115年身障生升四技二專-數學(C)詳解

115 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)

解答:$$\overline{AP} =3\overline{PB} \Rightarrow P=(3B+A)/4={1\over 4}[(30,-21)+(2,1)] ={1\over 4}(32,-20) =(8,-5)，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{\vec a=(2,1) \\\vec b=(m+1,n-1) \\\vec c=(m-2,n+2)} \Rightarrow \vec a+2\vec b-3\vec c=(2,1)+(2m+2,2n-2)-(3m-6,3n+6) \\=(2-m,-7-n) =0  \Rightarrow \cases{10-m=0\\-7-n=0} \Rightarrow \cases{m=10 \\ n=-7} \Rightarrow m+n=3，故選\bbox[red, 2pt]{(B)}$$

解答:$$\cases{A(1,-1) \\B(3,0) \\C(x,2) \\D(5,y)} \Rightarrow \cases{\overrightarrow{AB}=(2,1) \\\overrightarrow{AC}= (x-1,3) \\ \overrightarrow{AD} =(4,y+1) } \Rightarrow \cases{\displaystyle {2\over x-1} ={1\over 3} \\ \displaystyle {2\over 4}={1\over y+1}} \Rightarrow \cases{x=7\\ y=1}\Rightarrow x+y=8，故選\bbox[red, 2pt]{(C)}$$

解答:$$x^2+y^2-2x+8y+13=0 \Rightarrow (x^2-2x+1)+(y^2+8y+16)+13=1+16 \\ \Rightarrow (x-1)^2+(y+4)^2=4 \Rightarrow 圓心(1,-4) \Rightarrow {1\over 2}((1-4)+(5,8)) =(3,2)，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{前4項之和={1\over 2}(2a_1+3d)\cdot 4=28 \\ 前7項之和={1\over 2}(2a_1+6d)\cdot 7=84} \Rightarrow \cases{2a_1+3d=14 \\ a_1+3d=12} \Rightarrow \cases{a_1=2\\ d=10/3} \\ \Rightarrow 前10項之和={1\over 2}(2a_1+9d)\cdot 10 =5(4+30) =170，故選\bbox[red, 2pt]{(A)}$$

解答:$$A= \begin{bmatrix}a_{11}& a_{12}\\ a_{21}&  a_{22}\end{bmatrix} = \begin{bmatrix}-1& -3\\ 0& -2 \end{bmatrix} \Rightarrow A^2 = \begin{bmatrix}-1& -3\\ 0& -2 \end{bmatrix}  \begin{bmatrix} -1& -3\\ 0& -2 \end{bmatrix} = \begin{bmatrix}1& 9\\0 &4 \end{bmatrix} \Rightarrow \cases{c_{11}=1\\ c_{22}=4} \\ \Rightarrow c_{11}+c_{22} =1+4=5，故選\bbox[red, 2pt]{(D)}$$

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

解答:$$f(2)=f(4) \Rightarrow |2-c|=|4-c| \Rightarrow c=3 \Rightarrow f(x)=|x-3| \Rightarrow k=f(2)=1，故選\bbox[red, 2pt]{(A)}$$

解答:$${\sin \theta+\cos \theta\over \sin \theta-\cos \theta} ={\sin \theta/\cos \theta+1\over \sin \theta/\cos \theta-1} = {\tan \theta+1\over \tan \theta-1}=2 \Rightarrow \tan \theta+1=2\tan \theta-2 \Rightarrow \tan \theta=3\\，故選\bbox[red, 2pt]{(D)}$$

解答:$$f(x)=\cos^2 x+\cos x+2=(\cos x+{1\over 2})^2+{7\over 4} \Rightarrow \cases{M= (1+1/2)^2+7/4=4\\ m=7/4} \\ \Rightarrow M+m={23\over 4}，故選\bbox[red, 2pt]{(B)}$$

