107年台綜大轉學考-微積分B詳解

臺灣綜合大學系統107學年度學士班轉學生聯合招生考試

科目名稱：微積分B

解答： $$\lim_{x\to 0}{\sin(4x^3)\over x} =\lim_{x\to 0}{(\sin(4x^3))'\over (x)'} =\lim_{x\to 0}{12x^2\cos(4x^3)\over 1} = \bbox[red, 2pt] 0$$

解答： $$依題意:{d\over dt}V= -2 m^3/sec，再由相似三角形\Rightarrow {2\over 1} ={r\over h} \Rightarrow r=2h \Rightarrow V={1\over 3}\pi r^2h ={4\over 3}\pi h^3 \\ \Rightarrow {d\over dt}V= 4\pi h^2 {dh\over dt} \Rightarrow -2=4\pi \cdot ({1\over 2})^2 {dh\over dt}(此時r=1 \Rightarrow h=1/2) \Rightarrow {dh\over dt} = \bbox[red, 2pt]{-{2\over \pi} m/sec}$$

解答： $$u=\sqrt x \Rightarrow du ={1\over 2\sqrt x}dx = {1\over 2u}\,dx \Rightarrow dx= 2u\,du \Rightarrow \int \sin(\sqrt x)\,dx = \int 2u\sin u\,du\\ =2(\sin u-u\cos u)+C = \bbox[red, 2pt]{2(\sin \sqrt x -\sqrt x\cos \sqrt x)+C}$$

解答： $$f(x,y,z)=x^2+ 2y^2 +xy+e^z -2 \Rightarrow \cases{f_x=2x+y\\ f_y=4y+x\\ f_z= e^z} \\\Rightarrow (f_x(P),f_y(P), f_z(P))= (2,1, 1) \Rightarrow 切平面: 2(x-1)+y+z=0 \Rightarrow \bbox[red, 2pt]{2x+y+z= 2}$$

解答： $$\int {3x^2 +1\over x^4+x^2}\,dx =\int {1 \over  x^2} +{2 \over x^2+1}\,dx =\bbox[red,2pt]{ -{1\over x}+ 2\tan^{-1}x +C}$$

解答： $${\partial \over \partial x}f(0,0) = \lim_{t \to 0}{f(t,0)-f(0,0)\over t}= \lim_{t \to 0}{2t-0\over t}= \bbox[red, 2pt]2$$

解答： $$f(x)= \int_0^{x^3-3x} e^{\cos(t^2+1)}\,dt  \Rightarrow f'(x)=e^{\cos((x^3-3x)^2+1)}\cdot (3x^2-3) =g(x)\cdot (3x^2-3)\\由於g(x)= e^{\cos((x^3-3x)^2+1)} \gt 0,\forall x\in \mathbb{R},因此 f'(x)=0 \Rightarrow 3x^2-3=0 \Rightarrow x=\pm 1\\ 又f''(x)= g'(x)(3x^2-3)+ g(x)(6x) \Rightarrow \cases{f''(1)= 6g(6)\gt 0\\ f''(-1)=-6g(-1)\lt 0} \Rightarrow \bbox[red, 2pt]{x=1}有相對極小值$$

解答： $$f(x)={x^3\over 1+x^2} = {x^3\over 1-(-x^2)} = x^3(1-x^2 +x^4 -x^6 +x^8-\cdots) =x^3-x^5+x^7 -x^9+x^{11}-\cdots\\ = \bbox[red, 2pt]{\sum_{k=1}^\infty (-1)^{k+1}x^{2k+1}}$$

解答： $$\int_0^1 \int_x^1 \cos(y^2+1)\,dydx = \int_0^1 \int_0^y \cos(y^2+1)\,dxdy =\int_0^1 y\cos(y^2+1)\,dy \\=\int_1^2 {1\over 2}\cos u\,du \;(u=y^2+1) = \left.\left[ {1\over 2}\sin u \right] \right|_1^2 = \bbox[red, 2pt]{{1\over 2}(\sin 2-\sin 1)}$$

解答： $$令\cases{f(x,y,z)=2x+3y+ 5z\\ g(x,y,z)=x^2+y^2+z^2-19},則\cases{f_x= \lambda g_x\\ f_y=\lambda g_y\\ f_z= \lambda g_z\\ g=0} \Rightarrow \cases{2=\lambda (2x)\cdots(1)\\ 3=\lambda (2y)\cdots(2)\\ 5=\lambda (2z)\cdots(3) \\ x^2+y^2+z^2=19 \cdots(4)}\\ \cases{(1)\div (2)\\ (2)\div (3)} \Rightarrow \cases{x=2y/3\\ z=5y/3} 代入(4) \Rightarrow ({4\over 9}+1 +{25\over 9})y^2=19 \Rightarrow y^2 = {9\over 2} \Rightarrow y=\pm {3\over \sqrt 2} \\ \Rightarrow \cases{x=\pm 2/\sqrt 2\\ z= \pm 5/\sqrt 2} \Rightarrow 最大值f(2/\sqrt 2,3/\sqrt 2,5/\sqrt 2) ={4+9+25\over \sqrt 2} = \bbox[red, 2pt]{19\sqrt 2}$$

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