新竹市立香山高級中學114學年度第2次教師甄選
一、 單選題(每題 4 分，共計 60 分)
解答:$$假設二根為\alpha, \beta \Rightarrow \cases{\alpha+ \beta= -(m^2-4m+3)/(m-2) \\ \alpha\beta = -(6m^2-2)/(m-2)} \\ \Rightarrow \alpha^3 +\beta^3 = (\alpha+\beta)(\alpha^2+\beta^2-\alpha\beta ) = (\alpha+ \beta)((\alpha+\beta)^2 -3\alpha\beta ) = (\alpha+ \beta)^3-3\alpha \beta( \alpha+ \beta)=0 \\\Rightarrow -(m-3)^3(m-1)^3 = 3(6m^2-2)(m-2)(m-3)(m-1) \Rightarrow m=3為其中一根,故選\bbox[red, 2pt]{(C)}$$
解答:
$$假設\cases{\overline{BC}=a \\ \overline{AC}=b\\ \overline{PB}=x}, 由於\angle APB= \angle BPC= \angle APC \Rightarrow \angle APB= \angle BPC= \angle APC=120^\circ \\ \Rightarrow \cos \angle APC =\cos 120^\circ=-{1 \over 2}={10^2+6^2-b^2 \over 2\cdot 6\cdot 10} \Rightarrow b=14 \\ \Rightarrow \cos \angle BPC =-{1\over 2} ={x^2+36-a^2\over 12x} \Rightarrow a^2= x^2+6x+36 \cdots(1) \\ \triangle ABC面積={1\over 2}ab ={1\over 2}(60+6x+10x)\sin 120^\circ \Rightarrow 7a={\sqrt 3\over 4}(16x+60) \\ \Rightarrow a ={\sqrt 3\over 7}(4x+15) \Rightarrow a^2={3\over 49}(4x+15)^2=(1)=x^2+6x+36 \\ \Rightarrow x^2-66x+1089=0 \Rightarrow (x-33)^2=0 \Rightarrow \overline{PB}= x=33,故選\bbox[red, 2pt]{(E)}\\但公佈的答案是\bbox[cyan,2pt]{(D)}$$
解答:$$a={2\over \sqrt 5-1} ={\sqrt 5+1\over 2} \Rightarrow (2a-1)^2=(\sqrt 5)^2 \Rightarrow a^2=a+1 \\ \Rightarrow a^{2025}-a^{2024}-a^{2023}-8a+4 = a^{2025}-a^{2023}(a+1)-8a+4 =a^{2025}-a^{2023}\cdot a^2-8a+4 \\=-8a+4 =4(-2a+1)=4(-\sqrt 5-1+1) =-4\sqrt 5,故選\bbox[red, 2pt]{(B)}$$
解答:
$$假設\angle B=\angle C=2\theta \Rightarrow \cases{\angle ABD=\angle DBC=\theta\\ \angle A=\pi-4\theta \\ \angle BDC=\pi-3\theta} \Rightarrow \cases{ \triangle BDC:\displaystyle {\overline{BC} \over \overline{BD}}={\sin(\pi-3\theta) \over \sin 2\theta} ={\sin 3\theta\over \sin 2\theta} \\ \triangle ABD: \displaystyle {\overline{AD} \over \overline{BD}} ={\sin \theta\over \sin (\pi-4\theta)} ={\sin \theta\over \sin 4\theta}} \\ 又\overline{BC} =\overline{BD}+ \overline{AD} \Rightarrow {\overline{BC} \over \overline{BD}} =1+{\overline{AD} \over \overline{BD}} \Rightarrow {\sin 3\theta\over \sin 2\theta} =1+{\sin \theta\over \sin 4\theta} =1+{\sin \theta\over 2\sin 2\theta \cos 2\theta} \\ \Rightarrow 2\cos 2\theta \sin 3\theta=2\sin 2\theta \cos 2\theta+\sin \theta \Rightarrow \sin 5\theta+\sin \theta=\sin 4\theta+\sin \theta \\ \Rightarrow \sin 5\theta=\sin 4\theta \Rightarrow \theta=20^\circ \Rightarrow \angle A=180^\circ-4\cdot 20^\circ =100^\circ ={5\pi\over 9},故選\bbox[red, 2pt]{(C)}$$
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解題僅供參考，其他教甄試題及詳解
