[段考] 108上第1次段考-台中-曉明女中-高一(詳解)

108上第1次段考-台中-曉明女中-高一(詳解)

範圍：第一冊 龍騰單元4~單元7

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一、單一選擇題：(每題5分，共10分)

1. $(-3$，$-2)$與圓$C$圓心$(6$，$7)$距$=\sqrt{{{[6-(-3)]}^{2}}+{{[7-(-2)]}^{2}}}$$=\sqrt{182}$$=13.\cdots \cdots$$\Rightarrow$$(-3$，$-2)$到圓最近$=\sqrt{182}-r$，最遠$=\sqrt{182}+r$$\Rightarrow$$9.\cdots \cdots <$距離$<17.\cdots \cdots$，整數$\Rightarrow$$10\sim17$，共8種，兩側皆有$\Rightarrow$$8\times 2=16$，選(5)
   
   
   
   
   
2. $a={{10}^{21}}$，$b={{10}^{22}}$，$c={{10}^{23}}$$\Rightarrow$$log(a+b+c)$$=log[{{10}^{21}}(1+10+100)]$$=log{{10}^{23}}\times 1.21\approx 23$，選(3)
   
   

二、多重選擇題：(每題7分，錯一選項得4分，錯兩選項得2分，共28分)

1. ${{m}_{{{L}_{1}}}}>0$$\Rightarrow$$a>0$，${{m}_{{{L}_{2}}}}$、${{m}_{{{L}_{3}}}}<0$$\Rightarrow$$c<0$，$e<0$，${{L}_{1}}$之$y$截距$<0$$\Rightarrow$$b<0$，${{L}_{2}}$過原點$\Rightarrow$$d=0$，${{L}_{3}}$之$y$截距$=f$$\Rightarrow$$f>0$$\Rightarrow$(1)正確，(2)(3)錯誤。又${{m}_{{{L}_{1}}}}<1$、${{m}_{{{L}_{2}}}}<-1$$\Rightarrow$$\left| c \right|>\left| a \right|$。又${{m}_{{{L}_{2}}}}<{{m}_{{{L}_{3}}}}<-1$$\Rightarrow$最小，(5)正確，故選(1)(5)
   
   
2. (1)  $(7$，$-8)$、$(1$，$2)$皆過$5x+3y=11$，且$(13$，$-18)$亦過$5x+3y=11$$\Rightarrow$三點共線，無法作圓，不選。
   (2)  經配方，${{(x+2)}^{2}}+{{(y-4)}^{2}}=-60$，無圖形，不選。
   (3)  作圖觀察知有兩圓。
   
   
   
   (4)  令$P(x$，$y)$$\Rightarrow$$2\overline{PA}=3\overline{PB}$知$2\sqrt{{{(x+3)}^{2}}+{{(y-7)}^{2}}}=3\sqrt{{{(x-2)}^{2}}+{{(y-5)}^{2}}}$$\Rightarrow$$4{{x}^{2}}+24x+36+4{{y}^{2}}-28y+441=9{{x}^{2}}-36x+36+9{{y}^{2}}-90y+225$$\Rightarrow$$5{{x}^{2}}+5{{y}^{2}}-60x-62y-216=0$$\Rightarrow {{x}^{2}}+{{y}^{2}}-12x-\displaystyle{\frac{62}{5}}y-\displaystyle{\frac{216}{5}}=0$$\Rightarrow {{(x-6)}^{2}}+{{(y-\displaystyle{\frac{31}{5}})}^{2}}=\displaystyle{\frac{21}{5}}+36+\displaystyle{\frac{901}{25}}$$\Rightarrow$圓
   (5)  $y-2=\sqrt{9-{{x}^{2}}}$$\Rightarrow$${{(y-2)}^{2}}=9-{{x}^{2}}$$\Rightarrow$${{x}^{2}}+{{(y-2)}^{2}}=9$$\Rightarrow$又$y-2>0$$\Rightarrow$上半圓
   
   
3. 作不等式範圍並解出交點如下，
   
   
   
