105年警專35期乙組數學科詳解

\(\bbox[blue,2pt]解：\)$$a=\log _{ 2 }{ 3 } ,b=\log _{ 3 }{ 11 } \Rightarrow ab=\log _{ 2 }{ 11 } \\ \log _{ 22 }{ 33 } =\frac { \log { 33 }  }{ \log { 22 }  } =\frac { \log _{ 2 }{ 33 }  }{ \log _{ 2 }{ 22 }  } =\frac { \log _{ 2 }{ 3 } +\log _{ 2 }{ 11 }  }{ \log _{ 2 }{ 2 } +\log _{ 2 }{ 11 }  } =\frac { a+ab }{ 1+ab } $$

故選\(\bbox[red,2pt]{(A)}\)

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\(\bbox[blue,2pt]解：\)$$\overrightarrow { AB } \cdot \overrightarrow { BC } =\left| \overrightarrow { AB }  \right| \left| \overrightarrow { BC }  \right| \cos { \left( 180°-60° \right)  } =2\times 2\times \left( -\frac { 1 }{ 2 }  \right) =-2$$

故選\(\bbox[red,2pt]{(A)}\)

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\(\bbox[blue,2pt]解：\)$$\begin{cases} \left| x-3 \right| \le 2 \\ \left| x-1 \right| \le 2 \end{cases}\Rightarrow \begin{cases} 1\le x\le 5 \\ -1\le x\le 3 \end{cases}\Rightarrow 1\le x\le 3\\ \Rightarrow \left| x-2 \right| \le 1\Rightarrow \left( a,b \right) =\left( 2,1 \right) $$

故選\(\bbox[red,2pt]{(D)}\)

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\(\bbox[blue,2pt]解：\)$$\left( 2+\sqrt { 3 }  \right) x+\left( 1-\sqrt { 3 }  \right) y=-3\sqrt { 3 } \Rightarrow \left( 2x+y \right) +\left( x-y \right) \sqrt { 3 } =-3\sqrt { 3 } \\ \Rightarrow \begin{cases} 2x+y=0 \\ x-y=-3 \end{cases}\Rightarrow x=-1,y=2$$

故選\(\bbox[red,2pt]{(B)}\)

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\(\bbox[blue,2pt]解：\)$$f\left( x \right) =\left( 1-x \right) +{ \left( 1-x \right)  }^{ 2 }+{ \left( 1-x \right)  }^{ 3 }+\cdots +{ \left( 1-x \right)  }^{ 10 }\\ =\frac { \left( 1-x \right) \left[ 1-\left( 1-x \right) ^{ 10 } \right]  }{ 1-\left( 1-x \right)  } =\frac { \left( 1-x \right) -\left( 1-x \right) ^{ 11 } }{ x } \\ \Rightarrow xf\left( x \right) =\left( 1-x \right) -\left( 1-x \right) ^{ 11 }\Rightarrow x^{ 3 }的係數=C^{11}_8=165$$

故選\(\bbox[red,2pt]{(A)}\)

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\(\bbox[blue,2pt]解：\)

\(C^7_3\times C^4_2\times C^2_2=210\)，故選\(\bbox[red,2pt]{(C)}\)

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\(\bbox[blue,2pt]解：\)

假設正方形邊長為x，則\(\overline{BM}=\overline{MC}=\frac{x}{2}，且\overline{AC}=\sqrt{2}x, \overline{AM}=\frac{\sqrt{5}x}{2}\)，如上圖。
利用餘弦定理：$${ \overline { MC }  }^{ 2 }={ \overline { AM }  }^{ 2 }+{ \overline { AC }  }^{ 2 }-2{ \overline { AM }  }\cdot { \overline { AC }  }\cos { \angle MAC } \\ \Rightarrow \frac { x^{ 2 } }{ 4 } =\frac { 5x^{ 2 } }{ 4 } +2x^{ 2 }-\sqrt { 10 } x^{ 2 }\cos { \alpha  } \Rightarrow \cos { \alpha  } =\frac { 3 }{ \sqrt { 10 }  } $$

故選\(\bbox[red,2pt]{(B)}\)

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\(\bbox[blue,2pt]解：\)
相當於由\(\overrightarrow{AB}及\overrightarrow{AC}\)所展開的平行四邊形面積的4倍，即$$4\times \begin{vmatrix} 1 & 1 \\ 2 & 3 \\ -1 & 4 \\ 1 & 1 \end{vmatrix}=4\times \left( 3+8-1-2+3-4 \right) =4\times 7=28$$故選\(\bbox[red,2pt]{(A)}\)

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\(\bbox[blue,2pt]解：\)

