109年大學指考(補考)數學甲詳解

109學年度指定科目考試試題(補考)
數學甲

選擇題： 共 40 題，總分 100 分，每題 2.5 分

解：

$$f(x)=\begin{cases} 1+x, & x\le 1\\ 1, & x> 1\end{cases} \Rightarrow \cases{\lim_{x\to 1^+}f(x)=1\\ \lim_{x\to 1^-}f(x)=2} \Rightarrow \lim_{x\to 1}f(x)不存在\\ g(x)=\begin{cases} 1, & x\le 1\\ 3-x, & x> 1\end{cases} \Rightarrow \cases{\lim_{x\to 1^+}g(x)=2\\ \lim_{x\to 1^-}g(x)=1} \Rightarrow \lim_{x\to 1}g(x)不存在 \\f(x)+g(x) =\begin{cases} 2+x, & x\le 1\\ 4-x, & x> 1\end{cases} \Rightarrow \cases{\lim_{x\to 1^+}f(x)+g(x)=3\\ \lim_{x\to 1^-}f(x)+g(x)=3} \Rightarrow \lim_{x\to 1}f(x)+g(x)存在\\，故選\bbox[red, 2pt]{(4)}$$

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解：

$$s(t)=\int_0^t (-x^2+6x)\;dx \Rightarrow s'(t)=-t^2+6t  \Rightarrow s''(t)=-2t+6 =-2(t-3)\\  \Rightarrow s''(t)= \cases{正值(加速), t< 3\\ 負值(減速),t> 3} \Rightarrow a=3，故選\bbox[red, 2pt]{( 1)}$$

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解：

$$\log_2(x-1)=\log_4(25-y^2) = \log_2 \sqrt{25-y^2} \Rightarrow (x-1)^2=25-y^2 \Rightarrow (x-1)^2+y^2=5^2\\ \Rightarrow 為一圓，其圓心O(1,0),半徑r=5 \Rightarrow (x,y)=(6,0),(5,\pm 3)(-3,\pm 3),(4,\pm 3)\\ 但x-1>0,即(-3,\pm 3)不合，因此符合要求的(x,y)共有5組，故選\bbox[red, 2pt]{(2 )}$$

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解：

$$(1)\bigcirc:\cases{M(A)=A'\\ M(B)=B'\\ M(O)=O} \Rightarrow \triangle A'B'O面積\ne 0 \Rightarrow det(M)\ne 0 \Rightarrow M可逆 \\(2)\bigcirc:  \overrightarrow{OC} = 2\overrightarrow{OA} + 3\overrightarrow{OB} \Rightarrow M(C)=2M(A)+3M(B) \Rightarrow C'=2A'+3B' \Rightarrow \overrightarrow{OC'} = 2\overrightarrow{OA'} + 3\overrightarrow{OB'}\\(3)\times: \cases{M=\left[\matrix{1& 1\\ 0 & 1}\right]\\ A(1,0)\\ B(0,1) \\ \angle AOB=90^\circ} \Rightarrow \cases{A'=M(A)=(1,0) \\ B'=M(B)=(1,1)} \Rightarrow \angle A'OB'=45^\circ \ne \angle AOB\\ (4)\times: 同上例\Rightarrow \cases{\overline{OA} =\overline{OB}=1\\ \overline{OA'}=1\\ \overline{OB'}=\sqrt 2} \Rightarrow {\overline{OA} \over \overline{OB}}=1 \ne {1\over \sqrt 2} = {\overline{OA'} \over \overline{OB'}} \\(5)\bigcirc: \cases{M=\left[\matrix{a& b\\ c & d}\right]\\ A(x_1,y_1)\\ B(x_2,y_2)} \Rightarrow \cases{A'=M(A)=(ax_1+by_1,cx_1+dy_1) \\ B'=M(B)=(ax_2+by_2,cx_2+dy_2)} \\ \Rightarrow \cases{\triangle OAB={1\over 2}\begin{Vmatrix}x_1 & x_2\\ y_1 & y_2 \end{Vmatrix}\\ \triangle OA'B'={1\over 2}\begin{Vmatrix}ax_1+by_1 & ax_2+by_2\\ cx_1+dy_1 & cx_2+dy_2 \end{Vmatrix}} \Rightarrow \begin{Vmatrix}ax_1+by_1 & ax_2+by_2\\ cx_1+dy_1 & cx_2+dy_2 \end{Vmatrix}=|ad-bc|\begin{Vmatrix}x_1 & x_2\\ y_1 & y_2 \end{Vmatrix} \\ \Rightarrow \triangle OAB = \triangle OA'B'\times |det(M)|\\，故選\bbox[red, 2pt]{(1,2,5 )}$$

