112年政大轉學考-微積分(一)詳解

國立政治大學112學年度轉學生招生考試

科目:微積分(一)

系所別:應用數學系二年級

解答: $$\textbf{(a) }\lim_{x \to -\infty} {\sqrt{4x^2+x+7} \over 5x-1} =\lim_{x \to -\infty} {|2x|\sqrt{1+1/4x+7/4x^2} \over 5x-1} = \bbox[red, 2pt]{-{2\over 5}} \\\textbf{(b) }{x\over x+1} \le {x\over x-\sin x}\le {x\over x-1}, x\gt 0 \Rightarrow \lim_{x \to \infty} {x\over x+1}\le \lim_{x \to \infty} {x\over x-\sin x} \le \lim_{x \to \infty} {x\over x-1} \\\qquad \Rightarrow 1\le \lim_{x \to \infty} {x\over x-\sin x} \le 1 \Rightarrow \lim_{x \to \infty} {x\over x-\sin x}=\bbox[red, 2pt]1 \\\textbf{(c) } f(x,y) ={3y^2-x^2 \over x^2+y^2} \Rightarrow f(x,kx) ={(3k^2-1)x^2 \over (k^2+1)x^2} = {3k^2-1\over k^2+1} \text{ is dependent on }k \\\qquad \Rightarrow \lim_{(x,y)\to (0,0)} f(x,y) \bbox[red, 2pt]{\text{ the limit does not exist}}$$

解答: $$(\cos x)^{\sin y} = e^{\sin y \ln(\cos x)} =1 \Rightarrow \left(y'\cos y \ln(\cos x)- {\sin x\sin y\over \cos x} \right) e^{\sin y \ln(\cos x)}=0 \\ \Rightarrow y'\cos y \ln(\cos x)- {\sin x\sin y\over \cos x}=0 \Rightarrow y' \cos y\ln (\cos x)= \tan x \sin y \Rightarrow y'={\tan x \sin y\over \cos y \ln (\cos x)} \\ \Rightarrow y'= \bbox[red, 2pt]{{dy\over dx} ={\tan x\tan y \over \ln(\cos x)}}$$

解答: $$f(x)=x^3 \arctan(x) =\sum_{n=0}^\infty (-1)^n\cdot {1\over 2n +1}x^{2n+4} =x^4-{1\over 3}x^6+\cdots+ {1\over 61}x^{64}+ \cdots\\ \Rightarrow f^{[64]}(0) =\bbox[red, 2pt]{64!\over 61}$$

解答: $$(\rho,\theta,\phi) =(2,-{\pi\over 4}, {\pi\over 4}) \Rightarrow \cases{x =\rho\sin \phi \cos \theta = 2(\sqrt 2/2)(\sqrt 2/2) =1\\ y=\rho \sin \phi \sin \theta= 2(\sqrt 2/2)(-\sqrt 2/2)=-1\\ z=\rho \cos \phi =2\cdot (\sqrt 2/2) =\sqrt 2} \\  \Rightarrow \frac{\partial }{\partial \phi}F =\frac{\partial F}{\partial x} \cdot \frac{\partial x}{\partial \phi} +\frac{\partial F}{\partial y} \cdot \frac{\partial y}{\partial \phi} +\frac{\partial F}{\partial z} \cdot \frac{\partial z}{\partial \phi}  \\\qquad =\frac{\partial F}{\partial x} \cdot \rho\cos \phi \cos \theta +\frac{\partial F}{\partial y} \cdot \rho \cos \phi \sin \theta +\frac{\partial F}{\partial z} \cdot (-\rho \sin \phi) \\ \Rightarrow  \frac{\partial }{\partial \phi}F(1,-1,\sqrt 2) =1\cdot 2\cdot {\sqrt 2\over 2}\cdot {\sqrt 2\over 2} +2\cdot 2\cdot {\sqrt 2\over 2}\cdot {-\sqrt 2\over 2} +(-2)\cdot (-2)\cdot {\sqrt 2\over 2} =1-2+2\sqrt 2\\ = \bbox[red, 2pt]{-1+2\sqrt 2}$$

