國立臺灣師範大學 114學年度碩士班招生考試試題
科目:基礎數學
適用系所:數學系
Part I: Calculus(無計算過程或說明不給分,總分 50分)
解答:$$\textbf{(a) }f(x)= \left( {x+2\over x-1} \right)^x \Rightarrow \ln f(x)=x\ln{x+2\over x-1} \Rightarrow \lim_{x\to \infty} \ln f(x)=\lim_{x\to \infty} {(\ln{x+2\over x-1})' \over (1/x)'}\\\qquad = \lim_{x\to \infty} \left( {x-1 \over x+2}\times {3x^2\over (x-1)^2} \right)=3 \Rightarrow \lim_{x\to \infty} f(x)=e^3 \ne 0 \Rightarrow \int_2^\infty f(x)\,dx \text{ is divergent.} \\\textbf{(b) }\lim_{x\to \infty} {\int_2^x \left( {t+2\over t-1} \right)^t dt\over x} = \lim_{x\to \infty} {(\int_2^x \left( {t+2\over t-1} \right)^t dt)'\over (x)'} = \lim_{x\to \infty} { \left( {x+2\over x-1} \right)^x \over 1} = \bbox[red, 2pt]{e^3}$$
解答:$$\textbf{(a) } \cases{x\gt 0 \\ \ln x\gt 0 \Rightarrow x\gt 1} \Rightarrow \text{ domain of }f= \bbox[red, 2pt]{\{x\mid x\gt 1\}} \\\textbf{(b) }f'(x)={\ln(\ln x)\over x}+ \ln x \left( {1\over \ln x} \cdot {1\over x}\right) ={\ln(\ln x)+1\over x} \gt 0 \Rightarrow \ln(\ln x)+1 \gt 0 \Rightarrow \ln x \gt {1\over e} \\\qquad \Rightarrow x\gt e^{1/e} \Rightarrow \text{increasing interval: }\bbox[red, 2pt]{(e^{1/e}, \infty)} \\\textbf{(c) }\cases{f \text{ is strictly decreasing}, 1\lt x\lt e^{1/e} \\ f\text{ is strictly increasing, }e^{1/e}\lt x\lt \infty} \text{ and }\cases{f(1)=-1\lt 0\\ f(e^{1/e})=-{1\over e}-1 \lt 0 \\ f(\infty) \to \infty} \\\qquad \Rightarrow f\text{ has exactly one zero in its domain}$$
解答:$$x^2+y^2=1 \Rightarrow \cases{x=\cos t\\ y=\sin t} \Rightarrow x-y+z=1 \Rightarrow z=1-x+y=1-\cos t+\sin t \\ \Rightarrow f(x,y)=x+2y+3z=\cos t+2\sin t+3(1-\cos t+ \sin t) =3-2\cos t+5\sin t \\=3-\sqrt{29} \sin(t-\alpha) \Rightarrow \text{ maximum value: }\bbox[red, 2pt]{3+\sqrt{29}}$$
解答:
$$\text{Let }\cases{u=x+y\\ v=y-2x}, \text{ then }\cases{x=(u-v)/3\\ y=(2u+v)/3} \Rightarrow \left| {\partial(x,y) \over \partial(u,v)}\right| = \begin{vmatrix}1/3&-1/3\\2/3& 1/3 \end{vmatrix} ={1\over 3} \\ \text{The region of integration in the xy-plane is bounded by }x=0,y=0, \text{ and }x+y=1. \\ \cases{x=0 \Rightarrow u=v\\ y=0 \Rightarrow v=-2u\\ x+y=1 \Rightarrow u=1} \Rightarrow \int_0^1 \int_0^{1-x} \sqrt{x+y}(y-2x)^2 \,dydx = \int_0^1 \int_{-2u}^u \sqrt uv^2 \cdot {1\over 3}\,dv \,du \\=\int_0^1 \left. \left[ {1\over 9}\sqrt u v^3 \right] \right|_{-2u}^u \,du = \int_0^1 u^{7/2}\,du = \left. \left[ {2\over 9}u^{9/2} \right] \right|_0^1 = \bbox[red, 2pt]{2\over 9}$$
解答:
$$1-\cos \theta=1+\cos \theta \Rightarrow \cos \theta=0 \Rightarrow \theta={\pi\over 2},{3\pi\over 2} \\ \Rightarrow A= 2 \int_{\pi/2}^{\pi/2} {1\over 2}(1-\cos \theta)^2\,d\theta = \int_{\pi/2}^{\pi/2} \left( {3\over 2}-2\cos \theta+{1\over 2}\cos 2\theta \right)\,d\theta \\= \left. \left[ {3\over 2}\theta-2\sin \theta+{1\over 4}\sin 2\theta \right] \right|_{-\pi/2}^{\pi/2} = \left( {3\pi\over 4}-2 \right)- \left( -{3\pi\over 4}+2 \right) = \bbox[red, 2pt]{{3\pi\over 2}-4}$$
