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2019年9月6日 星期五

107年身心障礙學生大學甄試-數學甲-詳解


107學年度身心障礙學生升學大專校院甄試試題

甄試類(群)組別:大學組
考試科目(編號):數學甲
單選題,共 20 題,每題 5 分


$$f\left( x \right) =\left( x-1 \right) \left( x-2 \right) \left( x-3 \right) \Rightarrow \begin{cases} f\left( 1/2 \right) =(-1/2)(-3/2)(-5/2)=-15/8 \\ f\left( 3/2 \right) =(1/2)(-1/2)(-3/2)=3/8 \\ f\left( 5/2 \right) =(3/2)(1/2)(-1/2)=-3/8 \\ f\left( 9/2 \right) =(7/2)(5/2)(3/2)=105/8 \end{cases}\\ \Rightarrow f(9/2)>f(3/2)>f(5/2)>f(1/2),故選\bbox[red,2pt]{(D)} $$



$$\\ \begin{cases} \frac { 521+523 }{ 2 } >\sqrt { 521\times 523 } \Rightarrow \log { \frac { 521+523 }{ 2 }  } >\log { \sqrt { 521\times 523 }  }  \\ \frac { \log { 521 } +\log { 523 }  }{ 2 } >\sqrt { \log { 521 } \times \log { 523 }  }  \end{cases} \Rightarrow \begin{cases} a>b \\ b>c \end{cases}\Rightarrow a>b>c,故選\bbox[red,2pt]{(A)}$$


:$$ 第1袋抽1藍球的機率為1/2,第2袋抽1藍球的機率為1/3\\
兩袋都抽中藍球的機率為{1\over 2}\times {1\over 3}={1\over 6}\\
第1袋抽中藍球且第2袋沒抽中藍球的機率為{1\over 2}\times {2\over 3}= {1\over 3}\\
第1袋沒抽中藍球且第2袋抽中藍球的機率為{1\over 2}\times {1\over 3}= {1\over 6}\\
期望值為500\times{1\over 6}+200\times   ({1\over 3}+{1\over 6})= {500\over 6}+100   ={550\over 3},故選\bbox[red,2pt]{(B)}$$


解:
$$M\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}=M\Rightarrow \begin{bmatrix} a & b \\ c & d \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\Rightarrow \begin{cases} 2a+b=a \\ a+2b=b \\ 2c+d=c \\ c+2d=d \end{cases}\Rightarrow \begin{cases} a+b=0 \\ c+d=0 \end{cases}\\ \Rightarrow \begin{cases} 第1列相加=0 \\ 第2列相加=0 \end{cases},故選\bbox[red,2pt]{(D)}$$



:$$\begin{cases}P=(1,a)\\ Q=(7,a^7) \end{cases} \Rightarrow \overline{PQ}斜率=900 ={a^7-a \over 7-1} \Rightarrow a^7-a=5400\\
由\begin{cases}2^7=128\\ 3^7= 2187<5400 \\4^7= 16384 >5400 \end{cases} \Rightarrow 3< a< 4,故選\bbox[red,2pt]{(C)}$$


:$$P(a)至1,2,3的距離和為4 \Rightarrow  \begin{cases} a>3\\ a<1 \end{cases}  \Rightarrow  \begin{cases} (a-1)+(a-2)+(a-3)=4 \\  (1-a)+(2-a)  +(3-a)=4 \end{cases}  \\  \Rightarrow   \begin{cases} a={10\over 3}\\ a= {2\over 3} \end{cases}  \Rightarrow  |a-4| +|a-5| +a-6|=\begin{cases} 5 \\  13 \end{cases}  ,故選\bbox[red,2pt]{(A)}$$




:$$\begin{cases} f\left( x \right) +g\left( x \right) =2018x^{ 5 } \\ f\left( x \right) -g\left( x \right) =107x^{ 7 } \end{cases}\Rightarrow \begin{cases} f\left( x \right) =(2018x^{ 5 }+107x^{ 7 })\div 2=\frac { 107 }{ 2 } x^{ 7 }+1009x^{ 5 } \\ g\left( x \right) =(2018x^{ 5 }-107x^{ 7 })\div 2=-\frac { 107 }{ 2 } x^{ 7 }+1009x^{ 5 } \end{cases}\\ f\left( x \right)=g\left( x \right)q\left( x \right)+r\left( x \right)\Rightarrow q(x)為常數,r(x)為5次式,故選\bbox[red,2pt]{(A)}$$


