114年台大碩士班-基礎數學詳解

 國立臺灣大學114學年度碩士班招生考試

題號:289
科目:基礎數學

解答: $$f(x)={1\over 2-x}={1\over 2}\cdot {1\over 1-x/2}={1\over 2}\sum_{k=0}^\infty\left( {x\over 2}\right)^k \Rightarrow f^{n}(x)= \bbox[red, 2pt]{{1\over 2} \sum_{k=0}^\infty {(n+k)!\over 2^{n+k}\cdot k!}x^k}$$

解答: $$f(x)= {d\over dx} \left[x{d\over dx}(x-x^2) \right] = {d\over dx} \left[x(1-2x) \right] = {d\over dx} \left[x-2x^2 \right] =\bbox[red, 2pt]{1-4x}$$

解答: $$\lim_{x\to 5^+} {\sqrt{x^2-4x-5} \over x-5} =\lim_{x\to 5^+} {(\sqrt{x^2-4x-5})' \over (x-5)'} =\lim_{x\to 5^+} {x-2 \over \sqrt{x^2-4x-5}} =\bbox[red, 2pt] \infty$$

解答: $$f(x)=x^4-4x^3+1 \Rightarrow f'(x)=4x^3-12x^2 \Rightarrow f''(x)= 12x^2-24x \\f'(x)=0\Rightarrow 4x^2(x-3)=0 \Rightarrow x=0,3 \Rightarrow \cases{f''(0)=0\\ f''(3)=36\gt 0} \\ \Rightarrow \begin{cases} f'(x)\ge 0& x\ge 3\\ f'(x)\le 0& x\le 3\end{cases} \Rightarrow \begin{cases} f(x)遞增& x\ge 3\\ f(x)遞減& x\le 3\end{cases} \;又\cases{f(-1)=6\\ f(0)=1\\ f(3)=-26\\ f(5)=126}$$

解答: $$A= \begin{bmatrix} 4 & 2 & 2 \\2 & 4 & 2 \\2 & 2 & 4\end{bmatrix}  \Rightarrow \det(A-\lambda I) =-(\lambda-2)^2 (\lambda-8)=0 \Rightarrow \lambda =2,8\\ \lambda_1=2 \Rightarrow (A-\lambda_1 I) v=0 \Rightarrow  \begin{bmatrix}2 & 2 & 2 \\ 2 & 2 & 2 \\2 & 2 & 2 \end{bmatrix}  \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1+x_2+x_3=0 \\\qquad \Rightarrow v= x_2 \begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix} + x_3\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}, \text{ choose }v_1= \begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix}, v_2=\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix} \\ \lambda_2=8  \Rightarrow (A-\lambda_2 I) v=0 \Rightarrow  \begin{bmatrix}-4 & 2 & 2 \\ 2 & -4 & 2 \\2 & 2 & -4 \end{bmatrix}  \begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix} =0 \Rightarrow x_1=x_2=x_3  \\\qquad \Rightarrow v= x_3\begin{pmatrix}  1\\ 1\\ 1 \end{pmatrix} ,\text{ choose }v_3= \begin{pmatrix} 1\\ 1\\ 1 \end{pmatrix} \\ \Rightarrow \bbox[red, 2pt]{特徵值:2,8; 特徵向量:\begin{pmatrix} -1\\ 1\\ 0 \end{pmatrix}, \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}, \begin{pmatrix}  1\\ 1\\ 1 \end{pmatrix}}$$

解答: $$假設L為一對一，若\vec x\in NULL(L),則L(\vec x)= \vec 0. 由於L為線性轉換,因此L(\vec 0)= \vec0\\\qquad 也就是說L(\vec x) =L(\vec 0), 由於L為一對一，因此\vec x=\vec 0,即NULL(L)=\{\vec 0\}\\ 假設NULL(L)=\{\vec 0\}, 若L(\vec x)=L(\vec y) \Rightarrow L(\vec x)-L(\vec y) =L(\vec x-\vec y) =\vec 0 \Rightarrow \vec x-\vec y\in NULL(L) \\ \qquad \Rightarrow \vec x-\vec y=\vec 0 \Rightarrow \vec x=\vec y \Rightarrow L為一對一\\ 由上述可知: L\text{ is 1-1 iff }NULL(L)=\{\vec 0\} \quad \bbox[red, 2pt]{QED}$$

解答: $$\cases{P(A\to B)=1\\ P(B\to C) =P(B\to A) = P(B\to D)=1/3\\ P( C\to B)=1\\ P(D\to B) =P(D\to E)=1/2\\  P( E\to D)1} \Rightarrow 轉移矩陣A= \bbox[red, 2pt]{\begin{bmatrix}0 & 1/3&0 & 0 & 0 \\ 1& 0 & 1& 1/2& 0\\ 0 & 1/3& 0 & 0 & 0\\ 0 & 1/3& 0 & 0 & 1\\ 0 & 0& 0& 1/2& 0\end{bmatrix}}\\ \det(A-\lambda I) =0 \Rightarrow \lambda=0,\pm {\sqrt 3\over 3}, \pm 1 \Rightarrow \bbox[red, 2pt]{無穩定狀態}$$

