114豐原高中教甄-數學詳解

臺中市立豐原高級中等學校 114 學年度第 1 次教師甄選

第一部分：填充題

解答:$$f(x)=(x+1)^4p(x)+x^3-1 \Rightarrow f'(x)=4(x+1)^3p(x)+ (x+1)^4p'(x)+3x^2 \\ \Rightarrow g(x)=xf(x) \Rightarrow g'(x)=f(x)+xf'(x) \Rightarrow g'(-1) =f(-1)-f'(-1)=-2-3=-5 \\ \Rightarrow 一次近似估計: y=g'(-1)(x+1)+g(-1) =-5(x+1)+2  =\bbox[red, 2pt]{-5x-3}$$

解答:

$$假設\cases{\overline{AC} =\overline{AD}=a\\ \overline{AB}=b} \Rightarrow a^2+b^2=10^2=100\\ 在\triangle ABC中,依中線定理: \overline{AB}^2+ \overline{AC}^2=2(\overline{AM}^2+\overline{MB}^2) \Rightarrow b^2+a^2= 100=2(\overline{AM}^2+ 4^2) \\ \Rightarrow \overline{AM}^2 =34 \Rightarrow \overline{AM}= \bbox[red, 2pt]{\sqrt{34}}$$

解答:

$$\overline{AB}=\overline{AC} \Rightarrow \angle B =\angle C=\alpha \Rightarrow \cos \alpha={13^2+10^2-13^2 \over 2\cdot 13\cdot 10}={5\over 13} \Rightarrow \sin \alpha={12\over 13}\\ 又\overline{PB}^2+ \overline{PC}^2= \overline{BC}^2 \Rightarrow \angle BPC =90^\circ  \\取\angle PCB=\theta  \Rightarrow \cases{\sin \theta=4/5\\ \cos \theta=3/5}\Rightarrow \angle PBC =90^\circ-\theta \Rightarrow \cases{\angle ACP=\alpha-\theta\\ \angle ABP=\alpha-(90^\circ-\theta) =\alpha +\theta-90^\circ} \\ \Rightarrow {\triangle PAB\over \triangle PAC} ={8\cdot 13\sin (\alpha+\theta-90^\circ) \over 6\cdot 13 \sin(\alpha-\theta)} ={-4 \cos(\alpha+ \theta) \over 3\sin(\alpha-\theta)} =-{4\over 3}\cdot {\cos \alpha\cos \theta-\sin \alpha\sin \theta\over \sin \alpha\cos \theta-\sin \theta\cos \alpha} \\=-{4\over 3}\cdot {(5/13)\cdot (3/5)-(12/13)\cdot (4/5) \over (12/13)\cdot (3/5)-(4/5)\cdot (5/13)} =-{4\over 3}\cdot (-{33\over 16}) =\bbox[red, 2pt]{11\over 4}$$

解答:$$已知\cases{A(1)\\ B(z) \\C(z^2) \\D(z^3)} 及 \cases{\overline{PC} =2\overline{PA} \\ \overline{PD}= 5\overline{PB}} \Rightarrow \cases{P=(2A+C)/3=(2+z^2)/3\\ P=(5B+D)/6 =(5z+z^3)/6} \Rightarrow {2+z^2\over 3}={5z+z^3\over 6} \\ \Rightarrow z^3-2z^2+5z-4=0 \Rightarrow (z-1)(z^2-z+4)=0 \Rightarrow z={1\pm \sqrt{15}i\over 2} \Rightarrow (a,b)= \bbox[red, 2pt]{\left({1\over 2},{\sqrt{15}\over 2} \right)}$$

解答:

$$\cases{\angle ABO=90^\circ\\ \overline{OB}=圓半徑=3\\ \overline{OA}=5} \Rightarrow \overline{AB}=\overline{AC} =4\\令圓心O=(0,0)及\angle BAO= \theta \Rightarrow \cases{\cos \theta=4/5\\ \sin \theta=3/5} \Rightarrow \cos 2\theta={7\over 25} \\ {\overline{AP} \over \overline{BP}}有最大值\Rightarrow P=\overleftrightarrow{AC} \cap \overleftrightarrow{OB} \Rightarrow k={\overline{AP} \over \overline{AC}} ={4/\cos 2\theta\over 4}= \bbox[red, 2pt]{25\over 7}$$