   (1)  正確
   (2)  面積$\displaystyle{\frac{1}{2}}\cdot 4\cdot 4=8$，正確
   (3)  $y=-1$$\Rightarrow$$\left\{ \begin{array}{l} 2x-(-1)+1\ge 0 \\ 2x+(-1)-5\le 0 \\ \end{array} \right.$$\Rightarrow$$3\ge x\ge -1$$\Rightarrow$$x=-1$，$0$，$1$，$2$，$3$
   $y=0$$\Rightarrow$$\left\{ \begin{array}{l} 2x+1\ge 0 \\ 2x-5\le 0 \\ \end{array} \right.$$\Rightarrow$$\displaystyle{\frac{5}{2}}\ge x\ge -\displaystyle{\frac{1}{2}}$$\Rightarrow$$x=0$，$1$，$2$
   $y=1$$\Rightarrow$$\left\{ \begin{array}{l} 2x-1+1\ge 0 \\ 2x+1-5\le 0 \\ \end{array} \right.$$\Rightarrow$$2\ge x\ge 0$$\Rightarrow$$x=0$，$1$，$2$
   $y=2$$\Rightarrow$$\left\{ \begin{array}{l} 2x-2+1\ge 0 \\ 2x+2-5\le 0 \\ \end{array} \right.$$\Rightarrow$$\displaystyle{\frac{3}{2}}\ge x\ge \displaystyle{\frac{1}{2}}$$\Rightarrow$$x=1$
   $y=3$$\Rightarrow$$\left\{ \begin{array}{l} 2x-3+1\ge 0 \\ 2x+3-5\le 0 \\ \end{array} \right.$$\Rightarrow$$1\ge x\ge 1$$\Rightarrow$$x=1$
   $\Rightarrow$共$5+3+3+1+1=13$個
   (4)  ${{(x-3)}^{2}}+{{(y-2)}^{2}}$即$(x$，$y)$到$(3$，$2)$之距離平方，即$(3$，$2)$到$2x+y-5=0$之距離平方$={{(\displaystyle{\frac{\left| 6+2-5 \right|}{\sqrt{5}}})}^{2}}=\displaystyle{\frac{9}{5}}$
   (5)  即$(x$，$y)$到$(4$，$2)$之斜率最大，即$(4$，$2)$到頂點$(3$，$-1)$之斜率$=\displaystyle{\frac{2-(-1)}{4-1}}=\displaystyle{\frac{3}{3}}=1$
   
   
4. (1)  $1.253\times {{10}^{845}}$與$1.254\times {{10}^{845}}$皆為$846$位數$\Rightarrow$${{7}^{1000}}$為$846$位數，故(1)不選
   (2)  ${{49}^{50}}={{7}^{100}}$$\Rightarrow$${{(1.253\times {{10}^{845}})}^{\frac{1}{10}}}<{{7}^{100}}<{{(1.254\times {{10}^{845}})}^{\frac{1}{10}}}$$\Rightarrow$$1.\cdots \cdots \times {{10}^{84.5}}<{{7}^{100}}<1.\cdots \cdots \times {{10}^{84.5}}$$\Rightarrow$${{7}^{100}}$為$85$位數
   (3)  ${{(1.253\times {{10}^{845}})}^{\frac{1}{1000}}}<7<{{(1.254\times {{10}^{845}})}^{\frac{1}{1000}}}$$\Rightarrow$$1.\cdots \cdots \times {{10}^{0.845}}<7<1.\cdots \cdots \times {{10}^{0.845}}$$\Rightarrow$$0.845<log7<0.846$$\Rightarrow$${{(1.253\times {{10}^{845}})}^{-1}}>{{7}^{-1000}}>{{(1.254\times {{10}^{845}})}^{-1}}$$\Rightarrow$$\displaystyle{\frac{1}{1.253}}\times {{10}^{-845}}>{{7}^{-1000}}>\displaystyle{\frac{1}{1.254}}\times {{10}^{-845}}$$\Rightarrow$$0.8\cdots \times {{10}^{-845}}<{{7}^{-1000}}<0.8\cdots \times {{10}^{-845}}$$\Rightarrow$$8\times {{10}^{-846}}<{{7}^{-1000}}<9\times {{10}^{-846}}$$\Rightarrow$$846$位數
   (5)  ${{49}^{-100}}={{7}^{-200}}$$\Rightarrow$${{(1.253\times {{10}^{845}})}^{-\frac{1}{5}}}>{{7}^{-200}}>{{(1.254\times {{10}^{845}})}^{-\frac{1}{5}}}$$\Rightarrow$$1.\cdots \cdots \times {{10}^{-169}}>{{7}^{-200}}>1.\cdots \cdots \times {{10}^{-169}}$$\Rightarrow$小數點以下$169$位始不為$0$$\Rightarrow$小數點以下有$168$個$0$
   
   

三、填充題：全對才給分(共62分)