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解：

$$\cos {\pi \over 7}+i\sin{\pi \over 7} =e^{\pi i/7} \\(1)\times: e^{\pi i/7}\cdot e^{\pi i/7}= e^{2\pi i/7} =\cos {2\pi \over 7}+ i\sin{2\pi \over 7}\Rightarrow 有虛部\\(2)\bigcirc: e^{\pi i/7} \cdot e^{-\pi i/7} =e^0=1\Rightarrow 只有實部 \\(3)\bigcirc: -\sin{5\pi \over 14}+i\cos {5\pi \over 14} =i(\cos {5\pi \over 14}+i\sin {5\pi \over 14})=e^{\pi i\over 2}\cdot e^{5\pi i/14} = e^{{12\pi i\over 14}} \\\qquad \Rightarrow e^{{12\pi i\over 14}}\cdot e^{\pi i/7} =e^{\pi  i }=-1\Rightarrow 只有實部\\ (4) \times:\sin{\pi \over 7}+ i\cos{\pi \over 7} =i(\cos {\pi \over 7}-i\sin{\pi \over 7}) =e^{\pi i/2}\cdot e^{-\pi i/ 7} = e^{5\pi i/14} \\\qquad \Rightarrow e^{5\pi i/14}\cdot e^{\pi i/7} =e^{\pi i/2}=i \Rightarrow 只有虛部 \\(5)\times: \sin{\pi \over 7}- i\cos{\pi \over 7} =-i(\cos {\pi \over 7}+i\sin{\pi \over 7}) =e^{-\pi i/2}\cdot e^{\pi i/ 7} =e^{-5\pi i/ 14}\\\qquad \Rightarrow e^{-5\pi i/14}\cdot e^{\pi i/7} = e^{-3\pi i/14}\Rightarrow 有虛部\\，故選\bbox[red, 2pt]{(2,3 )}$$

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解：

$$(1)\bigcirc: 第2次需擲出1，機率為1/6\\(2)\bigcirc: 兩次就停止的情況:a+b=7 \Rightarrow (a,b)=(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)\\\qquad ，共有6種，機率為6/36=1/6\\ (3)\times: 擲三次的情況(5,a,b),需滿足\cases{a+b=7\\ a\ne 2}，共有6-1=5種情況(與(2)同,但扣除(2,5))\\\qquad，機率為5/36\\(4)\times:(1-6,a,b)各有5種，共6\times 5=30種，機率為30/6^3 < 36/6^3 =1/6\\(5)\times: 1-(擲2次就停止的機率)=1-1/6=5/6\\ ，故選\bbox[red, 2pt]{(1,2 )}$$

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解：

$$(1)\bigcirc: 符合均值定理\\(2)\times: \cases{f(1)>0\\ f(1.5)<0 \\ f(2)>0} \Rightarrow 存在c\in (1,1.5)滿足f(c)=0\\ (3)\times: \cases{f(1)<0 \\ f(2)< 0 \\ f(3)< 0},不一定存在c\in (1,3)滿足f(c)=0\\ (4)\bigcirc: 令g(t)=\int_0^t f(x)\;dx \Rightarrow \left(\int_0^1 f(x)\;dx\right)\left(\int_0^2 f(x)\;dx\right) = g(1)g(2) < 0\\\qquad \Rightarrow 存在c\in (1,2)滿足g(c)= \int_0^c f(x)\;dx =0\\(5)\times: f(x)=\cases{4x-7,x\ge 3/2\\ -4x+5,x\le 3/2}，如下圖，滿足\int_1^2 f(x)\;dx=0,但f(1)=f(2)=1\ne 0\\，故選\bbox[red, 2pt]{(1,4 )}$$

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解：

$$(2)\bigcirc: 2^a,2^b,2^c成等差\Rightarrow 2^a+2^c=2\cdot 2^b \Rightarrow 2^{100}(2^a+2^c)=2^{100}\cdot 2\cdot 2^b \\\qquad \Rightarrow 2^{a+100}+ 2^{c+100}=2\cdot 2^{b+100} \Rightarrow 2^{a+100},2^{b+100},2^{c+100}成等差\\ (4) \bigcirc: 2^a+2^c=2\cdot 2^b =2^{b+1}\Rightarrow a< b+1(因為2^c > 0)\\(5)\bigcirc: 2^b={2^a+2^c\over 2} \ge \sqrt{2^a\cdot 2^c} =\sqrt{2^{a+c}} =2^{a+c\over 2} \Rightarrow b\ge {a+c\over 2}\\，故選\bbox[red, 2pt]{(2,4,5 )}$$

