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2020年5月4日 星期一

109學年度四技二專統測--數學(C)詳解


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


22<0limx2x20(B)


{a=tan480=tan120<0b=sec135<0c=cos(60)>0{(b,c)=(,+)(a,b)=(,)(c,a)=(+,)(c,b)=(+,)(A)



f(x)=p(x)(x1)(x2+x+1)+3x2+5x2=p(x)(x1)(x2+x+1)+3(x2+x+1)+2x5=(x2+x+1)(p(x)(x1)+3)+2x52x5(B)



(A)×:10020=10020=80(B):39206070(C)×:(80)(D)×:39×90%=35.1PR904(B)



6×14=84(D)


cos21=cos(8059)=cos80cos59+sin80sin59=b1a2+a1b2(A)


(x4)225+(y+2)2144=1{C(4,2)a=12b=5(A)×:a2=b2+c2144=25+c2c=119=(4,2±119)(B)×:(4,2±12)(C)×:(4±5,2)(D):=2b2a=2×2512=256(D)


(3+i)z=23+2iz=23+2i3+i=(23+2i)(3i)(3+i)(3i)=4+43i4=1+3i=2(12+32i)=2(cos2π3+sin2π3)2π3(B)



{a5=41a13=73{a1+4d=41a1+12d=73{a1=25d=4{a19=25+18×4=97a20=25+19×4=1014203011100(B)



{A(2,2)B(9,3/2)C(9,1)D(2,8)ABCD=BCDEABE=(¯BC+¯DE)ׯBE2¯AEׯBE2=(5/2+19/2)×727/2×72=42494=1194(A)


cos3x12cos3x121=12cos3x1=0cos3x=123x=π3+nπ,n=0,1,2,...x=π9+nπ3,n=0,1,2,...x[0,2π]x=π9,49π,79π,119π,149π,179π6{a=6b=1ab=6(A)


10(1+3%)r=20(1+3%)r=21.03r=2rlog1.03=log2r=log2log1.03=0.3010.012823.520+23.5=43.544(C)


f(x)=x33x224x+32f(x)=3x26x24f




\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!


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)}。


\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)}。



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)}。


\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)}。



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)}。



\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)}


\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)}


(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)}


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)}



\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)}。




\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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