105年中壢高中教甄-數學詳解

國立中央大學附屬中壢高級中學105學年度第 1次教師甄選

一、填充題

解答： $$\sqrt{a_n} =2\sqrt{a_{n+1}}+\sqrt{a_na_{n+1}} \Rightarrow {\sqrt{a_n} \over \sqrt{a_na_{n+1}}} =2{\sqrt{a_{n+1}}\over \sqrt{a_na_{n+1}}} +{\sqrt{a_na_{n+1}} \over \sqrt{a_na_{n+1}}} \Rightarrow {1\over \sqrt {a_{n+1}}}=2{1\over \sqrt{a_n}}+1 \\ \Rightarrow b_{n+1}=2b_n+1，其中\cases{b_n={1\over \sqrt{a_n}}\\ b_1={1\over \sqrt{a_1}}=1} \\ \Rightarrow b_n=2b_{n-1}+1 = 2(2b_{n-2}+1)+1 =2^2b_{n-2}+2+1 =\cdots = 2^{n-1}b_1+2^{n-2}+2^{n-3}+\cdots +1\\ =1+2+2^2 +\cdots +2^{n-1} =2^n-1 \Rightarrow {1\over \sqrt{a_n}} =2^n-1 \Rightarrow a_n={1\over (2^n-1)^2} =\bbox[red, 2pt]{1\over 2^{2n}-2^{n+1}+1}$$

解答： $$令\alpha=x-[x]，則0\le \alpha \lt 1；將[x]=x-\alpha 代回原式\Rightarrow 4x^2-20(x-\alpha)+23=0 \\ \Rightarrow 4x^2-20x+23=-20\alpha \Rightarrow 4(x-{5\over 2})^2-2=-20\alpha \Rightarrow -20\lt 4(x-{5\over 2})^2-2 \le 0\\ \Rightarrow \cases{4(x-{5\over 2})^2\le 2\\ -18\lt 4(x-{5\over 2})^2} \Rightarrow 0\le 4(x-{5\over 2})^2\le 2 \Rightarrow {5\over 2}-{1\over \sqrt 2}\le x \le {5\over 2}+{1\over \sqrt 2} \Rightarrow [x]=1,2,3\\將[x]值代回原式\Rightarrow \cases{[x]=1   \\ [x]=2 \\ [x]=3} \Rightarrow \cases{4x^2+3=0(無解)\\ 4x^2-17=0\\ 4x^2-37=0} \Rightarrow x= \bbox[red, 2pt]{{\sqrt{17}\over 2},{\sqrt{37}\over 2}}$$

解答：

$$\sqrt{x^2+(mx-3m+2)^2} +\sqrt{x^2+(mx-3m+10)^2} =10\\\Rightarrow \sqrt{x^2+(mx-3m-(-2))^2} +\sqrt{x^2+(mx-3m-(-10))^2} =10\\上述方程式相當於求兩圖形\cases{直線L:y=m(x-3)\\ 橢圓\Gamma:\sqrt{x^2+(y+2)^2} +\sqrt{x^2+(y+10)^2}=10}的交點；\\而\Gamma:\cases{焦點F_1(0,-2)\\ 焦點F_2(0,-10)\\ 2a=10} \Rightarrow \cases{中心點O(0,-6)\\a=5\\ b=3\\ c=4} \Rightarrow \Gamma:{x^2\over 9}+{(y+6)^2\over 25}=1\\依題意有兩相異實根，即交點數為2；\\橢圓的切線方程式:y-k=m(x-h)\pm \sqrt{b^2m^2+a^2}，其中(h,k)為中心點\\ \Rightarrow y+6=mx\pm \sqrt{9m^2+25} 經過(3,0) \Rightarrow 6=3m\pm \sqrt{9m^2+25 } \Rightarrow 9m^2+25 =(3m-6)^2\\ \Rightarrow m={11\over 36}，另外一根m=\infty，即x=-3 \Rightarrow \bbox[red, 2pt]{m\gt {11\over 36}}，兩圖形有兩交點；$$

