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2024年4月3日 星期三

113年台師大附中教甄-數學詳解

國立臺灣師範大學附屬高級中學 113 學年度
第 1 次專任教師甄選數學科筆試

一、 選填題:(每題 5 分,共 90 分。 填在答案卡上, 分數或根式須以最簡形式回答,否則不予計分)

解答:x=8log2x9log3x4log2x+log0.50.25=2log2x33log3x22log2x2+2=x3x2x2+2x32x2x+2=0(x21)(x2)=0x=1,2(x=1,x>0)=3
解答:(12x)5(1+4x2)5(1+2x)5=(14x2)5(1+4x2)5=(116x4)5=5k=0C5k(16)kx4k{k=3a=C53(16)3=10163k=4b=C54(16)4=5164ba=5×16410×163=8
解答:E(X)=(a1++a12)÷12=13a1++a12=13×12=156Var(X)=E(X2)(E(X))252=E(X2)132E(X2)=194a21++a212=19412=232812i=1aibi=12i=1ai(ai135)=1512i=1(a2i13ai)=15(a21++a212)135(a1++a12)=152328135156=3005=60

解答:

x2+y2=10r=10=¯OP¯AO=62+72=85¯AP=8510=53PAO=θ{sinθ=10/85cosθ=53/85sin2θ=2sinθcosθ=23017APB:¯BPsin2θ=¯BAsinAPB¯BA¯BP=sinAPBsin2θ=sinAPB230/17=kksinAPB=1,k=17230=173060
解答::m=6,n=6,:(n1)m+(1)m(n1)=56+5=15630=30×521k=521
解答:
{A(1,7)B(7,1)O(0,0)P(x,y){PA=(x1,y7)PB=(x7,y+1)PAPB=(x1)(x7)+(y7)(y+1)=19(x4)2+(y3)2=6,Q(4,3),r=6¯OP=¯OQ+r=5+6
解答:[(3a2b+1)2+(2a+b2)2+(4a5b3)2][(2)2+12+12][2(3a2b+1)+(2a+b2)+(4a5b3)]2[(3a2b+1)2+(2a+b2)2+(4a5b3)2]6(7)2[(3a2b+1)2+(2a+b2)2+(4a5b3)2]496=496
解答:PLP(t+1,t+2,t){¯PA=(t+1)2+(t1)2(t3)2¯PB=t2+(t1)2+(t+2)2{¯PA=3(t1)2+8/3¯PB=3(t+1)2+2/3¯PA+¯PB=3(¯QC+¯QD),{Q(t,0)C(1,26/3)D(1,6/3)¯QC+¯QD=¯CD=10¯PA+¯PB=310=30

解答:H93(4!2!2!+5!2!3!+6!2!4!)H44H93=51155775=3135:35,,3,44:H44:,3593,H93=H44H93=57752:224!2!2!93H933:235!2!3!93H934:246!2!4!93H93H93(4!2!2!+5!2!3!+6!2!4!)H44H93
解答:α,α,β,β,γ,γ,cosα+cosβ+cosγ=16i=1cosθi=26i=12cosθi=46i=2cosθi=42cos120=5
解答:lim
解答:(x-1)f(x)= 4\int_1^x f(t)\,dt \Rightarrow \frac{\text{d} }{\text{d}x}( (x-1)f(x)) = \frac{\text{d} }{\text{d}x}\left(  4\int_1^x f(t)\,dt\right) \\ \Rightarrow f(x)+(x-1)f'(x)=4f(x) \Rightarrow f'(x)-{3\over x-1}f(x)=0 \\ 一階微分方程, 取積分因子I(x)=e^{\int -{3\over x-1}\,dx} ={1\over (x-1)^3}\\ \Rightarrow I(x)f'(x)-{3\over x-1 }I(x) f(x)=0 \Rightarrow {1\over (x-1)^3} f'(x) -{3\over (x-1)^4}f(x)=0\\ \Rightarrow \left({1\over (x-1)^3} f(x) \right)'=0 \Rightarrow {1\over (x-1)^3} f(x)= C \Rightarrow f(x)=C(x-1)^3 \\\Rightarrow f(0)=-C=-2 \Rightarrow C=2 \Rightarrow f(x)=2(x-1)^3 \Rightarrow f(5)=2\cdot 4^3=\bbox[red, 2pt]{128}

