62年大學聯考數學科試題(自)詳解

一、多重選擇題(占60分)

1. 令$z=cos\ \theta +\ i\ sin\ \theta$
   $\Rightarrow R+\ i\ S\ =\ \sum\limits_{k=1}^{2\ n}{{{z}^{k}}}=\displaystyle{\frac{1-{{z}^{2n}}}{1-z}}=\displaystyle{\frac{1-(cos\ 2\pi +i\ sin\ 2\pi )}{1-\left( cos\displaystyle{\frac{\pi }{n}}+i\ sin\displaystyle{\frac{\pi }{n}} \right)}}=0$
   $\Rightarrow R=S=0$，故選(A)(E)
   
   
2. $(1+x)y=x$$\Rightarrow$$xy-x+y=0$
   $\Rightarrow$$x(y-1)+1\cdot (y-1)=-1$
   $\Rightarrow$$(x+1)(y-1)=-1$為一雙曲線，其漸近線為$x+1=0$與$y-1=0$，則中心為$(-1，1)$
   故選(A)(B)(D)
   
   
3. $RS=\left[ \begin{matrix} 1 & 3 & 1 \\ 0 & -2 & 1 \\ 1 & 0 & 2 \\ \end{matrix} \right]\left[ \begin{matrix} -4 & x & 5 \\ 1 & 1 & -1 \\ 2 & 3 & -2 \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]$
   $\Rightarrow$$x=-6$
   因$RS=I$$\Rightarrow$$R={{S}^{-1}}$
   $\Rightarrow$$SR=I=RS$
   $\Rightarrow$${{R}^{2}}{{S}^{2}}=RRSS=RIS=RS=I$
   故選(B)(D)(E)
   
   
4. ${{n}^{4}}<{{10}^{6}}<{{(n+1)}^{4}}$ $\Rightarrow$ ${{n}^{2}}<{{10}^{3}}<{{(n+1)}^{2}}$ $\Rightarrow$ $n=31$、$32$、$33$、...，故選(D)(E)
   
   
5. $P={{\left( \displaystyle{\frac{1}{2}} \right)}^{5}}=\displaystyle{\frac{1}{32}}$$\Rightarrow$$E=\displaystyle{\frac{1}{32}}\cdot 5+\left( 1-\displaystyle{\frac{1}{32}} \right)\cdot S=0$$\Rightarrow$$S=\displaystyle{\frac{5}{31}}\approx 0.161$，故選(B)(E)
   
   
6. (A)  $sin\displaystyle{\frac{\pi }{4}}=cos\displaystyle{\frac{\pi }{4}}=\displaystyle{\frac{\sqrt{2}}{2}}$
   (B)  $sin\displaystyle{\frac{\pi }{8}}=+\sqrt{\displaystyle{\frac{1-cos\displaystyle{\frac{\pi }{4}}}{2}}}=\displaystyle{\frac{\sqrt{2-\sqrt{2}}}{2}}$
   (C)  $cos\displaystyle{\frac{\pi }{8}}=+\sqrt{\displaystyle{\frac{1+cos\displaystyle{\frac{\pi }{4}}}{2}}}=\sqrt{\displaystyle{\frac{2+\sqrt{2}}{2}}}$
   (D)  $sin\displaystyle{\frac{\pi }{16}}=\sqrt{\displaystyle{\frac{1-cos\displaystyle{\frac{\pi }{8}}}{2}}}=\sqrt{\displaystyle{\frac{2-\sqrt{2+\sqrt{2}}}{2}}}$
   (E)  $cos\displaystyle{\frac{\pi }{16}}=+\sqrt{\displaystyle{\frac{1+cos\displaystyle{\frac{\pi }{8}}}{2}}}=\sqrt{\displaystyle{\frac{2+\sqrt{2+\sqrt{2}}}{2}}}$
   故選(A)(B)(C)(D)(E)
   