1. $(\displaystyle{\frac{{{3}^{3}}}{{{2}^{3}}}})\times {{({{3}^{2}})}^{\frac{3}{2}}}\times {{(\displaystyle{\frac{{{3}^{4}}}{{{2}^{4}}}})}^{-\frac{1}{2}}}$$={{3}^{\frac{3}{4}}}\times {{2}^{-\frac{3}{4}}}\times {{3}^{3}}\times {{3}^{-2}}\times {{2}^{2}}$$={{2}^{\frac{5}{4}}}\times {{3}^{\frac{7}{4}}}$$\Rightarrow$$-\displaystyle{\frac{5}{4}}+\displaystyle{\frac{7}{4}}=\displaystyle{\frac{1}{2}}$
   
   
2. $\displaystyle{\frac{({{a}^{x}}+{{a}^{-x}})({{a}^{2x}}-1+{{a}^{-2x}})}{{{a}^{x}}+{{a}^{-x}}}}=9$$\Rightarrow$${{a}^{2x}}+{{a}^{-2x}}-10=0$，令${{a}^{2x}}=t$$\Rightarrow$$t+\displaystyle{\frac{1}{t}}-10=0$$\Rightarrow$${{t}^{2}}-10t+1=0$$\Rightarrow$$t=\displaystyle{\frac{10\pm \sqrt{96}}{2}}=5\pm 2\sqrt{6}$
   
   
3. $x={{10}^{-1.0899}}$、$123000={{10}^{5.0899}}$$\Rightarrow$$123000x={{10}^{4}}$$\Rightarrow$$12.3x=1$$\Rightarrow$$1\div 12.3\approx 0.081\cdots \approx 0.08$
   
   
4. 不能圍成三角形$\Rightarrow$三點共線、兩線平行。因${{L}_{1}}$、${{L}_{2}}$不平行$\Rightarrow$有「${{L}_{1}}$ $/\kern -0.8em /$ ${{L}_{2}}$」、「${{L}_{2}}$ $/\kern -0.8em /$ ${{L}_{3}}$」、「三線共點」。${{L}_{1}}$ $/\kern -0.8em /$ ${{L}_{3}}$則$\displaystyle{\frac{2}{a}}=\displaystyle{\frac{3}{1}}$$\Rightarrow$$a=\displaystyle{\frac{2}{3}}$；${{L}_{2}}$ $/\kern -0.8em /$ ${{L}_{3}}$則$\displaystyle{\frac{1}{a}}=\displaystyle{\frac{-4}{1}}$$\Rightarrow$$a=-\displaystyle{\frac{1}{4}}$；三線共點則解${{L}_{1}}$、${{L}_{2}}$交點得$(x$，$y=(\displaystyle{\frac{25}{11}}$，$\displaystyle{\frac{-2}{11}})$$\Rightarrow$$\frac{25}{11})a+\displaystyle{\frac{-2}{11}}=1$$\Rightarrow$$a=\displaystyle{\frac{13}{25}}$
   
   
5. 配方$\Rightarrow$${{(x+1)}^{2}}+{{(y-3)}^{2}}=-3k-3$$\Rightarrow$$-3k-3>0$$\Rightarrow$$k<-1$且$(-1$，$2)$代入$>0$$\Rightarrow$$1+4-2-12+3k+13>0$$\Rightarrow$$k>-\displaystyle{\frac{4}{3}}$。過$(8$，$4)$且與$x$軸，$y$軸相切
   
   
6. 圓心必為$(t$，$t)$，半徑為$\left| t \right|$$\Rightarrow$${{(x-t)}^{2}}+{{(y-t)}^{2}}={{t}^{2}}$$\Rightarrow$${{(8-t)}^{2}}+{{(4-t)}^{2}}={{t}^{2}}$$\Rightarrow$${{t}^{2}}-24t+80=0$$\Rightarrow$$t=4$或$20$$\Rightarrow$方程式為${{(x-4)}^{2}}+{{(y-4)}^{2}}=16$或${{(x-20)}^{2}}+{{(y-4)}^{2}}=400$
   
   
7. 令$P(a$，$b)$$\Rightarrow$$Q(2a$，$2b-5)$，又$Q$在$C$上$\Rightarrow$${{(2a-5)}^{2}}+{{(2b-5-3)}^{2}}=36$$\Rightarrow$${{(a+2)}^{2}}+{{(b-4)}^{2}}=9$$\Rightarrow$${{(x+2)}^{2}}+{{(y-4)}^{2}}=9$
   
