[段考] 108上第1次段考-台北-麗山高中-高一(詳解)

108上第1次段考-台北-麗山高中-高一(詳解)

範圍：翰林 第一冊1-1~2-2

　(※索取各種題目檔案請來信索取。)

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一、單選題(每題5分，共10分)

1. ${{3}^{k}}=a-1$，${{3}^{-k}}=-b+1$，因${{3}^{k}}$，${{3}^{-k}}$互為倒數$\Rightarrow$$\displaystyle{\frac{1}{a-1}}=-b+1$$\Rightarrow$$b-1=\displaystyle{\frac{-1}{a-1}}$$\Rightarrow$$b=1+\displaystyle{\frac{-1}{a-1}}$$=\displaystyle{\frac{a-2}{a-1}}$，選(5)
   
   
2. (1)  $0.3\overline{43}=a$$\Rightarrow$$100a=34.3\overline{34}$$\Rightarrow$$99a=34$$\Rightarrow$$a=\displaystyle{\frac{34}{99}}$，錯誤
   (2)  $\displaystyle{\frac{34}{99}}>\displaystyle{\frac{33}{99}}=\displaystyle{\frac{1}{3}}$，正確
   (3)  $0.\overline{34}=0.3434\cdots \cdots >0.343$，正確
   (4)  $0.\overline{34}=0.3434\cdots \cdots <0.35$，正確
   (5)  $0.3\overline{43}=0.34\overline{34}=0.\overline{34}$，正確
   故選(1)
   
   

二、填充題I(每題5分，共40分)

1. $a=\displaystyle{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}\cdot \displaystyle{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}}=5-2\sqrt{6}$
   $\displaystyle{\frac{1}{a}}=\displaystyle{\frac{1}{5-2\sqrt{6}}}=5+2\sqrt{6}$
   $\Rightarrow$$a+\displaystyle{\frac{1}{a}}=10$
   $\Rightarrow$${{(a+\displaystyle{\frac{1}{a}})}^{2}}={{a}^{2}}+2\cdot a\cdot \displaystyle{\frac{1}{a}}+\displaystyle{\frac{1}{{{a}^{2}}}}=100$
   $\Rightarrow$${{a}^{2}}+2+\displaystyle{\frac{1}{{{a}^{2}}}}=100$
   $\Rightarrow$${{a}^{2}}+\displaystyle{\frac{1}{{{a}^{2}}}}=98$
   
   
2. 由${{9}^{x}}={{15}^{xy}}$得$9={{15}^{y}}$；由${{25}^{y}}={{15}^{xy}}$得$25={{15}^{x}}$
   $9\times 25={{15}^{y}}\cdot {{15}^{x}}$$\Rightarrow$$225={{15}^{x+y}}$$\Rightarrow$$x+y=2$
   
   
3. 由定義${{10}^{k}}=3$，所求$={{1000}^{k}}+{{100}^{-k}}$
   $={{({{10}^{3}})}^{k}}+{{({{10}^{2}})}^{-k}}$$={{({{10}^{k}})}^{3}}+{{({{10}^{k}})}^{-2}}$$={{3}^{3}}+{{3}^{-2}}$$=27+\displaystyle{\frac{1}{9}}$$=27\displaystyle{\frac{1}{9}}$
   
   
4. 令長邊$=x$、寬$=y$$\Rightarrow$$(x+y)\cdot 2-3=33$$\Rightarrow$$2(x+y)=36$$\Rightarrow$$x+y=18$，
   所求$=xy$$\Rightarrow$$\displaystyle{\frac{x+y}{2}}\ge \sqrt{xy}$$\Rightarrow$$9\ge \sqrt{xy}$$\Rightarrow$$81\ge xy$，面積最大為$81$
   
   
5. $-2$與$5$之中點為$\displaystyle{\frac{3}{2}}$，距離為$7$$\Rightarrow$所求$-2\le x\le 5$之範圍即$\left| x-\displaystyle{\frac{3}{2}} \right|\le \displaystyle{\frac{7}{2}}$$\Rightarrow$$\left| -\displaystyle{\frac{2}{3}}x+1 \right|\le \displaystyle{\frac{7}{3}}$$\Rightarrow$$(-\displaystyle{\frac{2}{3}}$，$\displaystyle{\frac{7}{3}})$
   