解答:$$\vec a\cdot \vec b= |\vec a||\vec b| \cos 60^\circ=2\cdot 1\cdot {1\over 2}=1 \Rightarrow \left| 2\vec a-\vec b\right|^2 =(2\vec a-\vec b)\cdot (2\vec a-\vec b) =4|\vec a|^2-4\vec a\cdot \vec b+|\vec b|^2 \\=4\cdot 4-4\cdot 1+1=13 \Rightarrow \left| 2\vec a-\vec b\right|=\sqrt{13}，故選\bbox[red, 2pt]{(A)}$$

解答:$$x^2+mx+n=0 的兩根為2+3i, 2-3i \Rightarrow \cases{兩根之和=4=-m \Rightarrow m=-4\\ 兩根之積=13=n} \\ \Rightarrow m+n=-4+13=9，故選\bbox[red, 2pt]{(D)}$$

解答:$$A(x+2)+B(2x-1)= (A+2B)x+2A-B=x+17 \Rightarrow \cases{A+2B=1\\ 2A-B=17} \Rightarrow \cases{A=7\\ B=-3} \\ \Rightarrow A+B=7-3=4，故選\bbox[red, 2pt]{(D)}$$

解答:$$\cases{首項a_1:1/2\\ 公比:r} \Rightarrow {1\over 2}\cdot r^3=-{4\over 27} \Rightarrow r=-{2\over 3} \Rightarrow \cases{a_2 ={1\over 2}r\\ a_3={1\over 2}r^2} \Rightarrow a_2+a_3 ={1\over 2} \left( -{2\over 3} +{4\over 9}\right)=-{1\over 9}\\，故選\bbox[red, 2pt]{(C)}$$

解答:$$\cases{9人選3位有C^9_3=84種\\ 3位都是男性有C^5_3=10種\\ 3位都是女性有C^4_3=4種} \Rightarrow 兩種性別都有:84-10-4=70種，故選\bbox[red, 2pt]{(D)}$$

解答:$$(a+a^{-1})^3 =a^3+a^{-3}+3(a+a^{-1}) \Rightarrow 5^3=a^3+a^{-3}+3\cdot 5 \Rightarrow a^3+a^{-3}=110，故選\bbox[red, 2pt]{(C)}$$

解答:$$f(x)={3x-2\over x^2+1} \Rightarrow f'(x) ={3\over x^2+1} -{(3x-2)(2x) \over (x^2+1)^2} \Rightarrow f'(1)={3\over 2}-{2\over 4}=1，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{A(1,1,2)\\ B(2,-1,2) \\C(3,-1,5)} \Rightarrow \cases{\overrightarrow{AB} =(1,-2,0) \\ \overrightarrow{AC}=(2,-2,3)} \Rightarrow \overrightarrow{AB} \times \overrightarrow{AC}=(-6,-3,2) \\\Rightarrow E:-6(x-1)-3(y-1)+2(z-2)=0 \Rightarrow -6x-3y+2z+5=0 \\\Rightarrow d(P,E)={|-12-9+2+5|\over \sqrt{6^2+3^2+2^2}} =2，故選\bbox[red, 2pt]{(A)}$$

解答:$$\cases{y\ge0\\ 3x+2y-6 \le 0\\3x-2y+6\ge 0} 所圍區域頂點坐標\cases{A(-2,0) \\B(2,0)\\ C(0,3)} \Rightarrow \triangle ABC面積={1\over 2} \begin{vmatrix} -2& 0& 1\\2& 0& 1\\0&3& 1 \end{vmatrix}=6\\，故選\bbox[red, 2pt]{(B)}$$

解答:$$\int_{-1}^3 |x-2|\,dx =\int_{-1}^2 (2-x)\,dx +\int_2^3 (x-2)\,dx = \left. \left[ 2x-{1\over 2}x^2 \right] \right|_{-1}^2 + \left. \left[ {1\over 2}x^2-2x \right] \right|_2^3 \\={9\over 2}+{1\over 2}=5，故選\bbox[red, 2pt]{(C)}$$

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解題僅供參考，身障升大學歷年試題及詳解