109學年度科技校院四年制與專科學校二年制
統一入學測驗試題本數學(C)詳解
$$2^- -2 <0 \Rightarrow \lim_{x\to 2^-} \sqrt {x-2} \ne 0， 故選\bbox[red,2pt]{(B)}$$
解：$$\cases{a=\tan 480^\circ =\tan 120^\circ <0 \\b=\sec 135^\circ <0\\ c=\cos(-60^\circ)>0} \Rightarrow \cases{(b,c)=(-,+) \\ (a,b)=(-,-) \\ (c,a)=(+,-) \\ (c,b)=(+,-)}，故選\bbox[red,2pt]{(A)}。$$
解：$$f(x)=p(x)(x-1)(x^2+x+1)+ 3x^2+5x-2 = p(x)(x-1)(x^2+x+1)+ 3(x^2+x+1) +2x-5 \\= (x^2+x+1)(p(x)(x-1)+3)+2x-5 \Rightarrow 餘式為2x-5，故選\bbox[red,2pt]{(B)}。$$
解：$$(A) \times: 最高分100，最低分20 \Rightarrow 全距=100-20=80\\ (B)\bigcirc: 全班39人，中位數位於第20名，其得分在60-70之間\\ (C)\times: 標準差一定小於全距(80) \\(D)\times: 39\times 90\%= 35.1 \Rightarrow PR值高於90的為排名前4名\\，故選\bbox[red,2pt]{(B)}。$$
解：
$$6\times 14=84，故選\bbox[red,2pt]{(D)}。$$
解：$$\cos 21^\circ = \cos(80^\circ -59^\circ) = \cos 80^\circ\cos 59^\circ +\sin 80^\circ\sin 59^\circ =b\sqrt{1-a^2}+a\sqrt{1-b^2}，故選\bbox[red,2pt]{(A)}。$$
解：$${(x-4)^2 \over 25}+ {(y+2)^2 \over 144}=1 \Rightarrow \cases{中心C(4,-2) \\ a=12 \\b=5} \\ (A)\times: a^2=b^2+c^2 \Rightarrow 144=25+c^2 \Rightarrow c=\sqrt{119} \Rightarrow 焦距=(4,-2\pm \sqrt{119}) \\(B) \times:長軸頂點(4,-2\pm 12) \\(C) \times: 短軸頂點(4\pm 5,-2) \\(D)\bigcirc: 正焦弦長={2b^2 \over a}={2\times 25 \over 12} ={25\over 6}\\，故選\bbox[red,2pt]{(D)}。$$
解：$$(\sqrt 3+i)z = -2\sqrt 3+2i \Rightarrow z={-2\sqrt 3+2i \over \sqrt 3+i} ={(-2\sqrt 3+2i)(\sqrt 3-i) \over (\sqrt 3+i)(\sqrt 3-i)} = { -4+4\sqrt 3i\over 4} =-1+\sqrt 3 i \\= 2(-{1\over 2}+{\sqrt 3\over 2}i) = 2(\cos {2\pi \over 3} +\sin {2\pi \over 3}) \Rightarrow 主幅角為{2\pi \over 3}，故選\bbox[red,2pt]{(B)}。$$
解：$$\cases{a_5=41 \\ a_{13}=73} \Rightarrow \cases{a_1+4d=41 \\ a_1+12d=73 } \Rightarrow \cases{a_1=25 \\d=4} \Rightarrow \cases{a_{19}=25+18\times 4=97 \\ a_{20}=25+19\times 4=101} \\ \Rightarrow 4月20日至30日，共11天投球超過100個，故選\bbox[red,2pt]{(B)}。$$
解：
解：$$\cos 3x的最大值為1 \Rightarrow 2\cos 3x-1 的最大值為2-1=1\\又2\cos 3x-1=0 \Rightarrow \cos 3x={1\over 2} \Rightarrow 3x={\pi \over 3}+n\pi ,n=0,1,2,... \Rightarrow x={\pi \over 9}+{n\pi \over 3},n=0,1,2,...\\由於x\in [0,2\pi] \Rightarrow x= {\pi \over 9},{4\over 9}\pi,{7\over 9}\pi,{11\over 9}\pi,{14\over 9}\pi,{17\over 9}\pi，共6個解\\ 因此\cases{a=6 \\b=1} \Rightarrow ab=6，故選\bbox[red,2pt]{(A)}。$$