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解：

$$\begin{array}{}樣本& 機率 & 獎金 & 期望值\\\hline 二紅球 & C^4_2/C^{25}_2 & 450 & 450C^4_2/C^{25}_2 \\ 二藍球 & C^8_2 /C^{25}_2 & 450 & 450C^8_2/C^{25}_2 \\ 二白球 & C^{13}_2/C^{25}_2 & 450 &450C^{13}_2 /C^{25}_2\\\hdashline 一紅一藍& C^4_1C^8_1/C^{25}_2 & 75 & 75C^4_1C^8_1/C^{25}_2 \\一紅一白& C^4_1C^{13}_1/C^{25}_2 & 75 & 75C^4_1C^{13}_1/C^{25}_2 \\一藍一白& C^8_1C^{13}_1/C^{25}_2 & 75 & 75C^8_1C^{13}_1/C^{25}_2\\\hline \end{array} \\ \Rightarrow 期望值={450\over C^{25}_2}(C^4_2 +C^8_2+C^{13}_2) +{75\over C^{25}_2}(C^4_1C^8_1 +C^4_1C^{13}_1 +C^8_1C^{13}_1)\\ ={450\over 300}(6+28+78) +{75\over 300}(32+52+104)=168+47=\bbox[red, 2pt]{215}$$

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解：

$$令\cases{O為原點\\ A為圓心\\ B為切點} \Rightarrow \cases{O(0,0)\\ A(0,5k) \\ B(m/k,k)} \Rightarrow \overline{OA}^2 =\overline{AB}^2 +\overline{BO}^2 \Rightarrow 25k^2 = ({k^2\over m^2}+16k^2) + ({k^2\over m^2}+k^2) \\ \Rightarrow 8k^2 ={2\over m^2}k^2 \Rightarrow m^2={1\over 4} \Rightarrow m=\bbox[red, 2pt]{1\over 2} (負值違反m>0)$$

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解：

$$\cos \theta = {\overline{AB}^2 +\overline{AC}^2 -\overline{BC}^2 \over 2\times \overline{AB}\times \overline{AC}} ={\sin \theta +\sin \theta-\sin^2\theta \over 2\times \sqrt{\sin \theta}\times \sqrt{\sin \theta}} ={2\sin\theta -\sin^2\theta \over 2\sin \theta} =1-{1\over 2}\sin \theta\\ 又\cos^2\theta +\sin^2\theta =1 \Rightarrow (1-{1\over 2}\sin \theta)^2 +\sin^2\theta =1 \Rightarrow {5\over 4}\sin^2\theta-\sin \theta=0 \\\Rightarrow \sin\theta({5\over 4}\sin\theta-1)=0 \Rightarrow \sin \theta= {4\over 5} \Rightarrow \triangle ABC 面積= {1\over 2}\times \overline{AB}\times \overline{AC} \sin \theta \\ ={1\over 2}\sin^2 \theta ={1\over 2}\times {16\over 25} =\bbox[red, 2pt]{8\over 25}$$

