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2019年9月17日 星期二

106年身心障礙學生四技二專甄試-數學(B)-詳解


106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:四技二專組
考試科目(編號):數學(B)
單選題,共 20 題,每題 5 分


$$直線2x+4y+3=0的斜率為-{1\over 2},因此所求直線的斜率為2,可寫成y=2x+b;\\該直線過A(-4,3),即3=-8+b \Rightarrow b=11,方程式可寫成y=2x+11,即2x-y+11=0\\,故選\bbox[red,2pt]{(C)} $$



$$\begin{cases}A = (-2,1)\\ B=(1,-3) \\ C=(1,5)\end{cases} \Rightarrow \begin{cases}\overline{AB} =\sqrt{3^2+4^2}=5\\ \overline{AC}=\sqrt{3^2+4^2}=5 \\ \overline{BC}=5-(-3)=8 \end{cases} \Rightarrow \triangle ABC周長=5+5+8=18,故選\bbox[red,2pt]{(A)}$$


:$$2017^o = 2017^o-360^o\times 5 = 217^o \Rightarrow  180^o<217^o<270^o ,故選\bbox[red,2pt]{(C)}$$


解:
$$\begin{cases} \tan{\theta} = {\sin{\theta}>0 \over \cos{\theta}<0}<0 \\ \cot{\theta}= {\cos{\theta}<0 \over \sin{\theta}>0}<0 \end{cases} \Rightarrow (\tan{\theta},\cot{\theta})在第3象限,故選\bbox[red,2pt]{(C)}$$



:$$\begin{cases} A(2,4)\\ B(k,5) \\ C(5,1)\end{cases} \Rightarrow \begin{cases} \overrightarrow{AB}=(k-2,1) \\ \overrightarrow{AC}=(3,-3)\end{cases} \Rightarrow \overrightarrow{AB}\bot \overrightarrow{AC}\Rightarrow \overrightarrow{AB}\cdot \overrightarrow{AC} =0 \Rightarrow 3(k-2)-3=0 \Rightarrow k=3\\,故選\bbox[red,2pt]{(B)}$$


:$$a^x+a^{-x}=3 \Rightarrow (a^x+a^{-x})^2= a^{2x}+a^{-2x}+2=9 \Rightarrow a^{2x}+a^{-2x}=7,故選\bbox[red,2pt]{(D)}$$




:$$\log_{10}{5} +\log_{10}{12} +\log_{10}{15} -\log_{10}{9} =\log_{10}{5\times 12\times 15\over 9} =\log_{10}{100} =2,故選\bbox[red,2pt]{(B)}$$


:$$\log_2{(x-5)}+\log_2{(x+2)} =3 \Rightarrow \log_2{(x-5)(x+2)} =3 \Rightarrow (x-5)(x+2)=8\\ \Rightarrow x^2-3x-18=0 \Rightarrow (x-6)(x+3)=0 \Rightarrow x=6(x-5>0 \Rightarrow x=-3不合),故選\bbox[red,2pt]{( D)}$$



$$利用長除法可得4x^3+5x^2+6x+7 =(x^2+2x+3)(4x-3)+16  \Rightarrow \begin{cases} a=4\\b=-3\\ c=16 \end{cases}\\ \Rightarrow 6a+6b+c=24-18+16=22,故選\bbox[red,2pt]{(D)} $$



$$\begin{cases}a_1=14 \\ a_{10}=41 \Rightarrow a_1+9d=41 \Rightarrow d=3 \end{cases} \Rightarrow a_2+a_3+\cdots+a_9= {8(a_2+a_9)\over 2} \\= 4(2a_1+9d)=4(28+27)= 220,故選\bbox[red,2pt]{(A)}$$


:$$2017(x-1)=x-2017 \Rightarrow 2017x-2017=x-2017 \Rightarrow 2016x=0 \Rightarrow x=0,故選\bbox[red,2pt]{(D)}$$



$$\begin{vmatrix}x&1\\ 2&x \end{vmatrix} =\begin{vmatrix}8& x\\ x&2 \end{vmatrix} \Rightarrow x^2-2=16-x^2 \Rightarrow x^2=9 \Rightarrow x=\pm 3,故選\bbox[red,2pt]{(C)}$$




$$|x-3|\le 7 \Rightarrow -7\le x-3\le 7 \Rightarrow -4\le x\le 10 \Rightarrow x=-4,-3,\dots,10,共15個, 故選\bbox[red, 2pt]{(B)}$$




$$(2+1)\times (3+1)\times (5+1)= 3\times 4\times 6=72,故選\bbox[red,2pt]{(A)}$$



$$答對得a分,答錯扣2分,期望值為a\times {1\over 4}-2\times {3\over 4}=0 \Rightarrow a=6,故選\bbox[red,2pt]{(D)}$$



$$\cos{(106x)} =\cos{(2\times 53x)} = 1-2\sin^2{(53x)}= {1\over 9} \Rightarrow \sin^2{(53x)}= {4\over 9} ,故選\bbox[red,2pt]{(C)}$$



$$\cos{A}={\overline{AB}^2+ \overline{AC}^2-\overline{BC}^2 \over 2\times\overline{AB}\times \overline{AC}} \Rightarrow {1\over 4}= {1+4-\overline{BC}^2\over 4} \Rightarrow \overline{BC}^2=4 \Rightarrow \overline{BC}=2,故選\bbox[red,2pt]{(B)}$$




$${(x-1)^2\over a^2} - {(y-1)^2\over b^2}=1 \Rightarrow 漸近線: b(x-1)\pm a(y-1)=0 \Rightarrow \begin{cases} bx-ay-b+a=0 \\ bx+ay-b-a=0\end{cases} \\\Rightarrow \begin{cases} bx-ay-b+a=2x-y-1\Rightarrow \begin{cases} a=1 \\ b=2 \end{cases} \\ bx+ay-b-a=2x-y-1\Rightarrow \begin{cases} a=-1 \\ b=2\end{cases}\end{cases},故選\bbox[red,2pt]{(A)}$$


:$$f(x)=x^{2017}+{1\over x^{2017}} \Rightarrow f'(x)=2017x^{2016}-2017\times{1\over x^{2018}} \Rightarrow f'(1)=2017-2017=0\\,故選\bbox[red,2pt]{(D)}$$



$$\int_0^1{(x+1)(x+2)\,dx}= \int_0^1{x^2+3x+2\,dx}= \left. \left[ {1\over 3}x^3 +{3\over 2}x^2+2x \right]\right|_0^1 ={1\over 3}+{3\over 2}+2= {23\over 6}\\,故選\bbox[red,2pt]{(C)} $$


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