\(\overline{BC}=\overline{AB}\times\sqrt{3}\div 2=3\)，故選\(\bbox[red,2pt]{(B)}\)

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\(\bbox[blue,2pt]解：\)
直線L: x-2y+2=0，斜率為1/2。與此直線垂直的直線M 斜率為-2，因此M可寫成 y=-2x+b。
由於A經過M，所以\(0=-2\times 3+b\Rightarrow b=6\)，M: y=-2x+6。
兩直線L與M的交點B為 (2, 2)。
令對稱點座標為(m, n) 則\(\frac{m+3}{2}=2且\frac{n+0}{2}=2\)，可求得m=1,n=4。
故選\(\bbox[red,2pt]{(A)}\)

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\(\bbox[blue,2pt]解：\)

先求各直線的交點A、B、C、D如上圖，再將各點代入-2x+y找出最小值。
可求出B點代入-2x+y=-4+1=-3，為最小值，故選\(\bbox[red,2pt]{(C)}\)

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\(\bbox[blue,2pt]解：\)
令\(x=\sin{\theta},y=\cos{\theta}\Rightarrow -3x+4y=-3\sin{\theta}+4\cos{\theta} = 5\left( \frac{-3}{5}\sin{\theta}+\frac{4}{5}\cos{\theta}\right)\)
\(=5\sin{(\alpha+\theta)}\Rightarrow \)最小值為-5，故選\(\bbox[red,2pt]{(A)}\)

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\(\bbox[blue,2pt]解：\)
(A)對稱點為(-a, b, c)
(B) (a,b,c)到原點的距離\(\sqrt{a^2+b^2+c^2}\)
(C)(a, b, c)到x軸距離=\(\sqrt{b^2+c^2}\)

只有(D)是正確的，故選\(\bbox[red,2pt]{(D)}\)

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\(\bbox[blue,2pt]解：\)
E: 2016x+2017y+2018z=k，經過(0,0,0)，所以k=0，故選\(\bbox[red,2pt]{(C)}\)

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\(\bbox[blue,2pt]解：\)
解聯立方程組，可求得a=0, b=-2, c=-2，因此-8a-b-c=0+2+2=4，故選\(\bbox[red,2pt]{(B)}\)

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\(\bbox[blue,2pt]解：\)
星期三坐捷運的機率=星期二坐捷運且星期三坐捷運及星期二坐公車且星期三坐捷運
=\(0.3\times 0.3+0.7\times 0.4 = 0.09+0.28=0.37\)，故選\(\bbox[red,2pt]{(B)}\)

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\(\bbox[blue,2pt]解：\)
sin的週期為\(2\pi\)，所以x的週為\(\pi\)，故選\(\bbox[red,2pt]{(A)}\)

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\(\bbox[blue,2pt]解：\)
找抽樣誤差較大的，即p(1-p)要最大，故選\(\bbox[red,2pt]{(B)}\)

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\(\bbox[blue,2pt]解：\)
括號內的數字需小於1，只有(D)符合此條件，故選\(\bbox[red,2pt]{(D)}\)

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\(\bbox[blue,2pt]解：\)

令最大的正方形S1的邊長為a，第2二大的正方形邊長為b，如上圖。
由於\(\triangle EFG\)與\(\triangle   ABC\)相似，所以\(\overline{EF}:\overline{GF}= \overline{AB}: \overline{BC}=4:6  \Rightarrow  a-b:b=2:3 \Rightarrow  \) a:b=5:3\(\Rightarrow   S_1:S_2=a^2:b^2 =25:9\)
因此所有內接正方形面積的總和S=\(S_1+\frac{9}{25}S_1+{\left(\frac{9}{25}\right)}^2S_1+...   =   \frac{S_1}{1-\frac{9}{25}}\)
由\(\overline{AD}:\overline{ED}=\overline{AB}:\overline{BC}\Rightarrow   (4-a):a:4:6  \)
\( \Rightarrow   a=\frac{12}{5}\Rightarrow   S_1=a^2=\frac{144}{25}\Rightarrow   S= \frac{\frac{144}{25}}{1-\frac{9}{25}} = 9\)

故選\(\bbox[red,2pt]{(C)}\)

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\(\bbox[blue,2pt]解：\)
平方數: \(1^2, 2^2, ..., 9^2, 10^2\)共有10個；立方數: \(1^3, 2^3, 3^3, 4^3\)共有4個；
其中\(1^2=1=1^3, 4^3=64=8^2\)，有2個重複計算，因此共有100-10-4+2=88個，故選\(\bbox[red,2pt]{(B)}\)