   
7. 若${{a}_{n}}<2$$\Rightarrow$${{a}_{n+1}}=\sqrt{2+{{a}_{n}}}<\sqrt{2+2}=2$
   ${{a}_{n+1}}-{{a}_{n}}=\sqrt{{{a}_{n}}+2}-{{a}_{n}}=\displaystyle{\frac{{{a}_{n}}+a-{{a}_{n}}^{2}}{\sqrt{{{a}_{n}}+2}+{{a}_{n}}}}=\displaystyle{\frac{-({{a}_{n}}+1)({{a}_{n}}-2)}{\sqrt{{{a}_{n+2}}}+{{a}_{n}}}}>0$
   $\Rightarrow$$\left\langle {{a}_{n}} \right\rangle$遞增有界$\Rightarrow$$\underset{n\to \infty }{\mathop{lim}}\,{{a}_{n}}$存在，令$\underset{n\to \infty }{\mathop{lim}}\,{{a}_{n}}=L$，則$L=\sqrt{L+2}$$\Rightarrow$$L=2$
   ${{b}_{n}}-{{b}_{n+1}}=\sqrt{2-{{a}_{n-1}}}-\sqrt{2-{{a}_{n}}}=\displaystyle{\frac{-{{a}_{n-1}}+{{a}_{n}}}{\sqrt{2-{{a}_{n-1}}}+\sqrt{2-{{a}_{n}}}}}>0$
   $\Rightarrow$${{b}_{n}}$遞減
   $2cos\displaystyle{\frac{\pi }{{{2}^{n}}}}=\sqrt{2+2cos\displaystyle{\frac{\pi }{{{2}^{n-1}}}}}$$\Rightarrow$可令$2cos\displaystyle{\frac{\pi }{{{2}^{n}}}}={{a}_{n}}$且與上述${{a}_{n}}$相符
   $\Rightarrow$$\underset{n\to \infty }{\mathop{lim}}\,cos\displaystyle{\frac{\pi }{{{2}^{n}}}}=\underset{n\to \infty }{\mathop{lim}}\,\displaystyle{\frac{{{a}_{n}}}{2}}=\displaystyle{\frac{2}{2}}=1$
   故選(A)(B)(D)(E)
   
   
8. (A)  正確
   (B)  錯誤，$g$在$(\displaystyle{\frac{-\pi }{2}}$，$\displaystyle{\frac{\pi }{2}})$上的反函數是$f$
   (C)  錯誤，$g$在$\{x|x=\displaystyle{\frac{\pi }{2}}+k\pi,k\in$整數$\}$上沒有定義。
   (D)  正確
   (E)  錯誤，$f(g(x))=x+k\pi,k\in$整數，其中$x+k\pi\in(\displaystyle{\frac{-\pi }{2}}$，$\displaystyle{\frac{\pi }{2}})$，故選(A)(D)
   
   
9. 將$x=1$、$x=-2$、$y=0$、$y=3$、$xy+1=y$、$y=3-x^2$解交點座標並在坐標平面上作圖得
   其中$A(-2$，$3)$、$B(0$，$3)$、$C(\displaystyle{\frac{\sqrt{5}-1}{2}}$，$\displaystyle{\frac{3+\sqrt{5}}{2}})$、$D(1$，$3)$、$E(-2$，$0)$、$(\displaystyle{\frac{-1-\sqrt{5}}{2}}$，$\displaystyle{\frac{3-\sqrt{5}}{2}})$、$G(1$，$0)$、$H(2$，$-1)$
   分別取交集得到以下圖形：
   $P$：
   
   
   
   $Q$：
   
   
   
   $R$：
   
   
   
   $S$：
   
   
   
   $T$：
   
   
   
   $U$：
   
   
   
   $V=R\cap P$：
   
   故(A)錯誤、(B)(C)(D)正確，又因$V$都在$Q$內部，故面積$<9$，(E)正確。
   故選(B)(C)(D)(E)
   
   
10. $\theta =lo{{g}_{\frac{3}{2}}}4=\displaystyle{\frac{log\ 4}{log\ \displaystyle{\frac{3}{2}}}}=\displaystyle{\frac{log\ 4}{log\ 3-log\ 2}} \doteqdot \displaystyle{\frac{0.6020}{0.4771-0.3010}}\doteqdot 3.42$，故選(D)(E)
    
    
11. $\overline{OP}=\sqrt{5}<3$；$P$恰過直線$y-2=m(x+1)$，且$\overline{OP}$斜率為$-2$$\Rightarrow$$m=\displaystyle{\frac{1}{2}}$，故選(B)(C)
    
