115 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(B)
解答:$$\cases{\overrightarrow{OS} =(0,2) \\ \overrightarrow{OT}=(3,0)} \Rightarrow \overrightarrow{OS}-\overrightarrow{OT} =(-3,2) =\overrightarrow{OB}，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{兩根之和=\alpha+\beta+ \alpha-\beta=2\alpha =8\\ 兩根之積=(\alpha+\beta)(\alpha-\beta)= \alpha^2-\beta^2=5(\alpha-\beta)} \Rightarrow \alpha=4 \Rightarrow 16-\beta^2=5(4-\beta) \\ \Rightarrow \beta^2-5\beta+4=0 \Rightarrow (\beta-4)(\beta-1)=0 \Rightarrow \beta=1 (\beta \ne \alpha \Rightarrow \beta\ne 4) \Rightarrow \alpha\beta=4\cdot 1=4，故選\bbox[red, 2pt]{(A)}$$
解答:$$\cases{L_1:2x+3y =4 \Rightarrow L_1與x軸交於(2,0) \Rightarrow a=2\\ L_2與L_1平行\Rightarrow L_2的斜率=L_1的斜率m=-{2\over 3}} \Rightarrow (a,m) = \left( 2,-{2\over 3} \right)，故選\bbox[red, 2pt]{(A)}$$
解答:$$2x^2+7x+3=(2x+1)(x+3) \le 0 \Rightarrow -3\le x\le -{1\over 2} \Rightarrow \cases{a= -3\\ b= -1/2}，故選\bbox[red, 2pt]{(C)}$$
解答:$$ \cases{A(-1,0) \\B(0,-5) \\C(a,b) \\D(2,3)} \Rightarrow \cases{\overrightarrow{AD}=(3,3) \\ \overrightarrow{BC}=(a,b+5) } \Rightarrow \overrightarrow{AD} =\overrightarrow{BC} \Rightarrow \cases{a=3\\ b+5=3} \Rightarrow \cases{a=3\\ b=-2} \\\Rightarrow \overline{BC}中點= {1\over 2}(B+C)={1\over 2}(3,-7)，故選\bbox[red, 2pt]{(B)}$$
解答:$$假設餘式為ax+b \Rightarrow f(x)=x^4-3x^3+2x^2-x+1=(x-1)(x-2)Q(x)+ax+b \\ \Rightarrow \cases{f(1)=0=a+b\\ f(2)=-1=2a+b} \Rightarrow \cases{a=-1\\ b=1} \Rightarrow 餘式為-x+1，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{2026=360\times 5+226 \Rightarrow a=226^\circ-360^\circ=-134^\circ \\b=-115^\circ +360^\circ=245^\circ} \Rightarrow \cases{a+b=111^\circ\\ a-b=-379^\circ} \\ \Rightarrow \cases{\tan(a-b)= \tan(-379^\circ) \lt 0\\ \sin(a+b)=\sin (111^\circ) \gt 0} \Rightarrow 第二象限，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{A(0,0) \\B(4,0) \\C(-2,2\sqrt 3)} \Rightarrow \cases{\overrightarrow{AB}=(4,0) \\ \overrightarrow{AC}=(-2,2\sqrt 3)} \Rightarrow \cos \theta={\overrightarrow{AB} \cdot \overrightarrow{AC} \over |\overrightarrow{AB}||\overrightarrow{AC}|} ={-8\over 4\cdot 4}=-{1\over 2}，故選\bbox[red, 2pt]{(D)}$$
解答:$$L_1: 4x+(a+5)y-10=0通過(a,1) \Rightarrow 4a+(a+5)\cdot 1 -10=0 \Rightarrow 5a=5 \Rightarrow a=1 \\ L_3與L_2垂直\Rightarrow L_3斜率=2 \Rightarrow L_3: y=2(x-a)+1 \Rightarrow y=2(x-1)+1 \Rightarrow 2x-y-1=0\\，故選\bbox[red, 2pt]{(A)}$$
解答:$$沒抽中甲的機率p={2\over 3} \Rightarrow 連三次都沒抽中甲的機率=p^3={8\over 27}，故選\bbox[red, 2pt]{(C)}$$
解答:$$f(-2)=d= g(-2)=-32+12-4+1=-23 \Rightarrow f(-1)=4+b+c-23=g(-1)=-2 \\ \Rightarrow b+c=-2-4+23=17，故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{2x+3y=4\\ ax+by=c} 有無限多組解\Rightarrow {2\over a}={3\over b}={4\over c} ={1\over k} \Rightarrow \cases{a=2k\\ b=3k\\ c=4k} \\ (A) \times:1000ax+1000by=c \Rightarrow 2000x+3000y=4 \Rightarrow 2x+3y={1\over 250} \\\qquad \Rightarrow 兩平行線\cases{2x+3y=4\\ 2x+3y={1\over 250}}無解 \\(B)\times: ax+by=1000c \Rightarrow 2x+3y=4000\Rightarrow 兩平行線\cases{2x+3y=4\\ 2x+3y=4000}無解 \\ (C)\times: 1000ax+by=c \Rightarrow 2000x+3y=4 \Rightarrow \cases{2x+3y=4\\ 2000x+3y=4}有唯一解 \\(D)\bigcirc: 理由同(C)\\，故選\bbox[red, 2pt]{(D)}$$