   
6. 將$m=-1$，$n=4$代入得$4-(-1)=\displaystyle{\frac{5}{2}}log\displaystyle{\frac{{{L}_{-1}}}{{{L}_{4}}}}$
   $\Rightarrow$$\displaystyle{\frac{{{L}_{-1}}}{{{L}_{4}}}}={{10}^{2}}$$\Rightarrow$$100$倍
   
   
7. $a={{(\sqrt{5})}^{5\sqrt{5}}}$$={{({{5}^{\frac{1}{2}}})}^{5\sqrt{5}}}$$={{5}^{\frac{5}{2}\sqrt{5}}}$，所求$=\sqrt[5]{a}$$={{({{5}^{\frac{5}{2}\sqrt{5}}})}^{\frac{1}{5}}}$$={{5}^{\frac{1}{2}\sqrt{5}}}$$\approx {{({{10}^{0.699}})}^{\frac{1}{2}\sqrt{5}}}$，又$\sqrt{5}\approx 2.23$$\Rightarrow$${{({{10}^{0.699}})}^{\frac{1}{2}\sqrt{5}}}$$\approx {{10}^{0.699\cdot \frac{1}{2}\cdot 2.23}}$$={{10}^{0.779385}}$。
   又$6\approx {{10}^{0.7781}}$，$7\approx {{10}^{0.8451}}$，所求整數部分為$6$
   
   
8. ${{n}^{300}}<{{7}^{800}}$$\Rightarrow$$n<{{7}^{\frac{8}{3}}}$$\approx {{({{10}^{0.8451}})}^{\frac{8}{3}}}$$={{10}^{2.2536}}$，由計算機得${{10}^{2.2536}}\approx 179.3$$\Rightarrow$$n<179.3$$\Rightarrow$$n$最大$=179$
   
   

三、填充題II(每題6分，共30分)

9. 移項$\Rightarrow$${{x}^{2}}+2=2\sqrt{7}x$$\Rightarrow$${{({{x}^{2}}+2)}^{2}}={{(2\sqrt{7}x)}^{2}}$$\Rightarrow$${{x}^{2}}+4{{x}^{2}}+4=28{{x}^{2}}$$\Rightarrow$${{x}^{4}}-24{{x}^{2}}+4=0$$\Rightarrow$${{x}^{4}}-24{{x}^{2}}=-4$，所求$=-4+7=3$
   
   
10. $\sqrt{{{a}^{2}}-2a+1}=\left| a-1 \right|$；$\sqrt{49-14a+{{a}^{2}}}=\left| a-7 \right|$
    $\Rightarrow$原式為$\left| a-1 \right|+\left| a-7 \right|=12-\left| b-2 \right|-\left| b-4 \right|$
    $\Rightarrow$$\left| a-1 \right|+\left| a-7 \right|+\left| b-2 \right|+\left| b+4 \right|=12$
    又$\left| a-1 \right|+\left| a-7 \right|$之最小值$=6$、$\left| b-2 \right|+\left| b+4 \right|$之最小值$=6$
    $\Rightarrow$和$=12$$\Rightarrow$$\left| a-1 \right|+\left| a+7 \right|=6$且$\left| b-2 \right|+\left| b+4 \right|=6$；
    其中$1\le a\le 7$且$-4\le b\le 2$
    $\Rightarrow$$1\le {{a}^{2}}\le 49$且$0\le {{b}^{2}}\le 16$
    $\Rightarrow$${{a}^{2}}+{{b}^{2}}$之最大值為$49+16=65$
    