   
8. 因$SR$過$(3$，$0)$且$C$在$\overline{SR}$上$\Rightarrow$$SR$方程式為$x-3y=3$又$\overline{QR}\bot \overline{SR}$$\Rightarrow$$\overline{QR}$為$3x+y=6$，$\overline{PQ}$ $/\kern -0.8em /$ $\overline{SR}$且過$A$點$\Rightarrow$$\overline{PQ}$為$x-3y=-9$；$\overline{SP}$ $/\kern -0.8em /$ $\overline{QR}$且過$D$點$\Rightarrow$$\overline{SP}$為$3x+y=-12$$\Rightarrow$兩組對邊長為$\displaystyle{\frac{18}{\sqrt{10}}}$、$\displaystyle{\frac{\sqrt{12}}{10}}$$\Rightarrow$對角線長$=\sqrt{{{(\displaystyle{\frac{18}{\sqrt{10}}})}^{2}}+{{(\displaystyle{\frac{12}{\sqrt{10}}})}^{2}}}=\displaystyle{\frac{3}{5}}\sqrt{130}$
   
   
9. 將${{x}^{2}}+{{y}^{2}}-4y+1=0$配方$\Rightarrow$${{x}^{2}}+{{(y-2)}^{2}}=3$，所求$=$「${{x}^{2}}+{{y}^{2}}\ge 0$且${{x}^{2}}+{{(y-2)}^{2}}\le 3$」，或「${{x}^{2}}+{{y}^{2}}\le 0$且${{x}^{2}}+{{(y-2)}^{2}}\ge 3$」，即在「${{x}^{2}}+{{y}^{2}}=1$外部與${{x}^{2}}+{{(y-2)}^{2}}=3$內部」或「${{x}^{2}}+{{y}^{2}}=1$內部與${{x}^{2}}+{{(y-2)}^{2}}=3$外部」$\Rightarrow$作圖如下，所求為塗色範圍，長度標示如圖，中間空白部分$=$「左弓形」$+$「右弓形」
   
   
   
   左弓形$={{\sqrt{3}}^{2}}\pi \cdot \displaystyle{\frac{1}{6}}-\displaystyle{\frac{\sqrt{3}}{4}}\cdot {{\sqrt{3}}^{2}}=\displaystyle{\frac{\pi }{2}}-\displaystyle{\frac{3}{4}}\sqrt{3}$；右弓形$={{1}^{2}}\pi \cdot \displaystyle{\frac{1}{3}}-\displaystyle{\frac{\sqrt{3}}{4}}\cdot {{1}^{2}}=\displaystyle{\frac{\pi }{3}}-\displaystyle{\frac{\sqrt{3}}{4}}$，空白部分$=\displaystyle{\frac{5}{6}}\pi -\sqrt{3}$$\Rightarrow$塗色$=$兩圓面積和$-2\times$「空白部分」$=3\pi +\pi -2\cdot (\displaystyle{\frac{5}{6}}\pi -\sqrt{3})$$=\displaystyle{\frac{7}{3}}\pi +2\sqrt{3}$
   
   
10. ${{C}_{1}}$圓心$(-3$，$3)$、半徑$2$；${{C}_{2}}$圓心$(2$，$9)$、半徑$3$，作圖如下，作另一圓${{C}_{3}}$圓心$(-3$，$-3)$、半徑$2$。
    
    
    
    ${{C}_{3}}$即為${{C}_{1}}$對$x$軸的對稱圓。所求最小值即為${{C}_{2}}$上任意一點到${{C}_{3}}$上任意一點最小的距離，即$\overline{{{C}_{2}}{{C}_{3}}}-({{r}_{2}}+{{r}_{3}})=13-5=8$
    
    
11. $9$年衰變為$\displaystyle{\frac{16}{128}}=\displaystyle{\frac{1}{8}}$倍$\Rightarrow$半衰期$=9\cdot \displaystyle{\frac{1}{3}}=3$(年)
    
    
12. $7.6=\displaystyle{\frac{log(\displaystyle{\frac{{{E}_{1}}}{A}})}{log32}}$，$9.1=\displaystyle{\frac{log(\displaystyle{\frac{{{E}_{2}}}{A}})}{log32}}$$\Rightarrow$${{E}_{1}}$為墨西哥地震的能量、${{E}_{2}}$為蘇門答臘地震的能量$\Rightarrow$$\left\{ \begin{array}{l} 7.6\cdot log32=log{{E}_{1}}-logA \\ 9.1\cdot log32=log{{E}_{2}}-logA \\ \end{array} \right.$、兩式相減$\Rightarrow$$1.5log32=log\displaystyle{\frac{{{E}_{2}}}{{{E}_{1}}}}$$\Rightarrow$$log{{({{2}^{5}})}^{\frac{3}{2}}}=log\displaystyle{\frac{{{E}_{2}}}{{{E}_{1}}}}$$\Rightarrow$$\displaystyle{\frac{{{E}_{2}}}{{{E}_{1}}}}={{2}^{\frac{15}{2}}}=128\sqrt{2}$