115年成功大學企管碩士班-微積分詳解

國立成功大學115學年度碩士班招生考試試題

系所:企業管理系 
科目:微積分

Part A: Multi-Select/Multiple-Choice Question. Select all that apply.

解答:$$(B)\times: f(x)=g(x)=x-3 \Rightarrow \lim_{x\to 3}f(x)= \lim_{x\to 3}g(x)=0 \Rightarrow \lim_{x\to 3}{f(x)\over g(x)}= \lim_{x\to 3}{x-3\over x-3}= \lim_{x\to 3} 1=1 \text{ exists} \\ (C) \times: \cases{f(x)={1\over x-a} \\ g(x)=-{1\over x-a}} \Rightarrow \cases{\lim_{x\to a}f(x) \text{ does not exist} \\\lim_{x\to a}g(x) \text{ does not exist} } \Rightarrow \lim_{x\to a}[f(x)+g(x)] =\lim_{x\to a} 0=0\text{ exists} \\ \Rightarrow \text{True statements: }\bbox[red, 2pt]{(AD)}$$

解答:$$(A)\times: f(x)=(x-3)^4 \Rightarrow f''(3)=0 \Rightarrow  (3, 0)\text{ is a local minimum, not an inflection point.} \\(B)\times: f'(c)={f(6)-f(3) \over 6-3}=1 \Rightarrow \text{the Mean Value Theorem guarantees there is at least} \\\qquad\text{ one point where }f'(c)=1 \not \gt 1 \\(D)\times: f''(x)\gt 0 \Rightarrow \text{ strictly concave up everywhere} \Rightarrow \text{ the graph lies completely above} \\\qquad \text{ any of tangent lines} \Rightarrow f(x)\ge f(0)+f'(0)x \ge 0 \text{ as }x\to -\infty \Rightarrow f(x)\ge 0 \not \lt 0 \\ \Rightarrow \text{True statements: }\bbox[red, 2pt]{(C)}$$

解答:$$u=3x \Rightarrow du=3dx \Rightarrow \int_{b/3}^{a/3} g(3x)\,dx= \int_b^a g(u){1\over 3}du = {1\over 3}\int_b^a g(x)\,dx \Rightarrow \bbox[red, 2pt]{(C)}$$

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

Part B: No partial credit. Showing your work is not necessary, only the answer will be graded.

解答:$$x^2+2xy+4y^2=12 \Rightarrow 2x+2y+2xy'+8yy'=0 \Rightarrow y'= {-x-y\over x+4y} \\ \Rightarrow y'(2,1)={-2-1\over 2+4} =-{1\over 2} =a \Rightarrow y=ax+b \text{ passes (2,1)} \Rightarrow 1=2a+b \Rightarrow b=1-2a=2 \\ \Rightarrow (a,b)= \bbox[red, 2pt]{\left( -{1\over 2},2 \right)}$$

解答:$$\text{Let }\cases{x= \text{distance of the first car from the intersection } \\y= \text{distance of the second car from the intersection } \\ z=\text{distance between the two casrs}} \Rightarrow \cases{{dx\over dt}=40 \text{ mph} \\{dy\over dt}=30 \text{ mph}} \\ t={120\over 40}=3 \text{ hours} \Rightarrow y=30\times 3=90 \Rightarrow z^2=120^2+90^2=22500 \Rightarrow z=150\\ x^2+y^2=z^2 \Rightarrow 2x{dx\over dt}+2y{dy\over dt}=2z{dz\over dt} \Rightarrow 120\cdot 40+90\cdot 30=150\cdot {dz\over dt} \Rightarrow {dz\over dt}=\bbox[red, 2pt]{50 \text{ mph}}$$

解答:$$P_3(x)=f(0)+f'(0)x+ {f''(0)\over 2}x^2 +{f'''(0) \over 6}x^3 = \bbox[red, 2pt]{1+6x-2x^2+5x^3}$$

解答:$$\textbf{(A) }\cases{u=\ln (2x) \\ dv=dx} \Rightarrow \cases{du=dx/x \\v=x} \Rightarrow \int \ln(2x)\,dx =x\ln(2x)-\int 1\,dx= \bbox[red, 2pt]{x\ln(2x)-x+C} \\\textbf{(B) } u=x^2 \Rightarrow du=2xdx \Rightarrow \int x^3e^{x^2}\,dx = \int ue^u\cdot {1\over 2}du = {1\over 2} \left( ue^u-e^u \right)+C \\ \qquad = \bbox[red, 2pt]{{1\over 2}(x^2e^{x^2}-e^{x^2})+C} \\\textbf{(C) } {1\over x-1} \text{ is undefined at }x=1 \in[0,2] \Rightarrow \bbox[red, 2pt]{DNE} \\\textbf{(D) }u=\ln x \Rightarrow du={dx\over x} \Rightarrow \int_e^\infty {1\over x(\ln x)^2}\,dx =\int_1^\infty {1\over u^2}\,du = \left. \left[ -{1\over u} \right] \right|_1^\infty = \bbox[red, 2pt]1$$

Part C: Free-response questions. Show all your work for full credit.

解答:$$f(x)=x^3 \Rightarrow f'(x)=3x^2 \Rightarrow f(x) \approx L(x)=f(a)+f'(a)(x-a) \Rightarrow L(5.2)=f(5)+f'(5)\cdot(0.2)\\=125+75\cdot 0.2= \bbox[red, 2pt]{140}$$

解答:$$\int_0^2 \int_0^{4-x^2} {xe^{2y} \over 4-y}\,dydx = \int_0^4 \int_0^{\sqrt{4-y}}  {xe^{2y} \over 4-y}\,dxdy ={1\over 2} \int_0^4 e^{2y}\,dy = \bbox[red, 2pt]{{1\over 4}(e^8-1)}$$

解答:$$f(x,y)=x^3-3x-y^2+4y \Rightarrow \cases{f_x= 3x^2-3\\ f_y=-2y+4} \Rightarrow \cases{f_{xx}=6x\\ f_{xy}=0\\ f_{yy} =-2} \Rightarrow D(x,y)=f_{xx}f_{yy}-f_{xy}^2=-12x \\ \cases{f_x=0\\ f_y=0} \Rightarrow \cases{x=\pm 1\\ y=2} \Rightarrow \cases{D(1,2)=-12 \lt 0\\ D(-1,2)=12 \gt 0} \Rightarrow f_{xx}(-1,2)=-6\lt 0 \Rightarrow \text{relative max =}f(-1,2)=6 \\ \Rightarrow \bbox[red, 2pt]{\cases{\text{relative max: 6}\\ \text{relative min: none}}}$$

解答:$${dy\over dx}=x^3-x^3y=x^3(1-y) \Rightarrow \int {1\over 1-y}\,dx =\int x^3\,dx \Rightarrow -\ln|1-y|={1\over 4}x^4+c_1 \\ \Rightarrow 1-y=c_2e^{-x^4/4} \Rightarrow y=1-c_2e^{-x^4/4} \Rightarrow y(0)=1-c_2=3 \Rightarrow c_2=-2 \Rightarrow \bbox[red, 2pt]{y=1+2e^{-x^4/4}}$$

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解題僅供參考，碩士班歷年試題及詳解