114年海洋大學碩士班-微積分詳解

 國立臺灣海洋大學114學年度碩士班考試入學招生考試

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:

$$x^3-3x+3=x+3 \Rightarrow x^3-4x=0 \Rightarrow x(x-2)(x+2)=0 \Rightarrow x=-2,0,2\\ \Rightarrow \cases{f(-1)=6\\ g(-1)=2 \\f(1)=1\\ g(1)=4} \Rightarrow \cases{f(-1) \gt g(-1)\\ g(1) \gt f(1)} \Rightarrow \text{area =} \int_{-2}^0 \left(f(x)-g(x) \right)\,dx +\int_0^2 \left(g(x)-f(x) \right)\,dx \\=\int_{-2}^0 (x^3-4x)\,dx + \int_0^2 \left( -x^3+4x\right) \,dx =4+4 = \bbox[red, 2pt]8$$

解答:$$f(x)=xe^{-x^2} \Rightarrow f(-x)=-f(x) \Rightarrow f(x) \text{ is odd} \Rightarrow \int_{-\infty}^\infty xe^{-x^2} \,dx = \bbox[red, 2pt]0$$

解答:$$f(x,y)=e^{xy^2} \Rightarrow \cases{f_x=y^2 e^{xy^2} \\f_y=2xy e^{xy^2}} \Rightarrow \cases{f_{xx}=y^4 e^{xy^2} \\ f_{xy}= 2ye^{xy^2}+ 2xy^3 e^{xy^2} \\ f_{yy} =2xe^{xy^2}+ 4x^2y^2 e^{xy^2}} \Rightarrow  \bbox[red, 2pt]{\cases{f_{xx}=y^4 e^{xy^2} \\ f_{xy}= 2y(1+xy^2)e^{xy^2} \\ f_{yy} =2xe^{xy^2}(1+xy^2) e^{xy^2}}}$$

解答:$${x+y\over 2} \ge \sqrt{xy} \Rightarrow xy\le 10^2=100 \Rightarrow f(x,y)= x^2+xy+y^2 =(x+y)^2-xy \ge 20^2-100= \bbox[red, 2pt]{300}$$

解答:$${1\over (5-2)\cdot (4-1)} \iint f(x,y)\,dA ={1\over 9} \int_1^4 \int_2^5 3x^2y\,dx\,dy = {117\over 9} \int_1^4y\, dy={117\over 9}\cdot {15\over 2} = \bbox[red, 2pt]{ 585\over 6}$$

解答:$$\text{We need to find a number }\delta \gt 0 \text{ such that if }0\lt |x-10| \lt \delta, \text{ then }|\sqrt{19-x}-3|\lt 1 \\ \Rightarrow 2\lt \sqrt{19-x}\lt 4 \Rightarrow 3\lt x\lt 15 \Rightarrow |x-10|=5,7 \Rightarrow \text{choose } \bbox[red, 2pt]{\delta=5}$$

解答:$${x\over a}+{y\over b}=1 \Rightarrow L:bx+ay=ab \Rightarrow y=-{b\over a}x+b \Rightarrow \text{ the slope of }L:-{b\over a} \\ \Rightarrow L'\bot L \text{ and }L \text{ passes through origin }(0,0) \Rightarrow L': y={a\over b}x \Rightarrow P=L\cap L' \\ \Rightarrow  -{b\over a}x+b={a\over b}x \Rightarrow x={ab^2\over a^2+b^2} \Rightarrow y={a^2b\over a^2+b^2} \Rightarrow \text{ the closet point:}\bbox[red, 2pt]{ \left({ab^2\over a^2+b^2}, {a^2b\over a^2+b^2} \right)}$$

解答:$${\partial f\over \partial x}(0,0) =\lim_{h\to 0}{f(h,0)-f(0,0)\over h}  =\lim_{h\to 0}{\sin(h^3)/h^2-0\over h} = \lim_{h\to 0}{ \sin(h^3) \over h^3}  = \lim_{h\to 0}{ 3h^2\cos(h^3) \over 3h^2} =1 \\{\partial f\over \partial y}(0,0) =\lim_{h\to 0}{f(0,h)-f(0,0)\over h}  =\lim_{h\to 0}{\sin(h^4)/h^2-0\over h} = \lim_{h\to 0}{\sin(h^4)\over h^3} = \lim_{h\to 0}{4h^3 \cos(h^4) \over 3h^2} =0 \\ \Rightarrow \bbox[red, 2pt]{{\partial f\over \partial x}(0,0) =1 , {\partial f\over \partial y}(0,0) =0}$$

解答:$$\textbf{(a) } \cases{\lim_{x\to 1^+} \displaystyle {\sqrt{2x}(x-1)\over |x-1|} =\lim_{x\to 1^+}{\sqrt{2x}(x-1)\over x-1} = \lim_{x\to 1^+} \sqrt{2x}  = \sqrt 2 \\   \lim_{x\to 1^-}\displaystyle {\sqrt{2x}(x-1)\over |x-1|} =\lim_{x\to 1^-}{\sqrt{2x}(x-1)\over -(x-1)} =\lim_{x\to 1^-} -\sqrt{2x} =-\sqrt 2} \\\qquad \Rightarrow \lim_{x\to 1^+}{\sqrt{2x}(x-1)\over |x-1|} \ne \lim_{x\to 1^-}  {\sqrt{2x}(x-1)\over |x-1|} \Rightarrow \lim_{x\to 1 }{\sqrt{2x}(x-1)\over |x-1|} \bbox[red, 2pt]{\text{ does not exist}} \\\textbf{(b) }  \lim_{x\to 0^+} {(\ln x)^2 \over \ln(\sin x)} =  \lim_{x\to 0^+} {2(\ln x)/x \over \cos x/\sin x} = \lim_{x\to 0^+} {2  (\ln x) \over x\cot x} ={-\infty \over 1} =\bbox[red, 2pt]{-\infty}$$

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解題僅供參考，其他碩士班試題及詳解