解答:$$取g(x)=f(x)+13 \Rightarrow g(x)=0的三根為0,a,b \Rightarrow g(x)=x(x-a)(x-b) =x^3-(a+b)x^2+abx \\ \Rightarrow g'(x)=3x^2-2(a+b)x+ab \Rightarrow g''(x)=6x-2(a+b) \\ f(x)的對稱中心為(3,-15) \Rightarrow \cases{f(3)=-15\\ f''(3)=0} \Rightarrow \cases{g(3)=f(3)+13=-2=27-9(a+b)+3ab \\g''(3)=0=18-2(a+b)}\\  \Rightarrow \cases{a+b=9\\ ab= \bbox[red, 2pt]{52/3}}$$

解答:

$$A=\begin{bmatrix}4/5& -3/5\\ 3/5& 4/5 \end{bmatrix} 為一旋轉矩陣\Rightarrow \cases{\cos \theta=4/5\\ \sin \theta=3/5} \\P_1, P_2, P_3在半徑為r 的圓周上,令P_1=re^{i\alpha} \Rightarrow P_2=A(P_1)=re^{i(\theta+\alpha)} \Rightarrow P_3=A^2(P_1)=re^{i(2 \theta+ \alpha)} \\ \Rightarrow \triangle P_1P_2P_3 =\triangle OP_1P_2+ \triangle OP_2P_3-\triangle OP_1P_3 ={1\over 2}r^2 \sin \theta+{1\over 2}r^2 \sin \theta-{1\over 2}r^2\sin 2\theta \\=r^2({3\over 5}-{12\over 25}) ={3\over 25}r^2=9 \Rightarrow r=5\sqrt 3  \Rightarrow \overline{QP_1}的最大值= \overline{OQ}+r= \bbox[red, 2pt]{10+5\sqrt 3}$$

解答:$$ix^2-(i+1)x+\lambda=0 \Rightarrow i(x^2-x)+\lambda-x=0 \Rightarrow \cases{x^2-x=0\\ \lambda-x=0 } \\ \Rightarrow \lambda=x=0或1 \Rightarrow 只要\bbox[red, 2pt]{\lambda\ne0且 \lambda\ne 1},x就是虛根$$

解答:

$$假設\cases{\angle EAI=\angle DAI= \theta\\ \angle AEI=\alpha\\ \angle ADI=\beta\\  \overline{AI}=a\\ \overline{IE} =\overline{ID} =b} \Rightarrow \cases{\triangle AEI:\displaystyle {b\over \sin \theta} ={a\over \sin \alpha} \\ \triangle ADI: \displaystyle {b\over \sin \theta}={a\over \sin \beta}} \Rightarrow {a\over \sin \alpha}={a\over \sin \beta} \\ \Rightarrow \sin \alpha=\sin \beta \Rightarrow  \alpha+\beta=180^\circ (\because \alpha\ne \beta) \Rightarrow (\angle B+{1\over 2}\angle C) +({1\over 2} \angle B+\angle C) =180^\circ \\ \Rightarrow {3\over 2}(\angle B+\angle C)=180^\circ \Rightarrow \angle B+\angle C=120^\circ \Rightarrow \angle A=60^\circ \Rightarrow {\overline{BC} \over \sin \angle A} ={6\over \sqrt 3/2}=2R \Rightarrow R= \bbox[red, 2pt]{2\sqrt 3}$$

解答:$$考慮\cases{ a_n=2(a_{n+1})^2-1 \\ \cos 2\theta=2\cos^2 \theta-1} \Rightarrow a_1={1\over 2} =\cos {\pi\over 3} \Rightarrow a_2= \cos {\pi\over 3\cdot 2} \Rightarrow a_n= \cos {\pi\over 3\cdot 2^{n-1}} \\=1-{1\over 2!}\left( {\pi\over 3\cdot 2^{n-1}}\right)^2+ \cdots \Rightarrow \lim_{n\to \infty} 4^n(1-a_n) =\lim_{n\to \infty} 4^n({1\over 2}\left( {\pi\over 3\cdot 2^{n-1}}\right)^2-\cdots) \\= \lim_{n\to \infty}{2^{2n-1}} \cdot {\pi^2\over9\cdot 2^{2n-2}} =\bbox[red, 2pt]{{2\over 9}\pi^2}$$