解答:\omega=t(-\sqrt 3+4i)+(2-2t)i =-\sqrt 3 t+(2+2t)i \; 相當於直線L:2x+\sqrt 3y=2\sqrt 3\\  z^8=-{1\over 2}+{\sqrt 3\over 2}i = \cos({2\pi \over 3}+2k\pi) +i\sin ({2\pi \over 3}+2k\pi) \\ \Rightarrow z_i=\cos \left( {3k+1\over 12}\pi \right)+ i\sin \left( {3k+1\over 12}\pi \right), i=0,1,\dots,7\\ 此題相當於單位圓上8個點 P_i(\cos \left( {3k+1\over 12}\pi \right), \sin \left( {3k+1\over 12}\pi \right)),i=0-7,與L最近的距離\\考量L特性,只要計算第一象限的點就可以,即i=0,1\\ \cases{ P_0(\cos(\pi/12), \sin(\pi/12))=((\sqrt 6+\sqrt 2)/4,(\sqrt 6-\sqrt 2)/4 \\ P_1(\cos(\pi/3), \sin(\pi /3)) =(1/2,\sqrt 3/2)} \\\left| (\sqrt 6+5\sqrt 2)/4-2\sqrt 3\right| \gt \left| 5/2-2\sqrt 3\right|\Rightarrow 最短矩離=d(P_1,L) =\cfrac{\left| 5/2-2\sqrt 3\right|}{ \sqrt 7} \\ =\cfrac{2\sqrt 3- 5/2 }{ \sqrt 7} =\bbox[red, 2pt]{{4\sqrt{21}-5\sqrt 7 \over 14}}
解答:13x^2-10xy+13y^2-6x-42y-27=[x,y]\begin{bmatrix}13 & -5 \\-5 & 13   \end{bmatrix} \begin{bmatrix}x \\y \end{bmatrix} +[-6,-42] \begin{bmatrix}x \\y \end{bmatrix}-27 \\ A=\begin{bmatrix}13 & -5 \\-5 & 13   \end{bmatrix}  =\begin{bmatrix}\sqrt 2/2 & -\sqrt 2/2 \\\sqrt 2/2 & \sqrt 2/2 \end{bmatrix} \begin{bmatrix}8 & 0 \\0 & 18 \end{bmatrix} \begin{bmatrix}\sqrt 2/2 & \sqrt 2/2 \\-\sqrt 2/2 & \sqrt 2/2 \end{bmatrix} \\ \Rightarrow [-6,-42] \begin{bmatrix}\sqrt 2/2 & -\sqrt 2/2 \\\sqrt 2/2 & \sqrt 2/2 \end{bmatrix} =[-24\sqrt 2,-18\sqrt 2] \\ \Rightarrow 8x'^2+18y'^2-24\sqrt 2x'-18\sqrt 2y'-27=0 \Rightarrow 8(x'-3/\sqrt 2)^2+ 18(y'-1/\sqrt 2)^2=72 \\ \Rightarrow {(x-3/\sqrt 2)^2 \over 9}+{(y'-1/\sqrt 2)^2 \over 4}=1 \Rightarrow \cases{a=3\\b=2} \Rightarrow 正焦弦長={2b^2 \over a}=\bbox[red, 2pt]{8\over 3}
解答:P(X=k)=0.8^{k-1}\times 0.2 \Rightarrow \cases{P(X=10)=0.8^{9}\cdot 0.2\gt 0.02 \\ P(X=11)=0.8^{10}\cdot 0.2\gt 0.02 \\ P(X=12)=0.8^{11}\cdot 0.2 \approx 0.017} \\ \Rightarrow k=\bbox[red, 2pt]{12}\\ \log(0.8^{10}\cdot 0.2)=10\log 0.8+\log 0.2=10(3\log 2-1)+\log 2-1 =31\log 2-11\\\qquad =31\times 0.301-11=-1.669\\ \log 0.02=\log 2-2=0.301-2=-1.699\\ 因此\log(0.8^{10}\cdot 0.2) \gt \log 0.02 \Rightarrow 0.8^{10}\cdot 0.2 \gt 0.02
解答:\cases{x(y+z-x)=39-2x^2\\ y(z+x-y)=52-2y^2\\ z(x+y-z)=78-2z^2} \Rightarrow \cases{xy+xz=39-x^2 \\ yz+xy=52-y^2\\ xz+yz=78-z^2} \\三式相加 \Rightarrow 2(xy+yz+zx)=169-(x^2+y^2+z^2) \\\Rightarrow x^2+y^2+z^2-2(xy+yz+zx)=(x+y+z)^2 =169 \Rightarrow x+y+z=13 \\ \Rightarrow \cases{y+z-x=13-2x\\ z+x-y=13-2y\\ x+y-z=13-2z} 代回原式 \Rightarrow \cases{13x-2x^2=39-2x^2\\ 13y-2y^2=52-2y^2\\ 13z-2z^2=78-2z^2} \Rightarrow \cases{x=3\\y =4\\z= 6} \\ \Rightarrow abc=3 \cdot 4\cdot 6 =\bbox[red, 2pt]{72}
解答:
\cases{A(0,a)\\ B(\sqrt 3a,0)\\ C(0,0)\\ P(\cos \theta, \sin \theta) } \Rightarrow \cases{\overline{PA}^2=\cos^2\theta+(\sin\theta-a)^2=2\\ \overline{PB}^2= (\cos\theta-\sqrt 3a)^2+ \sin^2\theta= 10} \Rightarrow \cases{-2a\sin \theta+a^2=1\\ -2\sqrt 3a\cos \theta+3a^2=9} \\ \Rightarrow \cases{\sin\theta =(a^2-1)/2a\\ \cos\theta =(3a^2-9)/2\sqrt 3a} \Rightarrow \sin^2 \theta+\cos^2 \theta= {(a^2-1)^2 \over 4a^2} +{(3a^2-9)^2 \over 12a^2} =1 \\ \Rightarrow 3(a^2-1)^2+(3a^2-9)^2=12a^2 \Rightarrow a^4-6a^2+7=0 \Rightarrow a^2=3+\sqrt 2\\ \Rightarrow \triangle ABC={1\over 2} \cdot \overline{CA}\cdot \overline{CB}={1\over 2}\sqrt 3a^2 = \bbox[red, 2pt]{{3\sqrt 3+\sqrt 6\over 2}}


