115年板橋高中教甄-數學詳解

 新北市立板橋高級中學115學年度正式教師甄選

一、填充題(每題5分)

解答:$$105=3\times 5\times 7 \Rightarrow 歐拉函數\text{(Euler's Totient Function) }\\ \phi(105)=105\times  \left( 1-{1\over 3} \right)\times \left( 1-{1\over 5} \right) \times \left( 1-{1\over 7} \right) =48 \Rightarrow 2026=48\times 42+10 \\ \Rightarrow a_{2026}=105\times 42+(從 1 開始數起的第 10 個與 105 互質的數)\\ 與105互質的數:1,2,4,8,11,13,16,17,19,22 \Rightarrow 第10個是22 \Rightarrow a_{2026}=105\times 42+22= \bbox[red, 2pt]{4432}$$

解答:$$\cases{a=\log {8\over 25}=\log 8-\log 25=3\log 2-2\log 5=3(1-\log 5)-2\log 5=3-5\log 5\\b=\log 12=\log 3+2\log 2=\log3+2(1-\log 5)=\log 3-2\log 5+2} \\ \Rightarrow \cases{\log 5= {3\over 5}-{a\over 5} \\ \log 3=-{2\over 5}a+b-{4\over 5}} \Rightarrow \log 15= \log 3+\log 5=-{3\over 5}a+b-{1\over 5} =xa+yb+z \\ \Rightarrow (x,y,z) = \bbox[red, 2pt]{\left( -{3\over 5},1,-{1\over 5} \right)}$$

解答:$$由題意可知:最後一球是黑球，機率為{黑球\over 全部} ={5\over 12}\\3白球4紅球中，紅球是最後一球的機率為{4\over 7} \Rightarrow 符合要求的機率為{5\over 12}\times {4\over 7}= \bbox[red, 2pt]{5\over 21}$$

解答:$$取g(x)=f(x)-x-{1\over x} \Rightarrow h(x)=xg(x)=xf(x)-x^2-1 \Rightarrow x=1,2,\dots,11是h(x)=0的根\\ \Rightarrow h(x)=xf(x)-x^2-1=a(x-1)(x-2)\cdots(x-11) \Rightarrow h(0)=-1=-a\cdot 11! \Rightarrow a={1\over 11!} \\ \Rightarrow h(13)=13f(13)-170={1\over 11!}\cdot 12\cdot 11\cdots2 =12 \Rightarrow f(13)={182\over 13}=\bbox[red, 2pt]{14}$$

解答:

$$\angle ACB=90^\circ \Rightarrow 取\cases{C(0,0) \\ A(0,b)\\ B(-a,0)} \Rightarrow \overrightarrow{CD}=(\cos(90^\circ-\beta), \sin(90^\circ-\beta)) =(\sin \beta,\cos \beta)=({12\over 13},{5\over 13}) \\ \Rightarrow 取D=(12k,5k),又\cases{\overrightarrow{BA}=(a,b) \\ \overrightarrow{BD}=(12k+a,5k )} \Rightarrow \overrightarrow{BA}旋轉(-\alpha)變\overrightarrow{BD} \Rightarrow \cases{12k+a=a\cos \alpha+ b\sin \alpha \\ 5k=-a\sin \alpha+b\cos \alpha} \\ \Rightarrow \cases{12k+a=4a/5+3b/5\\ 5k=-3a/5+4b/5} \Rightarrow \cases{60k=-a+3b\\ 25k=-3a+4b} \Rightarrow \cases{a=33k\\ b=31k} \Rightarrow \tan \angle ABC={b\over a}= \bbox[red, 2pt]{31\over 33}$$

解答:$$|z|=1 \Rightarrow z=\cos \theta+ i\sin \theta \Rightarrow |z^3-3z-2|=|(z+1)^2(z-2)|= |z+1|^2 \cdot |z-2| \\=((\cos \theta+1)^2+\sin^2\theta) \cdot \sqrt{(\cos\theta-2)^2+\sin^2\theta} =(2+2\cos\theta)\cdot \sqrt{5-4\cos \theta}\\ 取x=\cos \theta \Rightarrow f(x)=(2+2x)^2\cdot (5-4x) \Rightarrow f'(x)=24(x+1)(1-2x) =0 \Rightarrow x=-1,{1\over 2} \\ \Rightarrow \cases{f(-1)=0\\f(1/2)=27\\f(1)=16} \Rightarrow f(x)最大值27 \Rightarrow |z^3-3z-2|最大值=\sqrt{27}= \bbox[red, 2pt]{3\sqrt 3}$$