:$$公司目前業績為a ,n年後的業績為a(1+0.4)^2;依題意{a(1+0.4)^n \over a}>100\\ \Rightarrow \log{(1+0.4)^n}>\log{100} \Rightarrow n>{2\over \log{1.4}} = {2\over \log{14/10}} = {2 \over \log{2}+\log{7}-1} \\= {2\over 0.301+0.8451-1} \approx 13.7,故選\bbox[red,2pt]{(B)}$$



$$全國被篩檢為陽性的比率為2\%\times 98\%+98\%\times 4\% = 5.88\%,\\已知全國有2\%人口感染此病,因此某甲實際感染此病的機率為{2\over 5.88} \approx 34\%,故選\bbox[red,2pt]{(A)} $$




$$令R=(x,y), 由\frac{\triangle POQ}{\triangle QOR}= {\overline{PQ} \over \overline{QR}}=2 \Rightarrow \left( {2x+5\over 3}, {2y-1 \over 3} \right) = (-2,7) \Rightarrow {2y-1 \over 3}=7 \\\Rightarrow y=11,故選\bbox[red,2pt]{(C)}$$


:$$\begin{cases} \vec{u}//(1,-2) \\ \vec{v}//(3,4) \end{cases} \Rightarrow \begin{cases} \vec{u}=(m,-2m) \\ \vec{v}=(3n,4n) \end{cases} \Rightarrow (\vec{u}+\vec{v})\bot (6,-5) \Rightarrow (m+3n,-2m+4n)\cdot (6,-5)=0 \\ \Rightarrow 6m+18n+10m-20n=0 \Rightarrow 8m=n \Rightarrow {|\vec{u}| \over |\vec{v}|} = {\sqrt{m^2+4m^2} \over \sqrt{9n^2+16n^2}} = {\sqrt{5m^2} \over \sqrt{25n^2}} ={\sqrt{5m^2} \over \sqrt{25\times 64m^2}}\\ = {\sqrt{5} \over 40},故選\bbox[red,2pt]{(D)}$$



$$依題意: 大樓高= \overline{FA}, 塔高=\overline{DC}, E為\overline{FD}中點,如上圖\\ 令\overline{FE}= \overline{ED}=a \Rightarrow \overline{DC}=\sqrt{3}a \Rightarrow \overline{BC}=\sqrt{3}a-50 \Rightarrow \overline{AB}=\sqrt{3}(\sqrt{3}a-50) \\ 由 \overline{FD}=\overline{AB} \Rightarrow \sqrt{3}(\sqrt{3}a-50)=2a \Rightarrow a=50\sqrt{3} \Rightarrow \overline{DC}=\sqrt{3}a=150,故選\bbox[red,2pt]{(C)}$$


:$$後5次需出現2正3反,機率為C^5_2(1/4)^2(3/4)^3 =10\times (1/16)\times (27/64) =135/512, 故選\bbox[red, 2pt]{(D)}$$


:$$假設平面E: ax+by+cz=0,其法向量為\vec{n}=(a,b,c)\\ \begin{cases} \vec{n} \cdot \vec{u}=0 \\ \vec{n} \cdot \vec{v}=0\end{cases} \Rightarrow \begin{cases} a+2b-c=0 \\ 2a-b=0\end{cases} \Rightarrow \begin{cases} b=2a \\ c=5a \end{cases}\\ \Rightarrow dist((1,1,1),E) = \left|  {a+b+c \over \sqrt{a^2+b^2+c^2}} \right| = {a+2a+5a \over \sqrt{a^2+4a^2+25a^2}} ={ 8\over \sqrt{30}},故選\bbox[red,2pt]{(D)}$$




