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2020年6月5日 星期五

100年 警專30期甲組數學科詳解


臺灣警察專科學校專科警員班三十期(正期學生組)
新生入學考試甲組數學科試題
壹、單選題


f(x)=x3+5x2+11x+10=(x+2)(x2+3x+5)x2+3x+5f(x)g(x)x2+3x+5g(x)=(x2+3x+5)(x2+1)a=3(x)(C)


12432143=(124214)(1242+124×214+2142)=90(1242+124×214+2142)3(C)


2x+3x212x+3x210x+5x20(x+5)(x2)0x2x50x>2x5(D)




±αx2α2()0{4α28=0α2=26=α2(a+α2)=06=2(a+2)a=5(B)



α,βx2+6x+4=0{α+β=6αβ=4{α<0β<0α×β=αβ(α+β)2=α+β2αβ=624=64=10(D)


678905log67890(51)=4=logxx5(1)log0.12345=1+log0.12345=log1.2345(2)(1)(2)logx=4+log1.2345=log12345x=12345(C)



tanAtanB=1sinAsinBcosAcosB=1cosAcosBsinAsinB=0cos(A+B)=0A+B=90C=18090=90(B)



12π<2<23π32<sin2<1(A)



:asinA=bsinB=csinC=2R{a=2RsinAb=2RsinBc=2RsinC3(ab+c)=6R(sinAsinB+sinC)=14(sinAsinB+sinC)R=73(B)





2πsinx=xsinx=x2π{y=sinxy=x2π{y=x2π12π=0.5π¯OA1π/2=2π¯OB15π/2=0.4π¯OB<y=x2π<¯OA3(C)



x2+y2+z22x+4y+2z19=0(x1)2+(y+2)2+(z+1)2=52{O(1,2,1)R=5{a=dist (O,x+y+z=0)=|1213|=23b=dist (O,z=1)=0c=dist (O,y=1)=|21|=3d=dist (O,x=2y)=|1+45|=5b<a<d<cc(C)



x2y+3z=4n=(1,2,3){(A):P=(57,187,157)(B):Q=(157,187,57)(C):R=(157,57,187)(D):S=(57,157,187){PA=(27,47,67)QA=(87,47,167)RA=(87,97,37)SA=(27,17,37){PAnQAnRAnSAn(A)





AM=tAP=t(3AB+4AC)=3tAB+4tAC3t+4t=1t=1/7AM=37AB+47AC(B)





O(0,0,0)E=154+1+4=5¯PQ=53()=2(A)



L:{x=1+3ty=25tz=3,tRLz=3z=0(xy)(C)


{xA(4,0)F1(0,4)O(4,4){b=¯OA=4c=¯OF1=4a2=b2+c2=16+16=32a=422a=82(D)


x1,x2,x3,x4,x5x1+x2+x3+x4+x5=110xi9x1+x2+x3+x4+x5=11H511=1365(11,0,0,0,0)5(10,1,0,0,0)5!3!=205+20=25(0xi9)136525=13401340100000=0.0134(B)



{P(a,b)Γb2=4a(a,b)ΓP(a,b)Γa216+b29=1(a,b)Γ(a,b)ΓΓ(B)


an+1=n+1nana30=3029a29=30292928a28=302929282827a27=30292928282721a1=30a1=30(D)




(){A(1,2)B(4,2)C(3,3)D(0,0)f(x,y)=x2y{f(A)=3f(B)=0f(C)=9f(D)=09(D)



[13252312]3r1+r2[132520112112]211r2[13252011]32r2+r1[101011]{a=1<0b=1<0(C)



()1(A)


=20%×0.9+80%×0.05=0.22(B)



{x+kyz=1(1)8x+3y6z=1(2)4x+y3z=1(3)(2)2×(3)y=3(1)(3){x+3kz=1(4)4x3z=4(5)x=2x=2(4)(5){3kz=1z=1+3k3z=12z=41+3k=4k=1(A)


f(x)=mx2+(m1)x+(m1)>0,x(m1)24m(m1)<03m22m1>0(3m+1)(m1)>0m>1m<13;y=f(x)m>0m<13(B)


