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2020年6月7日 星期日

99年警專29期甲組數學科詳解


臺灣警察專科學校專科警員班二十九期(正期學生組)
新生入學考試甲組數學科試題
壹、單選題


a2+a3+a10+a11=(a1+d)+(a1+2d)+(a1+9d)+(a1+10d)=4a1+22d=2(a1+11d)=2(a1+5d+a1+6d)=2(a6+a7)=48a6+a7=24(D)


f(x)=x2+2x5=(x22x+1)4=(x2)24f(2)=4(C)


g(x)=f(f(x))=(x32x2x+5)32(x32x2x+5)2(x32x2x+5)+5g(1)=332×323+5=27183+5=11(D)



:m2>m1>m3>m4(B)



log354+log362log32=log3(54×6÷22)=log381=4(A)


{a=31=2a0=b11=b1b=2ac=91=82c=8c=3a+b+c=2+2+3=7(A)



{logx=log2468=3xlogx=log0.1357x=0.1357×104=1357(D)



cos(1680)=cos1680=cos(360×4+240)=cos240=cos60=12(B)



{ADC:¯DCsin30=2RABC:¯ABsin45=2R4sin30=2R=¯ABsin45¯AB=4×sin45sin30=42(C)





xx=32π2×32π=3π(D)



{cosθ=3/5>0π<θ<2π32π<θ<2π34π<θ2<πcosθ2<0cosθ=cos(θ2+θ2)=2cos2θ21=35cosθ2=25(B)


z2z1=42(cosAOB+isinAOB)=2(cos60+isin60)(C)




|89+242+(3)2|=155=3(B)




3AD=2AB+ACAD=23AB+13AC¯BD:¯DC=1:2ABDABC=¯BD¯BC=13(A)



{A(1,2,1)B(0,1,1)C(1,0,0){AB=(1,3,0)AC=(2,2,1)n=AB×AC=(3,1,4)(B)


x2+y2+2(m+2)x2(m+3)y+3m2+2=0(x+m+2)2+(y(m+3))2=(m+2)2+(m+3)23m22(x+m+2)2+(y(m+3))2=(m5)2+363636=6(C)


x2+y26x+2ay+b=0(x3)2+(y+a)2=9+a2b{O(3,a)r=a2b+9{P(4,1)L:y=x21¯OP=dist(O,L)1+(a+1)2=|2a+15|(a+1)2+1=(2a+1)25a2+6a+9=0(a+3)2=0a=3;P16+124+2a+b=0b=13a+b=133=10(A)



(2,3,0)(2,3,a):(x2)2+(y3)2+(za)2=r2;(6,6,4)42+32+(4a)2=r2r2=a28a+41(1);r2=a2+(17)2(2);(1)(2)a28a+41=a2+17a=3r2=32+17=26r=26(B)


Cn2=55n(n1)2=55n2n110=0(n11)(n+10)=0n=11(D)



{2:C32=32:C52=102:C22=12:C102=452=3+10+145=1445(C)


a1=1a2=3141=23a3=215/3=35a4=4/57/5=47an=n2n1a8=815(C)



1x<x1xx<01x2x<0x(1x2)<0(C)


:(1,3),(1,5),(1,7),(1,9),(2,4),(2,6),(2,8),(3,5),(3,7),(3,9),(4,6),(4,8),(5,7),(5,9),(6,8),(7,9)1666/16=3/8(A)



|123320814287|=|10332481407|=284×14×33=28(166)=28×65=1820(B)


A=[3121]det(A)=3+2=5c=2/det(A)=2/5(D)


lim



f(x)=x^3-kx^2+x-3 \Rightarrow f'(x)= 3x^2-2kx+1 \Rightarrow f''(x)=6x-2k \Rightarrow f''(-2)=0\\ \Rightarrow -12-2k=0 \Rightarrow k=-6,故選\bbox[red,2pt]{(D)}



f(x)=x^3-3x^2-x+2 \Rightarrow f'(x)= 3x^2-6x-1 \Rightarrow f'(1)=3-6-1=-4,故選\bbox[red,2pt]{(C)}



\int_{-1}^3(3x^2-4x)\;dx = \left. \left[ x^3-2x^2\right] \right|_{-1}^3 =(27-18)-(-1-2) =9+3=12,故選\bbox[red,2pt]{(C)}


