112 學年度科技校院四年制與專科學校二年制
統 一 入 學 測 驗-數學(C)
$$,由上圖可知交接區在第二象限, 故選\bbox[red,2pt]{(B)}$$
解答:$$\cos \angle ACB={3^2+ 5^2-7^2 \over 2\cdot 3\cdot 5} =-{1\over 2} \Rightarrow \cos \angle ACD={1\over 2} \\\Rightarrow \overline{AD} =\overline{AC}\sin \angle ACD=5\cdot {\sqrt 3\over 2}, 故選\bbox[red,2pt]{(D)}$$
解答:$$C=AB = \begin{bmatrix} -1 & 1 & 0 \\ 1 & 0 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\ 1& 0\\ 1 & 1 \end{bmatrix} =\begin{bmatrix} 1 & -1\\ 0 & 1\end{bmatrix} \Rightarrow C^2 =\begin{bmatrix} 1 & -1\\ 0 & 1\end{bmatrix}\begin{bmatrix} 1 & -1\\ 0 & 1\end{bmatrix} =\begin{bmatrix} 1 & -2\\ 0 & 1\end{bmatrix} \\ \Rightarrow d_{12}=-2, 故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{x= \sqrt 2(\cos{\pi\over 4}+ i\sin {\pi\over 4}) =\sqrt 2e^{\pi i/4}\\ y=2(\cos{\pi\over 6 }+ i\sin{\pi \over 6}) =2e^{\pi i/6} \\ z=2(\cos {\pi\over 3}+i \sin{\pi \over 3}) =2e^{\pi i/3}} \Rightarrow {x^2y^4\over z^3} ={2e^{\pi i/2}\cdot 2^4 e^{2\pi i/3} \over 8e^{\pi i}} \\ =4e^{\pi i/6} =2^2(\cos {\pi\over 6}+i \sin{\pi \over 6}) , 故選\bbox[red,2pt]{(A)}$$
解答:$$\cases{A(1,2,3)\\ B(2,4,6)\\ C(3,5,4)} \Rightarrow \cases{\overrightarrow{AB} =(1,2,3)\\ \overrightarrow{AC} =(2,3,1)} \Rightarrow E_1法向量=\vec n_1= \overrightarrow{AB} \times \overrightarrow{AC}=(-7,5,-1)\\ 又E_2法向量\vec n_2= (1,k,-2),由於\vec n_1\bot \vec n_2 \Rightarrow \vec n_1\cdot \vec n_2=0 \Rightarrow -7+5k+2=0 \Rightarrow k=1, 故選\bbox[red,2pt]{(A)}$$
解答:$$將\cases{x=1\\ y=1\\ z=-1}代入\cases{ax+ by+cz=-2\\ bx+cy+az= -4\\ cx+ay+bz=6} \Rightarrow \cases{a+b-c=-2 \cdots(1)\\ b+c-a=-4 \cdots(2)\\ c+a-b=6 \cdots(3)}\\ 因此\cases{ (1)+(2) \Rightarrow 2b=-6 \Rightarrow b=-3\\ (2)+(3) \Rightarrow 2c=2 \Rightarrow c=1\\ (1)+(3) \Rightarrow 2a=4 \Rightarrow a=2} \Rightarrow \cases{(A)ab=-6 \ne 6\\ (B)bc=-3 \ne 3\\ (C)ac= 2\\ (D)abc=-6\ne 6}, 故選\bbox[red,2pt]{(C)}$$
解答:$$|Ax+6| \ge B \Rightarrow \cases{Ax+6 \ge B\\ Ax+6\le -B} \Rightarrow \cases{Ax\ge B-6\\ Ax\le -B-6}\\ \cases{若A\gt 0 \Rightarrow \cases{x\ge (B-6)/A=6\\ x\le (-B-6)/A=-2 } \Rightarrow \cases{A=-3\\ B=-18不合, 違反B\ge 0}\\ 若A\lt 0 \Rightarrow \cases{x\le (B-6)/A =-2\\ x\ge (-B-6)/A=6} \Rightarrow \cases{A=-3\\ B=12}} \\ \Rightarrow 2A+B=-6+12=6, 故選\bbox[red,2pt]{(C)}$$
解答:$$b+2i,-1+ai同為f(x)=0的根 \Rightarrow \cases{a=-2\\ b=-1} \Rightarrow x=-1\pm 2i \Rightarrow x^2+2x+5=0\\ \Rightarrow x^2+2x+5為f(x)的因式\\ 再由長除法可得x^3+4x^2 +9x+10 =(x^2+2x+5)(x+2), 故選\bbox[red,2pt]{(D)}$$
解答:$$2018至2023共5年 \Rightarrow 五年本利和=50(1+r)^5=60, 故選\bbox[red,2pt]{(B)}$$
解答:$$烏龜花了2000\div 250=8小時走完全程\\ 7小時後,兔子醒了,此時烏龜還需1小時到終點\\ 兔子睡醒後1小時跑了\int_0^1 v(t)dt =\int_0^1 27t^2+52t+1262\,dt=1297公尺\\ 離終點還有2000-600-1297=103公尺, 故選\bbox[red,2pt]{(C)}$$
解答:$$\lim_{n\to \infty}\left(\sqrt{n^2+8n-3} -\sqrt{n^2+2n+5} \right) \\=\lim_{n\to \infty}\left({ (\sqrt{n^2+8n-3} -\sqrt{n^2+2n+5})(\sqrt{n^2+8n-3} +\sqrt{n^2+2n+5}) \over \sqrt{n^2+8n-3} +\sqrt{n^2+2n+5}} \right)\\=\lim_{n\to \infty}\left({ 6n-8 \over \sqrt{n^2+8n-3} +\sqrt{n^2+2n+5}} \right) =\lim_{n\to \infty}\left({ 6-8/n \over \sqrt{1+8/n-3/n^2} +\sqrt{1+2/n+5/n^2}} \right) \\={6\over 1+1}=3, 故選\bbox[red,2pt]{(D)}$$