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\(\bbox[blue,2pt]解：\)$$(A)\sum_{k=1}^{n}{1}=1+1+..+1=n\\ (B)\sum_{k=1}^{3}{a_k}= a_1+a_2+a_3 \neq a_4+a_5+a_6=\sum_{k=4}^{6}{a_k}\\ (C)\sum_{k=1}^{n}{k(k+1)}= \sum_{k=1}^{n}{k^2+k)} =\sum_{k=1}^{n}{k^2}+\sum_{k=1}^{n}{k}\\ (D) \sum_{k=1}^{n}{k^2}=1^2+2^2+...+n^2\neq (1+2+...+n)^2={\left(\sum_{k=1}^{n}{k}\right)}^2$$

故選\(\bbox[red,2pt]{(C)}\)

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\(\bbox[blue,2pt]解：\)
(A)\(\vec{a}與\vec{b}\)不同方向，兩者不相等
(C)\(\vec{a}+\overrightarrow{AB}-\vec{b}=0\Rightarrow \overrightarrow{AB}=\vec{b}-\vec{a}\)
(D)\(\overrightarrow{DB}=\overrightarrow{DO}+\overrightarrow{DC}=\vec{a}+vec{b}\)
故選\(\bbox[red,2pt]{(B)}\)

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\(\bbox[blue,2pt]解：\)
由雙曲線方程式可知: a=2, b=\(2\sqrt{3}\)，因此c=4。又$$\begin{cases} \overline { PF_{ 1 } } +\overline { PF_{ 2 } } =16 \\ \overline { PF_{ 1 } } -\overline { PF_{ 2 } } =2a=4 \end{cases}\Rightarrow \begin{cases} \overline { PF_{ 1 } } =10 \\ \overline { PF_{ 2 } } =6 \end{cases}\\ \Rightarrow \triangle PF_{ 1 }F_{ 2 }之三邊分別為\overline { PF_{ 1 } } =10,\overline { PF_{ 2 } } =6,\overline { F_{ 1 }F_{ 2 } } =2c=8\\ \Rightarrow \triangle PF_{ 1 }F_{ 2 }為一直角三角形\Rightarrow 面積=6\times 8\div 2=24$$

故選\(\bbox[red,2pt]{(B)}\)

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\(\bbox[blue,2pt]解：\)
迴歸直線方程式為 \(y-\mu_y=b(x-\mu_x)\)，由題意可知: \(\mu_x=40\div 10=4, \mu_y=50\div 10=5\)，且直線過(3,0)，可推得\(0-5=b(3-4)\Rightarrow b=5\)，因此直線為 y-5=5(x-4)，即 y=5x-15，故選\(\bbox[red,2pt]{(D)}\)

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\(\bbox[blue,2pt]解：\)
\(\sum{x_i}=8,\sum{x^2_i}=8\)
平均數\(\mu_x = 8/10=0.8\),標準差=\(\sqrt{(8-10\times 0.8^2)\div 10}=\sqrt{1.6\div 10}=0.4\)

故選\(\bbox[red,2pt]{(D)}\)

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\(\bbox[blue,2pt]解：\)

\(\frac{3}{4}\cdot 8=6\)，故選\(\bbox[red,2pt]{(D)}\)

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\(\bbox[blue,2pt]解：\)$$P\left( B'|A' \right) =\frac { P\left( B'\cap A' \right)  }{ P\left( A' \right)  } =\frac { 1-P\left( A\cup B \right)  }{ 1-P\left( A \right)  } =\frac { 1-\left( P\left( A \right) +P\left( B \right) -P\left( A\cap B \right)  \right)  }{ 1-P\left( A \right)  } \\ =\frac { 1-\left( \frac { 3 }{ 4 } +\frac { 1 }{ 3 } -\frac { 3 }{ 4 } \times \frac { 1 }{ 3 }  \right)  }{ 1-\frac { 3 }{ 4 }  } =\frac { 1-\left( \frac { 13 }{ 12 } -\frac { 1 }{ 4 }  \right)  }{ \frac { 1 }{ 4 }  } =\frac { 2 }{ 3 } $$

故選\(\bbox[red,2pt]{(C)}\)

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\(\bbox[blue,2pt]解：\)$$\left( \log _{ 3 }{ 4 } +\log _{ 9 }{ 2 }  \right) \left( \log _{ 4 }{ 9 } +\log _{ 8 }{ 27 }  \right) =\left( 2\log _{ 3 }{ 2 } +\frac { 1 }{ 2 } \log _{ 3 }{ 2 }  \right) \left( \log _{ 2 }{ 3 } +\log _{ 2 }{ 3 }  \right) \\ =\left( \frac { 5 }{ 2 } \log _{ 3 }{ 2 }  \right) \left( 2\log _{ 2 }{ 3 }  \right) =5$$

