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2020年5月28日 星期四

109年 警專39期甲組數學科詳解




2a+b22a×b=2×18=62a+b12(B)


x=31(x+1)2=(3)2x2+2x2=0:f(x)=x4+3x3x+2=(x2+2x2)(x2+x)+x+2f(31)=0×(x2+x)+x+2=x+2=(31)+2=3+1(C)


x=1+i(x1)2=i2x22x+2=0:f(x)=x4x3+ax=(x22x+2)(x2+x)+(a2)x0(a2)x=0a=2(A)


log0.1(log3(x+2)){x+2>0log3(x+2)>0{x>2x+2>1x>1x>1(A)


{y=log2|x|y=2xy=log2xy=2xx=y;y=log2|x|Y(B)


log(13)50=50(log3)=50×0.4771=23.85523+1=240(B)



8(1,1,6),(1,2,5),(1,3,4),(2,2,4),(2,3,3);3,3!,3!,3,3212163=772(B)


8212(1,2,5),(2,2,4),(2,3,3)6+3+3=12;88=1221=47()



:cosθ=52+62722×5×6=15sinθ=265;:7265=2RR=3546=35624(C)





:cosA=cos120=32+62¯BC22×3×612=45¯BC236¯BC=37¯AEA¯BE¯EC=¯AB¯AC=36¯BE=13¯BC=7:cosEAB=cos60=32+¯AE2(7)22×3ׯAE12=2+¯AE26¯AE¯AE=21(D)



cosθ+cos2θ=0cosθ+2cos2θ1=0cosθ={1/21sinθ={3/20tanθ={30(A)




{¯BC=a¯AC=bD¯AB¯CD¯AB{¯AD=bcosBAC=1213b¯DB=acosABC=35a¯AB=¯AD+¯DB126=1213b+35a(1):asinBAC=bsinABCa5/13=b4/5b=5225a(1)35a+1213×5225a=1266325a=126a=126×2563=50(A)



{ax+y=2ax3y=11/3(a)×13=1a=3(B)



{A(2,2)O(0,2)OA=(2,4)A(2,4)2(x2)+4(y2)=0x+2y=6(C)




{¯BD:¯DC=2:3AD=25AC+35ABP¯ADAP=12ADAP=12(25AC+35AB)=15AC+310AB{x=3/10y=1/5x+y=1/2(D)


(ab)(ab)=|a|22ab+|b|225=42ab+16ab=52cosθ=ab|a||b|=5/28=516()



{a=(x,3)b=(2k,4k)ab|b|2×b=(2,4)2kx+12k20k2(2k,4k)=(2,4)6+x=10x=4(D)



L:x12=y+23=z4L(2t+1,3t2,4t)A(2,3,3):(2t1)2+(3t5)2+(4t3)2=29t258t+35t=1(2+1,32,4)=(3,1,4)(A)





abc(x)a+b+c(y)x×y122155123663612487561333721134128961446954223372122448322336848234249216244121012033319933412101203441811198444412481201080=1080120=9(C)



:恰有一次正面的情形:(正、負、負),(負,正,負)及(負,負,正),三種情形的機率和為3×p×(1p)2=4/9p=1/3故選(C)



{sinπ3=sin2π3=sin2.09=32=0.866sinπ4=sin3π4=sin2.35=22=0.707sin2.40.7(A)




f(x)=sinx3cosx=10(110sinx310cosx)110sinx310cosx=1f(x)10{sinx=110cosx=310(C)


{A(0,1)B(1,0)滿|zi|=|z+1|z¯ABP(x,y)|z22i|P(2,2)P(x,y)=(t,t)dis(P,(2,2))=(2t)2+(2+t)2=2t2+8t=08=22(B)


z=2+ai|z|=a2+4|z|cos7π6=2a2+4(32)=2a2+4=43a2=43a=23()()


ω=cos4π5+isin4π5ω5=1ω3+ω4++ω16=ω3ω171ω=ω3(ω5)3ω21ω=ω3ω21ω=ω2(1ω)1ω=ω2(A)



f(x)=x(x1)(x2)(x3)=(x2x)(x25x+6)f(x)=(2x1)(x25x+6)+(x2x)(2x5)f(3)=0+(93)(65)=6(D)


f(x)x=12f(1)=21+a+b+5=2a+b=4()