解答:$$\lim_{n\to \infty} {1\over n^2}\left( \sqrt{16n^2-1^2} + \sqrt{16n^2-3^2}+ \sqrt{16n^2-5^2}+ \cdots +\sqrt{16n^2-(2n-1)^2}\right) \\=\lim_{n\to \infty} \sum_{k=1}^n{1\over n^2}\left( \sqrt{16n^2-(2k-1)^2}\right) =\lim_{n\to \infty} \sum_{k=1}^{2n}{1\over n^2}\left( \sqrt{16n^2-k^2}\right) -\lim_{n\to \infty} \sum_{k=1}^{n}{1\over n^2}\left( \sqrt{16n^2-(2k)^2}\right) \\=\lim_{n\to \infty} \sum_{k=1}^{2n}{1\over n}\left( \sqrt{16-(k/n)^2}\right) -\lim_{n\to \infty} \sum_{k=1}^{n}{1\over n}\left( \sqrt{16-4(k/n)^2}\right) =\int_0^2\sqrt{16-x^2}\,dx-\int_0^1 \sqrt{16-4x^2}\,dx \\=2\sqrt 3+{4\over 3}\pi-(\sqrt 3+{2\over 3}\pi) =\bbox[red, 2pt]{\sqrt 3+{2\over 3}\pi}$$

解答:$$轉換矩陣A=\begin{bmatrix}0 & 1/3 & 1/3 & 1/3 \\ 1/3 & 0 & 1/3 & 1/3 \\1/3 & 1/3 & 0 & 1/3 \\1/3 & 1/3 &1/3   & 0\end{bmatrix} \\= \begin{bmatrix}1 & -1 & -1 & -1 \\ 1 & 1 & 0 & 0 \\1 & 0 & 1 & 0 \\1 & 0 &0   & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & -1/3 & 0 & 0 \\0 & 0 & -1/3 & 0 \\0 & 0 &0   & -1/3\end{bmatrix} \begin{bmatrix}1/4 & 1/4 & 1/4 & 1/4 \\ -1/4 & 3/4 & -1/4 & -1/4 \\-1/4 & -1/4 & 3/4 & -1/4 \\-1/4 & -1/4 &-1/4   & 3/4\end{bmatrix} \\ \Rightarrow A^n=\begin{bmatrix}1 & -1 & -1 & -1 \\ 1 & 1 & 0 & 0 \\1 & 0 & 1 & 0 \\1 & 0 &0   & 1 \end{bmatrix} \begin{bmatrix}1 & 0 & 0 & 0 \\ 0 & (-1/3)^n & 0 & 0 \\0 & 0 & (-1/3)^n & 0 \\0 & 0 &0   & (-1/3)^n\end{bmatrix} \begin{bmatrix} 1/4 & 1/4 & 1/4 & 1/4 \\ -1/4 & 3/4 & -1/4 & -1/4 \\-1/4 & -1/4 & 3/4 & -1/4 \\-1/4 & -1/4 &-1/4   & 3/4\end{bmatrix}  \\= \left[\begin{matrix}\frac{3\cdot \left(-1\right)^n+3^n}{4\cdot 3^n} & \frac{-\left(-1\right)^n+3^n}{4\cdot 3^n} & \frac{-\left(-1 \right)^n +3^n}{4\cdot 3^n} & \frac{-\left(-1\right)^n+3^n}{4\cdot 3^n} \\ \frac{-\left(-1\right)^n+3^n}{4\cdot 3^n} & \frac{3\cdot \left(-1\right)^n+3^n}{4\cdot 3^n} & \frac{-\left(-1\right)^n+3^n}{4\cdot 3^n} & \frac{-\left(-1\right)^n+3^n}{4 \cdot 3^n} \\\frac{-\left(-1\right)^n+3^n}{4\cdot 3^n} & \frac{-\left(-1 \right)^n+3^n}{4\cdot 3^n} & \frac{3\cdot \left(-1\right)^n+3^n}{4*3^n} & \frac{-\left(-1\right)^n+3^n}{4\cdot 3^n} \\\frac{-\left(-1\right)^n+3^n}{4\cdot 3^n} & \frac{-\left(-1\right)^n+3^n}{4\cdot 3^n} & \frac{-\left(-1\right)^n+3^n}{4*3^n} & \frac{3\cdot \left(-1\right)^n+3^n}{4\cdot 3^n} \end{matrix}\right] \Rightarrow A^n \begin{bmatrix}1\\0\\0\\0 \end{bmatrix} =\begin{bmatrix}\frac{3\cdot \left(-1 \right)^n+3^n}{4\cdot 3^n} \\ \frac{-\left(-1\right)^n+3^n}{4\cdot 3^n}\\ \frac{-\left(-1\right)^n+3^n}{4\cdot 3^n}\\ \frac{-\left(-1\right)^n+3^n}{4\cdot 3^n} \end{bmatrix} \\ \Rightarrow P_n= \frac{3\cdot \left(-1\right)^n+3^n}{4\cdot 3^n} \Rightarrow 4P_n={(-1)^n\over 3^{n-1}}+1 \Rightarrow 4P_{114}-1={1\over 3^{113}} \\ \Rightarrow \log_9(4P_{114}-1) =\bbox[red, 2pt]{-{113\over 2}}$$

