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2020年3月26日 星期四

109年身心障礙學生四技二專甄試-數學(A)-詳解


109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(A)
單選題,共 20 題,每題 5 分


$$x+3y+2=0 \Rightarrow 斜率為-{1\over 3}\\(A) 12x-4y-5=0 \Rightarrow 斜率為3 \\(B) 4x+12y+5=0 \Rightarrow 斜率為-{1\over 3} \\(C) 3x+y-2=0 \Rightarrow 斜率為-3 \\(D) x-3y+2=0 斜率為{1\over 3}\\,故選\bbox[red,2pt]{(A)} $$



$$小扇形面積=1=1^2\pi \times {\theta \over 2\pi} \Rightarrow \theta = 2 \Rightarrow 大扇形面積的弧長 = 2\theta =4,故選\bbox[red,2pt]{(D)}$$


:$$\sin \theta= {\sqrt 5\over 3} \Rightarrow \cos \theta = {2\over 3} \Rightarrow \tan \theta = {\sin \theta \over \cos \theta} = {\sqrt 5 \over 2},故選\bbox[red,2pt]{(B)}$$


解:
$$\triangle ABC面積= {1\over 2}\times 4\times 3 = {1\over 2}\times 5\times (A\text{至直線}\overline{BC}的距離)\\ \Rightarrow A\text{至直線}\overline{BC}的距離 = {12 \over 5}=2.4,故選\bbox[red,2pt]{(C)}$$



:$$\cases{\cos \alpha ={-1\over 2} \\ \cos \beta={\sqrt 3\over 2}} \Rightarrow \cases{\sin \alpha = {\sqrt 3\over 2} \\ \sin \beta ={1 \over 2}} \Rightarrow \sin (\alpha+\beta) = \sin \alpha \cos \beta + \sin \beta \cos \alpha \\ = {\sqrt 3\over 2}\times {\sqrt 3\over 2}+ {1\over 2}\times {-1\over 2} = {3\over 4}-{1\over 4} = {1\over 2},故選\bbox[red,2pt]{(B)}$$


:$$已知\cases{\vec a=(2,3) \\ \vec b=(x,9) \\\vec c=(3,y) }, 由 \cases{\vec a與\vec b平行 \\ \vec a 與 \vec c垂直} \Rightarrow \cases{{2\over x}={3\over 9} \\ 2\times 3+3y=0} \Rightarrow \cases{x=6 \\ y=-2} \Rightarrow x+y=4,故選\bbox[red,2pt]{(A)}$$


:$$p(1)=p(2) =p(3)=0 \Rightarrow 1,2,3為p(x)=0之三根 \Rightarrow p(x)=a(x-1)(x-2)(x-3);\\又p(4)=6 \Rightarrow p(4)=a\cdot (4-1)\cdot (4-2)\cdot (4-3) = 6a=6 \Rightarrow a=1 \\\Rightarrow p(x)=(x-1)(x-2)(x-3) = x^3-6x^2+11x-6,故選\bbox[red,2pt]{(A)}$$


:$$令f(x)=ax^2+bx+1,由x=-1為f(x)=0之一根可得a-b+1=0 \Rightarrow a-b=-1,故選\bbox[red,2pt]{( B)}$$



$$\log 2^{100} =100\log 2= 100\times 0.301=30.1 \Rightarrow 2^{100}為31位數,故選\bbox[red,2pt]{(C)} $$



$$a^2-2+a^{-2}= (a- {1\over a})^2 = ({1\over 2}-2)^2 = (-{3\over 2})^2 = {9\over 4},故選\bbox[red,2pt]{(B)}$$


:$$x^2+y^2+2x-19=0 \Rightarrow (x+1)^2+y^2=20 \Rightarrow \cases{圓心O(-1,0)\\ 半徑r=\sqrt{20}=2\sqrt 5}\\ \Rightarrow \overrightarrow{OP} =(4,2) 為切線的法向量,故選\bbox[red,2pt]{(A)}$$



$$\cases{A(1,1)\\ C(0,5)\\ f(x,y)=2x-3y} \Rightarrow \cases{f(A)=2-3=-1 <0 \\ f(C)=-15 <0} \Rightarrow A,C同側,故選\bbox[red,2pt]{(B)}$$




$$直線x-2y=4與坐標軸的交點為\cases{A(0,-2)\\ B(4,0)} \Rightarrow 原點O與A、B所圍\triangle 面積={1\over 2}\times 2\times 4=4\\, 故選\bbox[red, 2pt]{(D)}$$



$${-1\over 2} < x <3 \Rightarrow (x+{1\over 2})(x-3)<0 \Rightarrow (2x+1)(x-3)<0 \Rightarrow 2x^2-5x-3 <0\\ \Rightarrow \cases{a=-5\\ b=-3} \Rightarrow ab=15,故選\bbox[red,2pt]{(D)}$$


$$\cases{a_1=18 \\ a_5={32\over 9} = a_1r^4 =18r^4 \Rightarrow r^4={32 \over 9\times 18} = ({2\over 3})^4 \Rightarrow r=\pm {2\over 3}}\\ \Rightarrow a_3=a_1r^2 = 18\times (\pm{2\over 3})^2 =8,故選\bbox[red,2pt]{(C)}$$



$$\sum_{n=1}^6 3(2)^{n-1} = 3(1+2+ 2^2+\cdots +2^5)=3(2^6-1) = 3\times 63=189,故選\bbox[red,2pt]{(D)}$$



$$第1次失敗、第2次失敗、第3次成功的機率為0.2\times 0.2\times 0.8 =0.032,故選\bbox[red,2pt]{(A)}$$



$$\cases{第1顆抽中紅球且第2顆抽中白球的機率為{6\over 20} \times {14\over 19} ={84 \over 380} \\ 第1顆抽中白球且第2顆抽中紅球的機率為{14\over 20} \times {6\over 19} ={84 \over 380}} \\ \Rightarrow 抽中一白球一紅球的機率為{84 \over 380}+{84 \over 380} ={42 \over 95},故選\bbox[red,2pt]{(C)}$$




$$投資一萬元的期望報酬為2000\times 0.35-500 \times 0.65 = 375\\ \Rightarrow 投資十萬元的期望報酬為375\times 10=3750,故選\bbox[red,2pt]{(C)}$$


:$$原售價x,新售價為1.2x \Rightarrow \cases{\mu(1.2X)= 1.2\mu(X) \\ \sigma(1.2X) = 1.2\sigma(X)},故選\bbox[red,2pt]{(D)}$$



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