解答: $$\cases{T(x,y,z)=xz+yz\\ g(x,y,z)=x^2+y^2+z^2-1} \Rightarrow \cases{T_x =\lambda g_x\\ T_y= \lambda g_y\\ T_z= \lambda g_z\\ g=0} \Rightarrow \cases{z= \lambda(2x)\\ z=\lambda(2y)\\ x+y=\lambda(2z) \\ x^2+y^2+z^2=1} \Rightarrow \cases{x=y\\ z^2= 2x^2} \\ \Rightarrow x^2 +y^2+ z^2 =x^2+ x^2 +2x^2 =4x^2=1 \Rightarrow \cases{x=y=1/2 \\ x=y =-1/2} \Rightarrow z^2={1\over 2} \Rightarrow z=\pm {1\over \sqrt 2}\\ \Rightarrow \cases{(x,y,z)=(1/2,1/2,1/\sqrt 2)\\ (x,y,z)= (1/2,1/2,-1/\sqrt 2) \\ (-1/2,-1/2,1/\sqrt 2) \\(-1/2,-1/2,-1/\sqrt 2)} \Rightarrow \cases{T(1/2,1/2,1/\sqrt 2)= 1/\sqrt 2\\ T(1/2,1/2,-1/\sqrt 2)= -1/\sqrt 2\\ T(-1/2,-1/2,1/\sqrt 2)= -1/\sqrt 2\\ T(-1/2, -1/2, -1/\sqrt 2)= 1/\sqrt 2} \\ \Rightarrow \text{the hottest spot: } \bbox[red, 2pt]{\sqrt 2\over 2}$$


解答: $$f(x,y)=2x^2+y^2-4y=2x^2 +(y-2)^2-4 \ge -4 \Rightarrow \text{abs. min=}f(0,2)=-4 \\ \Rightarrow \cases{f_x=4x\\ f_y=2y-4} \Rightarrow \cases{f_{xx}=4\\ f_{xy}=0\\ f_{yy}=2} \Rightarrow d(x,y)=f_{xx}f_{yy}-(f_{xy})^2=8 \gt 0\\ \Rightarrow  \cases{f_{xx}\gt 0 \\ d\gt 0} \Rightarrow \text{maximum at boundary points of region }  R \\ \Rightarrow \cases{f(0,0)=0\\ f(4,4)=32\\ f(-4,4)=32} \Rightarrow \bbox[red, 2pt]{ \cases{\text{max: }32\\ \text{min: }-4}}$$


解答: $$e^x=1+x+{1\over 2!}x^2+{1\over 3!}x^3+ \cdots+ {1\over n!}x^n+ \cdots\\ \Rightarrow f'(x)=e^{x^2} =1+x^2+ {1\over 2!}x^4+ {1\over 3!}x^6 + \cdots+{1\over n!}x^{2n}+ \cdots  \\ \Rightarrow f(x)=c+x+{1\over 3}x^3+ {1\over 5\cdot 2!}x^5 +{1\over 7\cdot 3!}x^7+ \cdots+ {1\over (2n+1)n!} x^{2n+1}+ \cdots\\ \Rightarrow f(0)=c=2 \Rightarrow f(x)= 2+x+{1\over 3}x^3+ {1\over 5\cdot 2!}x^5 +{1\over 7\cdot 3!}x^7+ \cdots+ {1\over (2n+1)n!} x^{2n+1}+ \cdots \\ \Rightarrow f(1)=3+{1\over 3} +{1\over 5\cdot 2!}+ \cdots \Rightarrow f(1)\gt 3\\ \text{and }f(1)=2+1+{1\over 3}+ {1\over 5\cdot 2!}+ {1\over 7\cdot 3!} +\cdots \\\qquad \qquad \lt 2+1+ 1+{1\over 2!}+ {1\over 3!} + \cdots =2+e \Rightarrow f(1) \lt 2+e \\ \Rightarrow 3\lt f(1) \lt 2+e, \bbox[red, 2pt]{QED.}$$

 

解答: $$f(x)=a^x \Rightarrow f'(x)= {d\over dx} a^x =\lim_{h\to 0}{f(x+h)-f(x) \over h} =\lim_{h\to 0}{a^{x+h}-a^x \over h} \\= \lim_{h\to 0}{\frac{d }{dh}(a^{x+h}-a^x) \over \frac{d }{dh}h} = \lim_{h\to 0}{(\ln a) a^{x+h}\over 1} =(\ln a)a^{x}, \bbox[red, 2pt]{QED.}$$

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解題僅供參考，轉學考歷年試題及詳解