解答:

$$假設\cases{\overline{AB}= \overline{CD}=a\\ \overline{AD}= \overline{BC}=2b \\A(0,0,0)} \Rightarrow \cases{A(0,0,0) \\ B(a,0,0)\\ D(0,2b,0)\\ C(a,2b,0)} \Rightarrow \cases{E(0,b,\sqrt 3 b)\\F(a/2,2b,0) \\ G(0,b,0) } \Rightarrow \cases{\overrightarrow{BE}=(-a,b,\sqrt 3 b)\\ \overrightarrow{BG} =(-a,b,0) \\ \overrightarrow{BF} =(-a/2,2b,0)} \\ \Rightarrow \cos \angle EGB ={\overrightarrow{BE} \cdot \overrightarrow{BG}\over \overline{BE}\cdot \overline{BG}} \Rightarrow {\sqrt 3\over 2}={a^2+b^2\over \sqrt{a^2+4b^2} \cdot \sqrt{a^2+b^2}} =\sqrt{a^2+b^2\over a^2+4b^2} \\ \Rightarrow  a^2=8b^2 \Rightarrow a=2\sqrt 2 b \\又\vec n=\overrightarrow{BE} \times \overrightarrow{BF} =(-2\sqrt 3 b^2,-{\sqrt 3\over 2}ab,-{3\over 2}ab)\\ \Rightarrow 平面E=\triangle EFB: -2\sqrt 3b(x-a)-{\sqrt 3\over 2}ay-{3\over 2}az=0 \Rightarrow 4bx+ay+\sqrt 3az=4ab \\ \Rightarrow d(A,E)=2 \Rightarrow  {4ab\over \sqrt{4a^2+16b^2}}={8\sqrt 2b^2 \over \sqrt{48b^2}} ={4b\over \sqrt 6}=2 \Rightarrow b={\sqrt 6\over 2} \Rightarrow \overline{AD}=2b= \bbox[red, 2pt] {\sqrt 6}$$

解答:$$f(x)=(1+x)^{90} =\sum_{k=0}^{90} C^{90}_kx^k \Rightarrow g(x) =xf'(x) =90x(1+x)^{89}=\sum_{k=0}^{90} kC^{90}_kx^k \\ \Rightarrow h(x) = xg'(x) = 90x(1+x)^{89} +8010x^2(1+x)^{88} = \sum_{k=0}^{90} k^2C^{90}_kx^k  \\ \Rightarrow h({1\over 2}) =45\cdot ({3\over 2})^{89} +{4005\over 2} \cdot ({3\over 2})^{88} = \sum_{k=1}^{90} k^2C^{90}_k2^{-k} \\ \Rightarrow 2^{90}h({1\over 2}) =90\cdot 3^{89}+8010\cdot 3^{88} =\sum_{k=1}^{90} k^2C^{90}_k2^{90-k} \Rightarrow \sum_{k=1}^{90} k^2C^{90}_k2^{90-k}=920\cdot 3^{90} \\ \Rightarrow \log(900\cdot 3^{90}) \lt \log(920\cdot 3^{90}) \lt\log(1000\cdot 3^{90}) \Rightarrow 45.8932\lt\log(920\cdot 3^{90}) \lt 45.939 \\ \Rightarrow 920\cdot 3^{90}為\bbox[red, 2pt]{46}位數$$

解答:

$$\cases{只經過A點有24種\\只經過B點有14種\\ 只經過C點有24種\\ 只經過D點有14種} \Rightarrow 經過1點的機率P_1={24\cdot 2+14\cdot 2\over 210} ={76\over 210} \\ \cases{經過A\to B有21種\\ 經過A\to C有36種\\ 經過A\to D有6種\\ 經過B\to C有6種\\ 經過D\to C有21種 }\Rightarrow 經過2點的機率P_2={21+36+6+6+21\over 210} ={90\over 210} \\ \cases{經過A\to B\to C有9種\\ 經過A\to D\to C有9種} \Rightarrow 經過3點的機率P_3={18\over 210} \\ \Rightarrow 經過黑點個數的期望值:{76+2\cdot 90+3\cdot 18\over 210} =\bbox[red, 2pt]{31\over 21}$$