解答: $$\cases{u=\ln(x)\\dv =x^2\,dx} \Rightarrow \cases{du =dx/x\\ v=x^3/3} \Rightarrow \int  x^2\ln(x)\, dx= {1\over 3}x^3 \ln(x)-{1\over 3}\int x^2\,dx= {1\over 3}x^3 \ln(x)-{1\over 9}x^3+C \\ \Rightarrow \int_0^1 x^2\ln(x)\,d x= \left. \left[ {1\over 3}x^3 \ln(x)-{1\over 9}x^3 \right] \right|_0^1= \bbox[red, 2pt]{-{1\over 9}}$$

解答: $$f(x)=\ln(x^2+1) \Rightarrow f'(x)={2x\over x^2+1} \Rightarrow f''(x)={2\over x^2+1}-{4x^2 \over (x^2+1)^2} = \bbox[red, 2pt]{-2x^2+2\over (x^2+1)^2}$$

解答: $$\int f(x)\,dx =1 \Rightarrow \int_{-1}^1 k(1-x^2)\,dx = \left. \left[ k(x-{1\over 3}x^3)\right] \right|_{-1}^1 ={4\over 3}k=1 \Rightarrow k=\bbox[red, 2pt]{3\over 4}$$

解答: $$e^x =1+x+{x^2\over 2!} + \cdots \Rightarrow e^{-x^2} =1+(-x^2) +{(-x^2)^2\over 2!}+   \cdots = \bbox[red, 2pt]{1-x^2+{x^4\over 2!}- \cdots}$$

解答: $$A=\begin{bmatrix}3& 1& 2\\ 0& 4& 5\\ 1&-2 & 0 \end{bmatrix} \Rightarrow \det(A)=0+0+5-8-0+30= \bbox[red, 2pt]{27}$$

解答: $$B=\begin{bmatrix}6& -1\\ 2& 3 \end{bmatrix} \Rightarrow \det(B-\lambda I)= (\lambda-4)(\lambda-5)=0 \Rightarrow \lambda=4,5\\ \lambda_1=4 \Rightarrow (B-\lambda_1 I)v=0 \Rightarrow \begin{bmatrix}2& -1\\ 2& -1 \end{bmatrix}  \begin{bmatrix}x_1 \\ x_2 \end{bmatrix} =0 \Rightarrow 2x_1=x_2 \Rightarrow v= x_2 \begin{pmatrix}1/2\\ 1 \end{pmatrix}, \text{choose }v_1= \begin{pmatrix}1/2\\ 1 \end{pmatrix} \\ \lambda_2= 5  \Rightarrow (B-\lambda_2 I)v=0 \Rightarrow \begin{bmatrix}1& -1\\ 2& -2 \end{bmatrix}  \begin{bmatrix}x_1\\ x_2 \end{bmatrix} =0 \Rightarrow  x_1=x_2 \Rightarrow v= x_2 \begin{pmatrix}1\\ 1 \end{pmatrix}, \text{choose }v_2= \begin{pmatrix}1 \\ 1 \end{pmatrix} \\ \Rightarrow \cases{P=[v_1 v_2] =\begin{bmatrix}1/2& 1\\ 1& 1 \end{bmatrix} \\ D= \begin{bmatrix} \lambda_1& 0\\ 0& \lambda_2 \end{bmatrix} =\begin{bmatrix} 4& 0\\ 0& 5 \end{bmatrix} } \Rightarrow B=PDP^{-1} = \bbox[red, 2pt]{\begin{bmatrix}1/2& 1\\ 1& 1 \end{bmatrix}\begin{bmatrix} 4& 0\\ 0& 5 \end{bmatrix} \begin{bmatrix} -2& 2\\ 2& -1 \end{bmatrix} }$$

解答: $$A= \begin{bmatrix}1& 2\\ 2& 4 \end{bmatrix} \Rightarrow AA^T= \begin{bmatrix} 5& 10\\ 10& 20 \end{bmatrix}  \Rightarrow \text{row space } \{ \begin{bmatrix} 1\\ 2\end{bmatrix}\} \Rightarrow P= \begin{bmatrix} 1\\ 2\end{bmatrix} \\ \Rightarrow P^TAP =\begin{bmatrix} 1& 2\end{bmatrix} \begin{bmatrix} 1& 2\\ 2& 4 \end{bmatrix} \begin{bmatrix} 1\\ 2\end{bmatrix} =[25] \Rightarrow (P^TAP)^{-1}= \left[{1\over 25} \right] \\\Rightarrow A^+= P(P^TAP)^{-1 } P^T = \begin{bmatrix} 1\\ 2\end{bmatrix} \left[{1\over 25} \right]\begin{bmatrix} 1& 2\end{bmatrix} = \begin{bmatrix} 1/25\\ 2/25\end{bmatrix}  \begin{bmatrix} 1& 2\end{bmatrix} =\bbox[red, 2pt]{\begin{bmatrix} {1\over 25}& {2 \over 25} \\{2 \over 25}& {4 \over 25}\end{bmatrix}}$$

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解題僅供參考，碩士班歷年試題及詳解