113年海洋大學碩士班-微積分詳解

 國立臺灣海洋大學113學年度碩士班考試入學招生考試試題

考試科目：微積分
學系組名稱：運輸科學系碩士班不分組

解答: $$f'(x)=\lim_{h\to 0}{f(x+h)-f(x)\over h} = \lim_{h\to 0}{2(x+h)^2-5(x+h)-2x^2+5x\over h} \\ = \lim_{h\to 0}{4hx+2h^2-5h\over h}  = \lim_{h\to 0}(4x+2h-5)= \bbox[red, 2pt]{4x-5}$$

解答: $$f(x)={x^3-2\over |x|^3+1} \Rightarrow \cases{\lim_{x\to \infty} f(x)=1 \\ \lim_{x\to -\infty} f(x)=-1} \Rightarrow \text{ horizontal asymptotes: }\bbox[red, 2pt]{y=1,y=-1}$$

解答: $$y=\tan^2 (\sin^3 t) \Rightarrow {dy\over dt}=2\tan(\sin^3t) \sec^2(\sin^3 t) 3\sin^2 t\cos t = \bbox[red, 2pt]{6\tan(\sin^3t) \sec^2(\sin^3 t) \sin^2 t\cos t}$$

解答: $$f(x,y)=x\cos y+ye^x \Rightarrow \cases{\displaystyle\frac{\partial f}{\partial x} =\cos y+ye^x \\\displaystyle\frac{\partial f}{\partial y} =-x \sin y+e^x } \Rightarrow \bbox[red, 2pt]{ \cases{\displaystyle\frac{\partial^2 f}{\partial x^2} = ye^x \\\displaystyle\frac{\partial^2 f}{\partial x \partial y} =-\sin y+e^x \\\displaystyle\frac{\partial^2 f}{\partial y^2} =-x \cos y }}$$

解答: $$\int_0^2 {dx \over \sqrt{|x-1|}} =\int_0^1 {dx \over \sqrt{1-x}} +\int_1^2 {dx \over \sqrt{x-1}} = \left. \left[ -2\sqrt{1-x} \right] \right|_0^1 + \left. \left[ 2\sqrt{x-1} \right] \right|_1^2 \\=2+2= \bbox[red, 2pt]4$$

解答: $$\int_0^{\pi/4} \sqrt{1+\cos(4x)} \,dx =\int_0^{\pi/4} \sqrt{1+2\cos^2 (2x)-1} \,dx = \sqrt 2 \int_0^{\pi/4} \cos(2x) \,dx \\ =\sqrt 2 \left. \left[ {1\over 2} \sin(2x)\right] \right|_0^{\pi/4}  =\bbox[red, 2pt]{\sqrt 2\over 2}$$

解答: $$x=3\tan u \Rightarrow dx=3\sec^2 u\,du \Rightarrow I=\int{1\over x^2\sqrt{x^2+9}} \,dx = \int{3 \sec^2 u\over 9\tan^2 u \cdot 3 \sec u}\,du \\= \int{\sec u\over 9\tan^2 u} \,du = \int {\cos u\over 9\sin^2 u}\,du =-{1\over 9 \sin u}+c = \bbox[red, 2pt]{-{\sqrt{x^2+9} \over 9x}+c}$$

解答: $$\iint_R e^{x+2y}\,dA= \int_1^{\ln 2} \int_0^{\ln 3} e^x \cdot e^{2y}\,dx\, dy = \int_1^{\ln 2} 2e^{2y}\,dy= \bbox[red, 2pt]{ 4-e^2 }$$

解答: $$f(x)=2x-3x^{2/3}\\ \textbf{(a)}\; \text{domain of }f: \bbox[red, 2pt]{(-\infty, \infty)} \\\quad \lim_{x\to \infty}f(x)=\infty, \lim_{x\to -\infty}f(x)=-\infty \Rightarrow \bbox[red, 2pt]{\text{No asymptotes}} \\\textbf{(b)}\;f'(x)=2-2x^{-1/3}=2(1-{1\over \sqrt[3]x})=0 \Rightarrow x=1 \Rightarrow \cases{f(0)=0\\ f(1)=-1}\\\quad \Rightarrow \bbox[red, 2pt]{\text{critical points: }(0,0),(1,-1)}\\\textbf{(c)}\; \begin{cases} f'(x)\gt 0& x\lt 0\\ f'(x)\lt 0 & 0\lt x\le 1\\ f'(x) \gt 0& x\ge 1\end{cases} \Rightarrow \bbox[red, 2pt]{\cases{f \text{ is increasing, if }x\lt 0, x\ge 1\\ \text{f is decreasing, if }0\lt x\le 1}} \\\textbf{(d)}\;\text{from (c), we have} \bbox[red, 2pt]{\text{local minimum: -1; local maximum: 0}} \\\textbf{(e)}\;f''(x)={2\over 3x^{4/3}} \gt 0, \forall x\ne 0 \Rightarrow \bbox[red, 2pt]{f(x) \text{ is concave up }, x \in (-\infty, 0) \cup (0, \infty)} \\\textbf{(f)}:$$

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