P(BA)=P(B)P(A)P(A)=P(B)×P(A)P(A)=P(B)=34P(BC)=P(B)P(C)=34P(C)=15P(C)=415P((AB)C)=P(AB)P(C)=(P(A)+P(B)P(A)P(B))P(C)=(13+3414)415=29(B)



y=x44x+8y=4x34y



x^3-3x^2-4x+6 = (x-1)(x^2-2x-6)=0 \Rightarrow x=1,1\pm \sqrt 7\Rightarrow 3相異實根,故選\bbox[red,2pt]{(D)}



f(x)=x^3-3x+2 \Rightarrow f'(x)=3x^2-3 \Rightarrow f''(x)=6x\\ 令f'(x)=0 \Rightarrow 3x^2-3=0 \Rightarrow x=\pm 1 \Rightarrow \cases{f''(1)=6>0 \\f''(-1)=-6<0 } \Rightarrow \cases{f(1)=0 為極小值\\ f(-1)=4為極大值}\\,故選\bbox[red,2pt]{(C)}



f(x)=x^3+3x^2+px-4 \Rightarrow f'(x)=3x^2+6x+p \Rightarrow f''(x)=6x+6\\ 沒有極值代表 f'(x)=0無解 \Rightarrow 判別式36-12p\le 0 \Rightarrow  p\ge 3,故選\bbox[red,2pt]{(B)}

貳、多重選擇題


(A) \bigcirc: a_n= S_n-S_{n-1}=2n+1 \Rightarrow < a_n > 為等差數列,其公差為2 \\(B)\bigcirc:< a_k >為等比\Rightarrow a_k=a_1r^{k-1} \Rightarrow a_k^2 = a_1^2r^{2k-2} = a_1^2(r^2)^{k-1} \Rightarrow < a_k^2 >為等比數列,其公比為r^2 \\(C) \bigcirc: a_k=5k+3 \Rightarrow b_k=3a_k+2 = 15k+6+2= 15k+8 \Rightarrow < b_k > =< 3a_k+2 >為等差數列 \\(D) \bigcirc: 2a_{k+1} = a_k+a_{k+2} \Rightarrow a_{k+1} = (a_k+a_{k+2})\div 2 \Rightarrow < a_k> 為等差數列 (前後平均為中間項) \\(E) \times: \cases{< a_k> =1,1,1,1,\dots\\ < b_k> = 1,-1,1,-1,\dots} \Rightarrow < a_k+b_k >=2,0,2,0,\dots 非等比數列\\,故選\bbox[red,2pt]{(A,B,C,D)}



(A)\times: f(\alpha+\beta)= a^{\alpha+\beta} =a^\alpha\times a^\beta =f(\alpha)f(\beta)\ne f(\alpha)+f(\beta) \\(B)\times: 0.5^5 < 0.5^4 \Rightarrow 5 \not \lt 4 \\(C) \times: y=f(x)=a^x \ne 0,\forall x \in R \\ (D) \bigcirc: {1\over 2}(f(\alpha)+f(\beta))= {1\over 2}(a^\alpha +a^\beta)\ge \sqrt{a^\alpha \times a^\beta}= \sqrt{a^{\alpha+\beta}} =a^{\alpha+\beta \over 2} =f({\alpha+\beta \over 2}) \\(E) \bigcirc:\lim_{x\to -\infty}a^x=0 \Rightarrow y=0(即X軸)為漸近線\\,故選\bbox[red,2pt]{(DE)}


正弦定理:{a\over \sin A} ={b\over \sin B} ={c\over \sin C} =2R\Rightarrow a\cos A+b\cos B = 2R\sin A\cos A+ 2R\sin B\cos B \\ =R(\sin 2A+ \sin 2B) = 2R\sin(A+B)\cos (A-B) =2R\sin(\pi-C)\cos(A-B)\\ =2R\sin C\cos(A-B) = c\cos(A-B) =c\cos C \Rightarrow \cases{A-B=C =\pi-A-B\\ B-A=C =\pi -A-B} \Rightarrow \cases{A=\pi/2 \\ B=\pi/2} \\ \Rightarrow \triangle ABC 為直角\triangle,其中A為直角或B為直角,故選\bbox[red,2pt]{(A,C,D)}



