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2022年3月19日 星期六

108年全國科學班資格考-數學詳解

108 年度全國科學班聯合學科資格考試 數學科

一、 單一選擇題:

解答{A(3,3,0)B(3,0,3)C(0,3,3)ABCE:x+y+z=6{d(P(4,5,6),E)=9/6d(Q(13,14,15),E)=8/6d(R(log2(1/3),log31,log4(1/5)),E)=(6+log235)/3d(P,E)>d(R,E)>d(Q,E)p>r>q(4)
解答(1)×:{Γ1:x2/4y2/9=122=4x2/4y2/9=t22t=4t4t/4=tt(2)×:{Γ1:xy(3):a2b2a2+b2=4×94+9=3613(4)×:|¯PF1¯PF2|=4¯PF1¯PF2=±4(5)×:Γ1(3)

二、 多重選擇題:

解答(1)×:y=f(x)y(0,0)y(2):f(x)=f(x)=log10x2+1(3):x20log10x2+10f(x)0(4)×:f(cot50)=log10cot250+1=log10csc50<log10csc45=log102=120.301=0.15050.16(5):f(3)=log1010=12Q(235)

解答XB(n,p)EX=npp0.2()EX=5()np=5n=5÷0.2=25XB(25,0.2)(45)

解答z6+z4+z2+1=0(z21)(z6+z4+z2+1)=0z8=1{z8=18cosk4π+isink4π,k=07z2=12coskπ+isinkπ,k=0,1z6+z4+z2+1=06zk={cosk4π+isink4πk=13cos(k+1)4π+isin(k+1)4πk=46z1=12+12i,z2=i,z3=12+12i,z4=1212i,z5=i,z6=1212i{ak=12,0,12,12,0,12bk=12,1,12,12,1,12S={42,1,22,42,1,22}(1):(2)×:1(3)×:(4)×:z2=iiz2=1=c2+d2i{c2=1d2=03c2+d2=3S(5):z4=1212iiz4=1212i{c4=1/2d4=1/23c4+d4=2/2S(15)

三、 填充題:(每題 5 分)

解答
{¯PB=8¯PC=6¯BC=10¯BC2=¯PB2+¯PC2BPC=90;Q¯BC¯BC¯PQ¯PB¯PC=¯BC¯PQ¯PQ=8×6÷10=245¯BQ=82(24/5)2=325{B(0,0)C(10,0){A(5,12)P(32/5,24/5){u=AB=(5,12)v=AP=(7/5,36/5)w=AC=(5,12){|u|=13|v|=269/5|w|=13uv=397/5vw=467/5{PAB=12|u|2|v|2(uv)2PAC=12|v|2|w|2(vw)2PABPAC=|u|2|v|2(uv)2|v|2|w|2(vw)2=12116=114


解答
{A(0,0)B(4,2)d=¯AB=25{25cosθ=425sinθ=2{cosθ=2/5sinθ=1/5d=25{¯AD=2d=45¯AC=3d=215¯AF=d=25{D(45cos(60+θ),45sin(60+θ))C(215cos(30+θ),215sin(30+θ))F(25cos(60θ),25sin(60θ){D(423,2+43)C(63,3+23)F(23,231)=(63+2+3)×(2+43)=16+323
解答
{Γ1:y=|2x+3|Γ2:y=x2+4x+kk?Γ1(3/2,0)Γ20=946+kk=154y=2x+3Γ2:x2+4x+k=2x+3x2+2x+k3=0044(k3)=0k=4154<k<4

解答
AC¯BDEF¯BD=¯AB2+¯AD2=4+1=5¯BDׯAE=¯ABׯAD¯AE=25¯DE=¯AD2¯AE2=15E{E(0,0,0)F(525,0,0)C(525,25,0)A(0,¯AEcos60,¯AEsin60)=(0,15,35)¯AC=(525)2+(35)2+(35)2=215=1055

解答2:16216×1=6;3:1625326×5×1=30;4:6×5×5×1=1505:6×5×5×5×6=45006+30+120+2160=4686

三、 計算與證明:(每題 8 分)

解答{L:y=mx+nΓ:y=f(x)f(x)mxn=0a,b,cf(x)mxn=k(xa)(xb)(xc)f(x)=k(xa)(xb)(xc)+mx+nf(x)=k(x3(a+b+c)x2+(ab+bc+ca)xabc)+mx+nf(x)=k(3x22(a+b+c)x+ab+bc+ca)+mf

解答
假設此拋物線的準線為L,A在L的垂足為A'、B在L的垂足為B';\\ 另L'\parallel L,且L'經過拋物線焦點F,L'與\overline{AA'}的交點為C、與\overline{BB'}的交點為D,如上圖;\\ \cases{\overline{AF}=10\\ \overline{BF}=5 \\ \overline{AB} =5\sqrt 5} \Rightarrow \overline{AB}^2 =\overline{AF}^2 +\overline{BF}^2 \Rightarrow \angle AFB=90^\circ\\ \angle AFC+ \angle BFD = 90^\circ = \angle BFD+\angle DBF \Rightarrow \angle AFC=\angle DBF =\theta\\ 由拋物線定義可知:\cases{\overline{AF} =10= \overline{AA'} =10\sin \theta+2c\\ \overline{BF}=5 = \overline{BB'} =5\cos \theta+2c} \Rightarrow 10-10\sin\theta =5-5\cos \theta\\ \Rightarrow \cos\theta = 2\sin \theta -1 \Rightarrow (2\sin\theta-1)^2 +\sin^2\theta =1 \Rightarrow 5\sin^2\theta -4\sin\theta=0 \Rightarrow \sin\theta =4/5\\ \Rightarrow 10=10\cdot {4\over 5}+2c \Rightarrow 2c=2 \Rightarrow c=1 \Rightarrow 正焦弦長=4|c|= \bbox[red,2pt]{4}

