國立花蓮女中 113 學年度教師甄選
一、填充題（每題 5 分，共 35 分）
解答:$$x^2+y^2=1 \Rightarrow \cases{x= \cos \theta\\ y=\sin \theta} \Rightarrow (x+y)(x+3y) =x^2+4xy+ 3y^2= 1+4\cos \theta \sin \theta+ 2\sin^2 \theta \\=1+2\sin 2\theta +1-\cos 2\theta=2+2\sin 2\theta-\cos 2\theta=2+\sqrt 5 \sin(2\theta-\alpha) \\ \Rightarrow 最大值 \bbox[red, 2pt]{2+\sqrt 5}$$
解答:$$任取3球，剩下47球插入四個空隙，平均每個空隙可以插入{47\over 4}個球，因此第三個球的序位為:\\ {47\over 4}+1+{47\over 4}+1+{47\over 4}+1= \bbox[red, 2pt]{153\over 4}$$
解答:$$f(x)= x^5+x^4+x^3 +x^2+x+1 \Rightarrow (x-1)f(x)=x^6-1 \Rightarrow (x-1)(x^6+1) f(x) =x^{12}-1 \\ \Rightarrow p(x)f(x)+1=x^{12} \Rightarrow x^{12}除以f(x)的餘式為1 \Rightarrow \cases{x^{60}除以f(x)的餘式為1 \\ x^{48}除以f(x)的餘式為1 \\x^{36}除以f(x)的餘式為1 \\x^{24}除以f(x)的餘式為1 }\\ \Rightarrow f(x^{12}) =x^{60} +x^{48} +x^{36} +x^{24} + x^{12} +1 除以f(x)的餘式=1+1+1+1+1+1= \bbox[red, 2pt]6 $$
解答:$$\cases{f_x=0\\ f_y=0} \Rightarrow \cases{2x-y+2=0\\ -x+2y-3=0} \Rightarrow (x,y)= \bbox[red, 2pt]{(-{1\over 3},{4\over 3})}$$
解答:$$(x+\sqrt 2i)^2 =x^2+2\sqrt 2 xi-2 \Rightarrow (x+\sqrt 2i)^4 = (x^2+2\sqrt 2x i-2)^2 = x^4+ 4\sqrt 2x^3 i-12x^2-8\sqrt 2xi +4 \\\Rightarrow (x+\sqrt 2 i)^4-4-4i=0 \Rightarrow (z+\sqrt 2 i)^4=4+4i \Rightarrow |z+\sqrt 2i|^4=4\sqrt 2=2^{5/2} \Rightarrow |z+\sqrt 2i|= \bbox[red, 2pt]{2^{5/8}}$$
解答:$$\sin A=\sin(B+C) =\sin B\cos C+ \sin C\cos B=2\sin B\sin C \\ \Rightarrow {\sin B\cos C \over \cos B\cos C}+ {\sin C\cos B \over \cos B\cos C}= {2\sin B\sin C \over \cos B\cos C} \Rightarrow \tan B+\tan C=2\tan B\tan C\\ \Rightarrow \tan A= -\tan(B+C) ={\tan B+\tan C\over \tan B\tan C-1} ={2\tan B \tan C\over \tan B\tan C-1} \\ 因此取\tan B\tan C-1=m,則 \tan B\tan C=m+1 \tan A={2(m+1) \over m} \\\Rightarrow 欲求之\tan A\tan B\tan C= {2(m+1)^2\over m} =2m+ 4+{2\over m} \ge 4+ 2\sqrt{2m\cdot {2\over m}} = \bbox[red, 2pt]8$$
