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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(


(A)\times: 三點需不在同一直線上\\(D)\times: 該定點需不在該直線上\\其他皆正確,故選\bbox[red,2pt]{(BCE)}


y=[x]圖形在x為整數時不連續,因此y=x[x]也不連續,故選\bbox[red,2pt]{(ACDE)}


(A)\bigcirc: \lim_{n\to \infty} \sum_{k=1}^n ({1\over 2})^k =\lim_{n\to \infty} {{1\over 2}-{1\over 2^{k+1}}\over 1-{1\over 2}} =\lim_{n\to \infty} (1-{1\over 2^k})=1 \\(B) \times: y=(1+{1\over n})^n \Rightarrow \log y=n\log(1+{1\over n}) \Rightarrow \lim_{n\to \infty}n\log(1+{1\over n}) = \lim_{n\to \infty}{\log(1+{1\over n})\over 1/n}\\ = \lim_{n\to \infty}{(\log(1+{1\over n}))'\over (1/n)'} = \lim_{n\to \infty}{{1\over 1+{1\over n}}\cdot (-1/n^2)\over -1/n^2} = \lim_{n\to \infty}{1\over 1+{1\over n}}=1 \Rightarrow \lim_{n\to \infty}(1+{1\over n})^n=e^1 \\(C) \bigcirc: \lim_{n\to \infty}{7^n+3^n\over 7^n-3^n} =\lim_{n\to \infty}{1+(3/7)^n\over 1-(3/7)^n}=1 \\(D)\times: \sum_{k=1}^n{1\over n}\left(1+{k \over n}\right)^2 ={1\over n}\sum_{k=1}^n\left(1+{2k\over n}+ {k^2\over n^2} \right) ={1\over n}\left( n+{n(n+1) \over n} + {n(n+1)(2n+1)\over 6n^2} \right) \\ =1+{n(n+1) \over n^2} + {n(n+1)(2n+1)\over 6n^3} \Rightarrow \lim_{n \to \infty} \sum_{k=1}^n{1\over n}\left(1+{k \over n}\right)^2 =1+1+{1\over 3}= {7\over 3} \\(E)\times: \lim_{n\to \infty}{n+2\over n^2+2n+1} =\lim_{n\to \infty}{1+2/n\over n+2+1/n} =0\\,故選\bbox[red,2pt]{(AC)}


(A)\bigcirc: {a\over \sin A}= {b\over \sin B}={c\over \sin C}=2R \Rightarrow \cases{\sin A=a/2R \\ \sin B=b/2R \\ \sin C= c/2R},二邊之和大於第三邊\Rightarrow a+b> c\\ \qquad \Rightarrow {a+b \over 2R} > {c\over 2R} \Rightarrow \sin A+\sin B >\sin C\\(B)\times: 若A=B=C=60^\circ \Rightarrow \cases{\cos A+\cos B=1/2+1/2=1\\ \cos C=1/2} \cos A+\cos B \not < \cos C \\(C)\times: \sin A={1\over 2} \Rightarrow \angle A=30^\circ\;或\;150^\circ \\(D) \bigcirc: cos A={1\over 2} \Rightarrow \angle A= 60^\circ\;或300^\circ(不合)\\(E)\times: {a\over \sin A}= {b\over \sin B}={c\over \sin C}=2R \Rightarrow \cases{a=2R\sin A\\ b=2R\sin B\\ c=2R\sin C};\\\qquad 若\cases{a < R \\ b < R\\ c < R} \Rightarrow \cases{2R\sin A < R\\ 2R\sin B < R\\ 2R\sin C < R} \Rightarrow \cases{\sin A<1/2 \\ \sin B< 1/2 \\ \sin C< 1/2} \Rightarrow \cases{\angle A< 30^\circ \\\angle B< 30^\circ \\\angle C< 30^\circ }\\\qquad,但\angle A+\angle B+\angle C=180^\circ,所以不可能發生\triangle 三邊長皆小於外接圓半徑,故選\bbox[red,2pt]{(AD)}


(A)\bigcirc: \lim_{x\to 0}|x|= 0 \\(B)\times: \cases{\lim_{x\to 0+} {|x|\over x}=1 \\ \lim_{x\to 0-} {|x|\over x}=-1} \lim_{x\to 0} {|x|\over x}不收斂 \\(C)\times: 分母為0,\lim_{x\to 0}{|x|\over x^2}不存在 \\(D)\bigcirc: \lim_{x\to 1}{|x| \over x} =1 \\(E) \times: 分母為0,\lim_{x\to 0}{|x+1|\over x}不存在\\,故選\bbox[red,2pt]{(AD)}

今年送分五題......

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