Loading [MathJax]/jax/output/CommonHTML/jax.js

2024年3月25日 星期一

113年身障生升四技二專-數學(A)詳解

113 學年度身心障礙學生升學大專校院甄試

甄試類(群)組別:四技二專組-數學(A)

單選題,共 20 題,每題 

解答:x3nf(x)=(x3)2+n=x26x+9+n=x2+mx6{m=69+n=6n=15mn=6+15=9,(B)
解答:A(2,5)x+4y=82+2081+42=1417=k17k=14,(C)
解答:{f(x)x214f(x)x+15{f(2)=14f(1)=5{4+2b+c=141b+c=5{b=2c=6cb=62=4,(A)
解答:|2x+3|<55<2x+3<54<x<1(1)1<|2x+3|{2x+3>12x+3<1{x>1x<2(2)(1)(2){1<x<14<x<2|bd|=|1(2)|=3,(C)
解答:22x9×2x+8=0(2x8)(2x1)=0{2x=82x=1{x=3x=0a+b=3+0=3,(D)
解答:{a=log0.30.5=log100.5log100.3=log10510log10310=log1051log1031=1log1051log103b=log0.30.05=log100.05log100.3=log105100log10310=log1052log1031=2log1051log103c=log0.30.8=log100.8log100.3=log10810log10310=log1081log1031=1log1081log1032log105>1log105>1log108b>a>c,(B)
解答:x2+2x15=(x3)(x+5)<05<x<3x=4,3,2,1,0,1,2,7,(A)
解答:f(x,y)=x2y4{f(A)=14<0f(B)=4<0f(C)=1>0f(D)=1<0{A,B,DC,(C)
解答:3,(1,2):y2=3(x1)3xy1=0f(x,y)=3xy1{f(4,11)=0f(7,20)=0f(11,32)=0f(15,43)=10,(D)
解答:



解答:{a1=11a21=a1+20d=71a11=a1+10d=12(a1+a21)=41,(B)
解答:sinπ3+tan4π3cos2π3=32+312=323,(C)
解答:120:2,3,5,7,11,13,17,19,8,820=25=100×25=40,(B)
解答:6436344634363463642886,(C)
解答:,,C929=369=27,(A)
解答:π2<θ<πcosθ<05cos2θ2cosθ3=(5cosθ+3)(cosθ1)=0cosθ=35,(A)
解答:{a1=1a2=ra3=r2a4=r3a5=r4a6=r5=243r=3{a1=1a2=3a3=9a4=27a5=81a6=243(A)×:r=33(B)×:=36131=365182(C)×:a4>027(D):3+9+27+81=120,(D)
解答:42,1,(D)
解答:
x2+y24x+6y12=0(x2)2+(y+3)2=52{O(2,3)r=5(A)×:O=d(O,L)=6+12+232+42=45(B)×:d(O,L)=4<5(C):52=42+32=12×3×4=6=12(D)×d(O,L)=43(C)
解答:(AB)=AB={2,3},(B)

=================== END =======================

解題僅供參考,其他歷年試題及詳解



沒有留言:

張貼留言