解答:$$\vec n=(6,-2,-3) \Rightarrow |\vec n|= 7 \Rightarrow \cases{S_{\triangle OAB} =14= 2|\vec n| ={1\over 2} |\overrightarrow{OA} \times \overrightarrow{OB}|\\ S_{\triangle OBC} =7= |\vec n| ={1\over 2} |\overrightarrow{OB} \times \overrightarrow{OC}| \\S_{\triangle OCA} =42= 6|\vec n| ={1\over 2} |\overrightarrow{OC} \times \overrightarrow{OA}| } \\ \Rightarrow \cases{ \overrightarrow{OA} \times \overrightarrow{OB} =(a_2b_3-a_3b_2,a_3b_1-a_1b_3, a_1b_2-a_2b_1) =4\vec n= (24,-8,-12) \\ \overrightarrow{OB} \times \overrightarrow{OC} =(b_2c_3-b_3c_2,b_3c_1-b_1c_3, b_1c_2-b_2c_1) =2\vec n= (12,-4,-6) \\ \overrightarrow{OC} \times \overrightarrow{OA} =(c_2a_3-c_3a_2, c_3a_1-c_1a_3, c_1a_2-c_2a_1) =12\vec n= (72,-24,-36) } \\ \Rightarrow \cases{ \begin{vmatrix} a_1 & a_2\\ b_1& b_2 \end{vmatrix}= -12 \\ \begin{vmatrix} b_1 & b_2\\ c_1& c_2 \end{vmatrix}= -6 \\ \begin{vmatrix}c_1 & c_2\\ a_1& a_2 \end{vmatrix} = -36 } \Rightarrow \begin{vmatrix}4& a_1 & a_2\\ 5& b_1& b_2 \\ 2& c_1& c_2\end{vmatrix}= 4\begin{vmatrix}b_1 & b_2\\ c_1& c_2 \end{vmatrix}-5 \begin{vmatrix}a_1 & a_2\\ c_1& c_2 \end{vmatrix} +2\begin{vmatrix}a_1 & a_2\\ b_1& b_2 \end{vmatrix} \\=4\cdot (-6)-5\cdot (36)+2\cdot (-12) =\bbox[red, 2pt]{-228} $$
二、計算證明題（共 65 分）
解答:$$f(x)= 2x^3+3x^2-12x-21 \Rightarrow f'(x)=6(x+2)(x-1) \\ f'(x)=0 \Rightarrow \cases{x=-2\\ x=1} \Rightarrow \cases{f(-2) =-1\lt 0\\ f(1) =-28\lt 0} \Rightarrow y=f(x)圖形為左下右上，且極值均為負值 \\ 又\cases{f'(x)\gt 0, x\gt 2 \Rightarrow 當x\ge 2,f(x)為遞增\\ f(2) =-17 \lt 0},即當x\le 2,f(x) \le 0, \bbox[red, 2pt]{故得證}$$
解答:$$取h(x)=f(x)-g(x) =(x-a)(x-2021)+1-(x-b)(x-c)= -(a+2021-b-c)x+ 2021a-bc+1 \\又[f(2022)-g(2022)]^2 +[f(2023)-g(2023)]^2=0 \Rightarrow \cases{h(2022)=0\\ h(2023)=0} \Rightarrow h(x)=0\\ \Rightarrow \cases{a+2021-b-c=0\\ 2021a-bc+1=0} \Rightarrow a=b+c-2021={bc-1\over 2021} \Rightarrow (b-2021)(c-2021)=1\\ \Rightarrow \cases{ b=c =2022 \Rightarrow a= 2023\\b=c=2020 \Rightarrow a= 2019} \Rightarrow \cases{f(2024) =(2024-2023)(2024-2021)+1= 4\\ f(2024) =(2024-2019)(2024-2021) +1 =16} \\ \Rightarrow \bbox[red, 2pt]{f(2024)=4,16}$$
解答:$$\cases{若x=y=0 \Rightarrow f(0)=f(0)+f(0)-0 \Rightarrow f(0)=0\\ 若x=y=k \Rightarrow f(0)=f(k)+f(k)-2k^2 \Rightarrow f(k)=k^2} \Rightarrow \bbox[red, 2pt]{f(x)=x^2}$$
