110年身障生升四技二專-數學(C)詳解

110 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：$$\cases{A(2,0)\\ B(-4,3)\\ C(x,y)} \Rightarrow \cases{\overrightarrow{AC} =(x-2,y) \\\overrightarrow{BC} =(x+4,y-3)}\\ 3\overrightarrow{AC} =2\overrightarrow{BC} \Rightarrow (3x-6,3y) =(2x+8,2y-6) \Rightarrow \cases{3x-6=2x+8\\ 3y=2y-6} \Rightarrow \cases{x=14 \\y=-6} \\ \Rightarrow (x,y)=(14,-6)，故選\bbox[red,2pt]{(C)}$$

解答：$$\cases{A(2,-3) \\B(1,3)\\ C(-5,7)}，D=(B+C)\div 2 = (-2,5) \Rightarrow \overline{AD}斜率={5-(-3)\over -2-2}=-2，故選\bbox[red,2pt]{(A)}$$

解答：$$令f(x)=x^3+ax^2+bx+6 ，由於{f(x)\over x^2-2x-3} ={f(x)\over (x-3)(x+1)}為x的多項式，\\代表x-3與x+1均為f(x)的因式\\ \Rightarrow \cases{f(3)=0\\ f(-1)=0} \Rightarrow \cases{27+9a+3b+6=0\\ -1+a-b+6=0} \Rightarrow \cases{3a+b=-11 \\ a-b=-5} \Rightarrow \cases{a=-4\\ b=1} \\ \Rightarrow a+2b= -4+2=-2，故選\bbox[red,2pt]{(B)}$$

解答：$${x^2+5 \over x^3-1} ={A\over x-1} +{Bx+C\over x^2+x+1} \Rightarrow x^2+5=A(x^2+x+1)+(Bx+C)(x-1) \\= (A+B)x^2+ (A-B+C)x+A-C  \Rightarrow \cases{ A+B=1\\ A-B+C=0\\ A-C=5} \Rightarrow \cases{A=2\\ B=-1\\ C=-3} \\ \Rightarrow A+2B+3C= 2-2-9 =-9，故選\bbox[red,2pt]{(A)}$$

解答：$$3x^2-4x-15 \le 0 \Rightarrow (3x+5)(x-3)\le 0 \Rightarrow -5/3\le x \le 3 \\ \Rightarrow x=-1,0,1,2,3，共5個解，故選\bbox[red,2pt]{(C)}$$

解答：$$a_1\cdot a_2\cdot a_3\cdot a_4\cdot a_5 =a_1\cdot a_1r\cdot a_1r^2\cdot a_1r^3\cdot a_1r^4 = a_1^5r^{10} =(a_1r^2)^5 =(a_3)^5 =2^5 =32，故選\bbox[red,2pt]{(A)}$$

解答：$$\cases{\cos 233^\circ \lt 0 \\ \tan(-542^\circ)=\tan(360^\circ \times 2-542^\circ)= \tan 178^\circ \lt 0\\ -\sin 447^\circ =-\sin 87^\circ \lt 0\\ \sec 629^\circ= \sec 269^\circ \lt 0} \\ \Rightarrow \cases{\cos 233^\circ\tan(-542^\circ ) \gt 0\\ -\sin 447^\circ \sec 629^\circ \gt 0} \Rightarrow (\cos 233^\circ\tan(-542^\circ ), -\sin 447^\circ \sec 629^\circ)在第1象限\\，故選\bbox[red,2pt]{(A)}$$

解答：$$a+b-c:b+c-a:a+c-b= 1:3:5 \Rightarrow \cases{a+b-c= k\cdots(1)\\ b+c-a=3k \cdots(2)\\ a+c-b=5k\cdots(3)}\\ (1)+(2)+(3) \Rightarrow a+b+c=9k \cdots(4) \cases{(4)-(1) \Rightarrow 2c=8k\\ (4)-(2) \Rightarrow 2a=6k\\ (4)-(3) \Rightarrow 2b=4k} \Rightarrow \cases{a=3k\\ b=2k\\ c=4k} \\ \Rightarrow \sin A:\sin B:\sin C= a:b:c = 3:2:4，故選\bbox[red,2pt]{(D)}$$

解答：$${\sqrt 3-i\over \sqrt 3+i} ={(\sqrt 3-i)^2 \over (\sqrt 3+i)(\sqrt 3-i)} ={2-2\sqrt 3i \over 4} ={1\over 2}-{\sqrt 3\over 2}i = \cos {-\pi\over 3} +i\sin{-\pi\over 3} =e^{-{\pi \over 3}i} \\ \Rightarrow \left({\sqrt 3-i\over \sqrt 3+i} \right)^9 =\left(e^{-{\pi \over 3}i} \right)^9 =e^{-{3\pi  }i} =\cos (-3\pi) -i\sin(3\pi)=-1，故選\bbox[red,2pt]{(B)}$$

