Processing math: 100%

2021年9月11日 星期六

110年試辦考試(適用於108課綱)數學A詳解

110年試辦考試(適用於108課綱)數學A

第壹部分、選擇(填)題

一、單選題

解答{log(2+3)log(23)log(2+3)=log(12+3)=log(23)log(23)=log(123)=log(2+3)log2+log3=log233(3)
解答|x10|<|x60|<|x+10|{x10<x60<x+10if x60x10<60x<x+10if 10x6010x<60x<x+10if 10x1010x<60x<x10if x10{25<x<3525<x<35x=26,27,,349(2)
解答{8x+5y=14x+7y=6P(43,23)A(2,3)P(43,23)32/324/3=7/32/3=72(1)
解答{A(1,0,1)B(1,1,0)L:12x137=y+19=2z{AB=(0,1,1)L=u=(2,1,12)En=AB×u=(12,2,2)|n|=332{(1):cosθ=(1,0,0)n|n|=133(2):cosθ=(0,1,0)n|n|=433(3):(2),cosθ=433(4):cosθ=(1,1,0)n2|n|=366(5):cosθ=(0,1,1)n2|n|=866|cosθ|(5)
解答111610(3)

解答

{y=sin2x+cos2xy=sinx+1/2滿0x2πy=sin2x+cos2x=2(cos45sin2x+cos2xsin45)=2sin(2x+45){22πy=sinx+12{:3/2:1/22π()(4)

二、多選題


解答(1)×:23(2)×:10(3):(4):(5):{,,>(345)
解答

(1)×:{(2,3)y=f(x)y=f(x)=a(x2)2+3,a<0y=f(x)=a(x2)2+3=a(x+2)2+3(2):y=g(x)y=g(x)(3)×:{y=g(x)y=g(x)y(2,1)(2,1)(4)×:y=g(x)(2,1)g(x)=b(x2)3+c(x2)1f(x)+g(x)=b(x2)3+a(x2)2+c(x2)+2(2,2)(5):y=f(x)y=g(x)x<2(25)
解答(1):¯AC¯AB<¯BC32<aa>1(2)×:AB{cosA=13a212<0cosB=a254a<0{a>130<a<5a=2,B(3)×:¯AC>¯ABB>C(a)(4)×::3sinB=2R=22sinB=322>1(5):asinA=2R=4042a=4042sinA(15)
解答(1):(110)3(2)×:(110)31640(3)×:{1XX:XX=00991006YY:YY=004040(4):3601000×3601000×64010008.3%>5%(5)×:++=6401000+3601000×6401000+360210002×6401000=0.64(1+0.36+0.362)95.3%>93%(14)
解答(1):O,A,DOD=mOA,mkOA+(2k)OB=mOAk=2(2):CD=CO+OD=(OA+OB)+kOA+(2k)OB=(1k)(AO+OB)=(1k)ABCDAB(3)×:OC=OA+OB=(a1,a2)+(b1,b2)=(a1+b1,a2+b2)OAC=12|a1a2a1+b1a2+b2|=12|a1a2b1b2|=1(4):OD=kOA+(2k)OB=(ka1,ka2)+((2k)b1,(2k)b2)=(ka1+(2k)b1,ka2+(2k)b2)AD=((k1)a1+(2k)b1,(k1)a2+(2k)b2)=(k(b1a1)a1+2b1,k(b2a2)a2+2b2)ABD=12|ABAD|=12|b1a1b2a2k(b1a1)a1+2b1k(b2a2)a2+2b2|=12|b1a1b2a2a1+2b1a2+2b2|k(5)×:ACD=12|ACAD|=12|b1b2k(b1a1)a1+2b1k(b2a2)a2+2b2|k(124)
解答

L:x1/21=y1/32=z1/43QLQ=(t+1/2,2t+1/3,3t+1/4),tRLEFGH3t+14=1t=14Q=(34,56,1);LABCD3t+14=0t=112Q=(512,16,0)L(5)
解答log232=510005k=02k=1000(1+2+4+8+16+32)=1000×63=63
解答使1A使1=0.4×0.90.4×0.9+0.6×0.75=0.360.81=49
解答f(x)=x23x+3=(x32)2+34f(x)=f(32)=34ya3/4=278=(32)3a1/4=32a=(32)4=8116
解答
Γ:x2+y22x+6y+2=0(x1)2+(y+3)2=8{C(1,3)r=22;B(a,b)¯CAׯCB=2×(a1)2+(b+3)2=r2=8(a1)2+(b+3)2=(42)2BΓ:(x1)2+(y+3)2=32ABC¯CA:¯AB=2:(422)=1:3A=(3C+B)÷4{2=(3+a)÷42=(9+b)÷4{a=5b=1B=(5,1)
解答1{A(0,1,1)B(1,1,0)C(1,0,1)D(0,0,0){AB=(1,0,1)AC=(1,1,0)n=AB×AC=(1,1,1)AEAB=AEAC=0AEnEx1=y11=z11(An)E=(t,t+1,t+1),t<0(D,EABC){AE=(t,t,t)AD=(0,1,1)cosDAE=ADAE|AD||AE|=2t23t2=63

第貳部分、混合題或非選擇題

解答T=[abba]T(0,1)=(b,a)y=5x+13a=5b+13a+5b=13(2)
解答Py=x+1P(t,t+1)T(t,t+1)=P(atbtb,bt+at+a)Py=5x+13at+bt+a=5(atbtb)+13(4a6b)t+(13a5b)=0{4a6b=013a5b=0{a=3b=2

 解答{P(xp,yp)Q(xq,yq){P=T(P)=(axpbyp,bxp+ayp)Q=T(Q)=(axqbyq,bxq+ayq)¯PQ=(a(xpxq)b(ypyq))2+(b(xpxq)+a(ypyq))2=a2(xpxq)2+b2(ypyq)2+b2(xpxq)2+a2(ypyq)2=(a2+b2)(xpxq)2+(a2+b2)(ypyq)2=(a2+b2)((xpxq)2+(ypyq)2)=a2+b2ׯPQ¯PQ¯PQ=a2+b2,32+22=13

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