51|x3|dx=|31x3dx|+|53x3dx|=2+2=4(B)



f(x)=g(x)3x2=x3x2(x3)=0x=0,x=3;=|30f(x)g(x)dx|=|303x2x3dx|=|[x314x4]|30|=27814=274(C)


f(x)=x0(3x22t)dt=x3x2f(x)=3x22xf(x)=6x2f(x)=06x2=0x=1/3f(1/3)=1/271/9=2/27(1/3,2/27)()




|x2|+|x+2|4k4(CDE)



ff(x)=f(x)ax4+bx3+cx2+dx+e=ax4bx3+cx2dx+e{b=bd=db=d=0(BD)



x2+y26x6y+13=0(x3)2+(y3)2=5{O(3,3)r=53x4y+4=0=|912+432+42|=55<rPx0r+550x5+55x=0,1,2(ABC)




y=tanxπ(A):sinx2πsin(2x)π(B)×:cosx2πcos(x/2)4π(C)×:y=2sinxy=sinx2π(D)×:y=2secxy=secx2π(E):y=cotxy=tanx(AE)



(A):{A=[abcd]B=[efgh]{A+B=[a+eb+fc+gd+h]B+A=[e+af+bg+ch+d]A+B=B+A(B)×:{A=[0111]B=[1000]{AB=[0010]BA=[0100]ABBA(C)×:(A+B)(AB)=A2AB+BAB2A2B2(ABBA)(D):{A=[a11a12a21a22]B=[b11b12b21b22]C=[c11c12c21c22]{B+C=[b11+c11b12+c12b21+c21b22+c22]AB=[a11b11+a12b21a11b12+a12b22a21b11+a22b21a21b12+a22b22]AC=[a11c11+a12c21a11c12+a12c22a21c11+a22c21a21c12+a22c22]AB+AC=[a11(b11+c11)+a12(b21+c21)a11(b12+c12)+a12(b22+c22)a21(b11+c11)+a22(b21+c21)a21(b12+c12)+a22(b22+c22)]=A(B+C)(E)×:{A=[1001]B=[1001]{A2=[1001]B2=[1001]A2=B2,A±B(AD)


(A)×:(D)×:(BCE)


y=[x]xy=x[x](ACDE)


(A):limnnk=1(12)k=limn1212k+1112=limn(112k)=1(B)×:y=(1+1n)nlogy=nlog(1+1n)limnnlog(1+1n)=limnlog(1+1n)1/n=limn(log(1+1n))(1/n)=limn11+1n(1/n2)1/n2=limn11+1n=1limn(1+1n)n=e1(C):limn7n+3n7n3n=limn1+(3/7)n1(3/7)n=1(D)×:nk=11n(1+kn)2=1nnk=1(1+2kn+k2n2)=1n(n+n(n+1)n+n(n+1)(2n+1)6n2)=1+n(n+1)n2+n(n+1)(2n+1)6n3limnnk=11n(1+kn)2=1+1+13=73(E)×:limnn+2n2+2n+1=limn1+2/nn+2+1/n=0(AC)


(A):asinA=bsinB=csinC=2R{sinA=a/2RsinB=b/2RsinC=c/2Ra+b>ca+b2R>c2RsinA+sinB>sinC(B)×:A=B=C=60{cosA+cosB=1/2+1/2=1cosC=1/2cosA+cosBcosC(C)×:sinA=12A=30150(D):cosA=12A=60300()(E)×:asinA=bsinB=csinC=2R{a=2RsinAb=2RsinBc=2RsinC;{a<Rb<Rc<R{2RsinA<R2RsinB<R2RsinC<R{sinA<1/2sinB<1/2sinC<1/2{A<30B<30C<30A+B+C=180(AD)


(A):limx0|x|=0(B)×:{limx0+|x|x=1limx0|x|x=1limx0|x|x(C)×:0,limx0|x|x2(D):limx1|x|x=1(E)×:0,limx0|x+1|x(AD)

今年送分五題......

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