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

解答：$$\cases{a=\log_3 2\\ b=\log_3 7} \Rightarrow \log_6 28= {\log_3 28\over \log_3 6} = {2\log_3 2 +\log_3 7\over 1+\log_3 2} ={2a+b\over 1+a}，故選\bbox[red,2pt]{(D)}$$

解答：$$令\cases{7位同學總重a\\ 不正確的3位同學總重b=67.5\times 3\\ 正確的3位同學總重c=68.5\times 3} \\\Rightarrow (a+b)\div 10=64.8 \Rightarrow a= 648-b = 648-67.5\times 3\\ \Rightarrow 直正平均體重=(a+c)\div 10 =(648-67.5\times 3+68.5\times 3)\div 10=651\div 10=65.1\\，故選\bbox[red,2pt]{(C)}$$

解答：$$個位數是0的有5\times 4\times 3= 60個\\個位數是2的有4\times 4\times 3=48個(千位數不為0) \\個位數是4的有4\times 4\times 3=48個(千位數不為0) \\因此偶數共有60+48\times 2=156個，故選\bbox[red,2pt]{(D)}$$

解答：$$\cases{A:1R3W\\ B:3W}  \Rightarrow\cases{從A抽1R1W至B \stackrel{(2)}{\Rightarrow} \cases{A:2W\\ B:1R4W} ，再從B抽1R1W至A \stackrel{(4)}{\Rightarrow} \cases{A:1R3W\\ B:3W}\\從A抽2W至B \stackrel{(3)}{\Rightarrow}  \cases{A:1R1W\\ B:5W}，再從B抽2W至A \stackrel{(5)}{\Rightarrow} \cases{A:1R3W\\ B:3W}}\\\cases{  (2)\to (4)的機率={1\over 2}\times {4\over 10}={{1\over 5}}\\ (3)\to (5)的機率={1\over 2}\times 1={1\over 2}} \quad\Rightarrow 機率和={1\over 5}+{1\over 2}={7\over 10}，故選\bbox[red,2pt]{(D)}$$

解答：$$\begin{vmatrix} x^2 & 4& 5\\ 3x+1 & 7 & 3\\ 4x & 8 & 4\end{vmatrix} =\begin{vmatrix} x^2 & -6& 5\\ 3x+1 & 1 & 3\\ 4x & 0 & 4\end{vmatrix} = 4x^2-72x-20x+72x+24= 4x^2-20x+24=0\\ \Rightarrow x^2-5x+6=0 \Rightarrow (x-3)(x-2)=0 \Rightarrow \cases{a=3\\ b=2} \Rightarrow a^2+b^2 =9+4=13，故選\bbox[red,2pt]{(B)}$$

解答：$$\begin{vmatrix} 1 & -1& 1\\  1 & 1 & -2\\ 2 & 4 & k\end{vmatrix} =\begin{vmatrix} 1 & 0 & 0\\  1 & 2 & -3\\ 2 & 6 & k-2\end{vmatrix} = 2k-4+18=2k+14=0 \Rightarrow k=-7，故選\bbox[red,2pt]{(A)}$$

解答：$$x=-{1\over 3}y^2+2y-2 \Rightarrow 3x=-(y^2-6y+9)+3 \Rightarrow 3(x-1)=-(y-3)^2 \\ \Rightarrow (y-3)^2 =4\cdot (-{3\over 4})(x-1) \Rightarrow \cases{頂點坐標(1,3)\\ c=-3/4} \Rightarrow 焦點坐標(1-{3\over 4},3)=({1\over 4},3)，故選\bbox[red,2pt]{(C)}$$

解答：$$圓C:5x^2+5y^2-20x-10y+24=0 \Rightarrow 5(x^2-4x+4)+5(y^2-2y+1)=1\\ \Rightarrow (x-2)^2+(y-1)^2 ={1\over 5}\Rightarrow \cases{圓心O(2,1)\\ 半徑r={1\over \sqrt 5}}\\圓與直線交於相異兩點\Rightarrow d(O,L) \lt r \Rightarrow {|4-1-k|\over \sqrt 5} ={|3-k|\over \sqrt 5} \lt {1\over \sqrt 5} \Rightarrow 只有k=3符合條件\\，故選\bbox[red,2pt]{(C)}$$

解答：$$f(x)= (x^2-4x)^3 \Rightarrow f'(x)=3(x^2-4x)^2(2x-4) \Rightarrow f'(1)=3\cdot (-3)^2\cdot (-2)=-54，故選\bbox[red,2pt]{(B)}$$

解答：

$$\cases{紅色面積=\int_{-2}^2 4-x^2\;dx =\left. \left[ 4x-{1\over 3}x^3\right] \right|_{-2}^2 ={32\over 3}\\ 藍色面積=\int_2^3 x^2-4\;dx =\left. \left[  {1\over 3}x^3 -4x\right] \right|_{2}^3 ={7\over 3}} \\ \Rightarrow 紅+藍={39\over 3}=13，故選\bbox[red,2pt]{(D)}$$