114中央企管碩士班-微積分詳解

國立中央大學114學年度碩士班考試入學

系所: 企業管理學系 碩士班 工商管理丙組(一般生)
科目: 微積分

一、填充題:共8題,每題8分、總計64分。請在答案卷上列出題號並依序作答。

解答:$$\lim_{h \to 0} {3h\over \sqrt{2+h}-\sqrt 2} =\lim_{h \to 0} {3\over {1 \over 2\sqrt{2+h}}} = \lim_{h\to 0} 6\sqrt{2+h} = \bbox[red, 2pt]{6\sqrt 2}$$

解答:$$x^2-4 \gt 0 \Rightarrow (x+2)(x-2)\gt 0 \Rightarrow x\gt 2, x\lt -2 \\\Rightarrow \text{ domain of }f(x): \bbox[red, 2pt]{\{x \mid x\gt 2 \text{ or }x\lt -2, x\in \mathbb R\}}$$

解答:$$f(x)=e^x \sin(x) \Rightarrow f'(x)=e^x \sin (x)+ e^x \cos(x) \Rightarrow \cases{f(\pi)=0\\ f'(\pi) =-e^{\pi}} \\ \Rightarrow \text{ tangent line: } y=-e^\pi (x-\pi) \Rightarrow \bbox[red, 2pt]{y+e^\pi x=\pi e^\pi}$$

解答:$$g(x)=x^3 \Rightarrow \int_a^b g(f(x))f'(x)\,dx =\int_a^b (f(x))^3f'(x)\,dx = \left. \left[ {1\over 4}(f(x))^4\right] \right|_a^b ={1\over 4} \left(f(b)^4-f(a)^4 \right) \\={1\over 4}(3^4-2^4) = \bbox[red, 2pt]{65\over 4}$$

解答:$$\int_a^b f(x)\,dx \approx {b-a\over 2n} \left(f(x_0)+2f(x_1)+ \cdots+2f(x_{n-1}+ f(x_n)) \right) \\ \Rightarrow \int_0^2 x^2\,dx \approx{2-0\over 2\cdot 4}(f(0)+ 2f(0.5) +2f(1)+ 2f(1.5) +f(2)) \\={2\over 8}(0+2\cdot 0.25+ 2 \cdot 1+2\cdot 2.25+ 4)  = \bbox[red, 2pt]{11\over 4}$$

解答:$$\int_0^{10} f(x)^2 \pi \,dx  =\int_0^1 4x \pi \,dx + \int_1^{10} 4\pi \,dx =2\pi +36\pi= \bbox[red, 2pt]{38\pi}$$

解答:$$\int_0^2 \int_y^2 e^{x^2}\,dx dy = \int_0^2 \int_0^x e^{x^2}\,dy dx =\int_0^2 xe^{x^2 }\,dx = \left. \left[ {1\over 2}e^{x^2}\right] \right|_0^2 = \bbox[red, 2pt]{{1\over 2} (e^4-1)}$$

解答:$$f(x,y)=x^2-2xy+2y^2-3x+3y+1 \Rightarrow \cases{f_x=2x-2y-3\\ f_y=-2x+4y+3} \Rightarrow \cases{f_{xx} =2\\ f_{yy}=4\\ f_{xy} =-2} \\ \Rightarrow D(x,y)= f_{xx}f_{yy}-f_{xy}^2=4 \gt 0 \\ \cases{f_x=0\\ f_y=0} \Rightarrow \cases{2x-2y=3\\ -2x+4y=-3} \Rightarrow \cases{x=3/2\\ y=0} \Rightarrow D(3/2,0) =4\gt 0 \text{ and }f_{xx}(3/2,0)=2 \gt 0 \\ \Rightarrow \bbox[red, 2pt]{f(3/2,0) \text{ is a relative minimum}}$$

二、計算、證明題:共3題,每題12分、總計36分。請將題號標明清楚。

解答:$$f(x)=(1-x)^{-1} \Rightarrow f'(x)=(1-x)^{-2} \Rightarrow f''(x)=2(1-x)^{-3} \Rightarrow \cdots \Rightarrow f^{[n]}(x) =n! (1-x)^{-(n+1)} \\ \Rightarrow f^{[n]}(2)=n!(-1)^{n+1} \Rightarrow \sum_{n=0}^\infty {f^{[n]}(2) \over n!}(x-2)^n = \bbox[red, 2pt]{\sum_{n=0}^\infty (-1)^{n+1}(x-2)^n} \\ \Rightarrow  \lim_{n\to \infty} \left| {a_{n+1} \over a_n}\right| =\lim_{n\to \infty} \left| {(-1)^{n+2}(x-2)^{n+1} \over (-1)^{n+1} (x-2)^n}\right| =\lim_{n\to \infty} \left| x-2\right|  \\ \Rightarrow |x-2|\lt 1 \Rightarrow 1\lt x\lt 3 \Rightarrow \cases{x=1 \Rightarrow \sum (-1)^{n+1}(-1)^n = \sum(-1) \text{ divergent} \\ x=3 \Rightarrow \sum (-1)^{n+1} \text{ divergent}} \\ \Rightarrow \text{interval of convergence: }\bbox[red, 2pt]{(1,3)}$$

解答:

$$f(x)={x\over 1+x^2} \Rightarrow f'(x)={1-x^2\over (1+x^2)^2} \Rightarrow f''(x)={2x^3-6x\over (1+x^2)^3} \\ f'(x)=0 \Rightarrow x^2=1 \Rightarrow x=\pm 1 \Rightarrow \cases{f''(1)=-1/2 \lt 0\\\ f''(-1) 1/2 \gt 0} \Rightarrow \cases{f(1) \text{ maximum} \\f(-1) \text{ minimum}} \\ \Rightarrow \text{ area =} 2\int_0^1 {x\over 1+x^2}\,dx = 2 \left. \left[ {1\over 2}\ln(1+x^2) \right] \right|_0^1 = \bbox[red, 2pt]{\ln 2}$$

解答:$$\int_3^\infty {1\over x(\ln x)^2} \,dx = \int_{\ln 3}^\infty {1\over u^2}\,du = \left. \left[ -{1\over u} \right] \right|_{\ln 3}^\infty =\ln 3\\ \text{By intergal test, the series } \sum_{n=3}^\infty {1\over n(\ln n)^3} \bbox[red, 2pt]{\text{ convergent}}$$

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解題僅供參考，其他碩士班試題及詳解