108學年度指定科目考試試題
數學甲
第壹部分:選擇題一、單選題
解:$$丟第一次的期望值:800\times \frac{1}{4}+800\times\frac{1}{2}+400\times\frac{1}{4}= 200+400+100= 700\\丟第二次的期望值也是700,但需在第一次出現兩個正面(機率為1/4),\\也就是第二次的期望值為700\times \frac{1}{4}=175\\總期望值為700+175=875,故選\bbox[red,2pt]{(2)}$$
$$\frac{F_{13}}{F_{12}}=\frac{2^{2^{13}}+1}{2^{2^{12}}+1}\Rightarrow \log{\frac{F_{13}}{F_{12}}} =\log{\frac{2^{2^{13}}+1}{2^{2^{12}}+1}}\approx \log{\frac{2^{2^{13}}}{2^{2^{12}}}} =2^{13}\log{2}-2^{12}\log{2}\\ =\left(2^{13}-2^{12}\right) \log{2}= 2^{12}\log{2}=2^{12}\times 0.301=4096\times 0.301=1232.896,故選\bbox[red,2pt]{(5)}。$$
$$令塔高為h,A至塔底的距離為a=h\cot{14^o},B至塔底的距離為b=h\cot{18^o30'}\\直角\triangle ABC\Rightarrow a^2+b^2=65^2\Rightarrow h^2(\cot^2{14^o}+\cot^2{18^o30'} =65^2 \Rightarrow h^2(4.01^2+2.99^2)\\\approx h^2(4^2+3^2)=25h^2=65^2 \Rightarrow h=13\\ 離塔底越近仰角越大,C至直線\overline{AB}的距離x即為所求,即65x=ab\\ \Rightarrow 65x=h^2(4.01\times 2.99)\approx h^2(4\times 3)=13^2\times 12 \Rightarrow x=13\times 12\div 5=31.2,故選\bbox[red,2pt]{(3)}$$
二、多選題
若取出第1球是紅球,第2球取出是紅球的機率為1/5、若取出第1球不是紅球,第2球取出是紅球的機率為2/5,因此取出第二球為紅球的機率為\(\frac{1}{3}\times\frac{1}{5}+\frac{2}{3}\times\frac{2}{5}=5/15=1/3\)
(2)\(\times:\)第2次取出紅球的機與第1次是否取到紅球有相關,兩者不是獨立事件
(3)\(\times:\)取出第一顆為紅球,第二顆也能取白球或藍球,所以兩者不互斥
(4)\(\times:\)第一、二顆皆為紅球的機率為\(\frac{2}{6}\times\frac{1}{5}=\frac{1}{15}\);第一、二顆皆為白球的機率為\(\frac{3}{6}\times\frac{2}{5}=\frac{1}{5}\);兩者不同
(5)\(\bigcirc:\)前三顆皆為白球的機率\(\frac{3}{6}\times\frac{2}{5}\times\frac{1}{4}=\frac{1}{20}\);前三顆為異色的機率\(3!\times\frac{3\times 2\times 1}{6\times 5\times 4}=\frac{3}{10}>\frac{1}{20}\);
故選\(\bbox[red,2pt]{(1,5)}\)
解:
(1)\(\times\): 圖形可能如下,其\(a<0\)
(2)\(\bigcirc\):圖形在\(-2\le x\le 1\)內,斜率是遞增的,即\(f''>0\Rightarrow f''(0)>0\)
(3)\(\bigcirc\):圖形在\(x=0\),斜率是正數,即\(f'(0)>0\Rightarrow c>0\)
(4)\(\times\): 上圖只有一實根
(5)\(\bigcirc\):反曲點在上圖的A點處,其Y坐標為正;若a>0,反曲點仍在X軸之上
(2)\(\bigcirc\):圖形在\(-2\le x\le 1\)內,斜率是遞增的,即\(f''>0\Rightarrow f''(0)>0\)
(4)\(\times\): 上圖只有一實根
(5)\(\bigcirc\):反曲點在上圖的A點處,其Y坐標為正;若a>0,反曲點仍在X軸之上
故選\(\bbox[red,2pt]{(2,3,5)}\)
解:$$(1)\bigcirc :\left| 2\overrightarrow { OA } +3\overrightarrow { OB } \right| =\left| -4\overrightarrow { OC } \right| =4\left| \overrightarrow { OC } \right| =4\\ (2)\times :\overrightarrow { OA } \cdot \overrightarrow { OB } =\left| \overrightarrow { OA } \right| \left| \overrightarrow { OB } \right| \cos { \angle AOB } =\cos { \angle AOB } 可能為正\\ (3)\times :2\overrightarrow { OA } +3\overrightarrow { OB } +4\overrightarrow { OC } =0\Rightarrow 2\overrightarrow { OA } +3\overrightarrow { OB } =-4\overrightarrow { OC } \Rightarrow { \left( 2\overrightarrow { OA } +3\overrightarrow { OB } \right) }^{ 2 }=16{ \left| \overrightarrow { OC } \right| }^{ 2 }\\ \Rightarrow 4{ \left| \overrightarrow { OA } \right| }^{ 2 }+9{ \left| \overrightarrow { OB } \right| }^{ 2 }+12\overrightarrow { OA } \cdot \overrightarrow { OB } =16{ \left| \overrightarrow { OC } \right| }^{ 2 }\Rightarrow 13+12\overrightarrow { OA } \cdot \overrightarrow { OB } =16\\ \Rightarrow \overrightarrow { OA } \cdot \overrightarrow { OB } =\frac { 1 }{ 4 } =\left| \overrightarrow { OA } \right| \left| \overrightarrow { OB } \right| \cos { \angle AOB } =\cos { \angle AOB } \Rightarrow \cos { \angle AOB } =\frac { 1 }{ 4 } \\ 同理可得\cos { \angle BOC } =-\frac { 7 }{ 8 } ,\cos { \angle AOC } =-\frac { 11 }{ 16 } \Rightarrow \angle BOC>\angle AOC>\angle AOB\\ (4)\times :\cos { \angle AOB } =\frac { { \overline { OA } }^{ 2 }+{ \overline { OB } }^{ 2 }-{ \overline { AB } }^{ 2 } }{ 2\cdot { \overline { OA } }\cdot { \overline { OB } } } \Rightarrow \frac { 1 }{ 4 } =\frac { 2-{ \overline { AB } }^{ 2 } }{ 2 } \Rightarrow { \overline { AB } }^{ 2 }=\frac { 3 }{ 2 } \Rightarrow { \overline { AB } }<\frac { 3 }{ 2 } \\ (5)\bigcirc :\begin{cases} \cos { \angle AOB } =\frac { 1 }{ 4 } \\ \cos { \angle AOC } =-\frac { 11 }{ 16 } \end{cases}\Rightarrow \begin{cases} \sin { \angle AOB } =\frac { \sqrt { 15 } }{ 4 } \\ \sin { \angle AOC } =\frac { 3\sqrt { 15 } }{ 16 } \end{cases}\Rightarrow 3\sin { \angle AOB } =\frac { 3\sqrt { 15 } }{ 4 } =4\sin { \angle AOC } $$故選\(\bbox[red,2pt]{(1,5)}\)
三、選填題
解:
$$\begin{cases} A,B在y=\log _{ 2 }{ x } 上 \\ A,B連線斜率為2 \\ \overline { AB } =\sqrt { 5 } \end{cases}\Rightarrow \begin{cases} r=\log _{ 2 }{ a } \\ s=\log _{ 2 }{ b } \\ \frac { s-r }{ b-a } =2 \\ \sqrt { { (a-b) }^{ 2 }+{ (r-s) }^{ 2 } } =\sqrt { 5 } \end{cases}\Rightarrow \begin{cases} \frac { \log _{ 2 }{ b } -\log _{ 2 }{ a } }{ b-a } =2 \\ { (a-b) }^{ 2 }+{ (\log _{ 2 }{ a } -\log _{ 2 }{ b } ) }^{ 2 }=5 \end{cases}\\ \Rightarrow { (a-b) }^{ 2 }+{ (2(a-b)) }^{ 2 }=5\Rightarrow 5{ (a-b) }^{ 2 }=5\Rightarrow \begin{cases} a-b=1 \\ a-b=-1 \end{cases}\Rightarrow \begin{cases} \log _{ 2 }{ b } -\log _{ 2 }{ a } =-2 \\ \log _{ 2 }{ b } -\log _{ 2 }{ a } =2 \end{cases}\\\Rightarrow \begin{cases} \log _{ 2 }{ \left( b/a \right) } =-2 \\ \log _{ 2 }{ \left( b/a \right) } =2 \end{cases} \Rightarrow \begin{cases} b/a=1/4 \\ b/a=4 \end{cases}\Rightarrow \begin{cases} b=a/4 \\ b=4a \end{cases}\Rightarrow \begin{cases} a-b=a-a/4=3a/4=1 \\ a-b=a-4a=-3a=-1 \end{cases}\\ \Rightarrow \begin{cases} a=4/3 \\ a=1/3 \end{cases}\Rightarrow \begin{cases} b=1/3(不符b>a) \\ b=4/3 \end{cases} \Rightarrow (a,b)=\left( \bbox[red,2pt]{\frac { 1 }{ 3 } ,\frac { 4 }{ 3 }} \right) $$