    
11. $\displaystyle{\frac{1}{\sqrt{14-6\sqrt{5}}}}$$=\displaystyle{\frac{1}{\sqrt{14-2\sqrt{45}}}}$$=\displaystyle{\frac{1}{3-\sqrt{5}}}$$=\displaystyle{\frac{3+\sqrt{5}}{4}}$$=\displaystyle{\frac{5.2\cdots \cdots }{4}}$
    $\Rightarrow$$a=1$，$b=\displaystyle{\frac{3+\sqrt{5}}{4}}-1$$=\displaystyle{\frac{\sqrt{5}-1}{4}}$
    $\Rightarrow$$\displaystyle{\frac{1}{a+b}}+\displaystyle{\frac{1}{b}}$$=\displaystyle{\frac{1}{\displaystyle{\frac{3+\sqrt{5}}{4}}}}+\displaystyle{\frac{1}{\displaystyle{\frac{\sqrt{5}-1}{4}}}}$$=\displaystyle{\frac{4}{3+\sqrt{5}}}+\displaystyle{\frac{4}{\sqrt{5}-1}}$$=3-\sqrt{5}+\sqrt{5}+1$$=4$
    
    
12. $\left| \left| x \right|-5 \right|\le \displaystyle{\frac{13}{3}}$
    $\Rightarrow$$5-\displaystyle{\frac{13}{3}}\le \left| x \right|\le 5+\displaystyle{\frac{13}{3}}$
    $\Rightarrow$$\left| x \right|$整數部分$=1$，$2$，$3$，……，$9$，因此$x$共$9\times 2=18$個可能
    
    
13. $\left\{ \begin{array}{l} log(\sqrt{k}+1)=a \\ log(\sqrt{k}-1)=b \\ \end{array} \right.$，其中$a=b=3$
    $\Rightarrow$${{10}^{a}}=\sqrt{k}+1$，${{10}^{b}}=\sqrt{k}-1$
    $\Rightarrow$${{10}^{a}}\cdot {{10}^{b}}=(\sqrt{k}+1)(\sqrt{k}-1)=k-1$
    $\Rightarrow$${{10}^{a+b}}=k-1$
    $\Rightarrow$${{10}^{3}}=k-1$
    $\Rightarrow$$k=1001$
    
    

四、題組題(每題均需寫出演算過程或理由，否則將予扣分甚至零分。每一子題配分標於題末，共20分)

1. 原式$=(a+\displaystyle{\frac{1}{b}})(2b+\displaystyle{\frac{1}{2a}})$$=2ab+\displaystyle{\frac{1}{2}}+2+\displaystyle{\frac{1}{2ab}}$$=2ab+\displaystyle{\frac{1}{2ab}}+\displaystyle{\frac{5}{2}}$
   令$2ab=t>0$$\Rightarrow$$\displaystyle{\frac{t+\displaystyle{\frac{1}{t}}}{2}}\ge \sqrt{t\cdot \displaystyle{\frac{1}{t}}}=1$$\Rightarrow$$t+\displaystyle{\frac{1}{t}}\ge 2$$\Rightarrow$所求$\ge 2+\displaystyle{\frac{5}{2}}=\displaystyle{\frac{9}{2}}$
   其中$2ab=\displaystyle{\frac{1}{2ab}}$$\Rightarrow$${{(2ab)}^{2}}=1$$\Rightarrow$$2ab=\pm 1$，負不合$\Rightarrow$$ab=\displaystyle{\frac{1}{2}}$
   $\Rightarrow$$(1)$$\displaystyle{\frac{9}{2}}$、$(2)$$\displaystyle{\frac{1}{2}}$
   
   
2. (1)  ${{d}_{1}}$$=10log\displaystyle{\frac{{{10}^{-12}}}{{{10}^{-12}}}}$$=10log1=0$
   (2)  ${{d}_{2}}$$=10log\displaystyle{\frac{{{10}^{-3}}}{{{10}^{-12}}}}$$=10log{{10}^{9}}$$=10\cdot 9$$=90$
   (3)  千支齊鳴$\Rightarrow$強度$1000$倍$\Rightarrow$$70=10log\displaystyle{\frac{{{W}_{1}}}{{{10}^{-12}}}}$$\Rightarrow$${{W}_{1}}={{10}^{-5}}$
   $1000$倍$\Rightarrow$$1000{{W}_{1}}={{10}^{-2}}$$\Rightarrow$${{d}_{3}}=10log\displaystyle{\frac{{{10}^{-2}}}{{{10}^{-12}}}}$$=10log{{10}^{10}}=100$