\int_1^2 (f(x))^2\pi \;dx = \pi\int_1^2(2x+1)\;dx = \pi \left .\left[ x^2+x\right] \right|_1^2 =\pi(6-2)=4\pi,故選\bbox[red,2pt]{(D)}

貳、多重選擇題


f(x)=x^3+ax^2 +bx+c \Rightarrow \cases{f(-1)=-1 \\ f(2)=2 \\ f(4)=4} \Rightarrow \cases{-1+a-b+c=-1 \\ 8+4a+2b+c=2 \\ 64+16a+4b+c=4} \\ \Rightarrow \cases{a-b+c=0 \\ 4a+2b+c=-6 \\ 16a+4b+c=-60} \Rightarrow \cases{a=-5 \\ b=3 \\c=8} \Rightarrow f(x)=x^3-5x^2+3x+8 \Rightarrow \cases{f(0)=8>0 \\f(1)=7 > 0 \\f(3)=-1< 0\\ f(5)=23 > 0}\\ \Rightarrow \cases{(A)f(-\infty)f(0) < 0有實根 \\ (B)f(0)f(1) >0 \\(C) f(2)f(3)<0 有實根\\ (D) (2,3)有實根\Rightarrow (2,4)有實根 \\ (E)f(5)f(\infty)>0}$\\,故選\bbox[red,2pt]{(ACD)}



(A)\times: \tan \theta= 4/3 \\(B) \bigcirc:第3象限,\sin \theta <0 \\(C) \bigcirc:\cos (\theta+180^\circ) = -\cos \theta=3/5 \\(D) \bigcirc: \sin (90^\circ+\theta)=\cos \theta =-3/5 \\(E)\times: \sin(360^\circ+\theta) = \sin \theta= -4/5\\,故選\bbox[red,2pt]{(BCD)}


(A) \bigcirc: \vec a-2\vec b=(2,-3)-(8,16) =(-6,-19) \\(B) \times:\vec a\cdot \vec c = (2,-3)\cdot (2,-1)=4+3=6 \ne 1 \\(C) \bigcirc: \vec b\cdot \vec c=(4,8) \cdot (2,-1)=0 \Rightarrow \vec b\bot \vec c \\(D) \bigcirc: 依定義\\ (E) \bigcirc:\cos \theta ={\vec a\cdot \vec b\over |\vec a||\vec b|} ={8-24 \over \sqrt{13}\times \sqrt{80}}<0 \Rightarrow \theta > 90^\circ\\,故選\bbox[red,2pt]{(ACDE)}



\sqrt{(x-1)^2+(y-2)^2} +\sqrt{(x+1)^2+(y+2)^2} =6 \Rightarrow \cases{焦點F_1(1,2)\\ 焦點F_2(-1,-2)\\ 2a=6}\\ (A) \bigcirc: 中心位於\overline{F_1F_2}的中點\Rightarrow 中點坐標({1-1\over 2},{2-2\over 2})=(0,0) \\(B)\bigcirc: (1,2)及(-1,-2)為其兩焦點\\ (C)\bigcirc: 2a=6為長軸長\\ (D) \times: \overline{F_1F_2}不垂直也不在直線x=y上 \\(E)\bigcirc: 橢圓對稱兩焦點的連線\\,故選\bbox[red,2pt]{(ABCE)}





假設D為坐標原點(0,0,0),各頂點坐標如上圖;\overline{OD}=2 \Rightarrow \sqrt{1+1+a^2}=2 \Rightarrow a=\sqrt 2\\(A)\times: \cases{ \overrightarrow{AC} =(2,-2,0) \\\overrightarrow{BD} =(-2,-2,0)} \Rightarrow \overrightarrow{AC} \cdot\overrightarrow{BD} =-4+4=0 \ne 8 \\(B) \bigcirc: \cases{\overrightarrow{OA}-\overrightarrow{OC} =(-1, 1, -\sqrt 2)- (1,-1, -\sqrt 2) = (-2,2,0) \\\overrightarrow{CB}+\overrightarrow{CD} =(0,2,0)+(-2,0,0)=(-2,2,0)}\Rightarrow 兩者相同 \\(C) \bigcirc: \cases{\overrightarrow{OA}\cdot \overrightarrow{OD} =(-1, 1, -\sqrt 2)\cdot (-1,-1, -\sqrt 2) = 2 \\\overrightarrow{OB} \cdot \overrightarrow{OC} =(1,1,-\sqrt 2) \cdot (1,-1,-\sqrt 2)= 2}\Rightarrow 兩者相同\\(D) \times:\cases{\vec u=\overrightarrow{OA}\times \overrightarrow{OB} =(-1, 1, -\sqrt 2)\times (1,1, -\sqrt 2) = (0,-2\sqrt 2,-2) \\ \vec v=\overrightarrow{OB} \times \overrightarrow{OC} =(1,1,-\sqrt 2 \times (1,-1,-\sqrt 2)= (-2\sqrt 2,0,-2)} \\ \qquad \Rightarrow \cos \theta ={\vec u\cdot \vec v \over |\vec u||\vec v|} ={4\over 12} ={1\over 3} \Rightarrow \theta 為銳角 \\(E)\bigcirc: a=\sqrt 2\\,故選\bbox[red,2pt]{(BCE)}