Part II. Linear Algebra(總分 50分)
解答:$$\textbf{(a) }R = rref(A) \Rightarrow Rx=0 \Rightarrow \cases{x_1-x_3+x_5=0\\ x_2-2x_3+6x_5=0 \\ x_4+2x_5=0} \Rightarrow \cases{x_1=x_3-x_5\\ x_2=2x_3-6x_5 \\ x_4=-2x_5} \\ \Rightarrow x=\begin{bmatrix}x_3-x_5\\ 2x_3-6x_5\\x_3\\-2x_5\\ x_5 \end{bmatrix} =x_3 \begin{bmatrix}1\\2\\1\\0\\0 \end{bmatrix} + x_5 \begin{bmatrix}-1\\-6\\0\\-2\\-1 \end{bmatrix} \Rightarrow \text{ a basis of Null}(T): \bbox[red, 2pt]{\left \{ \begin{bmatrix}1\\ 2\\1\\0\\0 \end{bmatrix}, \begin{bmatrix}-1\\-6\\0\\-2\\-1 \end{bmatrix} \right\}} \\\textbf{(b) }\cases{ T(b_1) =e_1+ 6e_3 = \begin{bmatrix}1 \\0\\6 \end{bmatrix} \\T(b_2) =-e_1+2e_2 = \begin{bmatrix}-1\\2\\0 \end{bmatrix} \\T(b_4)= 3e_1- 4e_2 +5e_3= \begin{bmatrix}3\\ -4\\5 \end{bmatrix}} \Rightarrow A= \begin{bmatrix}1&-1 &a& 3&b\\0& 2& c &-4& d\\ 6& 0& e& 5& f \end{bmatrix} \xrightarrow{R_2/2\to R_2, R_3-6R_1\to R_3} \\ \begin{bmatrix}1&-1 &a& 3&b\\0& 1& c/2 &-2& d/2\\ 0& 6& e-6a& -13& f-6b \end{bmatrix} \xrightarrow{R_1+R_2\to R_1, R_3-6R_2\to R_3} \begin{bmatrix}1& 0 &a+c/2& 1&b+d/2\\0& 1& c/2 &-2& d/2\\ 0& 0& e-6a-3c& -1& f-6b-3d \end{bmatrix} \\\xrightarrow{R_1+R_3\to R_1, R_2-2R_2\to R_2} \begin{bmatrix}1& 0 &-5a-5c/2+e& 0&-5b-5d/2+f\\0& 1& 12a+13c/2-2e & 0& 12b+13d/2-2f\\ 0& 0& e-6a-3c& -1& f-6b-3d \end{bmatrix} \xrightarrow{-R_3\to R_3} \\ \begin{bmatrix}1& 0 &-5a-5c/2+e& 0&-5b-5d/2+f\\0& 1& 12a+13c/2-2e & 0& 12b+13d/2-2f\\ 0& 0& -e+6a+3c& 1& -f+6b+3d \end{bmatrix} = R= \begin{bmatrix}1&0&-1 &0 &1\\0&1& -2&0 &6 \\ 0& 0& 0& 1& 2\end{bmatrix} \\ \Rightarrow \cases{-5a-5c/2+e=-1\\ 12a+13c/2-2e=-2\\ -e+6a+3c=0} \Rightarrow \cases{a=1\\ c=-4\\ e=-6} \text{ and } \cases{-5b-5d/2+f=1\\ 12b+13d/2-2f=6\\ -f+6b+3d=2} \Rightarrow \cases{b=1\\ d=4\\ f=16} \\ \Rightarrow \bbox[red, 2pt]{A= \begin{bmatrix}1&-1 &1& 3&1\\0& 2& -4 &-4& 4\\ 6& 0& -6& 5& 16 \end{bmatrix}} \\\textbf{(c) }rref(A)=R= \begin{bmatrix}1&0&-1 &0 &1\\0&1& -2&0 &6 \\ 0& 0& 0& 1& 2\end{bmatrix} \Rightarrow \text{ the pivot columns are 1,2 and 4} \\\quad \Rightarrow M= [A_1\; A_2 \;A_3] = \begin{bmatrix}1&-1&3\\0& 2&-4\\ 6&0& 5 \end{bmatrix} \Rightarrow M^{-1} = \begin{bmatrix}-5&-2.5&1\\ 12& 6.5&-2\\ 6& 3& -1 \end{bmatrix} \\ \Rightarrow S= \begin{bmatrix}-5&-2.5&1\\ 12& 6.5&-2 \\0&0& 0\\ 6& 3& -1 \\0&0& 0\end{bmatrix} \Rightarrow AS= \begin{bmatrix}1&-1 &1& 3&1\\0& 2& -4 &-4& 4\\ 6& 0& -6& 5& 16 \end{bmatrix} \begin{bmatrix}-5&-2.5&1\\ 12& 6.5&-2 \\0&0& 0\\ 6& 3& -1 \\0&0& 0\end{bmatrix}=I_3 \\ \bbox[red, 2pt]{S\text{ is not unique.}}\text{ For example, we can construct matrix }K \text{ where the first column }v\in NULL(A)\\ \text{ and the others are zero. Say, }K= \begin{bmatrix}1&0&0\\ 2& 0& 0\\ 1&0&0\\ 0&0&0\\ 0&0&0 \end{bmatrix} \Rightarrow AK= 0 \Rightarrow A(S+K)=I+0=I \\ \Rightarrow S+K \text{ is another linear transformation. }\Rightarrow S \text{ is not unique.}$$