故選\(\bbox[red,2pt]{(A)}\)

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\(\bbox[blue,2pt]解：\)$$令f\left( x \right) =x^{ 3 }+2x^{ 2 }-3x-1=(x+2)^{ 3 }-4(x+2)^{ 2 }+(x+2)+5\\ \Rightarrow f\left( -2.01 \right) =(-0.01)^{ 3 }-4(-0.01)^{ 2 }+(-0.01)+5\\ =-0.000001-0.0004-0.01+5\approx -0.01+5=4.99$$

故選\(\bbox[red,2pt]{(C)}\)

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\(\bbox[blue,2pt]解：\)

(A)恆過(0,1)    (B)交Y軸於(0,1)   (E)若a<1為遞減函數

故選\(\bbox[red,2pt]{(CD)}\)

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\(\bbox[blue,2pt]解：\)
(A)  \(\log{X}=100\times   0.4771=47.71  \Rightarrow   x為48位數\)
(B)  \(\log{X}=100\times   0.4771=47.71\le 48\)
(C)    \(\log{X}=100\times   0.4771=47.71=47+0.71，又\log{5}<0.71<\log{6}\)，x的位數字是5
(D)  \(\log{X}=47.71\Rightarrow     \log{-X}=-47.71\le   -47\)
(E)\( \log{-X}=-47.71=-48+0.29\Rightarrow   首數-48\Rightarrow \)  小數位後48位不為零

故選\(\bbox[red,2pt]{(BC)}\)

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\(\bbox[blue,2pt]解：\)

(A)的確有三個交點，如上圖
(B)(C)(D)皆正確
(E)\(x^2向右移1單位變成(x-1)^2，再向下移2個單位變成(x-1)^2-2=x^2-2x-1\)
故選\(\bbox[red,2pt]{(ABCDE)}\)

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\(\bbox[blue,2pt]解：\)
(A)1-i為一根，則另一根為1+i
(B)(C)三次多項式已有2虛根，另一根為實根；2016+i非其實根，也不是1+i, 1-i，所以f(2016+i)不為0
(D)x=1-i\(\Rightarrow   x^2-2x+1=-1\Rightarrow   x^2-2x+2\)為其因式
(E)f(-1+2i)=\(-2+5i\Rightarrow   f(-1-2i)=\overline{f(-1-2i)}=\overline{-2+5i}=-2-5i\)

故選\(\bbox[red,2pt]{(AD)}\)

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\(\bbox[blue,2pt]解：\)
(A)正整數解的個數為\(H^4_{6-4}=C^5_2\)
(B)方法數為\(H^6_4=C^9_4\)
(C)\(C^9_6=C^9_3\)
(D)\( (1+x)^9=\sum _{ n=0 }^{ 9 }{ C^{ 9 }_{ n }x^{ n } } \Rightarrow  x^6係數=C^9_6=C^9_3\)
(E)\(H^4_6=C^9_6=C^9_3\)
故選\(\bbox[red,2pt]{(CDE)}\)

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\(\bbox[blue,2pt]解：\)
(A)\(\vec{a}\cdot\vec{a}=x_1^2+y_1^2\)
(E)\(|\vec{a}||\vec{b}| \ge |\vec{a}\cdot\vec{b}|\)
故選\(\bbox[red,2pt]{(BCD)}\)

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\(\bbox[blue,2pt]解：\)
(B)\(a^2=25\Rightarrow   a=5\Rightarrow   長軸長=2a=10\)
(C)\(b^2=16\Rightarrow  b=4\Rightarrow  短軸長=2b=8\)
(D)\(c^2=a^2-b^2=9\Rightarrow   c=3\)
故選\(\bbox[red,2pt]{(AE)}\)

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\(\bbox[blue,2pt]解：\)
(A)圓心(-2,0)    (B)半徑=2     (D)  (0,0)至(-2,0)的距離為2
故選\(\bbox[red,2pt]{(CE)}\)

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\(\bbox[blue,2pt]解：\)
(A)   舉一反例\(A=\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}\Rightarrow AA=\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}\)
(B)AB與BA不一定相等，所以此選項不正確
(C)\((AB)^2=ABAB\neq   A^2B^2\)
故選\(\bbox[red,2pt]{(DE)}\)

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\(\bbox[blue,2pt]解：\)
(A)不一定，只是有可能    (B)出現5次正面的機率=\(C^{10}_5\cdot   (\frac{1}{2})^{10}\)
(E)有可能，機率為\(\frac{1}{2^{10}}\)
故選\(\bbox[red,2pt]{(CD)}\)

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