解答：

$$\cases{A,B,C,D在同一圓上\\ \overline{AB}=\overline{BC}=\overline{AD}} \Rightarrow ABCD為一等腰梯形\Rightarrow 令\cases{A(-15/2,0) \\ B(15/2,0)\\ 圓心O(0,a)}，見上圖；\\作\overline{BP} \bot \overline{CD}及\overline{AQ}\bot \overline{CD} \Rightarrow \overline{CP}= (\overline{CD}-\overline{AB})\div 2=5/2；\\在直角\triangle BPC中，\overline{BP} = \sqrt{15^2-(5/2)^2} ={5\over 2}\sqrt{35} \Rightarrow C(10,{5\over 2}\sqrt{35});\\ 圓半徑r=\overline{OB} =\overline{OC} \Rightarrow \left({15\over 2}\right)^2+a^2 = 10^2+\left( {5\over 2}\sqrt{35}-a\right)^2 \Rightarrow a={3\over 2}\sqrt{35}\\ \Rightarrow r=\sqrt{\left({15\over 2}\right)^2+a^2} = 3\sqrt{15} \Rightarrow 正方形A'B'C'D'面積=(2r)\times (2r)= \left(6\sqrt{15}\right)^2 = \bbox[red, 2pt]{540}$$

解答： $$將\overleftrightarrow{AB}:y=m(x-4)+1代入xy=1 \Rightarrow mx(x-4)+x=1 \Rightarrow mx^2+(1-4m)x-1=0\\ \Rightarrow 兩根之和={4m-1 \over m} = 4\times 2(\because 兩根之平均=4) \Rightarrow m=-{1\over 4} \Rightarrow \overleftrightarrow{AB}:x+4y=8\\因此xy=1上的點C(t,1/t)至\overleftrightarrow{AB}的距離d={|t+4/t-8|\over \sqrt{17}} \le {|4-8|\over \sqrt{17}}=\bbox[red, 2pt]{4\sqrt{17}\over 17} \\(\because t+{4\over t}\ge 2\sqrt{t\cdot {4\over t}}=4)$$

解答： $$令a=10^{100}，則\left[\cfrac{10^{10000}}{10^{100}+1} \right] =\left[\cfrac{a^{100}}{a+1} \right] =\left[\cfrac{((a+1)-1)^{100}}{a+1} \right]\\ =\left[\sum_{k=0}^{100}C^{100}_k(a+1)^{k-1}(-1)^{100-k} \right] = \left[{1\over a+1} -C^{100}_1 + C^{100}_2(a+1)-\cdots +C^{100}_{100}(a+1)^{99}\right]\\ =\left[{1\over a+1} +很大的正整數 \right]= 很大的正整數=-C^{100}_1 + C^{100}_2(a+1)-\cdots +C^{100}_{100}(a+1)^{99}\cdots(1) \\ 由於C^{100}_m(a+1)^n= C^{100}_m(C^n_na^n +C^n_{n-1}a^{n-1}+\cdots+C^n_1a+1)=100k+C^{100}_m\\ 因此(1)=100K+ (C^{100}_{100}-C^{100}_{99}+\cdots +C^{100}_2-C^{100}_1+C^{100}_0)-C^{100}_0,m,n,k\in \mathbb{N}\\ =100K +(1+(-1))^{100}-1 =100k-1 \equiv 99 \mod 100,K\in \mathbb{N}\\ \Rightarrow \left[\cfrac{10^{10000}}{10^{100}+1} \right] \div 100的餘數為\bbox[red, 2pt]{99}$$

