2020年3月14日 星期六

96年大學指考數學甲詳解


96學年度指定科目考試試題
數學甲
第壹部分:選擇題
一、單選題


解:
z=cos2π7+isin2π7|1z|=|1cos2π7isin2π7|=(1cos2π7)2+(sin2π7)2=12cos2π7+cos22π7+sin22π7=22cos2π7=22(2cos2π71)=44cos2π7=4(1cos2π7)=4sin2π7=2sinπ7(1)


解:
f(x)=33xx2limx1f(x)=331=1limx1|f(x)|=limx1(f(x))limx1|f(x)|1x1=limx1f(x)1x1=limx1x2+3x4x1=limx1(x+4)(x1)x1=limx1(x+4)=5(4)



解:(0,0)10(5+5log2,5+152log2){5+5log2=loga1+loga3++loga95+152log2=loga2+loga4++loga10{5+5log2=log(a1a3a9)=log(a1a1r2a1r8)=log(a51r20)=5loga1+20logr5+152log2=log(a2a4a10)=log(a1ra1r3a1r9)=log(a51r25)=5loga1+25logr{5+5log2=5loga1+20logr(1)5+152log2=5loga1+25logr(2)(2)(1)52log2=5logrr5=25/2r=2(1)5+5log2=5loga1+10log2loga1=1log2a1=5(a1,r)=(5,2)(5)


二、多選題


解:(1):(2)×:0.0160(3):σ(x+5,y+5)σ(x+5)σ(y+5)=σ(x,y)σ(x)σ(y)=0.016(4):σ(100x,100y)σ(100x)σ(100y)=1002σ(x,y)100σ(x)100σ(y)=σ(x,y)σ(x)σ(y)=0.016(5):σ(xˉxsx,yˉysy)σ(xˉxsx)σ(yˉysy)=σ(xsx,ysy)σ(xsx)σ(ysy)=1sxsyσ(x,y)1sxsyσ(x)σ(y)=σ(x,y)σ(x)σ(y)=0.016(1,3,4,5)


解:P(x)=Q(x)(x3)+2(1)×:αQ(x)=0Q(α)=0P(α)=Q(α)(α3)+2=0αP(x)=0P(x)=0Q(x)=0(2)×:Q(x)=x4P(x)=x4(x3)+2x=0,3P(x)=2x=3(3):Q(x)Q(4)>0P(4)=Q(4)(43)+2=Q(4)+2>0P(4)0P(x)(x4)(4):Q(x)x3P(x)=Q(x)(x3)+2>0P(x)=03;P(x)53(5)×:Q(x)=(x+3)4+2P(x)=((x+3)4+2)(x3)+2=(x+3)4(x3)+2x4P(x)(x+3)(x3)2x42(3,4)


解:
{x32=y53(1)y53=z4(2)xa=z2(3)y+13=z2(4),(2)y53+2=(z4)+2y+13=z2(2)(4)(1)(2)(x,y,z)=(2t+3,3t+5,t+4),tR(1)×:(2t+3,3t+5,t+4)(3)2t+3a=t+2a=2t+3t+2=21t+22,tR(2):(2t+3,3t+5,t+4)(3)t=2a32a(x,y,z)(3):{x32=y53xa=z2(x,y,z)=(2t+3,3t+5,2t+3a+2),tR(4)×:{xa=z2y+13=z2(x,y,z)=(at,3t1,t+2),tR(5):{x32=y53y53=z4(x,y,z)=(2t+3,3t+5,t+4),tR(2,3,5)


解:{A=[1001]A2=[1001]=IB=[1/23/23/21/2]=[cosπ/3sinπ/3sinπ/3cosπ/3]逆時針旋轉60B6=I(1)×:{AB=[1001][1/23/23/21/2]=[1/23/23/21/2]BA=[1/23/23/21/2][1001]=[1/23/23/21/2]ABBA(2):{A2B=IB=BBA2=BI=BA2B=BA2(3)×:{A11B3=(A2)5A180=A[cosπsinπsinπcosπ]=[1001][1001]=[1001]B6A5=I(A2)2A=IA=A=[1001]A11B3B6A5(4):{AB12=A(B6)2=AI=AA7=(A2)3A=IA=AAB12=A7(5):(ABA)15=(ABA)(ABA)(ABA)=ABA2BA2BA2BA=ABIBIBIBA=AB15A(2,4,5)


解:(1)×:{limxy=limxy=(2)×:y=03x2+2=0(3):y=3x2+2>0(4):(a,b)b=a3+2a+3b=a32a3b+6=(a)3+2(a)+3(a,b+6)(5):=x=1x=0x3+2x+3dx=[14x4+x2+3x]|10=14+1+3>4(3,4,5)


三、選填題



5=5506k=1k50=521


解:
{A(2,7)B(1,6)O¯OB(4,3)¯OB:x+14=y63;O¯OBO(4t1,3t+6)¯OA=¯OB(4t+1)2+(3t1)2=(4t)2+(3t)22t+2=0t=1O(5,3)r=¯OB=(4)2+(3)2=5:(x+5)2+(y3)2=52x2+y2+10x6y+9=0{a=10b=6c=9


解:

x,y,;:{=4xy+x2=432=x2yx2+2xy+2xy33x2×2xy×2xyx2+4xy334x4y2432334x4y214434x4y214434x2y864x2y864x2=2xyx=2y=x2y=4y2y=4y3=864y=6x=2×6=12

第貳部份:非選擇題

解:
(1)f(x)=x36x2x+30=(x3)(x5)(x+2)f(x)=0x=3,5,2(2)cosACB=a2+b2¯AB22ab12=9+25¯AB230¯AB=7¯DE=¯AE+¯BD¯AB=5+37=1CDEABC=¯DE¯AB=17CDE=17ABC=17×12ׯBCׯACsinACB=114×15×32=15283


解:(1){A(2,7,15)B(1,16,3)C(10,7,3){AB=(3,9,12)AC=(12,0,12)n=AB×AC=(108,108,108)An:108(x+2)108(y7)108(z15)=0x+y+z=20(2)P(a,b,c){¯PA=¯PB=¯PCPABC{(a+2)2+(b7)2+(c15)2=(a1)2+(b16)2+(c3)2(a1)2+(b16)2+(c3)2=(a10)2+(b7)2+(c3)2a+b+c=20{a+3b4c=2ab=6a+b+c=20(a,b,c)=(3,9,8)



-- END   (僅供參考)  --

5 則留言:

  1. 第五題的詳解應該有錯喔 Q(x)必是四次式
    且係數均為正數,(2)Q(x)的常數項不可能為-4

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    1. 謝謝提醒,已修正替換另一個例子!

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  2. 算幾不等式 不是柯西不等式

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