解答:$$假設圓心O(\alpha,\beta) \Rightarrow \overline{OA} = \overline{OB} = \overline{OC}\\ \Rightarrow (\alpha-1)^2+ (\beta-2)^2 =(\alpha-2)^2+ (\beta+3)^2 =(\alpha-2)^2 +(\beta-7)^2 \\ \Rightarrow -2\alpha-4\beta+5 =-4\alpha+6\beta+13 =-4\alpha-14\beta+53 \Rightarrow \cases{\alpha=14\\ \beta=2} \\ \Rightarrow 圓半徑= \overline{OA}= \sqrt{13^2+0}=13 \Rightarrow 圓方程式:(x-14)^2 +(y-2)^2=13^2\\ 將D(a,-10)代入圓\Rightarrow (a-14)^2+ 144=169 \Rightarrow (a-14)^2= 25\\ \Rightarrow a-14=\pm 5 \Rightarrow a=19或9, 故選\bbox[red,2pt]{(A)}$$
解答:$$小明:30000+ 3000(x-1)\ge 80000 \Rightarrow x-1\ge {50\over 3}=16.6 \Rightarrow x=18\\ 小亮:30000 (1+3\%)^{y-1}\ge 80000 \Rightarrow (1+3\%)^{y-1} \ge{8\over 3} \\ \Rightarrow (y-1)\log 1.03 \ge 3\log 2-\log 3 = 3\cdot 0.301-0.4771 =0.4259 \\ \Rightarrow y-1 \ge {0.4259\over 0.0128(試題表頭資料)} \approx 33.3\Rightarrow y=35 \Rightarrow x-y=-17, 故選\bbox[red,2pt]{(D)}$$
解答:$$假設A(0,0)\Rightarrow \cases{O(2,0)\\ B(-6\cos 60^\circ, 6\sin 60^\circ) =(-3,3\sqrt 3)} \\\Rightarrow C=(-3+4\cos 60^\circ, 3\sqrt 3+4\sin 60^\circ) =(-3+2,3\sqrt 3+2\sqrt 3)=(-1,5\sqrt 3)\\ \Rightarrow \overline{OC}= \sqrt{3^2+(5\sqrt 3)^2} =\sqrt{84}= 2\sqrt{21}, 故選\bbox[red,2pt]{(B)}$$
解答:$$M與E成正比, 只有(A),(C)可選,又E\gt 0, 故選\bbox[red,2pt]{(A)}$$
解答:$$考慮均衡原則,主菜與湯只剩下8種組合:\\\qquad (牛,魚),(牛雞),(豬,魚),(豬,雞),(魚,牛),(魚,豬),(雞,牛),(雞豬)\\ 因此套餐共有(5飯)\times (8主菜與湯) \times (4飲料)=160種, 故選\bbox[red,2pt]{(B)}$$
解答:$$由題意可知此橢圓\cases{c=4\div 2=2\\ a=8\div 2=4} \Rightarrow b=\sqrt{4^2-2^2} = 2\sqrt 3, 故選\bbox[red,2pt]{(B)}$$
解答:$$直角\triangle ACB \Rightarrow \overline{AC} =\sqrt{20^2-10^2} =10\sqrt 3\\ 假設\overline{BC}中點O為圓心\Rightarrow \overline{OC}=\overline{OD}=5 \Rightarrow \cos \angle OCD={\overline{OC}^2 +\overline{CD}^2-\overline{OD}^2 \over 2\cdot \overline{OC}\cdot \overline{CD}} \\ \Rightarrow \cos 30^\circ={\sqrt 3\over 2} ={25+\overline{CD}^2-25\over 50\overline{CD}} \Rightarrow \overline{CD}=5\sqrt 3\\ 又\angle ACD=90^\circ \Rightarrow \overline{AD} =\sqrt{\overline{AC}^2 +\overline{CD}^2} =\sqrt{300+ 75}= 5\sqrt{15}, 故選\bbox[red,2pt]{(C)}$$
解答:$$g(x)=ax^3+bx 通過(1,0) \Rightarrow a+b=0\\ 又\cases{f(x)=\sqrt x-x\\ g(x)=ax^3+bx} \Rightarrow \cases{f'(x)={1\over 2\sqrt x}-1\\ g'(x)=3ax^2+b} \Rightarrow \cases{L_1斜率=f'(1)=-{1\over 2}\\ L_2斜率=g'(1) =3a+b=2a} \\ L_1\bot L_2 \Rightarrow -{1\over 2}\times 2a =-1 \Rightarrow a=1\Rightarrow b=-1 \Rightarrow ab=-1, 故選\bbox[red,2pt]{(D)}$$
解答:$$f(x)=x^3 +3x^2-72x-74 \Rightarrow f'(x)=3x^2+6x-72 \Rightarrow f''(x)=6x+6\\ 若f'(x)=0 \Rightarrow 3(x-4)(x+6)=0 \Rightarrow \cases{x=4\\ x=-6} \Rightarrow \cases{f''(4)=30\gt 0\\ f''(-6)=-30 \lt 0} \\ \Rightarrow \cases{f(x)的相對極大值發生於x=-6\\ f(x)的相對極小值發生於x=4}, 故選\bbox[red,2pt]{(D)}$$
解答:$$4秒旋轉2\pi \Rightarrow 4a =2\pi \Rightarrow a={\pi \over 2}, 故選\bbox[red,2pt]{(B)}$$
================ END =====================
解題僅供參考,其他歷屆試題及詳解
沒有留言:
張貼留言