解答:$$6^2=36 \equiv -13 \pmod{49} \Rightarrow 6^4 \equiv 22 \pmod{49} \Rightarrow 6^8 \equiv -6 \pmod{49} \\ \Rightarrow 6^{16} \equiv -13 \pmod{49} \Rightarrow 6^{32} \equiv 22 \pmod{49} \Rightarrow 6^{64} \equiv -6 \pmod{49} \\ \Rightarrow 6^{83} =6^{64} \times6^{16} \times 6^{2} \times6 \equiv (-6)\times (-13) \times (-13)\times 6 \equiv 41 \pmod{49} \\ 同理可得8^{83} \equiv 43 \pmod{49}\\ \Rightarrow \cases{6^{83} \equiv 41 \pmod{49} \\8^{83} \equiv 43\pmod{49}} \Rightarrow 6^{83}+8^{83} \equiv 41 +43=84 \equiv 35 \pmod{49},故選\bbox[red, 2pt]{(B)}$$
解答:$$S_n=S_m \Rightarrow a_1=r=1 \Rightarrow S_{m+n}=\log a_1^{m+n}r^{(m+n-1)(m+n)/2} =\log 1=0,故選\bbox[red, 2pt]{(A)}$$
解答:$$\tan(A+B) =\tan(\pi-C) =-\tan C=1 \Rightarrow {\tan A+\tan B\over 1-\tan A\tan B}=1 \\\Rightarrow \tan A+\tan B=1-\tan A\tan B \Rightarrow 1+\tan A+ \tan B+\tan A \tan B=2 \\ \Rightarrow(1+\tan A)(1+\tan B)=2，故選\bbox[red, 2pt]{(B)}$$
解答:$$n最高位數一定是8,個位數字一定是0， 因此只需考慮8X\cdots X0, X\in \{8,0\} \\ \Rightarrow 8880=15\times 592 \Rightarrow m+{n\over 15} =24+ 592= 616, 故選\bbox[red, 2pt]{(A)}$$
解答:$$\sqrt[4]x ={12\over 7-\sqrt[4]x} \Rightarrow \sqrt x-7\sqrt[4]x+12=0 \Rightarrow (\sqrt[4]x-4)(\sqrt[4]x -3)=0 \\ \Rightarrow \cases{\sqrt[4]x=4 \Rightarrow x=4^4=256\\ \sqrt[4] x=3 \Rightarrow x=3^4=81} \Rightarrow 256+81=337,故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{A(2,2,0) \\B(-1,0,2) \\C(0,4,3)} \Rightarrow \cases{\vec u=\overrightarrow{AB} =(-3,-2,2) \\ \vec v= \overrightarrow{AC} =(-2,2,3)} \Rightarrow \vec u\times \vec v=(-10,5,-10) \Rightarrow |\vec u\times \vec v|=15 \\ \Rightarrow \triangle ABC面積={1\over 2}\cdot 15, 故選\bbox[red,2pt]{(B)}$$
解答:
$$假設\cases{\overline{AE}=a\\ \overline{AF}=b} \Rightarrow \cases{\overline{EC}=15-a\\ \overline{BF}=15-b} \\ \cases{\overline{AF} \parallel \overline{DE} \\ \overline{AE} \parallel \overline{DF}} \Rightarrow \triangle FBD \sim \triangle ECD (AAA) \Rightarrow {\overline{BF} \over \overline{EC}} ={ \overline{DF} \over \overline{DE}} \Rightarrow {15-b\over 15-a} ={a\over b} \\ \Rightarrow 15b-b^2=15a-a^2 \Rightarrow a^2-b^2-15a+15b=0 \Rightarrow (a-b)(a+b)-15(a-b)0 \\ \Rightarrow (a-b)(a+b-15)=0 \Rightarrow \cases{a=b\\ a+b=15} \Rightarrow a+b=15 \Rightarrow AEDF周長=2(a+b)=30\\, 故選\bbox[red, 2pt]{(D)}$$
解答:$$\cases{\log_a x=24\\ \log_b x=40\\ \log_{abc}x =12} \Rightarrow \cases{\log x/\log a=24\\ \log x/\log b=40\\ \log x/\log abc=12} \Rightarrow \log x=24\log a=40\log b=12\log abc \\ 24\log a=40\log b \Rightarrow \log b={3\over 5}\log a \Rightarrow 12\log abc =12(\log a+\log b+\log c) =12(\log a+{3\over 5}\log a+\log c) \\=12({8\over 5}\log a+\log c) =24\log a \Rightarrow \log c={2\over 5}\log a \Rightarrow \log_c x={\log x\over \log c} = {\log x\over (2/5)\log a} ={5\over 2}\log_a x \\={5\over 2}\cdot 24=60, 故選\bbox[red, 2pt]{(E)}$$
