114年清華計量財金碩士班-微積分詳解

 國立清華大學 114學年度碩士班考試入學試題

系所班組別:計量財務金融學系乙組(財務工程組)
考試科目:微積分

解答:

$$3x=4-x^2 \Rightarrow x^2+3x-4 =0 \Rightarrow (x+4)(x-1) =0 \Rightarrow \min\{3x,4-x^2\} = \begin{cases} 3x & 0\le x\le 1\\ 4-x^2& 1\le x\le 2\end{cases} \\ \Rightarrow \int_0^2 \min\{3x,4-x^2\}\,dx = \int_0^13x\,dx +\int_1^2 (4-x^2) \,dx = \left. \left[ {3\over 2} x^2\right] \right|_0^1+ \left. \left[ 4x-{1\over 3}x^3 \right] \right|_1^2 \\={3\over 2}+{5\over 3} =\bbox[red, 2pt]{19\over 6}$$

解答:$$A_3 =(1,4) \cup [4,5) =(1,5) \Rightarrow I(A_3) =\int_1^5 e^{-x}\,dx =e^{-1} -e^{-5} \\ I((1,4)) +I([4,5)) = \int_1^4 e^{-x} \,dx +\int_4^5 e^{-x}\,dx = (e^{-1}-e^{-4}) +(e^{-4}-e^{-5}) =e^{-1}-e^{-5} \\ \Rightarrow \bbox[red, 2pt]{I(A_3) = I((1,4)) +I([4,5)) }$$

解答:$$f'(0) =\lim_{h\to 0} {f(0+h)-f(0) \over h}  =\lim_{h\to 0} {h^2 \sin(1/h)-0 \over h} =\lim_{h\to 0}  \left( h\sin{1\over h} \right) \\ \Rightarrow \lim_{h\to 0} -|h| \le \lim_{h\to 0}  \left( h\sin{1\over h} \right) \le \lim_{h\to 0} |h| \Rightarrow 0 \le \lim_{h\to 0}  \left( h\sin{1\over h} \right) \le 0 \Rightarrow \lim_{h\to 0}  \left( h\sin{1\over h} \right) =0 \\ \Rightarrow f'(0)=0 \Rightarrow \bbox[red, 2pt]{f \text{ is differentiable at }x=0}$$

解答:$$\text{Let }P(n) \text{ be the statement: }\log(x^n) =n\log x, x\gt 0\\ \text{Proof by Induction: }\\\text{ for }n=1 \Rightarrow \log x^1=\log x \Rightarrow P(1) \text{ is true}\\ \text{Assume }P(k)\text{ is true ,} k\in \mathbb N. \text{ That is, }\log x^k =k\log x \\ n=k+1 \Rightarrow  \log x^{k+1} =\log x\cdot x^{k} =\log x+\log x^k =\log x+k\log x=(k+1)\log x \Rightarrow P(k+1) \text{ is true.}\\ \text{By the principle of mathematical induction, the statement }\log x^n=n\log x \text{ holds, }\forall n\in \mathbb N\quad \bbox[red, 2pt]{Q.E.D.}$$

解答:$$\text{Let }g(x)=f(x)-x \Rightarrow g(x) \text{ is continuous on }[a,b] \Rightarrow \cases{g(a)=f(a)-a \ge 0\\ g(b)=f(b)-b\le 0} \Rightarrow \cases{g(a)\ge 0\\ g(b)\le 0} \\ \textbf{Case I }g(a)=0 \Rightarrow f(a)-a=0 \Rightarrow f(a)=a, \text{ and we are done.}\\ \textbf{Case II }g(b)=0 \Rightarrow f(b)-b=0 \Rightarrow f(b)=b, \text{ and we are done.} \\ \textbf{Case III }\cases{g(a)\gt 0\\ g(b)\lt 0} \Rightarrow \text{ By the Intermediate Value Theorem, there exists }c\in (a,b )\\ \qquad \text{ such that }g(c)=0 \Rightarrow f(c)-c=0 \Rightarrow f(c)=c \\ \text{In all cases, there exists at least one point }c\in [a,b] \text{ such that }f(c)=c. \qquad \bbox[red, 2pt]{Q.E.D.}$$

