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2022年3月30日 星期三

111年身障生升四技二專-數學(B)詳解

111 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別:四技二專組-數學(B)

單選題,共 20 題,每題 

解答:$$|2x+3|\lt 7 \Rightarrow -7\lt 2x+3\lt 7 \Rightarrow -5\lt x\lt 2 \Rightarrow x=-4,-3,\dots, 1,共6個整數解,故選\bbox[red, 2pt]{(C)}$$
解答:$$直線L的斜率=1\\ \cases{(A) x-y=2 \Rightarrow 斜率=1\\ (B)y=2 \Rightarrow 斜率=0 \lt 1\\ (C)2x-y=-1 \Rightarrow 斜率=2 \gt 1\\ (D)x+y=1 \Rightarrow 斜率=-1\lt 1},故選\bbox[red, 2pt]{(C)}$$
解答:$$\sum_{n=1}^5 a_n =\sum_{n=1}^5 (2n+3) =2 \sum_{n=1}^5 n+ 3\sum_{n=1}^5 1 =2\times 15+ 3\times 5=45,故選\bbox[red, 2pt]{(D)}$$
解答:$$資料有由小至大排序:2, 7, 7, 8 ,10,16,20,30,35 \Rightarrow \cases{Me=排序第5的數=10\\ R=最大-最小=35-2=33},故選\bbox[red, 2pt]{(C)}$$
解答:$$\overline{BC} =\sqrt{41^2-40^2}=\sqrt{81} =9 \Rightarrow \cases{\sin A=9/41\\ \cos A=40/41\\ \sin B=40/41\\ \cos B=9/41},故選\bbox[red, 2pt]{(D)}$$
解答:$$各組的圓皆相等,皆是圓心O(0,0),半徑r=1\\ (A)d(O,L)= 1/\sqrt 2\lt r\\ (B)d(O,L)=1=r\\(C) d(O,L)=1=r \\ (D)d(O,L)=2/\sqrt 2 \gt r \\ 只有(A)符合d(O,L)\lt r,故選\bbox[red, 2pt]{(A)}$$
解答:$$(A)\times: \cases{f(6)=3^6 =81\\f(2)+f(3)= 3^2+ 3^3 =9+27=36} \Rightarrow f(6)\ne f(2)+f(3) \\(B)\times: f(-1)=3^{-1}= {1\over 3}\gt 0\\ (C)\times: f(x) \gt 0 \Rightarrow (x,f(x))不過第四象限\\(D)\bigcirc: f(-\infty)= {1\over \infty} \to 0,故選\bbox[red, 2pt]{(D)}$$
解答:$$2x+y=6與x軸、y軸分別交於P(3,0)及Q(0,6) \Rightarrow \triangle OAB=3\times 6 \div 2=9,故選\bbox[red, 2pt]{(C)}$$
解答:$$R={P+Q\over 2} \Rightarrow \cases{0=(-2+a)\div 2\\ -1= (3+b)\div 2} \Rightarrow \cases{a=2\\ b=-5} \Rightarrow 2x-5y=10的x截距={10\over 2}=5,故選\bbox[red, 2pt]{(A)}$$
解答:$$x^2-2x-5 =(x-(1+\sqrt 6))(x-(1-\sqrt 6))\gt 0 \Rightarrow x\gt 1+\sqrt 6或x\lt 1-\sqrt 6,故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{\sin (-7\pi/3) = \sin(-2\pi-\pi/3) =\sin(-\pi/3) \lt 0\\ \tan(10\pi/3) = \tan(2\pi+4\pi/3) =\tan(4\pi/3) \gt 0} \Rightarrow P在第二象限,故選\bbox[red, 2pt]{(B)}$$
解答:$$利用長除法:f(x)=(x-7)(5x^4-x^3+2x^2)-2 \Rightarrow f(7)=-2,故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{個位數字:0,2,4,有3種選擇\\ 十位數字:0-5,有6種選擇\\ 百位數字:1-5,有5種選擇} \Rightarrow 共有3\times 6\times 5=90種偶數,故選\bbox[red, 2pt]{(C)}$$
解答:$$\left( \log_5 25\right)x^2 +\left(\log_3 1\right)x -\left( \log_7 \sqrt 7\right)=0 \Rightarrow 2x^2+0-{1\over 2}=0 \Rightarrow x^2={1\over 4} \Rightarrow x=\pm {1\over 2},故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{A(-1,2)\\ B(3,5)} \Rightarrow \overrightarrow{AB} =(4,3) \Rightarrow \vec u={\overrightarrow{AB}  \over |\overrightarrow{AB} |} =({4\over 5},{3\over 5}),故選\bbox[red, 2pt]{(A)}$$
解答:$$\left({x-1\over x^2-2x-3} \times {6x^2-18x\over 2x^2-x-1}\div {1\over 4x+2}\right)+ {1\over x} =\left({x-1\over (x-3)(x+1) } \times {6x(x-3)\over (2x+1)(x-1)}\times 2(2x+1)\right)+ {1\over x} \\= {12x\over x+1} +{1\over x} ={12x^2+x+1\over x(x+1)},故選\bbox[red, 2pt]{(C)}$$
解答:$$令\cases{首項a_1\\ 公比r},則\cases{a_1+a_3={5\over 4} \\a_2+a_4=-{5\over 8}} \Rightarrow \cases{a_1+a_1r^2= a_1(1+r^2 )={5\over 4} \\a_1r+a_1r^3= a_1r(1+r^2)=-{5\over 8}} ,兩式相除\Rightarrow {1\over r}=-2 \\ \Rightarrow r=-{1\over 2} \Rightarrow a_1(1+r^2) =a_1(1+{1\over 4})={5\over 4} \Rightarrow a_1= 1 \Rightarrow a_5= a_1r^4 = {1\over 2^4}={1\over 16},故選\bbox[red, 2pt]{(D)}$$
解答:$$\log_{10} (2x^3)= 4-\log_{10} 5 =\log_{10} 10^4-\log_{10} 5= \log_{10}{10^4\over 5} \Rightarrow 2x^3={10^4\over 5}  \Rightarrow x^3= 10^3 \Rightarrow x=10\\,故選\bbox[red, 2pt]{(B)}$$
解答:$$判別式\gt 0 \Rightarrow 36-4k \gt 0 \Rightarrow k\lt 9,故選\bbox[red, 2pt]{(B)}$$
解答:$$A\cap B=\{(2,3),(3,2)\} \Rightarrow P(A\cap B)={2\over 6\times 6} ={1\over 18},故選\bbox[red, 2pt]{(D)}$$

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