解答:$$2^a-3^b=1的明顯解為(a,b)=(1,0),(2,1),現在要證明沒有其他非負整數數解\\ a=0 \Rightarrow 3^b=0 無解 \\ a\ge 3 \Rightarrow 2^a=8k, k\in \mathbb N,而3^b \text{ mod }8 =3,1循環 \Rightarrow (3^b+1) \text{ mod }8 =4,2循環 \Rightarrow 3^b 不是8的倍數 \\\qquad \Rightarrow 無解\\因此，恰有兩組非負整數數解, \bbox[red, 2pt]{故得證}$$
解答:$$m^3+ n^3 = (m+n)^3-3mn(m+n) \Rightarrow m^3+n^3+93mn-31^3=0 \\ \Rightarrow (m+n)^3-31^3+93mn-3mn(m+n)=0 \\ \Rightarrow (m+n-31)((m+n)^2+ 31^2+ 31(m+n)) -3mn(m+n-31) =0\\ \Rightarrow (m+n-31) (m^2-mn+n^2+31^2+31m+31n) =0 \\ \Rightarrow {1\over 2}(m+n-31)(m^2-2mn+n^2+ m^2+62m+31^2+ n^2+62n+31^2) =0 \\ \Rightarrow {1\over 2}(m+n-31)((m-n)^2 +(m+31)^2 +(n+31)^2)=0 \\ \Rightarrow \cases{ m+n=31 有H^2_{31} =32組\\ m=n=-31有1組} \Rightarrow 共有\bbox[red, 2pt]{33}組$$
解答:$$nf(n) =(n-1)f(n-1)+f(n-2) \Rightarrow n(f(n)-f(n-1)) =-(f(n-1)-f(n-2)) \\ \Rightarrow {f(n)-f(n-1) \over f(n-1)-f(n-2)} =-{1\over n}, 因此取g(n)=f(n)-f(n-1),則{g(n) \over g(n-1)} =-{1\over n} \\ \Rightarrow {g(n) \over g(n-1)}\cdot {g(n-1) \over g(n-2)} \cdot {g(n-2) \over g(n-3)} \cdots{g(3) \over g(2)} =(-1)^{n-2}\cdot {1\over n(n-1)(n-2) \cdots 3} \\ \Rightarrow {g(n)\over g(2)} =(-1)^{n-2} \cdot {2\over n!} \Rightarrow {g(n)\over -{1\over 2}+1}=(-1)^{n-2} \cdot {2\over n!} \Rightarrow g(n)= (-1)^{n-2}\cdot {1\over n!} \\ \Rightarrow g(n)+g(n-1)+ \cdots +g(2) =[f(n)-f(n-1)] +[f(n-1)-f(n-2)] + \cdots+[f(2)-f(1)] \\=f(n)-f(1)= \sum_{k=2}^n (-1)^{k-2}\cdot {1\over k!} \Rightarrow f(n)= -1+ \sum_{k=2}^n (-1)^{k-2}\cdot {1\over k!} \\ \Rightarrow \bbox[red, 2pt]{f(n) =-{1\over 2}+ {1\over 3!}-{1\over 4!} +\cdots (-1)^{n}\cdot {1\over n!}}$$
解答:$$假設第k次抽到SSR的機率為p_k,則p_1= p_2=\cdots = p_{99} =p\\ \Rightarrow \cases{p_{100}= p+(1-p)^{100} \\ p_{101} =p+p_1(1-p)^{100} \\ p_{102} =p+p_2(1-p)^{100} \\\cdots\\ p_{250}=p+p_{150}(1-p)^{100}} \Rightarrow 期望值=p_1+\cdots+ p_{250} =250p + (1-p)^{100}(1+p_1+ \cdots +p_{150}) \\= 250p+(1-p)^{100}(1+p +p +\cdots +p ) +(1-p)^{100}(p_{100} +p_{101} +\cdots +p_{150}) \\=250p+ (1-p)^{100}(1+99p) +(1-p)^{100}(51p+ (1+50p)p^{100}) \\= \bbox[red, 2pt]{250p+(1-p)^{100}(1+ 150p) +(1+50p)(1-p)^{200}}$$
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