解答:$$\textbf{(i) }  \cases{\displaystyle\lim_{x\to 0^-}F(x) = \lim_{x\to 0^-} {1-\epsilon\over 1+e^{-x}} = {1-\epsilon\over 2} \\[1ex] \displaystyle \lim_{x\to 0^+} F(x) = \lim_{x\to 0^+} \left( \epsilon+{1-\epsilon\over 1+e^{-x}} \right) = \epsilon+{1-\epsilon\over 2}} \\\quad \text{To be continuous, we need }\epsilon+{1-\epsilon\over 1+e^{-x}} ={1-\epsilon\over 1+e^{-x}} \Rightarrow \epsilon=0 \not \in (0,1) \Rightarrow \bbox[red, 2pt]{F(x) \text{ is not continuous}} \\\textbf{(ii) } \cases{\displaystyle \lim_{x\to 0^+} F(x) = \epsilon+{1-\epsilon \over 2} \\ F(0)=\epsilon+ \displaystyle {1-\epsilon \over 2}} \Rightarrow  \lim_{x\to 0^+} F(x) =F(0) \Rightarrow \bbox[red, 2pt]{F(x) \text{ is right-continuous}}$$

解答:$$f(x+h)-f(x) = {f(x+h)-f(x)\over h}\cdot h \Rightarrow \lim_{h\to 0} (f(x+h)-f(x) ) =\lim_{h\to 0} \left( {f(x+h)-f(x)\over h}\cdot h \right) \\=  \left( \lim_{h\to 0}{f(x+h)-f(x)\over h} \right)\cdot  \left( \lim_{h\to 0} h \right) =f'(x) \cdot 0=0 \Rightarrow \lim_{h\to 0} (f(x+h)-f(x) ) =0 \\ \Rightarrow \lim_{h\to 0} f(x+h)=f(x)  \Rightarrow f \text{ is continuous at }x. \quad \bbox[red, 2pt]{Q.E.D.}$$

解答:$$\textbf{Case I }x=0 \Rightarrow f_n(0) =\max\{0,1\} =1 \Rightarrow \lim_{n\to \infty} f_n(0)=1 \\ \textbf{Case II }x\ne 0 \Rightarrow f_n(x) =\max\{0, \text{negative number}\} =0 \Rightarrow \lim_{n\to \infty} f_n(x)=0 \text{ for all }x\ne 0 \\ \Rightarrow \lim_{n\to \infty } f_n(x) =f(x)= \begin{cases} 1& \text{ if }x=0\\ 0& \text{ if }x\ne 0\end{cases} \\ \lim_{x\to 0}f(x) \ne f(0) \Rightarrow f(x) \text{ is discontinuous at }x=0 \\ \Rightarrow \bbox[red, 2pt]{f_n(x) \text{ does not converge to a continuous function}}$$

解答:$$\text{Let }S=\{x_1,x_2,x_3,\dots \}.\text{ By the Completeness Axiom of the real numbers, every }\\ \text{nonempty set of real numbers that is bounded above has a least upper bound. That is,} \\ \text{we can let }L= \text{sup}(S) \\ \text{Pick an }\epsilon \gt  0 \Rightarrow L-\epsilon \text{ is not an upper bound  } \Rightarrow x_N \gt L-\epsilon \cdots(1) \\ \{x_n\}_0^\infty \text{ is increasing } \Rightarrow x_n\gt x_N, \text{ for }n\gt N \Rightarrow L\gt x_n\gt x_N \cdots(2) \\ \text{From (1) and (2), we have } L-\epsilon \lt x_n\lt L \Rightarrow |x_n-L|\lt \epsilon \text{ for all }n\gt N \\\Rightarrow \lim_{n\to \infty} x_n =L,  \text{that is, the limit exists. }\bbox[red,2pt]{Q.E.D.}$$

解答:$$\text{Let }U(x)= C+\int_a^x f(t)g(t)\,dt \ge C \gt 0, \text{ for all }x\in [a,b]  \Rightarrow U(a)=C\\ \Rightarrow U'=f(x)g(x) \le U(x)g(x) \quad (\because U(x) \ge f(x)) \Rightarrow g(x)\ge {U'(x)\over U(x)} \\ \Rightarrow \int_a^x g(t)\,dt \ge \int_a^x {U'(t) \over U(t)}\,dt = \ln U(x)-\ln U(a) =\ln U(x)-\ln C = \ln {U(x) \over C} \\ \Rightarrow {U(x) \over C} \le e^{\int_a^x g(t)dt} \Rightarrow U(x)\le C e^{\int_a^x g(t)dt}\quad \bbox[red, 2pt]{Q.E.D.}$$

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解題僅供參考，碩士班歷年試題及詳解