解：$$10(1+3\%)^r=20 \Rightarrow (1+3\%)^r=2 \Rightarrow 1.03^r=2 \Rightarrow r\log 1.03=\log 2 \Rightarrow r={\log 2 \over \log 1.03} = {0.301 \over 0.0128}\\ \approx 23.5 \Rightarrow 20歲+23.5年=43.5歲，即44歲 ，故選\bbox[red,2pt]{(C)}$$
解：$$f(x)= x^3-3x^2-24x+32 \Rightarrow f'(x)=3x^2-6x-24 \Rightarrow f''(x)=6x-6\\ f'(x)=0 \Rightarrow 3x^2-6x-24=0 \Rightarrow x^2-2x-8=0 \Rightarrow (x-4)(x+2)=0 \Rightarrow f(4),f(-2)有極值;\\ 又\cases{f''(4)=24-6>0 \Rightarrow f(4)=-48為極小值\\ f''(-2) =-12-6<0 \Rightarrow f(-2)=60為極大值};\\ \cases{-2\in [-3,3] \\f(-3)=50 \\ f(3)=-40} \Rightarrow \cases{ f(-2)=60=m \\ f(3)=-40=n} \Rightarrow m-n=100，故選\bbox[red,2pt]{(C)}。$$
解：$$\begin{array}{} x & y& z \\\hline 3-6 & 1 & 2 \\ 3-6 & 2 & 1\\ \end{array} \Rightarrow 出現BG有4\times {4!\over 2} (x,x,y,z排列數)=4\times 12=48種情形\\，丟4顆骰子有6^4情形，因此機率為{48 \over 6^4} = {1\over 27}，故選\bbox[red,2pt]{(C)}。$$註:(1,1,1,2)或(2,2,2,1)不是BG，要重新擲骰子!
解：$$將P代入曲線\Rightarrow k+k^2+2-4k+k-1=0 \Rightarrow k^2-2k+1=0 \Rightarrow (k-1)^2=0 \Rightarrow k=1\\ k=1代回原曲線\Rightarrow x^2+y^2+2x-4y+1-1=0 \Rightarrow (x+1)^2 +(y-2)^2 = 5為一圓，故選\bbox[red,2pt]{(A)}。$$
解：$$\log_{10-x^2}(x^2+3x+2)有意義 \Rightarrow \cases{10-x^2>0 \\ 10-x^2 \ne 1 \\ x^2+3x+2 >0 } \Rightarrow \cases{\sqrt{10}>x >-\sqrt{10} \\ x\ne \pm 3 \\ (x+{3\over 2})^2 -{1\over 4}>0 } \\\Rightarrow \cases{x=-3,-2,-1,0,1,2,3 \\ x\ne \pm 3 \\ x+{3\over 2} > {1\over 2},x+{3\over 2} <- {1\over 2}} \Rightarrow \cases{x=-2,-1,0,1,2 \\ x>-1, x<-2} \Rightarrow x=0,1,2，共3個，故選\bbox[red,2pt]{(A)}。$$
解：
$$f(x)=\begin{cases} 2x-1 & x>2 \\ x^2-2x+3 & x\le 2\end{cases} \Rightarrow f'(x)=\begin{cases} 2 & x>2 \\ 2x-2 & x\le 2\end{cases} \Rightarrow 2=2\times 2-2 \Rightarrow f'(2)=2\\，故選\bbox[red,2pt]{(B)}。$$