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解：

(1)$$E之法向量\vec n=\vec u\times \vec v = (2,0,1)\times (0,1,1)=(-1,-2,2)，又(0,0,0)在E上 \\\Rightarrow E:-x-2y+2z=0 \Rightarrow x+2y-2z=0 \Rightarrow \bbox[red, 2pt]{\cases{b=2\\ c=-2 \\d=0}}$$(2)$$\cases{A(a_1,a_2,a_3) \\ E:x+2y-2z=0\\ \vec u=(2,0,1)} \Rightarrow \overline{AA'}直線方程式:{x-a_1\over 1} ={y-a_2\over 2} ={z-a_3\over -2}\\ \Rightarrow 線上的點可表示成(a_1+t,a_2+2t,a_3-2t)\\ \Rightarrow 求直線與E的交點: (a_1+t)+2(a_2+2t)-2(a_3-2t)=0 \Rightarrow t= (2a_3-2a_2-a_1)/9 \\ \Rightarrow A'=(a_1+{2a_3-2a_2-a_1\over 9},a_2+{4a_3-4a_2-2a_1\over 9},a_3+{-4a_3+4a_2+2a_1\over 9}) \\=( {2a_3-2a_2+8a_1\over 9},{4a_3+5a_2-2a_1\over 9},{5a_3+4a_2+2a_1\over 9})\\ 同理可求得B'=({2b_3-2b_2+8b_1\over 9},{4b_3+5b_2-2b_1\over 9},{5b_3+4b_2+2b_1\over 9})\\ 因此\cases{\overrightarrow{AB} =(b_1-a_1,b_2-a_2, b_3-a_3) \\ \overrightarrow{A'B'}= ({2(b_3-a_3)-2(b_2-a_2)+8(b_1-a_1)\over 9},{4(b_3-a_3)+5(b_2-a_2)-2(b_1-a_1)\over 9},{5(b_3-a_3)+4(b_2-a_2)+2(b_1-a_1)\over 9})} \\ \Rightarrow \cases{\overrightarrow{AB} \cdot \vec u=2(b_1-a_1)+(b_3-a_3) \\ \overrightarrow{A'B'}\cdot \vec u= {4(b_3-a_3)-4(b_2-a_2)+16(b_1-a_1)\over 9} +{5(b_3-a_3)+4(b_2-a_2)+2(b_1-a_1)\over 9}= (b_3-a_3)+2(b_1-a_1)} \\ \Rightarrow \overrightarrow{AB} \cdot \vec u=\overrightarrow{A'B'} \cdot \vec u\\ \bbox[blue,2pt] {另解}:\\\overrightarrow{A'B'} \cdot \vec u =(\overrightarrow{A'A}+\overrightarrow{AB}+\overrightarrow{BB'}) \cdot \vec u=\overrightarrow{A'A} \cdot \vec u+\overrightarrow{AB}\cdot \vec u +\overrightarrow{BB'} \cdot \vec u =\vec 0+ \overrightarrow{AB}\cdot \vec u +\vec 0= \overrightarrow{AB}\cdot \vec u\\，\bbox[red,2pt]{故得證} $$(3)$$\cases{\overrightarrow{A'B'}\cdot \vec u=\overrightarrow{AB}\cdot \vec u =5\\ (\alpha \vec u+\beta \vec v)\cdot \vec u =\alpha|\vec  u|^2+\beta( \vec u\cdot \vec v )=5\alpha+\beta} \Rightarrow 5\alpha+\beta =5 \cdots(1) \\\cases{\overrightarrow{A'B'}\cdot \vec v=\overrightarrow{AB}\cdot \vec v =2\\ (\alpha \vec u+\beta \vec v)\cdot \vec v =\alpha(\vec u\cdot \vec v)+\beta |\vec v|^2 =\alpha+2\beta} \Rightarrow \alpha+2\beta =2 \cdots(2)\\ 由(1)及(2) \Rightarrow \bbox[red, 2pt]{\cases{\alpha=8/9\\ \beta=5/9}}$$

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解：

(1)$$f(x)={1\over 3}f'(x)(x+k) \Rightarrow x^3+bx^2+cx+d = {1\over 3}(3x^2+2bx+c)(x+k) \\= x^3+(k+{2\over 3}b)x^2 +({2\over 3}bk+{1\over 3}c)x + {1\over 3}kc \Rightarrow \cases{k+{2\over 3}b=b\\ {2\over 3}bk+{1\over 3}c=c \\ {1\over 3}kc=d } \Rightarrow \cases{\bbox[red,2pt]{b=3k}\\ c=3k^2 \\d=k^3 }$$(2)$$f'(x)=3x^2+2bx+c=0 \Rightarrow 3x^2+6kx+3k^2=0 \Rightarrow 判別式36k^2-36k^2=0 \Rightarrow f'(x)=0有重根$$(3)$$f(-1)=0 \Rightarrow -1+b-c+d=0 \Rightarrow 3k-3k^2+k^3=1 \Rightarrow (k-1)^3=0 \Rightarrow k=1 \\ \Rightarrow f(x)=x^3+3x^2+3x+1\Rightarrow \int_0^1 f(x)\;dx = \left.\left[ {1\over 4}x^4 +x^3+{3\over 2}x^2+x\right] \right|_0^1 \\ ={1\over 4} +1+{3\over 2}+1 = \bbox[red, 2pt]{15\over 4}$$

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