解答： $$\lim_{n\to \infty}\left( \sqrt[n]{1+3^{2n}} +\sqrt[n]{3^{2n}+5^{2n}} + \cdots +\sqrt[n]{(2m-1)^{2n}+(2m+1)^{2n}} \right) \\ = \lim_{n\to \infty}\left(3^2 \sqrt[n]{({1\over 3})^{2n}+1} +5^2\sqrt[n]{({3\over 5})^{2n}+1} + \cdots +(2m+1)^2\sqrt[n]{({2m-1\over 2m+1})^{2n}+1} \right) \\ =3^2+5^2+\cdots +(2m+1)^2 =\sum_{k=1}^m(2k+1)^2 =4\sum_{k=1}^m k^2 +4\sum_{k=1}^m k+\sum_{k=1}^m 1\\ ={2\over 3}m(m+1)(2m+1)+2m(m+1)+m\\ 因此原式\Rightarrow \lim_{ m\to\infty}{{2\over 3}m(m+1)(2m+1)+2m(m+1)+m\over m^3} =\bbox[red,2pt]{4\over 3}$$

解答：

$$\left[\cfrac{x+1}{2}\right]=k,k\in \mathbb{Z} \Rightarrow k\le \cfrac{x+1}{2}\lt k+1 \Rightarrow x \in [2k-1,2k+1) \Rightarrow x介於兩個連續奇數之間; \\\log_2 \left\{ \log_2 \left( \left| x-\left[\cfrac{x+1}{2} \right]\right|\right)\right\} \lt 0 \Rightarrow 0\lt \log_2 \left( \left| x-\left[\cfrac{x+1}{2} \right]\right|\right) \lt 1 \\ \Rightarrow 1 \lt \left| x-\left[\cfrac{x+1}{2} \right]\right|\lt 2 \Rightarrow \cases{ 1\lt x-\left[\cfrac{x+1}{2}\right] \lt 2 \\ -2\lt x-\left[\cfrac{x+1}{2}\right] \lt -1} \Rightarrow \cases{ x+1\lt  \left[\cfrac{x+1}{2}\right] \lt x+2 \\ x-2\lt \left[\cfrac{x+1}{2}\right] \lt x-1}\\ \Rightarrow 所圍區域為四條水平線(見上圖)，即\bbox[red, 2pt]{-4\lt x\lt -2或 2\lt x\lt 4，但x\ne \pm 3\qquad}$$

解答： $$(a,b,c)代表3個箱子內的物品數，由於箱子相同，不計算a,b,c的排列\\(9,0,0)\to 分法=1\\(8,1,0)\to C^9_8=9,(7,2,0)\to C^9_7=36,(6,3,0)\to C^9_6=84, (5,4,0)\to C^9_5=126；\\(7,1,1)\to C^9_7 =36, (6,2,1)\to C^9_6C^3_2=252, (5,3,1)\to C^9_5C^4_3=504,(5,2,2)\to C^9_5C^4_2\div 2=378;\\(4,4,1)\to C^9_4C^5_1\div 2=315, (4,3,2)\to C^9_4C^5_3=1260; (3,3,3)\to C^9_3C^6_3\div 3!=280 \\ 共有1+9+36+84+126 +36+252+504+378+315 +1260 +280= \bbox[red, 2pt]{3281}種分法$$

解答：

$$\triangle OAB:正弦定理 \Rightarrow {|\vec z|\over \sin \alpha} ={|\vec w|\over \sin (\pi-\theta)} ={|\vec w|\over \sin \theta} \Rightarrow \sin\theta = {|\vec w|\over |\vec z|}\sin \alpha ={1\over 10}\sin \alpha \\ \Rightarrow \sin ^2\theta={\sin^2\alpha \over 100} \le {1\over 100} \Rightarrow \tan^2\theta ={\sin^2\theta \over \cos^2\theta} ={\sin^2\theta \over 1-\sin^2\theta}= {1 \over 1-\sin^2\theta} -1\le {1\over 1-1/100}-1\\= \bbox[red,2pt]{1\over 99}$$