$$\begin{cases}L_1: 4x-3y=0\\ L_2:4x-3y=10 \\ L_3: y=mx \end{cases} \Rightarrow dist(L_1,L_2)=\overline{OE}={ 10\over \sqrt{3^2+4^2}}=2,如上圖;\\
直角\triangle OEC \Rightarrow \overline{OC}^2=\overline{OE}^2+\overline{EC}^2 \Rightarrow (2\sqrt{5})^2 =2^2+ \overline{EC}^2 \Rightarrow \overline{EC}=4 \\
\tan{\angle OCE} = {\overline{OE}\over \overline{EC}} ={1\over 2}\Rightarrow \begin{cases}{m-4/3 \over 1+4m/3}=1/2\\ {4/3-m\over 1+4m/3}=1/2 \end{cases} \Rightarrow \begin{cases}m=11/2\\ m=1/2 \end{cases} \Rightarrow 11/2+1/2=6,故選\bbox[red,2pt]{(D)}$$




$$令L_1與L_2的交點為A(x,y),如上圖\\ L_1斜率為-1,且過P點 \Rightarrow L_1: y+1=-1(x-1) \Rightarrow L_1: x+y=0 \Rightarrow A=(-y,y)\\ 由於\overline{AP} =\overline{AQ} \Rightarrow (-y-1)^2+(y+1)^2 = (-y-3)^2+(y-5)^2 \Rightarrow 8y=32 \Rightarrow y=4,故選\bbox[red,2pt]{(A)}$$




$$\begin{cases} \angle NMQ=\theta\Rightarrow \angle PMN=2\theta \\ \overline{QN}=a\Rightarrow \overline{MN}=1.7a \end{cases},如上圖;\\
\begin{cases} \cos{\theta}= {\overline{MN}^2 +\overline{MQ}^2-\overline{QN}^2 \over 2\times\overline{MN}\times \overline{MQ}} = {(1.7a)^2+a^2-a^2 \over 2\times 1.7a\times a} ={1.7\over 2} \\
\cos{2\theta}={\overline{MN}^2 +\overline{MP}^2-\overline{PN}^2 \over 2\times\overline{MN}\times \overline{PM}} = {(1.7a)^2+b^2-b^2 \over 2\times 1.7a\times b} ={1.7a\over 2b} \end{cases} \Rightarrow \cos{2\theta}=2\cos^2{\theta}-1 \\ \Rightarrow  {1.7a\over 2b}=2\times \left({1.7\over 2}\right)^2-1 ={0.89\over 2} \Rightarrow {\overline{MN} \over \overline{PN}}={1.7a\over b}={0.89\over 2}\times 2=0.89,故選\bbox[red,2pt]{(B)}$$



:$$\begin{cases}(A)z^2= -24+70i \\ (B)(1+2i)z= -9+17i\\ (C) 2iz=-14+10i \\(D) 5-7i\end{cases}\\ \Rightarrow  \begin{cases} |z-(-z)|=|10+14i|\\ |(1+2i)z-z|=|(-9+17i)-(5+7i)|=|-14+10i|\end{cases} \\ \Rightarrow |z-(-z)|=|(1+2i)z-z| \Rightarrow (1+2i)z為另一頂點,故選\bbox[red,2pt]{(B)}$$


:$$\begin{cases} x-2y+z=4\cdots(1) \\ -y+z=3\cdots(2)\\ x-z=-1\cdots(3) \\-x+z=2\cdots(4)\end{cases} \Rightarrow 由(1)及(2)可得x-z=-2\cdots(5)\\
若(3)與(5)同時存在,則無解,故選\bbox[red,2pt]{(C)}$$






$$\overline{PQ}=\overline{AE}=1 \Rightarrow \overline{AB}= {\sqrt{2}\over 2}\\

\overline{PQ}=\overline{AC}=1 \Rightarrow m^2+4m^2=1 \Rightarrow m^2={1\over 5} \Rightarrow m={\sqrt{5}\over 5}\\

\overline{PQ}=\overline{AD}=1 \Rightarrow \overline{AD}^2 = \overline{AC}^2+ \overline{CD}^2 = 5m^2 +m^2 =1 \Rightarrow m^2={1\over 6} \Rightarrow m={\sqrt{6}\over 6}\\
,故選\bbox[red,2pt]{(B)} $$


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