國立清華大學 114學年度碩士班考試入學試題
系所班組別:科技管理研究所乙組(0551)
考試科目(代碼):微積分(5101)
解答:$$\textbf{(a) }\lim_{n\to \infty} {1\over n}\left( 3^{1/n} +3^{2/n}+ \cdots+3^{n/n} \right) =\lim_{n\to \infty} \sum_{k=1}^n \left( {1\over n}3^{k/n} \right) =\int_0^1 3^x\,dx = \left. \left[ {3^x\over \ln 3} \right] \right|_0^1 = \bbox[red, 2pt]{2\over \ln 3} \\\textbf{(b) } \cases{u=\ln x\\ dv=dx/\sqrt x} \Rightarrow \cases{du=dx/x\\ v=2\sqrt x} \Rightarrow \int_1^9 {\ln x\over \sqrt x}\,dx = \left. \left[ 2\sqrt x \ln x \right] \right|_1^9-\int_1^9 {2\sqrt x\over x}\,dx \\\qquad = 6\ln 9-\int_1^9 {2\over \sqrt x}\,dx =6\ln 9- \left. \left[ 4\sqrt x \right] \right|_1^9 =6\ln 9-4\sqrt 9+4 = \bbox[red, 2pt]{12\ln 3-8} \\\textbf{(c) } \int_0^1 \int_0^{1-x^2} {xe^{2y} \over 1-y}\, dy dx = \int_0^1 \int_0^{\sqrt{1-y}} {xe^{2y} \over 1-y}\,dx dy =\int_0^1 {1\over 2}e^{2y}\,dy = \bbox[red, 2pt]{e^2-1\over 4} \\\textbf{(d) } \text{Let }\cases{P(x,y) \text{ on the hyperbola} \\Q(5,0)} \Rightarrow \overline{PQ}^2 =f(x)= (x-5)^2+y^2= (x-5)^2+(x^2-4) \\ \qquad =2x^2-10x+21 \Rightarrow f'(x)=4x-10=0 \Rightarrow x={5 \over 2} \Rightarrow y^2=x^2-4={9\over 4} \Rightarrow y=\pm {3\over 2} \\\qquad \Rightarrow \text{the closet points: } \bbox[red, 2pt]{({5\over 2}, \pm{3\over 2})} \\\textbf{(e) }{dy\over dx}+3x^2y=6x^2 \Rightarrow e^{x^3} {dy\over dx}+ 3x^2 e^{x^3}y= 6x^2 e^{x^3} \Rightarrow \left( e^{x^3}y \right)'=6x^2e^{x^3} \\ \qquad \Rightarrow e^{x^3 }y=\int 6x^2 e^{x^3}\,dx = 2e^{x^3}+c_1 \Rightarrow y=2+c_1e^{-x^3} \Rightarrow y(0)=2+c_1=0 \Rightarrow c_1=-2 \\\qquad \Rightarrow \bbox[red, 2pt]{y=2-2e^{-x^3}} \\\textbf{(f) } xy+xz^3=2yz \Rightarrow y+z^3+3xz^2z_x =2yz_z \Rightarrow z_x= {y+z^3\over 2y-3xz^2} \Rightarrow z_x(1,1,1)={2\over -1} = \bbox[red, 2pt]{-2}$$
解答:$$\text{Suppose }f \text{ has more than one fixed point. Let's say }a \text{ and }b \text{ are two fixed points, where }a\lt b.\\ \text{Then we have }\cases{f(a)=a\\f(b)=b}. \text{ By the Mean Value Theorem, there exists }c \in (a,b) \text{ such that} \\ f'(c)={f(b)-f(a) \over b-a} ={b-a\over b-a}=1 \\ \text{It contradicts that }f'(x)\ne 1 \text{ for all }x\in \mathbb R. \text{ We can conclude that }f\text{ has at most one fixed point.}$$
解答:$$\lim_{x\to x_0} {f(x) \over g(x)} =\lim_{x\to x_0} {f(x) -f(x_0)\over g(x)-g(x_0)} =\lim_{x\to x_0} {(f(x) -f(x_0))/(x-x_0)\over (g(x)-g(x_0)) /(x-x_0)} = {\displaystyle \lim_{x\to x_0}{f(x)-f(x_0)\over x-x_0} \over \displaystyle \lim_{x\to x_0} {g(x)-g(x_0) \over x-x_0}} \\= \lim_{x\to x_0}{f'(x_0) \over g'(x_0)} \Rightarrow \lim_{x\to x_0} {f(x) \over g(x)} =\lim_{x\to x_0} {f'(x_0) \over g'(x_0)} \qquad \bbox[red, 2pt]{Q.E.D.}$$