二、計算題

解答：

(1) $$x^2f(x)={3\over 5}x^5+{1\over 2}ax^4 -{1\over 3}x^3+ 2\int_0^xtf(t)\;dt \\兩邊微分\Rightarrow 2xf(x)+x^2f'(x) =3x^4+2ax^3-x^2 +2xf(x)\\ \Rightarrow f'(x)=3x^2+2ax-1 \Rightarrow f(x)=x^3+ax^2-x+C,C為常數\\ f(0)=0 \Rightarrow C=0 \Rightarrow \bbox[red,2pt]{f(x)=x^3+ax^2-x}$$ (2) $$f(x)=x^3+ax^2-x=0 \Rightarrow x(x^2+ax-1)=0 \Rightarrow x=0,\alpha,\beta,其中\cases{\alpha={-a-\sqrt{a^2+4}\over 2}\\ \beta={-a+\sqrt{a^2+4}\over 2}} \\ \Rightarrow \cases{\alpha^2+\beta^2= (\alpha+\beta)^2-2\alpha\beta=a^2+2 \\ \alpha^3+\beta^3 =(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)= -a^3-3a\\ \alpha^4+\beta^4= (\alpha^2+ \beta^2)^2 -2\alpha^2\beta^2=(a^2+2)^2-2 =a^4+4a^2+2} \\ \Rightarrow S(a)= \int_\alpha^0 f(x)\;dx -\int_0^\beta f(x)\;dx =\left .\left[ {1\over 4}x^4+{1\over 3}ax^3-{1\over 2}x^2 \right]\right|_\alpha^0 -\left .\left[ {1\over 4}x^4+{1\over 3}ax^3-{1\over 2}x^2 \right]\right|_0^\beta \\ =-{1\over 4}\alpha^4-{1\over 3}a\alpha^3+{1\over 2}\alpha^2 -{1\over 4}\beta^4-{1\over 3}a\beta^3+{1\over 2}\beta^2 =-{1\over 4}(\alpha^4+\beta^4)-{1\over 3}a(\alpha^3+\beta^3)+{1\over 2}(\alpha^2+\beta^2) \\ =-{1\over 4}(a^4+4a^2+2)-{1\over 3}a(-a^3-3a)+{1\over 2}(a^2+2) = \bbox[red, 2pt]{{1\over 12}a^4+{1\over 2}a^2+{1\over 2}}$$ 3 $$S'(a)=0 \Rightarrow {1\over 3}a^3+ a=0 \Rightarrow a(a^2+3)=0 \Rightarrow 當a=0時，S(a)有最小值S(0)=\bbox[red,2pt]{1\over 2}$$

解答： $$\omega = \cos{2\pi\over 7}+i\sin{2\pi \over 7} \Rightarrow \omega^7=1  \\ (2-\omega)(2-\omega^3)(2-\omega^5) =(4-2\omega-2\omega^3 +\omega^4)(2-\omega^5)= 8-2\omega-\omega^2-4\omega^3+2\omega^4-4\omega^5+2\omega^6\\ \Rightarrow  實部=8-2\cos{2\pi\over 7}-\cos{4\pi\over 7}-4\cos{6\pi\over 7} +2\cos{8\pi\over 7} -4\cos{10\pi\over 7} +2\cos{12\pi\over 7}\\ = 8-2\cos{2\pi\over 7}-\cos{4\pi\over 7}+4\cos{\pi\over 7}  -2\cos{\pi\over 7} -4\cos{4\pi\over 7} +2\cos{2\pi\over 7}=8+2\cos{\pi\over 7}-5\cos{4\pi \over 7}\\ \Rightarrow 虛部=-2\sin{2\pi\over 7}-\sin{4\pi\over 7}-4\sin{6\pi\over 7} +2\sin{8\pi\over 7} -4\sin{10\pi\over 7} +2\sin{12\pi\over 7} \\= -2\sin{2\pi\over 7}-\sin{4\pi\over 7}-4\sin{\pi\over 7} -2\sin{ \pi\over 7} +4\sin{4\pi\over 7} -2\sin{2\pi\over 7} \\ =-6\sin{\pi\over 7}-4\sin{2\pi\over 7}+3\sin{4\pi\over 7} \\因此欲求之解為\bbox[red,2pt]{8+2\cos{\pi\over 7}-5\cos{4\pi \over 7}+i(-6\sin{\pi\over 7}-4\sin{2\pi\over 7}+3\sin{4\pi\over 7})}$$
 

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