解：$$\alpha,\beta 為x^2+5x+k=0之二根 \Rightarrow \cases{\alpha+\beta=-5 \cdots(1)\\ \alpha\beta=k \cdots(2)}\\依題意\cases{f(x)除以x-\alpha的餘式為 -1 \\ f(x)除以x-\beta的餘式為2} \Rightarrow \cases{ f(\alpha)= -1 \\ f(\beta)=2} \Rightarrow \cases{ 2\alpha^2+7\alpha+5=-1 \\ 2\beta^2+7\beta+5=2} \\ \Rightarrow \cases{2\alpha^2+7\alpha+6=0 \\ 2\beta^2+7\beta+3=0} \Rightarrow \cases{(2\alpha+3)(\alpha+2)=0 \\(2\beta+1)(\beta+3)=0} \Rightarrow \cases{\alpha=-3/2, -2 \\ \beta=-1/2,-3} \Rightarrow 由(1)可知\cases{\alpha=-2 \\ \beta=-3}\\\Rightarrow 代入(2):k=\alpha\beta=(-2)\times (-3)=6，故選\bbox[red,2pt]{(C)}。$$
解：$$假設\cases{每台A機器原先每日產能為a\\ 每台B機器原先每日產能為b} \Rightarrow 5a+3b=11070\cdots(1)\\ 因應疫情\cases{每台A機器每日產能增為1.5a\\ 每台B機器每日產能增為1.5b} \Rightarrow (5+3)\times 1.5a+(3+9)\times 1.5b=42120 \\ \Rightarrow 2a+3b=7020 \cdots(2);由(1)-(2)\Rightarrow 3a=4050 \Rightarrow a=1350，故選\bbox[red,2pt]{(A)}。$$
解：$$\begin{vmatrix}a_1-2b_1-3c_1 & a_1-2c_1 & a_1 \\ a_2-2b_2-3c_2 & a_2-2c_2 & a_2 \\a_3-2b_3-3c_3 & a_3-2c_3 & a_3 \end{vmatrix} \stackrel{C_2-C_3}{=} \begin{vmatrix}a_1-2b_1-3c_1 & -2c_1 & a_1 \\ a_2-2b_2-3c_2 & -2c_2 & a_2 \\a_3-2b_3-3c_3 & -2c_3 & a_3 \end{vmatrix} =-2\begin{vmatrix}a_1-2b_1-3c_1 & c_1 & a_1 \\ a_2-2b_2-3c_2 & c_2 & a_2 \\a_3-2b_3-3c_3 & c_3 & a_3 \end{vmatrix}\\ =2\begin{vmatrix}a_1-2b_1-3c_1 & a_1 & c_1 \\ a_2-2b_2-3c_2 & a_2 & c_2 \\a_3-2b_3-3c_3 & a_3 & c_3 \end{vmatrix}\stackrel{C_1-C_2+3C_3}{=} 2\begin{vmatrix}-2b_1 & a_1 & c_1 \\ -2b_2 & a_2 & c_2 \\-2b_3 & a_3 & c_3 \end{vmatrix} =-4\begin{vmatrix}b_1 & a_1 & c_1 \\ b_2 & a_2 & c_2 \\b_3 & a_3 & c_3 \end{vmatrix} \\ =4\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\a_3 & b_3 & c_3 \end{vmatrix} \Rightarrow \begin{vmatrix}a_1-2b_1-3c_1 & a_1-2c_1 & a_1 \\ a_2-2b_2-3c_2 & a_2-2c_2 & a_2 \\a_3-2b_3-3c_3 & a_3-2c_3 & a_3 \end{vmatrix}=8=4\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\a_3 & b_3 & c_3 \end{vmatrix} \\ \Rightarrow \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\a_3 & b_3 & c_3 \end{vmatrix}=2，故選\bbox[red,2pt]{(C)}$$