解答:$$假設\cases{P(\cos \theta, \sin \theta)\\A(2,0) \\B(x,y)} \Rightarrow \cases{\overrightarrow{PA}=(2-\cos \theta, -\sin \theta) \\ \overrightarrow{PB}=(x-\cos \theta,y-\sin \theta)} \Rightarrow \overrightarrow{PB} =\overrightarrow{PA}逆時針旋轉90^\circ \\ \Rightarrow(x-\cos \theta,y-\sin \theta) =(\sin \theta, 2-\cos\theta) \Rightarrow \cases{x=\cos \theta+\sin \theta\\ y=2-\cos\theta+\sin \theta} \Rightarrow \cases{\cos \theta=(x-y+2)/2\\ \sin \theta=(x+y-2)/2} \\ \Rightarrow {1\over 4}((x-y+2)^2+ (x+y-2)^2)=1 \Rightarrow 2x^2+2(y-2)^2=4 \Rightarrow \bbox[red, 2pt]{x^2+(y-2)^2=2}$$

解答:$$f(x)=x^3+3x+{12\over x}+{4 \over x^3} \Rightarrow f'(x)=3x^2+3-{12\over x^2}-{12\over x^4} \\ f'(x)=0 \Rightarrow x^6+x^4-4x^2-4=0 \Rightarrow (x^4-4)(x^2+1)=0 \Rightarrow x=\sqrt 2 \\ \Rightarrow f(\sqrt 2) =2\sqrt 2+3\sqrt 2+6\sqrt 2+\sqrt 2= \bbox[red, 2pt]{12\sqrt 2}$$

解答:$$相關係數 r_{X,Y}= {Cov(X,Y) \over \sigma_X \sigma_Y} \Rightarrow {1\over 2}={Cov(X,Y) \over 1\cdot 2} \Rightarrow Cov(X,Y)=1\\ Z= X+Y \Rightarrow Cov(X,Z) = Cov(X,X+Y) =Cov(X,X)+ Cov(X,Y) =Var(X)+1=1+1=2 \\ \Rightarrow Var(Z) =Var(X+Y)=Var(X)+Var(Y)+ 2Cov(X,Y)=1+4+2=7 \Rightarrow \sigma_Z=\sqrt 7 \\ \Rightarrow r_{X,Z} ={Cov(X,Z) \over \sigma_X \sigma_Z}={2 \over 1\cdot \sqrt 7} = \bbox[red, 2pt]{{2\sqrt 7\over 7}}$$

解答:$$f(x)=-8x^3+33x^2-18x+ \int_0^x f(t)\,dt \Rightarrow f(0)=0\\ \Rightarrow f'(x)=-24x^2+66x-18+f(x) \Rightarrow  f'(x)-f(x)=-24x^2+66x-18\; 一階微分方程 \\ \Rightarrow e^{-x}(f'(x)-f(x))= (e^{-x} f(x))' =e^{-x}(-24x^2+66x-18) \\ \Rightarrow e^{-x}f(x)= \int e^{-x}(-24x^2+66x-18) \,dx =e^{-x}(24x^2-18x)+C \Rightarrow f(x)= 24x^2-18x+Ce^{-x} \\f(0)=0 \Rightarrow C=0 \Rightarrow f(x)= \bbox[red, 2pt]{24x^2-18x}$$

