114年沙崙高中教甄-數學詳解

 臺南市立沙崙國際高級中學高中部114學年度第1次教師甄選

一、選擇題：5 題(10%) 

解答:$$y=f(x) =(x-3)(x-3-a)(x-3-2a) \Rightarrow \cases{\bigcirc:a=3 \Rightarrow f(x)=(x-3)(x-6)(x-9) \\ \bigcirc: a=2 \Rightarrow f(x)=(x-3)(x-5)(x-7) \\\bigcirc: a=1 \Rightarrow f(x)=(x-3)(x-4)(x-5) \\\bigcirc: a=-1 \Rightarrow f(x)=(x-3)(x-2) (x-1) \\\times: a=-2 \Rightarrow f(x) =(x-3)(x-1)(x+1) \\ \times:a=-3 \Rightarrow f(x)=(x-3)x(x+3)}\\，故選\bbox[red, 2pt]{(E)}$$

解答:$$\sin 18^\circ+\sin x^\circ =\cos 12^\circ \Rightarrow \sin x^\circ = \cos 12^\circ-\sin 18^\circ = \cos 12^\circ-\cos 72^\circ \\= -2 \sin42^\circ \sin (-30^\circ) = \sin 42^\circ  \Rightarrow x=42，故選\bbox[red, 2pt]{(D)}$$

解答:$$(B)\times: f(x)=(x-3)^3 \Rightarrow f'(x)=3(x-3)^2 \Rightarrow f'(3)是=0最最小值不是最大值 \\(D)\times: f(x)=x^2(x-1) =x^3-x^2\Rightarrow f'(x)= 3x^2-2x =x(3x-2)=0有兩相異實根,\\ \qquad 但f(x) =0僅有二實根0,1 \\(E) \times: f(x)=x^3+1 \Rightarrow f'(x)=3x^2=0有兩相等實根,x=0; \\\qquad 但f(x)=0的根為-1, {1 \pm \sqrt 3i\over 2} \Rightarrow 沒有重根\\，故選\bbox[red, 2pt]{(AC)}$$

解答:$$L:{x\over 2}={y\over -1} ={z\over 3} \Rightarrow L: \{(2t,-t,3t) \mid t\in \mathbb R\} \\(A) \times: P=(12,0,0) \Rightarrow d(P,L) =\sqrt{(2t-12)^2+t^2+9t^2} =\sqrt{14t^2-48t+144} \\ \qquad \Rightarrow \left.d(P,L) \right|_{t=12/7} ={12\over 7}\sqrt{35} \ne 3\sqrt{10}  \\ (B) \times: P=(4,-4,-4) \Rightarrow d(P,L) =\sqrt{(2t-4)^2+(-t+4)^2+(3t+4)^2} =\sqrt{14t^2+48} \\\quad \Rightarrow \left. d(P,L)\right|_{t=0} = \sqrt{48} \ne 3\sqrt {10} \\(C) \bigcirc: P=(0,-9,-3) \Rightarrow d(P,L) = \sqrt{4t^2+ (-t+9)^2+ (3t+3)^2} =\sqrt{14t^2+90} \\\quad \Rightarrow \left. d(P,L)\right|_{t=0} = \sqrt{90} =3\sqrt{10} \\(D) \times: P=(0,-3,-9) \Rightarrow d(P,L) =\sqrt{4t^2+(-t+3)^2+ (3t+9)^2} =\sqrt{14t^2+48t+90} \\\quad \Rightarrow \left. d(P,L)\right|_{t=-12/7} = \sqrt{342\over 7} \ne 3\sqrt{10} \\(E) \bigcirc: P=(8,1,-5) \Rightarrow d(P,L) = \sqrt{(2t-8)^2+ (-t-1)^2+ (3t+5)^2} = \sqrt{14t^2+90} \\\quad \Rightarrow \left. d(P,L)\right|_{t=0} = \sqrt{90} =3\sqrt{10}\\，故選\bbox[red, 2pt]{(CE)}$$