(A)\times: 含糖量平均約為40,最大值不到70,最小值為30,與平均值的差距都不到40,因此標準差小於40\\ (B)\bigcirc: 低於50的有6個樣本,剛好50的有5個樣本,因此中位數落在50\\ (C)\bigcirc: 由圖形知:熱量越高,則含糖量越高,兩者為正向關\\ (D)\times: 正比必須剛好一直線\\ (E)\bigcirc: 正相關\Rightarrow 斜率大於0\\故選\bbox[red,2pt]{(BCE)}



(A)\bigcirc: 符合算幾不等式{a+b\over 2} > \sqrt{ab}(此時a\ne b) \\(B)\bigcirc: \cases{\sqrt{10}+\sqrt{20} \approx 3.X+4.X =7.X\\ \sqrt{30}=5.X} \Rightarrow \sqrt{10}+\sqrt{20} > \sqrt{30} \\(C) \bigcirc: \log 10+\log 20= \log 200 > \log 30 \\(D) \bigcirc: 理由同(A) \\(E) \bigcirc: \cases{(10^2+20^2)/2=500/2=250 \\ ({10+20 \over 2})^2 =15^2=225} \Rightarrow {10^2+20^2\over 2} >({10+20 \over 2})^2\\,故選\bbox[red,2pt]{(ABCDE)}


(B)\times: det(5M)=5^3det(M)\\其餘皆正確,故選\bbox[red,2pt]{(ACDE)}


g(x)=x^3+ax^2+bx+c \Rightarrow g'(x)=3x^2+2ax+b \Rightarrow g''(x)=6x+2a\\ g(-2)及g(4)有極值 且g(-2)=29\Rightarrow \cases{g'(-2)=0 \\ g'(4)=0 \\ g(-2)=29} \Rightarrow \cases{12-4a+b=0 \\ 48+8a+b=0 \\ -8+4a-2b+c=29} \\ \Rightarrow \cases{a=-3\\b=-24 \\c=1} \Rightarrow g(x)=x^3-3x^2-24x+1\\(A)\times: g(-2)=29 \ne 0 \Rightarrow -2不是g(x)=0的根\\ (B)\bigcirc: \cases{g(-2)為極大值\\ g(4)為極小值\\ g(x)為三次式} \Rightarrow g(x)在區間(-2,4)遞減\\ (C)\times: g''(x)=0 \Rightarrow 6x-6=0 \Rightarrow x=1 \Rightarrow g(x)在x>1凹向上 \\(D)\bigcirc: 由上述聯立方程組可知:c=1 \\(E) \times: g(\infty)=\infty \not \le -79\\,故選\bbox[red,2pt]{(BD)}註: 公布的答案是ABD


(A) \times: f(x)=x^3-ax^2=x^2(x-a) \Rightarrow f(x)=0至少有重根0 \\(B)\bigcirc: f'(x)=3x^2-2ax \Rightarrow f''(x)=6x-2a \Rightarrow f''(1)=0 \Rightarrow 6-2a=0 \Rightarrow a=3\\ (C) \bigcirc: \cases{f'(0)=0\\ f''(0)=-2a=-6} \Rightarrow f(0)為極大值 \\(D)\times: f'(a)=f'(3)=27-18 \ne 0 \Rightarrow f(a)非極值\\ (E)\bigcirc: f(0)=0為極大值\Rightarrow 在x\in [0,3],f(x)為遞減且f(x)<0,因此所圍面積= -\int_0^a f(x)\;dx\\故選\bbox[red, 2pt]{(BCE)}


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