解答:$$假設\cases{S_0:目前沒有任何符合目標進度\\ S_1:目前最後一次擲出的點數為1\\ S_2:目前最後兩次擲出的點數為1,2\\ S_3:目前最後三次擲出的點數為1,2,3} \\及E_i為在S_i下達成目標還要再投擲的期望次數,i=0,1,2,3; 欲求E_0 \\ \Rightarrow \cases{E_0=1+{1\over 4}E_1+{3\over 4}E_0 \\E_1= 1+{1\over 4}E_2+{1\over 4}E_1+ {1\over 2}E_0\\ E_2=1+{1\over 4}E_3+ {1\over 4}E_1+ {1\over 2}E_0 \\E_3=1+{1\over 4}\cdot 0+{1\over 4}E_1 +{1\over 2}E_0} \Rightarrow E_0= \bbox[red, 2pt]{256}$$

解答:$$假設\cases{A(0,0,0) \\B(3,0,0) \\C(3,4,0) \\D(0,4,0) \\E(1,0,0)} \Rightarrow B'(0,y,h) \Rightarrow \cases{\overline{B'E}^2= \overline{BE}^2\\ \overline{B'C}^2= \overline{BC}^2} \Rightarrow \cases{y^2+h^2 =3 \\-8y+19=7} \Rightarrow \cases{y=3/2\\ h=\sqrt 3/2}\\ \Rightarrow 四角錐B-AECD體積={1\over 3}\times S_{AECD}\times h={1\over 3}\times 8\times {\sqrt 3\over 2}= \bbox[red, 2pt] {4\sqrt3 \over 3}$$

解答:$$\cases{x+5=y+z\\ z^2+xy=3z-5} \Rightarrow \cases{x-y=z-5 \\ xy=-z^2+3z-5} \Rightarrow (x-y)^2=x^2+y^2-2xy =(z-5)^2 \\ \Rightarrow x^2+y^2=(z-5)^2+2xy=(z-5)^2+2(-z^2+3z-5) \\\Rightarrow 欲求x^2+y^2+xy= f(z)= (z-5)^2+3(-z^2+3z-5) =-2z^2-z+10 \\ 考慮以x和(-y)為兩根的二次方程式g(u)\Rightarrow \cases{兩根之和=x-y=z-5\\ 兩根之積=-(xy)=z^2-3z-5} \\ \Rightarrow  g(u)=u^2-(z-5)u+(z^2-3z-5)=0有實根\Rightarrow (z-5)^2-4(z^2-3z-5)\ge 0 \\ \Rightarrow 3z^2-2z-5\le 0\Rightarrow (3z-5)(z+1)\le 0 \Rightarrow -1\le z\le {5\over 3} \\ f(z)=-2z^2-z+10=-2(z+{1\over 4})^2+{81\over 8} \Rightarrow \cases{最大值f(-1/4)=81/8\\ 最小值f(5/3)=25/9} \Rightarrow  \bbox[red, 2pt]{\cases{最大值=81/8\\ 最小值=25/9}}$$

解答:

$$\textbf{(1) }取M=(0,m), m\gt 0 \Rightarrow 過M的直線L: y=kx+m,並假設L與\Gamma:y=x^2交於\cases{A(x_1,y_1)\\ B(x_2,y_2)}\\ 將L:y=kx+m代入 \Gamma \Rightarrow kx+m=x^2 \Rightarrow x^2-kx-m=0 \Rightarrow \cases{兩根之和=k=x_1+x_2\\ 兩根之積=-m= x_1x_2} \\ \Rightarrow \overrightarrow{OA} \cdot \overrightarrow{OB} =x_1x_2 +y_1y_2 =x_1x_2+ x_1^2x_2^2=2 \Rightarrow -m+m^2=2 \Rightarrow (m-2)(m+1)=0 \\ \Rightarrow m=2 \;(\because m\gt 0) \Rightarrow L:y=kx+2 必通過(0,2), \bbox[red, 2pt]{故得證} \\\textbf{(2) }F=(0,1/4) 並假設B(t,t^2),t\gt 0; 由於x_Ax_B=-2 \Rightarrow A(-2/t,4/t^2) \\\Rightarrow \triangle OAB ={1\over 2} \begin{Vmatrix} -2/t& 4/t^2 & 1\\ t& t^2&1 \\ 0& 0& 1 \end{Vmatrix}  ={1\over 2} \left| -2t-{4\over t}\right|=t+{2\over t} \\ \Rightarrow \triangle OBF ={1\over 2}\times {1\over 4} \times t={t\over 8} \Rightarrow 四邊形OABC面積= \triangle OAB+\triangle OBC = \triangle OAB+\triangle OBF\\={9t\over 8}+{2\over t} \ge 2 \sqrt{{9t\over 8}\cdot {2\over t}} = \bbox[red, 2pt]3$$