    
12. $S$：$\left[ \ (\ x\ +\ y\ )\ +\ 1\ \right]\left[ \ (\ x\ +\ y\ )\ +\ 3\ \right]\le 0$$\Rightarrow$兩平行線間
    $T$：${{\left( x+1 \right)}^{2}}+{{\left( y+1 \right)}^{2}}\le 1$$\Rightarrow$圓內
    $\Rightarrow$$R=S\cap T$如圖：
    
    故$R$面積為「圓面積」$-$「$2\times \text{(}\displaystyle{\frac{1}{4}}\text{弓形)}$」$=\pi -2\cdot \left(\displaystyle{\frac{1}{4}}\pi -\displaystyle{\frac{1}{2}} \right)= \displaystyle{\frac{\pi }{2}}+1\doteqdot 2.57$，故選(E)
    
    
13. 由行列式得${{\left( x-cos\ \theta \right)}^{2}}+si{{n}^{2}}\theta =0$$\Rightarrow$${{x}^{2}}-2\cdot cos\ \theta \ x+1=0$$\Rightarrow$可令
    $\left\{ \begin{aligned} & \alpha =cos\ \theta +i\ sin\ \theta \\ & \beta =cos\ \theta -i\ sin\ \theta =\bar{\alpha } \\ \end{aligned} \right.$$\Rightarrow$${{\alpha }^{n}}+{{\beta }^{n}}={{\alpha }^{n}}+{{\left( {\bar{\alpha }} \right)}^{n}}=2\ cos\ n\theta$，故選(D)
    
    
14. 由$log\ {{g}_{n}}<log\ {{10}^{-3}}$$\Rightarrow$ $log\ {{g}_{n}}<-3$$\Rightarrow$ $log\ 400+n\left( log\ 3-log\ 4 \right)<-3$ $\Rightarrow$$2.602+n\left( 0.4771-0.602 \right)<-3$ $\Rightarrow$ $n>44.85$，故選(A)
    
    
15. 作$(0$，$1)$對於$x$軸的對稱點得到$(0$，$-1)$，如圖：
    
    再由中垂線性質可知：$(4$，$5)$到$(0$，$-1)$之距離即「$(0$，$1)$到$(x$，$0)$之距離$+$$(x$，$0)$到$(4$，$5)$之距離」，又因三角不等式，可得該最小距離為$(0$，$-1)$與$(4$，$5)$距離$\Rightarrow$$(0$，$-1)$與$(4$，$5)$之直線方程式為$3x-2y=2$，與$x$軸交於$\left( \displaystyle{\frac{2}{3}}，0 \right)$$\Rightarrow$$x=\displaystyle{\frac{2}{3}}$，故選(C)
    
    
16. 令$\displaystyle{\frac{3x+1}{x-3}}=t$ $\Rightarrow$ $t+5\cdot \displaystyle{\frac{1}{t}}-6=0$ $\Rightarrow$ ${{t}^{2}}-6t+5=0$ $\Rightarrow$0$\left( t-1 \right)\left( t-5 \right)=0$ $\Rightarrow$ $t=1，5$ $\Rightarrow$ $\displaystyle{\frac{3x+1}{x-3}}=1$，$5$ $\Rightarrow$ $x=-2$，$8$ $\Rightarrow$ $x$的解集合為$\{-2$，$8\}$
    故選(B)
    
    
17. 令$x=\sqrt{2}+i\sqrt{3}$ $\Rightarrow$ $x-\sqrt{2}=i\sqrt{3}$ $\Rightarrow$ ${{x}^{2}}-2\sqrt{2}x+2=-3$ $\Rightarrow$ ${{x}^{2}}+5=2\sqrt{2}x$ $\Rightarrow$ ${{x}^{4}}+10{{x}^{2}}+25=8{{x}^{2}}$ $\Rightarrow$ ${{x}^{4}}+2{{x}^{2}}+25=0$ $\Rightarrow$ 選(B)
    
    
18. $\left( 7-2 \right)\times 180{}^\circ =900{}^\circ$，故選(D)
    
    
19. $xy$的$n$次式必以${{x}^{a}}{{y}^{b}}$項組成，其中$0\le a+b\le n$ $\Rightarrow$ $0\le a+b\le n$的非負整數解為$H_{n}^{3}=C_{n}^{n+2}=C_{2}^{n+2}=\displaystyle{\frac{\left( n+1 \right)\left( n+2 \right)}{2}}$，故選(E)
    