解答:$$\det(A-\lambda I) =(\lambda+1)^2 \Rightarrow (A+I)^2 =0 \\ \Rightarrow f(x)=x^n =(x+1)^2 p(x)+ ax+b \Rightarrow f'(x)=nx^{n-1} =2(x+1)p(x)+(x+1)^2p'(x)=a \\ \Rightarrow \cases{f(-1)=(-1)^n=-a+b\\ f'(-1)=n(-1)^{n-1}=a} \Rightarrow \cases{a= n(-1)^{n-1} \\b=n(-1)^{n-1}+(-1)^n} \\ \Rightarrow A^n =n(-1)^{n-1}A+ \left( n(-1)^{n-1}+(-1)^n \right)I \Rightarrow \cases{a_n=n(-1)^{n-1} \\b_n=n(-1)^{n-1}+(-1)^n} \\(A) \times: \det(A^2+A) =\det(A(A+I)) =\det(A) \det(A+I) =\det(A)\cdot 0=0 \Rightarrow 無反方陣\\ (B)\times: a_{2n}=(2n)(-1)^{2n-1} =-2n\Rightarrow {a_{2(n+1)} \over a_{2n}} ={-2(n+1)\over -2n} ={n+1\over n}非常數 \\(C)\bigcirc: b_n=(-1)^{n-1}(n-1) \Rightarrow |b_n|-|b_{n-1}|=n-1-(n-2)=1 \Rightarrow \{|b_n|\}為等差數列 \\(D) \bigcirc: c_n= a_n-b_n= -(-1)^n=(-1)^{n+1} \Rightarrow {c_{n+1} \over c_n} =-1 \Rightarrow \{a_n-b_n\}為等比數列\\ (E)\bigcirc: c_n= a_n^2-b_n^2 =(a_n+b_n)(a_n-b_n)=(2n(-1)^{n-1}+(-1)^n) \cdot (-1)^{n+1}=2n-1 \\\qquad \Rightarrow c_{n+1}-c_n=2 \Rightarrow \{a_n^2-b_n^2\}為等差數列\\ ，故選\bbox[red, 2pt]{(CDE)}$$

二、非選題：10 題(40%) 

解答:$$假設費氏數為ab, 1\le a,b\le 9 且a,b\in \mathbb N\Rightarrow a+b\le 9 \Rightarrow 共有H^3_7 =C^9_7 =36\\ 再加上ab=10,20,\dots,90，共9個，一共有36+9= \bbox[red, 2pt]{45} 個費氏數$$

解答:

$$假設O為正六邊形的重心\Rightarrow O=B逆時鐘旋轉60^\circ = \begin{bmatrix}1/2& -\sqrt 3/2\\ \sqrt 3/2& 1/2 \end{bmatrix} \begin{bmatrix}4\\ 2 \end{bmatrix} = \begin{bmatrix}2-\sqrt 3\\ 1+2\sqrt 3 \end{bmatrix} \\ \Rightarrow O(2-\sqrt 3, 1+2\sqrt 3) =\overline{AD}的中心 \Rightarrow D=2O=(4-2\sqrt 3,2+4\sqrt 3) \Rightarrow \overline{QR}=2+4\sqrt 3\\ \overline{BO}的中點 = \overline{AC}中點 \Rightarrow B+O=A+C \Rightarrow C=B+O-A = (6-\sqrt 3,3+2\sqrt 3)\\ F=B逆時鐘旋轉120^\circ = \begin{bmatrix}-1/2& -\sqrt 3/2\\ \sqrt 3/2& -1/2 \end{bmatrix} \begin{bmatrix} 4\\ 2 \end{bmatrix} = \begin{bmatrix}-2-\sqrt 3\\ -1+2\sqrt 3 \end{bmatrix} \Rightarrow F(-2-\sqrt 3, -1+2\sqrt 3) \\ \Rightarrow \overline{PQ} =C的x坐標-F的x坐標=6-\sqrt 3+2+\sqrt 3=8 \Rightarrow PQRS面積= \overline{PQ} \cdot \overline{QR} \\= 8\cdot (2+4\sqrt 3) =\bbox[red, 2pt]{16+32\sqrt 3}$$

解答:

$$假設\cases{A(0,0) \\ \overleftrightarrow{AB}為x軸}, 又{\overline{AD}\over \overline{CD}} ={1\over 3} \Rightarrow \cases{\overline{AD} =k\\ \overline{CD}=3k} \Rightarrow \overline{AC}=4k  \\ 取D'為D在x軸的投影點 \Rightarrow D'=(k\cos 60^\circ, 0)=(k/2,0) \\ \triangle ABD: {\overline{AD} \over \sin \angle ABD} = {\overline{BD} \over \sin \angle A} =2R \Rightarrow {k \over \sin \angle ABD} = {\overline{BD} \over \sqrt 3/2} =2 \Rightarrow \cases{\overline{BD}= \sqrt 3 \\  \sin \angle ABD={k\over 2}} \\ \Rightarrow \overline{BD'}= \overline{BD} \cos \angle ABD = \sqrt 3 \cdot {\sqrt{4-k^2} \over 2} ={\sqrt 3\over 2}\cdot \sqrt{4-k^2} \Rightarrow \overline{AB}={k\over 2}+{\sqrt 3\over 2}\cdot \sqrt{4-k^2} \\ \Rightarrow \overline{AB}+ \overline{AC} =f(k)={9k\over 2}+{\sqrt 3\over 2}\cdot \sqrt{4-k^2} \Rightarrow f'(k)={9\over 2}-{\sqrt 3\over 2}\cdot {k\over \sqrt{4-k^2}} =0 \\ \Rightarrow k^2={27\over 7} \Rightarrow k={3\over 7} \sqrt{21} \Rightarrow f({3\over 7} \sqrt{21}) ={27\over 14}\sqrt{21}+ {\sqrt 3\over 2}\cdot \sqrt{1\over 7} = \bbox[red, 2pt]{2\sqrt{21}}$$

