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#9

##### 引用:

9.

C: x² + y² - 2x + 3y = 0 ...(1)

C 的圓心Q: (1, - 3/2)

(x-a)*(x-1) + (y-b)*(y+3/2) = 0 ...(2)

⇒ (a, b) = (3, -4)

x² + y² - 2x + 3y + k*(4x - 5y - 18) = 0

⇒ k = -1/2

⇒ S 的圓心: (2, -11/4) 為 P，Q 的中點

⇒ P 之坐標 (3, -4)

$$\displaystyle \lim_{\theta\to0}\frac{\tan\theta - \sin\theta}{\theta^3}=\lim_{\theta\to0}\frac{\sin\theta}{\theta^3}\cdot\frac{1-\cos\theta}{\cos\theta}$$

$$\displaystyle=\lim_{\theta\to0}\frac{\sin\theta}{\theta^3}\cdot\frac{1-\cos\theta}{\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta}=\lim_{\theta\to0}\frac{\sin^3\theta}{\theta^3}\cdot\frac{1}{\cos\theta\left(1+cos\theta\right)}$$

$$\displaystyle=1^3\cdot\frac{1}{1\cdot\left(1+1\right)}=\frac{1}{2}$$

\begin{align} & \frac{\tan \theta -\sin \theta }{{{\theta }^{3}}} \\ & =\frac{\sin \theta -\sin \theta \cos \theta }{{{\theta }^{3}}\times \cos \theta } \\ & =\frac{\sin \theta \times 2{{\sin }^{2}}\frac{\theta }{2}}{{{\theta }^{3}}\times \cos \theta } \\ & =\frac{\sin \theta }{\theta }\times {{\left( \frac{\sin \frac{\theta }{2}}{\frac{\theta }{2}} \right)}^{2}}\times \frac{1}{2\cos \theta } \\ & \\ & \underset{\theta \to 0 }{\mathop{\lim }}\,\frac{\tan \theta -\sin \theta }{{{\theta }^{3}}}=1\times 1\times \frac{1}{2}=\frac{1}{2} \\ \end{align}

sinA=A-1/3!*A^3+.....

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##### 引用:

g112大，我覺得應該是高雄聯招也看計算過程，可能答案對但計算過程有誤，所以沒給分(我的情況就是這樣，但我知道那7分是被扣定了)，可以拿題目一起討論看看!

##### 引用:

$$\displaystyle \lim_{n \to \infty}\left(\frac{1}{\sqrt{n}+\sqrt{n+2}}+\frac{1}{\sqrt{n+1}+\sqrt{n+3}}+\frac{1}{\sqrt{n+2}+\sqrt{n+4}}+\ldots\right)$$
$$\displaystyle =\lim_{n \to \infty}\frac{1}{n}\left(\frac{1}{\sqrt{1}+\sqrt{1+\frac{2}{n}}}+\frac{1}{\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{3}{n}}}+\frac{1}{\sqrt{1+\frac{2}{n}}+\sqrt{1+\frac{4}{n}}}+\ldots \right)$$
$$\displaystyle =\int_0^1 \frac{1}{2\sqrt{1+x}}dx=\sqrt{1+x}|_0^1=\sqrt{2}-1$$

##### 引用:

4575 ...

[attach]4578[/attach]補充板上尚未討論的題目，

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[解]

$$\matrix{C_2^{101}&+&C_2^{100}&=\frac{101 \times 100}{2}+\frac{100\times 99}{2}=10000種 \cr (奇+奇)&&(偶+偶)&}$$

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[解]
$$\displaystyle R=\int_0^2 x^3 dx=\frac{x^4}{4}\Bigg\vert\;_0^2=4$$

$$\displaystyle U_{n} = \sum_{k=1}^{n} \frac{2}{n} \cdot \left( \frac{2k}{n} \right)^{3} = \sum_{k=1}^{n} \frac{16}{n^4} \cdot k^3 = \frac{16}{n^4} \cdot \left[ \frac{n(n+1)}{2} \right]^2 = \frac{ 4n^4 + 8n^3 + 4n^2 }{n^4} = 4 + \frac{8n+4}{n^2}$$

$$\displaystyle \Rightarrow \; | U_{n} - R | = \left| \frac{8n+4}{n^2} \right| < \frac{1}{100} \quad \Rightarrow \quad n^2 > 800n + 400 \quad \Rightarrow \quad (n^2 - 800n + 400^2) > 400 + 400^2$$

$$\displaystyle \Rightarrow \; (n - 400)^2 > 160400 \quad \Rightarrow \quad n - 400 > 400. \cdots \quad \Rightarrow \quad n > 800$$，最小自然數$$\displaystyle n = 801$$。

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