解答:$$假設\cases{\angle PAB=\alpha\\ \angle PBA=\beta} \Rightarrow \cases{\angle PDC =\alpha \\ \angle PCD=\beta} \Rightarrow \cases{\angle APE=90^\circ -\alpha\\ \angle APD=\alpha+\beta \\ \angle DPH=90^\circ-\beta} \Rightarrow \angle APE+ \angle APD+ \angle DPH =180^\circ\\ \Rightarrow E,P,H共線, 同理可得F,P,G共線\\ 假設\angle APB= \angle DPC=\theta \Rightarrow \triangle PAB={1\over 2} \overline{PA}\cdot \overline{PB}\sin \theta={1\over 2} \overline{AB} \cdot \overline{PE} \Rightarrow \overline{PE} ={\overline{PA} \cdot \overline{PB} \sin \theta\over \overline{AB}} \\ 正弦定理:2R_G= \overline{AB\over \sin \theta} \Rightarrow \overline{PG} =R_G= {\overline{AB} \over 2\sin \theta}\\ 同理可得: \overline{PF}={\overline{PD}\cdot \overline{PD} \sin \theta\over \overline{CD}} ,\overline{PH}=R_H ={\overline{CD} \over 2\sin \theta} \\ \Rightarrow \overline{PE}\cdot \overline{PH} =   {\overline{PA} \cdot \overline{PB} \sin \theta\over \overline{AB}}   \cdot {\overline{CD} \over 2\sin \theta} = {\overline{PA} \cdot \overline{PB}\cdot \overline{CD} \over 2\overline{AB}} , \overline{PF}\cdot \overline{PG}={\overline{PC}\cdot \overline{PD} \cdot \overline{AB} \over 2\overline{CD}}\\ 由於\triangle PAB \sim \triangle PDC \Rightarrow {\overline{CD} \over \overline{AB}} ={\overline{PC} \over \overline{PB}}={ \overline{PD} \over \overline{PA}}=k \Rightarrow {\overline{PC} \over \overline{PB}} \cdot { \overline{PD} \over \overline{PA}} =k^2 = \left( {\overline{CD} \over \overline{AB}} \right)^2 \\ \Rightarrow \overline{PE} \cdot \overline{PH}= \overline{PF}\cdot \overline{PG},依圓冪定理逆定理, E,F,G,H共圓$$

解答:$$f(x+5)-f(x)\le 3(x+3)=3x+ 9 \Rightarrow \cases{f(x+15)-f(x+10) \le 3(x+10)+9=3x+39\\ f(x+10)-f(x+5) \le 3x+24\\f(x+5)-f(x) \le 3x+9} \\ \Rightarrow 三式相加: f(x+15)-f(x)\le 9x+72, 又題意:f(x+15)-f(x)\ge 9(x+8) =9x+72 \\ \Rightarrow 9x+72\le f(x+15)-f(x)\le 9x+72 \Rightarrow f(x+15)-f(x)=9x+72 \\ \Rightarrow f(x+5)-f(x)=3x+9 \Rightarrow f(x)為二次式\Rightarrow f(x)=ax^2+ bx+c \\ \Rightarrow a(x+5)^2+b(x+5)+c-(ax^2+bx+c)= 3x+9 \Rightarrow \cases{a=3/10 \\ b=3/10\\ c=1013} \\ \Rightarrow f(x)={3\over 10}x^2+{3\over 10}x+1013 ={3\over 10}x(x+1) +1013 \Rightarrow {f(2025)\over 2026}={{3\over 10}\cdot 2025\cdot 2026+1013 \over 2026} \\={3\over 10}\cdot 2025+0.5 =\bbox[red, 2pt]{608}$$

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