(A)\times: \text{dist}(A,E)= \left|{6+9-1\over \sqrt{4+9+36}} \right|={14\over 7}=2 \\(B)\bigcirc: 經過A且方向向量為(2,3,-6)的直線L:{x-3\over 2}= {y-3 \over 3} ={z\over -6}\Rightarrow L上的點(2t+3,3t+3,-6t)與E相交\\ \qquad \Rightarrow 2(2t+3)+3(3t+3)-6(-6t)=1 \Rightarrow t=-{2\over 7} \Rightarrow L與E的交點為(-{4\over 7}+3,-{6\over 7}+3,{12\over 7}) \\ \qquad =({17\over 7},{15\over 7},{12\over 7})即為正射影點\\ (C)\bigcirc: \cases{A(3,3,0) \\ A在E的正射影點P({17\over 7},{15\over 7},{12\over 7})\\ A對於E的對稱點A'(x,y,z)} \Rightarrow P=(A+A')\div 2 \Rightarrow \cases{{17\over 7} =(3+x)/2\\{15\over 7} =(3+y)/2\\ {12\over 7}=(0+z)/2} \Rightarrow \cases{x={13\over 7}\\ y={9\over 7} \\ z={24\over 7}} \\(D) \times: 由(E)知: \triangle ABC \bot E \Rightarrow \triangle ABC在E上的投影為一直線\\(E) \bigcirc: \cases{\overrightarrow{AB}= (-2,-1,2)\\ \overrightarrow{AC}=(1,-1,2)} \Rightarrow \overrightarrow{AB} \times \overrightarrow{AC} = (0,6,3) \Rightarrow (\overrightarrow{AB} \times \overrightarrow{AC})\cdot (2,3,-6)=0 \\\qquad \Rightarrow \theta=90^\circ \Rightarrow \sin \theta=1\\,故選\bbox[red,2pt]{(BCE)}





假設5個不同禮品分別為A、B、C、D、E;\\(A)\times: 每個禮品可以有3種分法,共有3^5=243種分法\\ (B)\times: 甲恰得A,其4個禮品各有2種分法,共有2^4=16種分法;\\ \qquad 甲恰得B、C、D、E也各有16種分法,因此共有16\times 5=80種分法;\\(C)\bigcirc: 理由同(B),乙恰得一件也是有80種分法\\ (D)\bigcirc: 甲沒有禮品,5個禮品各有2種分法,因此共有2^5=32種分法;\\\qquad因此甲至少得一件的分法=全部分法減去甲沒有禮物的分法=243-32=211 \\(E)\bigcirc: 全部-(甲沒禮物-乙沒禮物-丙沒禮物)+(甲乙都沒禮物+乙丙都沒禮物+甲丙都沒禮物)\\\qquad = 243-(32+32+32)-(1+1+1) =243-96+3= 150\\,故選\bbox[red,2pt]{(CDE)}




(A)\times: 有可能,機率為{1\over 2^{10}}\\ (B)\bigcirc: 10\times {1\over 2}=5 \\(C) \times: 5個0與5個1的排列數為{10! \over 5!5!},因此機率為{10! \over 5!5!}\times {1\over 2^{10}}\ne {1\over 2}\\ (D)\bigcirc:6個1與4個0的排列數與 6個0與4個1的排列數相同\\ (E)\bigcirc: 無論第幾次,出現正面或反面的機率都是1/2\\故選\bbox[red,2pt]{(BDE)}



