2021年3月24日 星期三

110年身障生升大學-數學乙詳解

110 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別:大學組數學乙

解答a8(3a1)<0.58.53<aa>4.9a=5(A)

解答{aiˉabiˉbciˉc{aibici{ai=ˉa+3bi=ˉb+1ci=ˉc5ai+bi+ci=(ˉa+ˉb+ˉc)1(D)

解答(C)

解答|8146|=48+4=52(C)

解答{g(1)=1,g(2)=2,g(3)=3h(1)=2,h(2)=4,h(3)=6{p(x)=g(x)xq(x)=h(x)2x{x=1,2,3q(x)=0x=1,2,3p(x)=0{p(x)=g(x)x=a(x1)(x2)(x3),aq(x)=h(x)2x=b(x1)(x2)(x3),b{g(x)=a(x1)(x2)(x3)+xh(x)=b(x1)(x2)(x3)+2xf(x)=g(x)(x1)h(x)(x2)=a(x1)2(x2)(x3)+x(x1)b(x1)(x2)2(x3)2x(x2)=a(x1)2(x2)(x3)b(x1)(x2)2(x3)+x2x2x2+4x=a(x1)2(x2)(x3)b(x1)(x2)2(x3)x2+3x=a(x1)2(x2)(x3)b(x1)(x2)2(x3)x(x3)x3f(x)(D)

解答{loga=m+klogb=n+k,m>n>0,m,nZ0<k<1{a=10m+kb=10n+k(A):ba=10mn>1bab(B)×:ab=10m+n+2km+nab(C)×:{m=2n=1a+b=102+k+101+k=10(101+k+10k)10ka+b10(D)×:(C)ab=10(101+k10k)(C)(A)

解答{11X11Y11Z(B)

解答(1,1)LOAOBOC(1,1)La=b=c(D)

解答(|x|12)(|x|20)<2012<|x|<20|x|=13,14,,19x=±13,±14,,±1914(B)

解答{A:log22log4=log4×log22=log4=2log20.602B:log2(2×2log2)=log22+log22log2=1+log2=1.301C:log2412log2=log22log2=log2=0.301D:log24log2=log222log2=2log2=aa2=4log2=1.204a<1.2B(B)

解答2:(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(4,6)(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)18a=1836=123:(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3),(6,6)12b=1236=13(A)

解答(A)×:{M1=[abcd]M2=[cdab]M112M2det(M1)det(M2)(B)×:{{2x+3y=102x+3y=8{2x+3y=102x+3y=10(D)×:(B)(C)

解答L:{x=23ty=3+4t(3,4)(D)(3,4)(D)

解答
xy{x+y=18000x7000yy2xf(x,y)=10x+8yx+y=1800y=2x(600,1200)(0,1800){f(600,1200)=15600f(0,1800)=14400f(x,y)15600=1.56(C)

解答(,)=(2,1),(2,1),(1,2)C52C31C32C21C11C22=10×3×3×2=180(C)


解答12xy(20,20)xyr1>r2m1>m2(A)
解答(B){u=a10bv=b{|u|2=|a|220ab+100|b|2|v|2=|b|2uv=ab10|b|2|u|2|v|2(uv)2=|a|2|b|220ab|b|2+100|b|4((ab)220ab|b|2+100|b|4)=|a|2|b|2(ab)2(B)


解答=3000×1100+1000×2100+500×2100+300×10100+100×85100=175(A)×:(B)×:{=(2×500+10×300+85×100)÷97=12500÷97=(3×500+8×300+86×100)÷97=12500÷97(C)×:{=(3000+85×100)÷86=11500÷86=(5000+85×50)÷86=92500÷86(D):{=(10×300+85×100)÷96=11500÷95=(10×150+85×150)÷95=14250÷95(D)

解答g(x)=(2x+3)f(x)+1:g(a)=0(2a+3)f(a)+1=0f(a)=12a+3f(0)=2f(x)=xp(x)+f(0)f(a)=ap(a)+f(0)af(a)f(0)a(12a+32)a4a+72a+3a=1(a=221)(B):g(x)7g(a=2)0
解答{abc1.5a+b+0.5c=10,a,b,cZ3a+2b+c=20abc6102520344053506270818090100112+3+5+6+8+9+11=44(C)


2 則留言:

  1. 您好:請問一下,為甚麼第11題中的a=1/36=1/2呢?

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