解:
第貳部分:非選擇題
解:
(1) $$\overrightarrow{OA}\cdot\overrightarrow {OP}= \left|\overrightarrow{OA}\right|\left|\overrightarrow{OP}\right|\cos{\angle AOP}=\sqrt{1+2+1}\cdot 2\cdot \frac{1}{2}=\bbox[red,2pt]{2}$$
(2)$$假設P=(x,y,z) \Rightarrow \overrightarrow{OP}\cdot \overrightarrow{OA}=2 \Rightarrow (x,y,z) \cdot (1,\sqrt{2},1) =x+\sqrt{2}y+z=2 \\\Rightarrow E:\bbox[red,2pt]{x+\sqrt{2}y+z=2}$$
(3)$$令\vec{u}=\overrightarrow{OA}\times\overrightarrow{OB} =(1,\sqrt{2},1)\times (2,0,0)=(0,2,-2\sqrt{2})\\ \Rightarrow 該直線方向向量即為\vec{u}=\bbox[red,2pt]{(0,2,-2\sqrt{2})}$$
(4)$$Q=(x,y,z)\Rightarrow \begin{cases} \overrightarrow{OQ}\cdot \overrightarrow{OA}=2\cdot 2\cdot \cos{60^o}=2 \\ \overrightarrow{OQ}\cdot \overrightarrow{OB}=2\end{cases} \Rightarrow \begin{cases} x+\sqrt{2}y+z=2 \\ 2x=2 \Rightarrow x=1\end{cases}\\ 任找一點符合以上兩式,如(1,0,1),再由直線向量(0,2,-2\sqrt{2})可得Q=(1,2t,-2\sqrt{2}t+1)\\ \Rightarrow |\overrightarrow{OQ}|=2\Rightarrow \sqrt{1+4t^2+1-4\sqrt{2}t+8t^2}=2 \Rightarrow 12t^2-4\sqrt{2}t-2=0\\ \Rightarrow \begin{cases} t={\sqrt{2}\over 2}\\t={-\sqrt{2}\over 6}\end{cases}\Rightarrow \bbox[red,2pt]{\begin{cases} Q=(1,\sqrt{2},-1)\\Q=(1,-{\sqrt{2}\over 3},{5\over 3})\end{cases}}$$
解:
(1) $$xf(x)=3x^4-2x^3+x^2+\int_1^x{f(t)\,dt}\Rightarrow 1\cdot f(1)=3-2+1+0 \Rightarrow f(1)=\bbox[red,2pt]{2}$$(2)$$xf(x)=3x^4-2x^3+x^2+\int_1^x{f(t)\,dt}\Rightarrow f(x)+xf'(x)=12x^3-6x^2+2x+f(x)\\\Rightarrow f'(x)=(12x^3-6x^2+2x)/x =\bbox[red,2pt]{12x^2-6x+2}$$(3)$$f(x)=\int{f'(x)\,dx}=\int{12x^2-6x+2\,dx}=4x^3-3x^2+2x+C\\ 由f(1)=2\Rightarrow 4-3+2+C=2\Rightarrow C=-1 \Rightarrow f(x)=\bbox[red,2pt]{4x^3-3x^2+2x-1}$$(4)$$令g(a)=\int_0^a{f(x)\,dx}= \int_0^a{4x^3-3x^2+2x-1\,dx}=\left.\left[x^4-x^3+x^2-x\right]\right|_0^a\\ =a^4-a^3+a^2-a=a^3(a-1)+a(a-1)=(a^3+1)(a-1)\Rightarrow g(1)=0\\ \Rightarrow g'(a)=f(a)=4a^3-3a^2+2a-1=4a\left(a^2-\frac{3}{4}a+\frac{1}{2}\right)-1\\=4a\left[\left(a-\frac{3}{8}\right)^2+\frac{1}{2}-\frac{9}{64}\right]-1 = 4a\left[\left(a-\frac{3}{8}\right)^2+\frac{23}{64}\right]-1\\\Rightarrow g'(1)=4\left(\frac{48}{64}\right)-1>0且g'(a)>0 對a\ge 1\\ 也就是說對a\ge 1,g(a)為嚴格遞增,即 g(m)< g(n), for\; 1\le m< n\\ 因此,由\begin{cases}g(1)=0\\g(a)為連續且嚴格遞增,對a\ge 1\end{cases}\Rightarrow 存在唯一的a> 1,使用g(a)=1,即\int_0^a{fx)\,dx}=1$$
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第六題的選項5舉例那邊的 (b_{n})^{2}=4-\frac{1}{n+0.5}
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1→4, 已修訂,謝謝!
刪除非選一 (4) 倒數第三個"⇒"後面的6應為12
回覆刪除謝謝提醒,已修訂
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