(A)\times: 三個未知數卻只有二個方程式,有無限多組解 \\(B)\times: 理由同(A) \\(C) \times: \begin{bmatrix} -2 & 1 & 0 & -5\\ 0 & 2 & 3& 1\\ 0 & 4 & 6& 4\end{bmatrix} \xrightarrow{-2r_2+r_3} \begin{bmatrix} -2 & 1 & 0 & -5\\ 0 & 2 & 3& 1\\ 0 & 0 & 0& 2\end{bmatrix} \Rightarrow 0=2 \Rightarrow 無解 \\(D) \bigcirc: \begin{bmatrix} 1 & 0 & 2 & 4\\ 0 & -3 & 1& 2\\ 0 & 1 & 0& 3\end{bmatrix} \Rightarrow \cases{x+2z=4 \\ -3y+z=2 \\ y=3}\Rightarrow \cases{x=-18 \\ z=11 \\ y=3} \\(E)\bigcirc: \begin{bmatrix} 1 & 1 & 0 & 5\\ 0 & 2 & 2& -6\\ 3 & 0 & 3 & -6\end{bmatrix} \xrightarrow{r_2/2,r_3/3} \begin{bmatrix} 1 & 1 & 0 & 5\\ 0 & 1 & 1& -3\\ 1 & 0 & 1 & -2\end{bmatrix} \Rightarrow \cases{x+y=5\\ y+z=-3\\ x+z=-2} \Rightarrow x+y+z=0 \Rightarrow \cases{z=-5\\ x=3\\ y=2}\\,故選\bbox[red,2pt]{(DE)}


(A)\bigcirc: f(x)=\begin{cases} x & x\ge 0\\ -x & x<0 \end{cases} \Rightarrow f'(x)=\begin{cases} 1 & x\ge 0\\ -1 & x<0 \end{cases} \Rightarrow 1\ne -1 \\(B)\bigcirc: f(x)=[x] \Rightarrow \cases{\lim_{x\to 0^+}f(x)=0 \\ \lim_{x\to 0^-}f(x)=-1} \Rightarrow f(x)在x=0不連續\\ (C) \times:f(x)=\begin{cases} x^2 & x\ge 0\\ -x^2 & x<0 \end{cases} \Rightarrow f'(x)= \begin{cases} 2x & x\ge 0\\ -2x & x<0 \end{cases} \Rightarrow f'(0)=0 \\(D)\bigcirc: f(x)=\sqrt x \Rightarrow f'(x)={1\over 2\sqrt x}  \Rightarrow f'(0) 不存在\\ (E)\times: f(x)={x^2 \over x+1} \Rightarrow f'(x)={2x \over x+1}- {x^2 \over (x+1)^2} \Rightarrow f'(0)=0\\,故選\bbox[red,2pt]{(ABD)}


f(\infty)>0 \Rightarrow a >0\\ f(0)=d>0\\由圖形知三實根為二正一負,而三根之和為正,即-{b\over a}>0 \Rightarrow b<0\\ 由圖形知f'(x)=0有相異二正根 \Rightarrow 3ax^2+2bx+c=0之二根為正值 \\\Rightarrow \cases{二根之積為正,即{c\over 3a}>0 \Rightarrow c>0 \\ 判別式4b^2-12ac>0 \Rightarrow b^2-3ac>0}\\,故選\bbox[red,2pt]{(ACDE)}


(A)\times: \int_0^2 f(t)\;dt為一常數,設為a \Rightarrow f(x)=x^2 +x\int_0^2 f(t)dt +1=x^2+ax+1 \\\qquad \Rightarrow \int_0^2 f(x)\;dx= \left. \left[ {1\over 3}x^3+{1\over 2}ax^2+x \right] \right|_0^2 = {8\over 3}+2a+2 = {14 \over 3}+2a \\ \qquad\Rightarrow \int_0^2 f(t)\;dt=a= {14 \over 3}+2a \Rightarrow a=-{14\over 3} \\(B)\bigcirc: 理由同(A)\\ (C)\bigcirc: f(x)=x^2 -{14\over 3}x +1 \Rightarrow f(3)=9-14+1=-4\\ (D)\times: f(-1)=1+{14\over 3}+1 = {20\over 3} \ne {8\over 3}\\ (E)\bigcirc:f(x)的 判別式 {196 \over 9}-4 >0 \Rightarrow 有相異實根\\故選\bbox[red, 2pt]{(BCE)}


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