選擇題：（共 34 分）

解答:$$\textbf{(1) }設 X_i  表示第 i  次試驗的結果 \Rightarrow  X_i = \begin{cases} 1, & \text{若第 } i \text{ 次試驗成功（機率為 } p \text{）} \\ 0, & \text{若第 } i \text{ 次試驗失敗（機率為 } 1-p \text{）} \end{cases} \\ \Rightarrow \cases{E[X_i]=1 \cdot p+0\cdot(1-p) =p \\ E[X_i^2]  =1^2\cdot p+0^2\cdot(1-p)=p} \Rightarrow Var(X_i)= E[X_i^2] -(E[X_i])^2 =p-p^2 =p(1-p) \\ \Rightarrow  E[X] = E\left[\sum_{i=1}^{n} X_i\right] = \sum_{i=1}^{n} E[X_i] = \sum_{i=1}^{n} p = np \\ \Rightarrow  \text{Var}(X) = \text{Var}\left(\sum_{i=1}^{n} X_i\right) = \sum_{i=1}^{n} \text{Var}(X_i) = \sum_{i=1}^{n} p(1-p) = np(1-p) \\\qquad  \Rightarrow \bbox[red, 2pt]{\cases{E[X]=np\\ Var(X)=np(1-p)}} \\\textbf{(2) }設\cases{設 Y_1 表示達到第 1 次成功所需的試驗次數\\設 Y_2 表示從第 1 次成功後，再達到第 2 次成功額外所需的試驗次數} \Rightarrow \cases{Y_1  \sim Geo(p) \\Y_1\sim Geo(p)}\\ \Rightarrow E[Y_1]=E[Y_2]={1\over p} \Rightarrow E[Y]=E[Y_1+Y_2]= E[Y_1]+E[Y_2]= \bbox[red, 2pt]{2\over p}$$

解答:$$\textbf{(1) }小天透過消去法分別求出  -1 \le x \le 5  與  -4 \le y \le 5 \\ 這種個別求範疇的作法只是必要條件，將可行解區域從原本的「平行四邊形」\\放寬成了一個較大的「矩形」。而且未考慮是否 x  與  y  可以同時取到極值\\ 小天算出來的極值來自\cases{x=5\\ y=5} \Rightarrow x+2y=15, 不符原條件x+2y\le 9 \\ \textbf{(2) 方法一: }8x+y=a(x+2y)+b(2x-y) =(a+2b)x+(2a-b) y \Rightarrow \cases{a+2b=8\\ 2a-b=1} \Rightarrow \cases{a=2\\ b=3} \\ \qquad \Rightarrow \cases{-6\le 2(x+2y)\le 18 \\ -3\le 3(2x-y) \le 24} \Rightarrow 兩式相加:-9\le 8x+y\le 42 \Rightarrow \cases{最大值:42\\ 最小值:-9} \\ \textbf{方法二: }\cases{  x + 2y = -3\\x + 2y = 9 \\ 2x - y = -1\\ 2x - y = 8} 所圍區域頂點\cases{A(-1,-1) \\B(1,-2) \\C(5,2) \\D(1,4)},令f(x,y)=8x+y,則\cases{f(A)=-9\\ f(B)=6\\ f(C)=42\\ f(D)=12} \\ \qquad  \Rightarrow \cases{最大值:42\\ 最小值:-9}$$

解答:$$\textbf{方法1: }\tan \theta ={6x\over x^2+16}={6\over x+{16\over x}},利用算幾不等式:x+{16\over x}\ge 2\sqrt{x\cdot {16\over x}} =8 \\ \qquad \tan \theta\ge {6\over 8}={3\over 4} \Rightarrow 最大值{3\over 4},此時x={16\over x} \Rightarrow x=4 \\ \textbf{方法2: }令k=\tan \theta ={6x\over x^2+16} \Rightarrow  k x^2 - 6x + 16k = 0 有實數解\Rightarrow \Delta=36-64k^2\ge 0 \\\qquad 64k^2\le 36 \Rightarrow 0\lt k\le {3\over 4}\Rightarrow 最大值{3\over 4} \Rightarrow  \frac{3}{4}x^2 - 6x + 12 = 0 \Rightarrow x^2 - 8x + 16 = 0 \\\qquad \Rightarrow  (x-4)^2 = 0 \Rightarrow  x = 4 \\ \textbf{方法3: } \tan\theta = f(x) = \frac{6x}{x^2+16} \Rightarrow f'(x)=  \frac{6(16 - x^2)}{(x^2+16)^2} =0 \Rightarrow x^2=16 \Rightarrow x=4 \\\qquad \Rightarrow \tan \theta=f(4)={24\over 32} ={3\over 4} \\ \textbf{方法4: } \frac{1}{\tan\theta} = \frac{x^2 + 16}{6x} = \frac{1}{6} \left( x + \frac{16}{x} \right) ={1\over 6} \left[ \left( \sqrt x -{4\over \sqrt x}\right)^2+8 \right] \ge {8\over 6} ={4\over 3} \Rightarrow \tan \theta={3\over 4} \\ \qquad 此時\sqrt x={4\over \sqrt x} \Rightarrow x=4$$

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