解：$$\cases{A(1,1) \\B(5,-2) \\C(5,2) \\ \overrightarrow{DC}=(x,y)} \Rightarrow \cases{\overrightarrow{AB}=(4, -3) \\\overrightarrow{AC}=(4,1)\\ \overrightarrow{DC}=(x,y)} \Rightarrow \overrightarrow{AD}={\overrightarrow{AC} \cdot \overrightarrow{AB}\over |\overrightarrow{AB}|}\times{\overrightarrow{AB} \over |\overrightarrow{AB}|} \\= {(4,1)\cdot (4,-3) \over \sqrt{4^2+3^2}}\times {(4,-3) \over \sqrt{4^2+3^2}} = {13\over 25}(4,-3) \\ \Rightarrow D={13\over 25}(4,-3)+(1,1) = \left( {77 \over 25}, -{14 \over 25}\right) \Rightarrow \overrightarrow{DC}=(5-{77 \over 25},2+{14 \over 25}) =({48 \over 25},{64\over 25})=(x,y)\\ \Rightarrow x+y={48 \over 25}+{64\over 5} ={112\over 25} ，故選\bbox[red,2pt]{(D)}$$
解：$$(ax+1)^4 =a^4x^4+4a^3x^3+ 6a^2x^2+4ax+1 \Rightarrow \cases{4a^3>a^4 \\ 4a^3>6a^2 \\ 4a^3 > 4a \\ 4a^3>1} \Rightarrow \cases{a^3(a-4)<0 \\ 2a^2(3-2a)<0 \\ 4a(1-a^2)<0 \\ a^3>1/4} \\ \Rightarrow \cases{0< a< 4 \\ 3/2 < a \\ 1>a>0, -1< a< 0, a > 1 \\ a> 1/\sqrt[3]4} \Rightarrow 3/2 < a< 4 ，故選\bbox[red,2pt]{(D)}$$
解：$$餘弦定理: \cos \angle A= {\overline{AB}^2 +\overline{AC}^2 -\overline{BC}^2 \over 2\times \overline{AB} \times \overline{AC}} \Rightarrow \cos 120^\circ ={3^2+ 6^2- \overline{BC}^2 \over 2\times 3\times 6} \Rightarrow -{1\over 2}= {45-\overline{BC}^2 \over 36} \\\Rightarrow \overline{BC}^2 = 45+18=63 \Rightarrow \overline{BC}=3\sqrt 7;\\ 又 \overline{AD} 為角平分線 \Rightarrow \overline{BD}:\overline{DC} =\overline{AB}:\overline{AC} = 3:6=1:2 \Rightarrow \overline{DC} = {2\over 3}\overline{BC} = {2\over 3} \times 3\sqrt 7=2\sqrt 7\\，故選\bbox[red,2pt]{(C)}$$
解：
$$\cases{D(-1,2) \\ G(-3,2)} \Rightarrow 新直線需能區分D與G\\(A) f(x,y)=3x+4y+2 \Rightarrow f(D)f(G)>0\\ (B) f(x,y)=3x+4y-6 \Rightarrow f(D)f(G)>0\\ (C) f(x,y)=6x+8y+3 \Rightarrow f(D)f(G)>0\\ (D)f(x,y)=6x+8y-2 \Rightarrow f(D)f(G)=8\times (-4)<0\\，故選\bbox[red,2pt]{(D)}。$$
解：
$$\cases{ \int_{-1}^1 f(x)\;dx=藍色+綠色=5\\ 綠色= {1\over 2}\times 1 \times {1\over 2}={1\over 4}\\ 黃色={3\over 2}\times 3 \times {1\over 2}={9\over 4}} \Rightarrow \int_{-1}^1 (f(x)-g(x))\;dx =藍色+黃色 =5-{1\over 4}+{9\over 4}=7\\，故選\bbox[red,2pt]{(D)}$$
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