113年身障生升四技二專-數學(C)詳解

113 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(C)

單選題，共 20 題，每題 5 分

解答: $$\cases{x=(1+9)/2=5\\ y=(5+3)/2=4} \Rightarrow x+y=5+4=9，故選\bbox[red, 2pt]{(B)}$$

解答: $$\left({8\over 125} \right)^{-2/3} =\left(\left({2\over 5} \right)^3\right)^{-2/3} = \left({2\over 5} \right)^{-2} =\left({5\over 2} \right)^2={25\over 4}，故選\bbox[red, 2pt]{(D)}$$

解答: $$\sqrt{50}-\sqrt{1\over 2}-\sqrt 8 =5\sqrt 2-{\sqrt 2\over 2}-2\sqrt 2={5\over 2}\sqrt 2，故選\bbox[red, 2pt]{(C)}$$

解答: $$\sum_{k=3}^5(2^k+k-3)=2^3+2^4+2^5+3+4+5-3-3-3=56+12-9=59，故選\bbox[red, 2pt]{(B)}$$

解答: $$\cases{\vec a=(5,4)\\ \vec b=(2,0)} \Rightarrow \vec a-\vec b=(3,4) \Rightarrow |\vec a-\vec b|=\sqrt{3^2+4^2}=5，故選\bbox[red, 2pt]{(C)}$$

解答: $$\sin \theta+\cos\theta=0 \Rightarrow \sin \theta \cos\theta \lt 0，故選\bbox[red, 2pt]{(D)}$$

解答: $${5\over x+2}+ {1\over x-2}={5(x-2)+(x+2)\over (x+2)(x-2)} ={6x-8\over x^2-4} ={6\over x} \\ \Rightarrow x(6x-8)=6(x^2-4) \Rightarrow 6x^2-8x=6x^2-24 \Rightarrow 8x=24 \Rightarrow x=3\\ \Rightarrow {x+2\over x-2}={5\over 1}=5，故選\bbox[red, 2pt]{(B)}$$

解答: $$L_1\parallel L_2且 L_3\parallel L_4 \Rightarrow d(L_1,L_2)=d(L_3,L_4) \Rightarrow {5\over 5} ={|4-k| \over 5} \Rightarrow k=9，故選\bbox[red, 2pt]{(A)}$$

解答: $$x,x+10,x+100成等比 \Rightarrow (x+10)^2 =x(x+100) \Rightarrow x^2+20x+100 =x^2+100x \\ \Rightarrow 80x=100 \Rightarrow x={5\over 4} \Rightarrow 公比r={x+10\over x} =1+{10\over x}=1+{10\over 5/4}=9，故選\bbox[red, 2pt]{(A)}$$

解答: $$\cases{2男1女: C^6_2C^4_1= 60\\ 1男2女: C^6_1C^4_2= 36} \Rightarrow 共60+36=96種選法，故選\bbox[red, 2pt]{(D)}$$

解答: $$\tan \theta={5\over 12} \Rightarrow \cases{\sin \theta=5/13\\ \cos \theta=12/13} \Rightarrow \sin 2\theta=2\sin \theta\cos \theta=2\cdot {5\over 13}\cdot {12\over 13}={120\over 169}，故選\bbox[red, 2pt]{(A)}$$

解答: $$|x-1|=8 \Rightarrow \cases{x-1=8\\ x-1=-8} \Rightarrow \cases{x=9\\ x=-7} \Rightarrow 9-7=2，故選\bbox[red, 2pt]{(A)}$$

解答: $$k\vec a+\vec b與\vec c垂直 \Rightarrow (k\vec a+\vec b)\cdot \vec c=0 \Rightarrow ((k,-k,2k)+(2,1,3))\cdot (1,1,1)=0\\ \Rightarrow (k+2,1-k,2k+3)\cdot (1,1,1)= 2k+6=0 \Rightarrow k=-3，故選\bbox[red, 2pt]{(C)}$$

解答: $$\begin{bmatrix}1 & 0 \\2 & 3 \end{bmatrix} \begin{bmatrix}2 & 1 \\1 & 2 \end{bmatrix} =\begin{bmatrix}2+0 & 1+0 \\4+3 & 2+6 \end{bmatrix} =\begin{bmatrix}2 & 1 \\7 & 8 \end{bmatrix}，故選\bbox[red, 2pt]{(D)}$$

解答:

$$假設直線L: x+2y=k,則\cases{A(2,0)\\ L與x=2交於B(2,(k-2)/2)\\ L與y軸交於C(0,k/2) \\原點O(0,0)} \\ 梯形OABC面積＝4 \Rightarrow {(\overline{AB}+ \overline{OC})\times 2\over 2} ={k-2\over 2}+{k\over 2}=4 \Rightarrow 2k-2=8 \Rightarrow k=5，故選\bbox[red, 2pt]{(B)}$$

解答:

$$x^2+4x+y=0 \Rightarrow \cases{A(-4,0)\\ B(0,0)\\ V(-2,4)} \Rightarrow \triangle AVB面積={1\over 2}\times 4\times 4=8，故選\bbox[red, 2pt]{(D)}$$

解答: $$2x^2-10x+11 =2(x-{5+\sqrt 3\over 2}) (x-{5-\sqrt 3\over 2}) \lt 0 \Rightarrow {5-\sqrt 3\over 2}\lt x\lt {5+\sqrt 3\over 2} \\\Rightarrow 1.64 \lt x\lt 3.37 \Rightarrow x=2,3,共2個整數解，故選\bbox[red, 2pt]{(C)}$$

解答: $$h(x)=f(g(x)) \Rightarrow h'(x)=f'(g(x))g'(x) \Rightarrow h'(2)=f'(g(2))g'(2) = f'(1)\cdot 8=4\cdot 8=32\\，故選\bbox[red, 2pt]{(C)}$$

解答: $$\int_0^5 f(x)\,dx =\int_0^2 f(x)\,dx +\int_2^5 f(x)\,dx  \Rightarrow 7=2+ \int_2^5 f(x)\,dx \Rightarrow \int_2^5 f(x)\,dx=7-2=5\\，故選\bbox[red, 2pt]{(A)}$$

解答: $$x^2+y^2-2x-4y-4=0 \Rightarrow (x-1)^2+(y-2)^2=3^2 \Rightarrow \cases{圓心O(1,2) \\ 圓半徑r＝3} \\ 相切\Rightarrow d(O,L)=r \Rightarrow {|3-8+k| \over \sqrt{3^2+4^2}} ={|k-5| \over 5} =3 \Rightarrow |k-5|=15 \Rightarrow k=20，故選\bbox[red, 2pt]{(B)}$$

================ END =================

解題僅供參考，其他歷年試題及詳解