    
20. 化簡$\sqrt{1+sin\ \alpha }-\sqrt{1-sin\ \alpha }$ $=\sqrt{co{{s}^{2}}\displaystyle{\frac{\alpha }{2}}+2\cdot sin\displaystyle{\frac{\alpha }{2}}\cdot cos\displaystyle{\frac{\alpha }{2}}+si{{n}^{2}}\displaystyle{\frac{\alpha }{2}}}-\sqrt{co{{s}^{2}}\displaystyle{\frac{\alpha }{2}}-2\cdot sin\displaystyle{\frac{\alpha }{2}}\cdot cos\displaystyle{\frac{\alpha }{2}}+si{{n}^{2}}\displaystyle{\frac{\alpha }{2}}}$ $=\sqrt{{{\left( cos\displaystyle{\frac{\alpha }{2}}+sin\displaystyle{\frac{\alpha }{2}} \right)}^{2}}}-\sqrt{{{\left( cos\displaystyle{\frac{\alpha }{2}}-sin\displaystyle{\frac{\alpha }{2}} \right)}^{2}}}$ $=\left| cos\displaystyle{\frac{\alpha }{2}}+sin\displaystyle{\frac{\alpha }{2}} \right|-\left| cos\displaystyle{\frac{\alpha }{2}}-sin\displaystyle{\frac{\alpha }{2}} \right|$
    因$0<\alpha <\displaystyle{\frac{\pi }{2}}$ $\Rightarrow$ $0<\alpha <\displaystyle{\frac{\pi }{4}}$ $\Rightarrow$ $cos\displaystyle{\frac{\pi }{4}}>sin\displaystyle{\frac{\pi }{4}}>0$ $\Rightarrow$ $\left| cos\displaystyle{\frac{\alpha }{2}}+sin\displaystyle{\frac{\alpha }{2}} \right|-\left| cos\displaystyle{\frac{\alpha }{2}}-sin\displaystyle{\frac{\alpha }{2}} \right|$ $=\left( cos\displaystyle{\frac{\alpha }{2}}+sin\displaystyle{\frac{\alpha }{2}} \right)-\left( cos\displaystyle{\frac{\alpha }{2}}-sin\displaystyle{\frac{\alpha }{2}} \right)$ $=2sin\displaystyle{\frac{\alpha }{2}}$，故選(A)
    
    
21. 在平面上作$x=2$、$x=5$、$x+y=8$、$x+3y=5$的圖形，並將$2\le x\le 5$、$x+y\le 8$、$x+3y\le 5$取交集得到下圖，其中出現四個交點$\left( 2，6 \right)$、$\left( 2，1 \right)$、$\left( 5，3 \right)$、$\left( 5，0 \right)$
    
    由頂點法可得：將$\left( 5，3 \right)$代入會有極大值$16$，故選(D)
    
    
22. 由$t=cos\ 2\theta =co{{s}^{2}}\theta -si{{n}^{2}}\theta$得$4\left( co{{s}^{6}}\theta -si{{n}^{6}}\theta \right)$ $=4\left( co{{s}^{2}}\theta -si{{n}^{2}}\theta \right)\left( co{{s}^{4}}\theta +co{{s}^{2}}\theta \cdot si{{n}^{2}}\theta +si{{n}^{4}}\theta \right)$ $=4t\left[ {{\left( co{{s}^{2}}\theta +si{{n}^{2}}\theta \right)}^{2}}-co{{s}^{2}}\theta \cdot si{{n}^{2}}\theta \right]$ $=4t\left( {{1}^{2}}-\displaystyle{\frac{1}{4}}\cdot 4\cdot co{{s}^{2}}\theta \cdot si{{n}^{2}}\theta \right)$ $=4t\left[ 1-\displaystyle{\frac{1}{4}}{{\left( sin\ 2\theta \right)}^{2}} \right]$ $=4t\left[ 1-\displaystyle{\frac{1}{4}}\left( 1-co{{s}^{2}}2\theta \right) \right]$ $=4t-t+{{t}^{3}}$ $={{t}^{3}}+3t$，故選(A)