解答:$$\textbf{(a) }a_1=a_2=\cdots=a_n=0 \Rightarrow 0T(v_1)+ 0T(v_2)+ \cdots+ 0T(v_n)=0 \\ \qquad \Rightarrow (0,\dots,0) \in \mathcal N\\ \cases{(a_1,a_2,\dots, a_n) \in \mathcal N \Rightarrow a_1T(v_1)+a_2T(v_2)+ \cdots+ a_nT(v_n)=0 \\ (b_1, b_2, \dots, b_n) \in \mathcal N \Rightarrow b_1T(v_1) +b_2T(v_2)+ \cdots+ b_nT(v_n)=0} \\ \quad \Rightarrow (a_1+b_1)T(v_1)+(a_2+b_2)T(v_2)+ \cdots+(a_n +b_n)T(v_n) =0 \\\quad \Rightarrow (a_1+b_1, a_2+b_2, \dots,a_n+b_n) \in \mathcal N \\ \cases{c\in \mathbb R\\(a_1,a_2,\dots, a_n) \in \mathcal N} \Rightarrow c (a_1T(v_1)+a_2T(v_2)+ \cdots+a_nT(v_n)) =c\cdot 0=0 \\\quad \Rightarrow (ca_1, ca_2, \dots, ca_n) \in \mathcal N \\ \Rightarrow \mathcal N \text{ is a subspace} \\ \text{Any vector }x\in \mathbb R^n \Rightarrow x=a_1v_1+a_2v_2+ \cdots+a_nv_n \\ (a_1,\dots,a_n)\in \mathcal N \text{ iff }T(x)=0 \Rightarrow dim(\mathcal N) =dim(ker(T)) =n-rank(T)=n-r \\ \Rightarrow dim(\mathcal N)=n-r. \bbox[red, 2pt]{QED.} \\\textbf{(b) }|I\cap R| \le dim(R)=r \Rightarrow |I\cap R| \le r \quad \bbox[red, 2pt]{QED.}$$
解答:$$\textbf{(a) }\begin{vmatrix}5& 2& 0 & \cdots&0\\ 3&5& 2& \cdots& 0\\ 0& 3& 5& \cdots&0\\ \vdots& \vdots& \vdots& \ddots& \vdots\\ 0&0&0& \cdots& 5 \end{vmatrix} =5\begin{vmatrix} 5& 2& \cdots& 0\\ 3& 5& \cdots&0\\ \vdots& \vdots& \ddots& \vdots\\ 0&0& \cdots& 5 \end{vmatrix} -2\begin{vmatrix} 3& 2& \cdots& 0\\ 0& 5& \cdots&0\\ \vdots& \vdots& \ddots& \vdots\\ 0&0& \cdots& 5 \end{vmatrix} =5\begin{vmatrix} 5& 2& \cdots& 0\\ 3& 5& \cdots&0\\ \vdots& \vdots& \ddots& \vdots\\ 0&0& \cdots& 5 \end{vmatrix} -6\begin{vmatrix} 5& \cdots&0\\ \vdots& \ddots& \vdots\\ 0& \cdots& 5 \end{vmatrix} \\ \Rightarrow \det(A_n)=5\det(A_{n-1})-6\det(A_{n-2}) \equiv a_n=5a_{n-1}-6a_{n-2}, \text{ where }a_n= \det(A_n) \\ \Rightarrow a_1= 5, a_2= \begin{vmatrix}5&2\\3&5 \end{vmatrix} =19 \\ r^2-5r+6=0 \Rightarrow (r-2)(r-3)=0 \Rightarrow r=2,3 \Rightarrow a_n =c_12^n+c_23^n \\\Rightarrow \cases{n=1 \Rightarrow 5=2c_1+3c_2\\ n=2 \Rightarrow 19=4c_1+ 9c_2} \Rightarrow \cases{c_1=-2\\ c_2=3} \Rightarrow a_n=\det(A) = \bbox[red, 2pt]{3^{n+1}-2^{n+1}} \\\textbf{(b) }n=3 \Rightarrow A= \begin{bmatrix}5&2& 0\\3&5& 2\\ 0&3&5 \end{bmatrix} \Rightarrow \det(A-\lambda I) =(5-\lambda)^3-12(5-\lambda) =(5-\lambda)(\lambda^2-10\lambda+13)=0 \\ \quad \Rightarrow \lambda=5,5\pm 2\sqrt 3 \Rightarrow \text{ three distinct real eigenvalues }\Rightarrow \bbox[red, 2pt]{A \text{ is diagonalizable}}$$
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解題僅供參考，碩士班歷年試題及詳解
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