解答:

|\vec a|=1 \Rightarrow 假設A(1,0); \\又|\vec a-\vec b|={1\over 2} \Rightarrow B在以A為圓心,半徑為1/2的圓上\\ |5\vec a-\vec c|=1 \Rightarrow C在以O_1(5,0)為圓心,半徑為1的圓上\\ \vec a和\vec d的夾角為{\pi\over 4}\Rightarrow D在直線L: x=y上\\ 欲求\overline{BD}+\overline{CD}的最小值,因此將L為對稱軸,圓O_2為圓O_1的對稱圓,\\則\overline{AO_2}與圓A的交點即為B,與圓O_2的交點即為C的對稱點C'\\ 因此\overline{BD}+\overline{CD}的最小值=\overline{AO_2}-\overline{O_2C'}-\overline{AB}=\sqrt{26}-1-{1\over 2}= \bbox[red, 2pt]{-{3\over 2}+\sqrt{26}}

二、 證明題:(共 10 分。請用黑色或藍色原子筆寫在作答卷上,須詳細過程,否則酌予扣分)

解答:\left({1\over a^3(b+c)} +{1\over b^3(a+c)} +{1\over c^3(a+b)}  \right) (a(b+c)+ b(a+c)+c(a+b)) \ge ({1\over a}+{1\over b}+{1\over c})^2 \\ \Rightarrow \left({1\over a^3(b+c)} +{1\over b^3(a+c)} +{1\over c^3(a+b)}  \right) (2(ab+bc+ca) \ge ({ab+bc+ca \over abc} )^2 \\ \Rightarrow {1\over a^3(b+c)} +{1\over b^3(a+c)} +{1\over c^3(a+b)}\ge {ab+bc+ca\over 2} \ge {3\over 2} \left( \because {ab+bc+ca\over 3} \ge \sqrt[3]{abc}=1\right) \\ \Rightarrow {1\over a^3(b+c)} +{1\over b^3(a+c)} +{1\over c^3(a+b)}\ge {3\over 2}\;\bbox[red, 2pt]{QED}
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解題僅供參考,其他歷年試題及詳解




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