解答:$$\sin^2 \theta_1 + \sin^2 \theta_2+ \cdots + \sin^2 \theta_{2025}=1= \sin^2 \theta_1+ \cos^2 \theta_1 \\ \Rightarrow \sin^2 \theta_2+ \cdots + \sin^2 \theta_{2025}= \cos^2 \theta_1 \\ 柯西不等式:\cos^2 \theta_1 = (\sin^2 \theta_2+ \cdots + \sin^2 \theta_{2025})(1^2 +1^2+ \cdots+ 1^2) \ge (\sin \theta_2+ \sin \theta_2+ \cdots+ \sin \theta_{2025})^2 \\ \Rightarrow \cos \theta_1 \ge {\sin \theta_2+\cdots+ \sin \theta_{2025} \over \sqrt{2024}} \Rightarrow \cos \theta_2\ge {\sin \theta_1+ \sin \theta_{3}+\cdots+ \sin \theta_{2025} \over \sqrt{2024}} \\ \Rightarrow \cdots \Rightarrow \cos \theta_{2025}\ge{\sin \theta_1+ \sin \theta_{2}+\cdots+ \sin \theta_{2024} \over \sqrt{2024}} \\ \Rightarrow \cos \theta_1+ \cos \theta_2+ \cdots+ \cos \theta_{2025} \ge {2024\over \sqrt{2024}}(\sin \theta_1+ \sin \theta_2+ \cdots+ \sin  \theta_{2025}) \\ \Rightarrow {\sin \theta_1+ \sin \theta_{2}+\cdots+ \sin \theta_{2025} \over \cos \theta_1+ \cos \theta_2+ \cdots + \cos \theta_{2025}}\le {\sqrt{2024}\over 2024} = \bbox[red, 2pt]{\sqrt{506} \over 1012}$$

第一部分：計算證明題

解答:$$\textbf{(1)}由切點可知拋物線圖形對稱y軸且凹向下 \Rightarrow \Gamma_1: y=ax^2+b, a\lt 0 \\將切點(-\sqrt{10},\sqrt{10})代入\Gamma_1 \Rightarrow 10a+b=\sqrt{10} \\將\Gamma_1代入 圓C \Rightarrow x^2+(ax^2+b)^2=20 \Rightarrow a^2x^4+(2ab+1)x^2+b^2-20=0 \\ \Rightarrow 判別式: (2ab+1)^2-4a^2(b^2-20)=0 \Rightarrow 80a^2+4ab+1=0 \Rightarrow 80a^2+4a(\sqrt{10}-10a)+1=0\\ \Rightarrow 40a^2+4\sqrt{10}a+1=0 \Rightarrow a={-4\sqrt{10}\over 80}=-{\sqrt{10}\over 20} \Rightarrow b=\sqrt{10}+{\sqrt{10}\over 2} ={3\over 2}\sqrt{10} \\ \Rightarrow \Gamma_1: \bbox[red, 2pt]{y=-{\sqrt{10}\over 20}x^2+{3\over 2}\sqrt{10}} \\\textbf{(2)}旋轉矩陣T=\begin{bmatrix} \cos \theta& -\sin \theta\\ \sin \theta& \cos \theta\end{bmatrix}  \Rightarrow T\begin{bmatrix} \sqrt{10} \\ \sqrt{10} \end{bmatrix} =\begin{bmatrix} 2 \\ 4 \end{bmatrix}  \Rightarrow \cases{\cos \theta=3/\sqrt{10} \\ \sin \theta =1/\sqrt{10}} \\ \Gamma_1頂點P(0,3\sqrt{10}/2) \Rightarrow \Gamma_2頂點=T(P)= \begin{bmatrix} -3/2 \\ 9/2\end{bmatrix} \Rightarrow \Gamma_2頂點= \bbox[red, 2pt]{(-{3\over 2},{9\over 2})}$$

解答:

$$假設\cases{\overline{BC}=a\\ \overline{AC}=b \\ \overline{AB}=c \\ \angle AHF= \alpha =\angle CHD\\ \angle AHE= \beta= \angle BHD\\ \angle CHE=\gamma= \angle BHF} \Rightarrow \triangle AFH \sim \triangle ADB (AA) \Rightarrow \angle B=\alpha\\\Rightarrow {\overline{AH} \over \overline{DH}} ={\overline{AF}/ \sin \alpha \over \overline{CD}/ \tan \alpha} ={b\cos A/\sin \alpha\over b\cos C/\tan \alpha}= {\cos A\over \cos C\cos \alpha} ={\cos A\over \cos C\cos B} ={-\cos(B+C) \over \cos B\cos C} \\ ={\sin B\sin C-\cos B\cos C\over \cos B\cos C} =\tan B\tan C-1 \Rightarrow {\overline{AH} \over \overline{DH}} ={\cos A\over \cos C\cos B}  =\tan B\tan C-1\quad \bbox[red, 2pt]{QED}$$

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解題僅供參考，其他教甄試題及詳解