解答:$$-1\le \cos(1/x) \le 1 \Rightarrow -|\sin x|\le \sin x\cos(1/x) \le |\sin x| \\ \lim_{x\to 0} |\sin x|=\lim_{x\to 0} -|\sin x|=0 \Rightarrow \lim_{x\to 0} \sin x\cos(1/x) =0 \Rightarrow \bbox[red, 2pt]{f \text{ is continuous at }x=0} \\ f'(0) =\lim_{h\to 0} {f(0+h)-f(0) \over h}=\lim_{h\to 0} {\sin h \cos(1/h) \over h} =\lim_{h\to 0} \left( {\sin h\over h} \right)\cos(1/h) \\ \text{Since }\lim_{h\to 0} \left( {\sin h\over h} \right) =1\text{ and }\lim_{h\to 0} \cos(1/h) \text{ does not exist} \Rightarrow f'(0) \text{ does not exist} \\ \Rightarrow \bbox[red, 2pt]{f \text{ is not differentiable at }x=0}$$
解答:$$a_n ={(1-x)^n \over n(\ln n)^2} \Rightarrow \lim_{n \to \infty} \left| {a_{n+1} \over a_n}\right| =\lim_{n \to \infty} \left| {(1-x)^{n+1} \over (n+1)(\ln n+1)^2} \cdot {n(\ln n)^2 \over (1-x)^n }\right| \\ =|1-x|\lim_{n \to \infty} \left| {n \over n+1} \cdot \left( {\ln n \over \ln(n+1) } \right)^2 \right| =|1-x|\lt 1 \Rightarrow 0\lt x\lt 2 \\ x=0 \Rightarrow a_n={1\over n(\ln n)^2 } \Rightarrow \int_2^\infty {1\over t(\ln t)^2}\,dt = \left. \left[ -{1\over \ln t} \right] \right|_2^\infty ={1\over \ln 2 } \Rightarrow \text{ the series converges at }x=0\\ x=2 \Rightarrow a_n= {(-1)^n\over n(\ln n)^2} \text{ decreases monotonically to zero} \Rightarrow \text{ the series converges at }x=2 \\ \text{The series converges for all } \bbox[red, 2pt]{0\le x\le 2}$$
解答:$$\textbf{Find Critical Points in the interior:} \\ \qquad \cases{f_x=y=0\\ f_y=x=0\\ f_z=2z=0} \Rightarrow \text{ critical point }(0,0,0) \Rightarrow f(0,0,0)=0 \\\textbf{Find Extrema on the Boundary:} \\ \qquad \cases{f(x,y,z)=xy+ z^2\\ g(x,y,z) =x^2+y^2+z^2-1} \Rightarrow \cases{y=\lambda(2x) \\ x=\lambda(2y)\\ 2z=\lambda(2z)} \\ \qquad \textbf{Case I }z=0 \Rightarrow x^2+y^2=1 \Rightarrow \cases{x=\cos \theta\\y=\sin \theta} \Rightarrow f=\cos \theta\sin \theta={1\over 2} \sin 2\theta \\\qquad \quad \Rightarrow \text{extrema: }\pm {1\over 2} \\ \qquad \textbf{Case II }\lambda=1 \Rightarrow \cases{y=2x\\ x=2y} \Rightarrow \cases{x=0\\ y=0} \Rightarrow z=\pm 1 \Rightarrow f(0,0,\pm 1)=1 \\ \text{Finally, we have } \bbox[red, 2pt]{\cases{ \max(f)=1\\ \min(f)=-1/2}}$$
解答:$$\textbf{(a) }A(t) = \int_0^t e^{-x} \,dt =1-e^{-t} \Rightarrow \lim_{t\to \infty} A(t)= \bbox[red, 2pt]1\\\textbf{(b) } V(t)= \int_0^t 2\pi x\cdot e^{-x}\,dx = 2\pi(1-e^{-t}-te^{-t}) \Rightarrow \lim_{t\to \infty} {V(t) \over A(t)} =\lim_{t\to \infty} {2\pi(1-e^{-t}-te^{-t})\over 1-e^{-t}} =\bbox[red, 2pt]{2\pi} \\ \textbf{(c) } \lim_{t\to 0^+} {V(t) \over A(t)} =\lim_{t\to 0^+} {2\pi(1-e^{-t}-te^{-t})\over 1-e^{-t}} =\lim_{t\to 0^+} {2\pi te^{-t} \over e^{-t}} =\bbox[red, 2pt]0$$
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解題僅供參考，碩士班歷年試題及詳解
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