解答:$$假設\int_1^2 f(x) \,dx =k \Rightarrow  \int_1^2 f(x)\,dx = \int_1^2 \left( 4x^3+3x^2-2kx +3 \right) \,dx \\ \Rightarrow k= \left. \left[ x^4+x^3-kx^2+3x \right] \right|_1^2 =25-3k \Rightarrow k= \bbox[red, 2pt]{25\over 4}$$

解答:$$n=1 \Rightarrow a_1^2 ={1\over 2}a_1a_2 \Rightarrow a_2=2a_1 ={2\over 3} \\ n=2 \Rightarrow a_1a_2+a_2a_1={1\over 2}(a_1a_2+ a_2a_3) \Rightarrow a_3={3\over 3} \\ n=3 \Rightarrow a_1a_3+a_2a_2+a_3a_1 = {1\over 2}(a_1a_2+ a_2a_3+ a_3a_4) \Rightarrow a_4={4\over 3} \\ \cdots \\n=n \Rightarrow a_n={n\over 3} \Rightarrow a_{2025}={2025\over 3} =\bbox[red, 2pt]{675}$$

解答:

$$f(x)=|x^2-3x-4|+x+1 =|(x+1)(x-4)|+x+1 \Rightarrow f(-1)=0為最小值\\ \Rightarrow  f(x)=\begin{cases} f_1(x)=x^2-2x-3& x\le -1\\ f_2(x)=-x^2+4x+5& -1\le x\le 4\end{cases} \Rightarrow M(a)=f_1(a)=f_2(a+1)=M(a+1) \\ \Rightarrow a^2-2a-3=-(a+1)^2+4(a+1)+5 \Rightarrow 2a^2-4a-11=0 \Rightarrow a={2+\sqrt{26} \over 2} \\ \Rightarrow f_1({2+\sqrt{26} \over 2})= \bbox[red, 2pt]{5\over 2}$$

解答:$$取z=x+y \Rightarrow 欲求\overline{PQ} +\overline{PR}的最小值, 其中\cases{P(x,y,z) \\Q(6,5,2)\\ R(8,2,1)} \\平面E:x+y-z=0 \Rightarrow 取f(x,y,z)=x+y-z \Rightarrow \cases{f(Q)=9\\ f(R)=9} \Rightarrow Q,R在E的同側\\ 欲求Q在對稱平面E的對稱點Q': 過Q且方向向量為(1,1,-1)之直線L:{x-6 \over 1}= {y-5\over 1} ={z-2\over -1} \\ \Rightarrow  \overline{QQ'}中點S(t+6,t+5,-t+2) \in L \Rightarrow S\in E \Rightarrow 3t+9=0 \Rightarrow t=-3 \Rightarrow S(3,2,5) \\ \Rightarrow Q'=2S-Q=(6,4,10)-(6,5,2)=(0,-1,8) \\ \Rightarrow \overline{PQ} +\overline{PR}的最小值=\overline{Q'R} =\sqrt{64+9+49} =\bbox[red, 2pt]{\sqrt{122}}$$

解答:$$f(x+2)除以(x^2-3)的餘式為(bx+5) \Rightarrow f(x+2)= (x^2-3)p(x)+bx+5\\ 取y=x+2 \Rightarrow f(y)=((y-2)^2-3)p(y-2)+b(y-2)+5 =(y^2-4y+1)p(y-2)+by+5-2b \\ \Rightarrow f(x)=(x^2-4x+1)p(x-2)+bx+5-2b \\ \Rightarrow f(2+\sqrt 3)=0+b(2+\sqrt 3)+5-2b =5+b\sqrt 3=a-2\sqrt 3 \Rightarrow (a,b)= \bbox[red, 2pt]{(5,-2)}$$

解答:$$5 顆紅球隨機排成一列，將15顆白球插入6個間隙，平均每個間隙可插入{15\over 6}=2.5顆白球\\ 抽獎結束代表所有紅球(5顆)及紅球之前的白球(12.5顆)都被抽走，共抽走了17.5顆球\\ 抽17.5顆球需花費100\times 17.5=1750元，獲利100+200+\cdots+500=1500元\\ 因此期望值=1500-1750=\bbox[red, 2pt]{-250}$$

解答:$$f(x)=1024^{\sin x} +1024^{\cos x} \ge 2\sqrt{1024^{\sin x+\cos x}} =2 \sqrt{2^{10( \sin x+\cos x)}} = 2^{5(\sin x+\cos x)+1} \\ \qquad =2^{5\sqrt 2 \sin(x+\pi/4)+1} \ge 2^{1-5\sqrt 2}\\ k=2^{1-5\sqrt 2} \Rightarrow {\log k\over \log 2}=1-5\sqrt 2 \Rightarrow \log k\approx 0.301 (1-5\cdot 1.414) =-1.82707 \\ \Rightarrow k=10^{-1.82707} \Rightarrow 10^{-2}\lt k\lt 10^{-1} \Rightarrow (m,n)= \bbox[red, 2pt]{(1,-2)}$$

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