解答:$$原點(0,0)符合兩不等式\cases{26x+15y\le 2026\\ 15x+26y\le 2026} \Rightarrow (B)與(C)是錯誤的\\ 又兩直線\cases{26x+15y= 2026\\ 15x+26y= 2026}斜率均為負值，，故選\bbox[red, 2pt]{(A)}$$
解答:$$(A)\bigcirc: 由圖(二)可知正確\\ (B)\bigcirc:1970-2025出生人數最多是在1976年,約43萬人;最少是2025年,107812人;兩者相距大於30萬 \\(D)\bigcirc:由表(一)可知逐年減少\\，故選\bbox[red, 2pt]{(C)}$$
解答:$$C^7_2= 21，故選\bbox[red, 2pt]{(C)}$$
解答:$$10000\times {1\over 1000}+5000\times {2\over 1000}+1000\times {5\over 1000}+ 500\times {1\over 100} =10+10+5+5=30 \\ \Rightarrow 30-50\times \left( 1-{1\over 1000}-{2\over 1000}-{5\over 1000}-{1\over 100} \right) =30-49.1 \approx -20，故選\bbox[red, 2pt]{(C)}$$
解答:$$\cases{\tan(1847^\circ=360^\circ\times 5+47^\circ) =\tan 47^\circ=1.072\\ \tan(2126^\circ= 360^\circ\times 5+326^\circ)= \tan (-34^\circ) =-0.675\\ \tan(2205^\circ= 360^\circ\times 6+45^\circ) =\tan 45^\circ=1} \\ \Rightarrow \tan(2026^\circ=360^\circ\times 5+226^\circ) =\tan (226^\circ =180^\circ+46^\circ)=\tan 46^\circ \Rightarrow \tan 45^\circ \lt\tan 46^\circ \lt \tan 47^\circ \\ \Rightarrow 1\lt \tan 2026^\circ \lt 1.072，故選\bbox[red, 2pt]{(B)}$$
解答:$$\angle ACB=180^\circ-60^\circ-75^\circ=45^\circ \Rightarrow 正弦定理:{\overline{AC} \over \sin 75^\circ} ={\overline{AB} \over \sin 45^\circ} \Rightarrow {\overline{AC} \over (\sqrt 6+\sqrt 2)/4} ={100 \over \sqrt 2/2} \\ \Rightarrow \overline{AC} ={25(\sqrt 6+\sqrt 2) \over \sqrt 2/2} ={50(\sqrt 6+\sqrt 2) \over \sqrt 2} =50(\sqrt 3+1)，故選\bbox[red, 2pt]{(A)}\\註:試題有提供參考數據:\sin 75^\circ={\sqrt 6+\sqrt 2\over 4}$$
解答:$$假設\overline{BC}=x \Rightarrow 餘弦定理: \cos 60^\circ={8^2+x^2-7^2\over 2\cdot 8\cdot x} \Rightarrow {1\over 2}={x^2+15\over 16x} \Rightarrow x^2-8x+15=0 \\ \Rightarrow (x-5)(x-3)= 0 \Rightarrow x=5，故選\bbox[red, 2pt]{(A)}$$
解答:$$五張卡片都顯示「無」的機率為{1\over 2^5}={1\over 32} \Rightarrow 至少有一張「有」的機率=1-{1\over 32}={31\over 32} \\ 總共有2^5=32種可能，故32\times {31\over 32}=31，故選\bbox[red, 2pt]{(B)}$$
解答:$$\cases{行程I費用=x+x-3000=2x-3000\\ 行程II費用=x+2000+ (x+2000)\times 0.8=1.8x+3600} \Rightarrow 2x-3000=1.8x+3600 \\ \Rightarrow 0.2x=6600 \Rightarrow x=6600\times 5=33000，故選\bbox[red, 2pt]{(B)}$$
解答:$$A={110\over 8}=13.75 \Rightarrow C=14\\ B={\log 3\over \log(1+{8\over 100})} ={\log 3\over \log{108\over 100}} ={\log 3\over \log 108-2}={\log 3\over \log(3^3\cdot 2^2)-2} ={\log 3\over 3\log 3+2\log 2-2} \\={0.4771\over 3\cdot 0.4771+2\cdot 0.301-2} ={4771\over 333} \approx 14.3 \Rightarrow D=15 =C+1，故選\bbox[red, 2pt]{(C)}$$
解答:$$圓:x^2-8x+(y-7)^2=a \Rightarrow (x-4)^2+(y-7)^2=a+16 \Rightarrow \cases{圓心P(4,7)\Rightarrow h=4 \\圓半徑r=\sqrt{a+16}} \\ 圓與L相切 \Rightarrow d(P,L)=r \Rightarrow \sqrt{a+16}=\sqrt 2 \Rightarrow a=-14 \Rightarrow h+a=4-14=-10，故選\bbox[red, 2pt]{(C)}$$
解題僅供參考，其他統測試題及詳解
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