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# System prepended metadata

title: 台大二階筆試-數學考古2013

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# 台大二階筆試-數學考古2013
## 2013台大電機
### 1.
#### $a_1=3,a_2=5,a_n=a_{n-1}+2a_{n-2},a_1+a_2+...a_n=?$
#### ans:
#### $a_n-1=(a_{n-1}-1)+2(a_{n-2}-1)\longrightarrow b_n=b_{n-1}+2b_{n-2}$
#### $b_n=\alpha2^n+\beta(-1)^n,$代入得$b_n=2^n,a_n=2^n+1$
#### $a_1+a_2+...a_n=2^1+2^2+...2^n+n=2^{n+1}+n-2$
### 2.
#### $x^3-7x^2+cx-8=0$三根成等比，求三根與c
#### ans:
#### 三根$a,ar,ar^2,a(ar)(ar^2)=8\longrightarrow ar=2$
#### $a+ar+ar^2=7\longrightarrow a=1,r=2,c=14$
### 3.
#### 兩隊勝率0.5，五戰三勝，比到第五局(兩勝兩負)機率為何?四人家庭，已知有一男一女，求男女各半機率?
#### ans:
#### $\frac{1}{16}C^4_2=\frac{3}{8},\frac{1}{2}$
### 4.
#### 編號1-10的球各兩顆，球抽三顆組合數
#### ans:
#### 三異加上兩同一異$C^{10}_3+10\times9=120+90=210$
### 5.
#### $13x^2-10xy+13y^2-6x-42y-27=0$為何種圖形?標準化後為何?標準化對應的矩陣為何
#### ans:
#### $A=\begin{bmatrix}13-\lambda& -5\\-5& 13\end{bmatrix},det(A)=0\longrightarrow \lambda=8,18,$可解出兩向量滿足$Av=\lambda v$，並可組成一旋轉矩陣
#### 可轉為$\begin{bmatrix}x& y\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}8& 0\\0& 18\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}+\begin{bmatrix}-6& -42\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}-27$
#### $\begin{bmatrix}x'\\ y'\end{bmatrix}\begin{bmatrix}8& 0\\0& 18\end{bmatrix}\begin{bmatrix}x'\\ y'\end{bmatrix}+\begin{bmatrix}-6& -42\end{bmatrix}\begin{bmatrix}\frac{1}{\sqrt{2}}& \frac{-1}{\sqrt{2}}\\\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}x'\\ y'\end{bmatrix}-27$
#### $8(x'-\frac{3\sqrt{2}}{2})^2+18(y'-\frac{\sqrt{2}}{2})^2=72 \longrightarrow \frac{X^2}{9}+\frac{Y^2}{4}=1,\begin{bmatrix}X\\ Y\end{bmatrix}=\begin{bmatrix}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\\frac{-1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}+\begin{bmatrix}\frac{-3}{\sqrt{2}}\\ \frac{-1}{\sqrt{2}}\end{bmatrix}$
### 6.
#### $a=1+i,b=-3+i,$兩複數極式?ab、a+b極式?最小a的次方數使其為正實數?最小a+b的次方數使其為正實數?首項係數1、以a,b為根的四次方程式?
#### ans:
#### (1)$a=\sqrt{2}(cos\frac{\pi}{4}+isin\frac{\pi}{4}),b=\sqrt{10}(cos\theta+isin\theta)$
#### (2)$ab=2\sqrt{5}(cos(\frac{\pi}{4}+\theta)+isin(\frac{\pi}{4}+\theta)),a+b=2\sqrt{2}(cos\frac{3\pi}{4}+isin\frac{3\pi}{4})$
#### (3)8(4)8(5)利用虛根成對及韋達定理$(x^2-2x+2)(x^2+6x+10)$
### 7.
#### $A(2,0,3)B(3,1,-2)C(4,0,-1),\overrightarrow{AB}\cdot\overrightarrow{AC}=?\angle BAC=?$
#### $\overrightarrow{AD}$為$\overrightarrow{AB}$在$\overrightarrow{AC}$上的正射影，點D座標?ABC平面方程式?
#### ans:
#### (1)4(2)$cos^{-1}\frac{\sqrt{6}}{3}(3)\sqrt{3}\times\frac{\sqrt{6}}{3}=\sqrt{2},(2,0,-3)+(1,0,1)=(3,0,-2)$
#### (4)$(1,1,1)\times(2,0,2)=(2,0,-2),$代入點可得平面$x-z=5$