解答:$$\lim_{h\to 0} {\int_2^{2+h} \sqrt{1+t^2}\,dt \over h} =\lim_{h\to 0} { \sqrt{1+(2+h)^2} \over 1} = \sqrt 5,故選\bbox[red, 2pt]{(D)}$$
二、 多選題(每題 8 分，共計 40 分;每題有 5 個選項，其中至少有一個是正確的選項。各題的選項獨立判斷，所有選項均答對者，得該題全部的分數；答錯 k 個選項者，得該題(5-2k)/5的分數，但分數低於零分或所有選項均未作答，該題以零分計算。)
解答:$$f(x)g(x)=(x+1)(x^3-4x^2+x+6) =(x+1)^2(x-2)(x-3) \\ \Rightarrow \cases{f(x)=x^2+ax+b =(x+1)(x-r) \\ g(x) =x^2+bx+c=(x+1)(x-s)} \Rightarrow r,s \in \{2,3\} \\ \Rightarrow \cases{r=2,s=3 \Rightarrow \cases{f(x)=x^2-x-2=x^2+ax+b\\ g(x)=x^2-2x-3= x^2+bx+c} \Rightarrow \cases{a=-1\\ b=-2\\ c=-3} \\ r=3,s=2 \Rightarrow \cases{f(x)=x^2-2x-3=a^2+ax+b \Rightarrow b=-2\\ g(x)= x^2-x-2 =x^2+bx+c \Rightarrow b=-1} 不合} \\(A) \times: a=-1\ne -2\\ (B)\bigcirc: a=-1\\ (C)\times: a+b=-3\ne -4 \\(D)\bigcirc: a+c=-4 \\(E) \bigcirc: a+b+c=-6 \\ 故選\bbox[red, 2pt]{(BDE)}$$
解答:$$假設\cases{\overline{AC}=a-1\\ \overline{BC} =a \\ \overline{AB}=a+1} \Rightarrow \cases{\angle B=\theta\\ \angle C=2\theta\\ \angle A=\pi-3\theta} \Rightarrow {a-1\over \sin \theta} ={a+1\over \sin 2\theta} ={a+1\over 2\sin \theta\cos \theta} \Rightarrow \cos \theta={a+1\over 2(a-1)} \\ 餘弦定理: \cos \theta={a^2+(a+1)^2-(a-1)^2\over 2a(a+1)} ={a+4\over 2(a+1)} ={a+1\over 2(a-1)} \\ \Rightarrow (a+1)^2=(a+4)(a-1) \Rightarrow a=5 \Rightarrow 三邊長分別為4,5,6 \Rightarrow \cases{最小邊長4\\ 周長=15}, 故選\bbox[red, 2pt]{(BD)}$$
解答:$$5(a^2+ab+b^2) =7(a+2b) \Rightarrow 5a^2+(5b-7)a+5b^2-14b=0 \\ (A)\times: a+b=-1 \Rightarrow b=-1-a \Rightarrow 5a^2+12a+19=0 \Rightarrow 無整數解\\ (B)\bigcirc: 顯然a=b=0符合條件 \Rightarrow a+b=0\\ (C) \bigcirc: a+b=2 \Rightarrow b=2-a \Rightarrow 5a^2-3a-8=0 \Rightarrow (5a-8)(a+1)=0 \Rightarrow a=-1 \Rightarrow b=3 \\(D) \bigcirc: 同理, a=1,b=2 \Rightarrow a+b=3符合條件\\ (E)\times:a+b=4 \Rightarrow a=4-b \Rightarrow 5a^2-13a+24=0 \Rightarrow 無實數解\\ 故選\bbox[red, 2pt]{(BCD)}$$
解答:$$f(x)=x^{2025}(x^2+ax+b)=(x-2)^2 p(x)+2^{2025}(x-2) \\ \Rightarrow f'(x)=2025x^{2024}(x^2+ax+b)+x^{2025}(2x+a) =2(x-2)p(x)+(x-2)^2p'(x)+2^{2025} \\ \Rightarrow \cases{f(2)=2^{2025}(4+2a+b)=0\\ f'(2)=2025\cdot 2^{2024}(4+2a+b)+2^{2025}(4+a)=2^{2025}} \Rightarrow \cases{2a+b +4=0\\ 2025(4+2a+b)+2(4+a)=2} \\ \Rightarrow \cases{a=-3\\ b=2} \Rightarrow \cases{a+b=-1\\ a^2+b^2=13},故選\bbox[red, 2pt]{(AE)}$$
解答:$$(A)\times: k\lt n \Rightarrow V\not \subseteq span(\{v_1,v_2, \dots,v_k\}) \\(B)\times: A=[0] \Rightarrow \{Av_1,Av_2,\dots, Av_n\ =\{\vec 0\}非基底 \\(D)\times: \cases{V=\{(a,0) \mid a\in R\} \\W=\{(0,a)\mid a\in R\}} \Rightarrow dim(V)=dim(W)=1,但V\ne W\\ 故選\bbox[red, 2pt]{(CE)}$$
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解題僅供參考，其他教甄試題及詳解
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