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2023年3月28日 星期二

112年身障升大學-數學A詳解

112 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別:大學組-數學 A

單選題,共 20 題,每題 

解答2,6,104,r=4a3=a1r2=242=32(D)
解答{y>x2y3xx2<y3xx2<3xx(x+3)<03<x<0{x=11<y3x=24<y6(x,y)=(1,2),(1,3),(2,5),(2,6),(D)
解答{12515x125+1510015y100+15{110x14085y115110115xy140855xy550|xy|55(A)
解答f(x)=(x1)(x2+x+1)p(x)+x2+bx+c=(x1)(x2+x+1)p(x)+(x2+x+1)+(b1)x+(c1)=(x2+x+1)((x1)p(x)+1)+(b1)x+(c1)f(x)x2+x+12x+1(b1)x+(c1)=2x+1{b=3c=2f(1)=1+b+c=1+3+2=6(C)
解答cosA=82+72¯BC2287=113¯BC211260<A<12012<cosA<1212<113¯BC2112<1256<113¯BC2<5657<¯BC2<16957<¯BC<138¯BC<13¯BC=8,9,10,11,12(B)
解答logab(a,b)=(2,4),(2,8),(2,16),(3,9),(4,16)(a,b)20×19=380=5380=176(B)
解答{logx=1+2alogy=2+aa=logx12=logy2logx=2logy3logx=logy2log103=logy2103x=y2103y2=1000x(D)
解答{P(,)=1/16P(,)=9/16{10/16=6/164001016+200616=250+75=325(C)
解答{¯OP=13OP(0,1)=2{a2+b2=13b=2a2=9a=3(a=3,P)OP(2,1)=(3,2)(2,1)=62=8(A)
解答A:{x=ny=10nxyra>0B:{x=ny=log10n=nx=yrb=1C:{x=ny=1/7nxyrc<0rc<ra<rb(C)
解答(cosθ+1)2+(sinθ2)2=6cos2θ+2cosθ+1+sin2θ4sinθ+4=62cosθ4sinθ=0tanθ=sinθcosθ=24=12(A)
解答{ab{1a+15b2ab+153a+2b+10{x(a+15)+by=519ax+y(b+15)=504x(a+2)+y(b+10)=500{ax+by+15x=519(1)ax+by+15y=504(2)ax+by+2x+10y=500(3){(1)(3)13x10y=19(2)(3)2x+5y=4{x=3y=22x+y=6+2=8(C)
解答(1,0),(0,3)Ly=3x3Cx2+y26x+8y+k=0(x3)2+(y+4)2=25k{O(3,4)r=25kCLd(O,L)=r10=25kk=15(D)
解答(3,1){f(x)=a(x3)3+b(x3)1g(x)=c(x3)3+d(x3)1{f=2g=1{a=2b=1{f(x)=2(x3)3+b(x3)1g(x)=(x3)3+d(x3)1(5,3)f(5)=g(5)=3{f(5)=16+2b1=3g(5)=8+2d1=3{b=6d=6{f(x)=2(x3)36(x3)1g(x)=(x3)3+6(x3)1(A)×:{f(0)=54+181<0g(0)=27181>f(0)
解答\begin{vmatrix}a & 0 &  1\\ b& \log 2 & \log 5 \\ 1& 0 & 1\end{vmatrix} =a\log 2-\log 2=(a-1)\log 2 = 0 \Rightarrow a=1,故選\bbox[red,2pt]{(B)}
解答E與2x-2y+z=3不相交 \Rightarrow 相互平行 \Rightarrow E:2x-2y+z= k\\ \Rightarrow\cases{d((1,-2,3), E)= {|9-k|\over 3} \\ d((-5,4,1), E)= {|-17-k|\over 3}} \Rightarrow |9-k|=|-17-k| \Rightarrow  k=-4\\ \Rightarrow E:2x-2y+z+4=0 \Rightarrow d((3,-2,1),E) ={|6+4+1+4|\over 3} =5,故選\bbox[red,2pt]{(A)}
解答\cos A= {\overline{AB}^2+\overline{AC}^2- \overline{BC}^2\over 2\overline{AB}\cdot \overline{AC}} ={1+1-1/4\over 2\cdot 1\cdot 1} ={7\over 8} \\ 又\cos A= {\overrightarrow{AB} \cdot \overrightarrow{AC} \over |\overrightarrow{AB} || \overrightarrow{AC}|} \Rightarrow {7\over 8}={\overrightarrow{AB} \cdot \overrightarrow{AC} \over 1\cdot 1} \Rightarrow \overrightarrow{AB} \cdot \overrightarrow{AC}={7\over 8}\\ \cases{\overrightarrow{DC} =\overrightarrow{DA} +\overrightarrow{AC}=-{1\over 2}\overrightarrow{AB}+ \overrightarrow{AC} \\ \overrightarrow{EB} =\overrightarrow{EA} +\overrightarrow{AB} =-{1\over 2}\overrightarrow{AC} +\overrightarrow{AB}} \Rightarrow \overrightarrow{DC} \cdot \overrightarrow{EB} =\left(-{1\over 2}\overrightarrow{AB}+ \overrightarrow{AC} \right)\cdot \left( -{1\over 2}\overrightarrow{AC} +\overrightarrow{AB}\right) \\={1\over 4} \overrightarrow{AB} \cdot \overrightarrow{AC} -{1\over 2}-{1\over 2}+\overrightarrow{AB} \cdot \overrightarrow{AC} ={5\over 4}\overrightarrow{AB} \cdot \overrightarrow{AC}-1 ={5\over 4}\cdot {7\over 8}-1 = {3\over 32},故選\bbox[red,2pt]{(D)}
解答A\begin{bmatrix} 1 \\0\end{bmatrix}= \begin{bmatrix} \sqrt 3/2 \\1/2 \end{bmatrix} = \begin{bmatrix} \cos (\pi/6) \\ \sin(\pi/6)\end{bmatrix} \Rightarrow A=\begin{bmatrix} \cos (\pi/6) & -\sin(\pi/6) \\ \sin (\pi/6) & \cos (\pi/6) \end{bmatrix} \Rightarrow A^6=\begin{bmatrix} \cos (\pi) & -\sin(\pi) \\ \sin (\pi) & \cos (\pi) \end{bmatrix} \\=\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \Rightarrow A^6\begin{bmatrix} -3 \\2 \end{bmatrix} =\begin{bmatrix} 3 \\-2 \end{bmatrix}=\begin{bmatrix} a \\b \end{bmatrix} \Rightarrow a+b=3-2=1,故選\bbox[red,2pt]{(B)}
解答全部有C^4_2\cdot C^5_3=60種組合\\豆腐和波菜一起被選有:C^4_2 \cdot C^3_1=18種組合\\ 牛肉和韭菜一起被選有:C^3_1\cdot C^4_2=18種組合\\豆腐和波菜一起被選且牛肉和韭菜有:C^3_1=3種組合\\ 因此符合要求的選法有60-18-18+3=27種,故選\bbox[red,2pt]{(C)}
解答\cases{\overrightarrow{AB}=(1,0,0)\\ E之法向量\vec n=(1,-1,\sqrt 6)} \Rightarrow \cos \theta ={\overrightarrow{AB} \cdot \vec n\over |\overrightarrow{AB}||\vec n|} =\sqrt{1\over 8} \Rightarrow \sin \theta =\sqrt{7\over 8} \\ \Rightarrow 投影長=|\overrightarrow{AB}|\sin \theta = 1\cdot \sqrt{7\over 8}=\sqrt{7\over 8},故選\bbox[red,2pt]{(D)}
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