解答
圓C向右平移2單位成為圓C':(x-3)^2+ (y-2)^2=8,又L':y=m(x+3) \Rightarrow mx-y+3m=0\\ L'與圓C'相切,也就是d((3,2),L')=\sqrt 8 \Rightarrow \left|{ 3m-2+3m\over \sqrt{m^2+1}} \right| =\sqrt 8 \Rightarrow {(6m-2)^2\over m^2+1} =8\\ \Rightarrow 7m^2-6m-1=0 \Rightarrow (7m+1)(m-1)=0  \\\Rightarrow \cases{m=1 \Rightarrow L':y=x+3 切點為B(1,4) \\ m=-1/7 \Rightarrow L':x+7y+3=0切點為B'(13/5,-4/5)}\\ B'在直線x=2的右邊,因此B'不在對摺後的圓周上,也就是m=-1/7不合\\也就是L':\bbox[red,2pt]{y=x+3} \\ \bbox[blue,2pt]{註}:公布的答案包含x+7y=3


解答
解答(1)f(x)=\log_2 (x+k) \Rightarrow \cases{f(0)=\log_2 k\\ f(2)=\log_2 (k+2) \\ f(6)=\log_2(k+6)} \Rightarrow 2f(2)= f(0)+ f(6) \\\Rightarrow 2\log_2(k+2) = \log_2 k+\log_2 (k+6) \Rightarrow \log_2 (k+2)^2 = \log_2 k(k+6) \Rightarrow (k+2)^2 = k(k+6)\\ \Rightarrow 2k=4 \Rightarrow \bbox[red,2pt]{k=2}\\(2)\;a,b,c成等比\Rightarrow b^2=ac \Rightarrow f(a)+ f(c)= \log_2(a+2) +\log_2 (c+2) =\log_2 (a+2)(c+2)\\ =\log_2 (ac+ 2(a+c)+4) =\log_2(b^2+ 2(a+c)+4) \gt \log_2 (b^2+4b+4)(註) = \log_2(b+2)^2=2f(b)\\ \Rightarrow f(b)+f(c) \gt 2f(b),\bbox[red,2pt]{故得證}\\(註:a+c \gt 2\sqrt{ac}=2b)

四、 數學寫作能力題: (每題 5 分, 共計 10 分)

解答(1)X\sim B(n,p) \Rightarrow P(X=x) = \bbox[red,2pt]{C^n_x p^x(1-p)^{n-x},x=0,1,\dots, n} \\(2)期望值EX= \sum_{x=0}^n x C^n_x p^x(1-p)^{n-x} = \sum_{x=0}^n x \cdot {n!\over x!\cdot (n-x)!} (1-p)^n \left({p\over 1-p}\right)^x  \\= (1-p)^n \sum_{x=1}^n   {n\cdot (n-1)!\over (x-1)!\cdot (n-x)!}  \left({p\over 1-p}\right)^x = n(1-p)^n \left({p\over 1-p}\right)\sum_{x=1}^n   {  (n-1)!\over (x-1)!\cdot (n-x)!}  \left({p\over 1-p}\right)^{x-1} \\ = n(1-p)^n \left({p\over 1-p}\right)\sum_{x=1}^n  C^{n-1}_{x-1}  \left({p\over 1-p}\right)^{x-1} = n(1-p)^n \left({p\over 1-p}\right)\sum_{k=0}^n  C^{n-1}_{k}  \left({p\over 1-p}\right)^{k} \\ =n(1-p)^n \left({p\over 1-p}\right)\left(1+{p\over 1-p}\right)^{n-1} =n(1-p)^{n-1}p \left({1 \over 1-p}\right)^{n-1} =\bbox[red,2pt]{np}
解答(1)A= \bbox[red, 2pt]{\begin{bmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{bmatrix}}\\(2)假設\cases{P(x,y) \\P'(x',y')} 且\overline{OP}與x軸的夾角為\alpha,則L與\overline{OP}的夾角為\theta-\alpha \Rightarrow L與\overline{OP'}的夾角為\theta-\alpha \\ \Rightarrow \overline{OP}與\overline{OP'}的夾角為2(\theta-\alpha) \Rightarrow \overline{OP'}與x軸的夾角為2(\theta-\alpha)+\alpha = 2\theta -\alpha \\ \Rightarrow \cases{(x,y)= (\overline{OP}\cos \alpha, \overline{OP}\sin \alpha) \\ (x',y')= (\overline{OP'}\cos(2\theta-\alpha), \overline{OP'}\sin(2\theta-\alpha))} ,由於 \overline{OP}=\overline{OP'} \\ \Rightarrow \cases{x'=\overline{OP}\cos(2\theta-\alpha) =\overline{OP} \cos 2\theta \cos \alpha +\overline{OP} \sin 2\theta \sin \alpha =x\cos 2\theta+ y\sin 2\theta\\ y'=\overline{OP}\sin(2\theta-\alpha)= \overline{OP}\sin 2\theta \cos \alpha -\overline{OP}\sin \alpha \cos 2\theta =x \sin 2\theta-y\cos 2\theta} \\ \Rightarrow \begin{bmatrix}x' \\ y'\end{bmatrix} =\begin{bmatrix}\cos 2\theta & \sin 2\theta \\ \sin 2\theta & -\cos 2\theta \end{bmatrix} \begin{bmatrix}x \\ y\end{bmatrix} \Rightarrow A\begin{bmatrix}x \\ y\end{bmatrix} =\begin{bmatrix}x' \\ y' \end{bmatrix},\bbox[red, 2pt]{故得證}
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