115年政大應數二轉學考-微積分一詳解

 國立政治大學 115 學年度學士班轉學生招生考試

考試科目 微積分 (一): 以微分為主 
系所別 應用數學系二年級

解答:$$\textbf{(a) } f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}= \lim_{x \to 0} \frac{x + x^2 \sin(1/x) - 0}{x}  = \lim_{x \to 0} \left( 1 + x \sin\left(\frac{1}{x}\right) \right) =1+0=1\\ \Rightarrow \bbox[red, 2pt]{\text{$f'(0)$ exists and $f'(0)=1$}} \\ \textbf{(b) }x\ne 0 \Rightarrow  f'(x) = \frac{d}{dx} \left( x + x^2 \sin\left(\frac{1}{x}\right) \right)   = 1 + 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \\ \Rightarrow \lim_{x \to 0} f'(x) = \lim_{x \to 0} \left( 1 + 2x \sin\left(\frac{1}{x}\right) - \cos\left(\frac{1}{x}\right) \right) \text{ does not exist, since } \lim_{x \to 0} \cos(1/x) \text{ does not exist.} \\ \Rightarrow \bbox[red, 2pt]{\text{$f'$ is NOT continuous at $x=0$}}$$

解答:$$\textbf{(a) }\bbox[red, 2pt]{true}: \text{Suppose }\lim_{x\to 1}{f(x) \over g(x)} =L \lt \infty \Rightarrow  \lim_{x \to 1} f(x) = \lim_{x \to 1} \left( \frac{f(x)}{g(x)} \cdot g(x) \right)  \\ = \left( \lim_{x \to 1} \frac{f(x)}{g(x)} \right) \cdot \left( \lim_{x \to 1} g(x) \right) =L\cdot 0 =0\ne 3 \Rightarrow \text{contradiction} \\ \textbf{(b) }\bbox[red, 2pt]{false}: \cases{f(x)=x-1\\ g(x)=x-1} \Rightarrow \cases{\lim_{x\to 1}f(x)=0 \\\lim_{x\to 1}g(x)=0 } \Rightarrow \lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{x - 1}{x - 1} =   1 \lt \infty$$

解答:$$\cases{\lim_{x\to 0} x^2+2x-1 \lt 0 \Rightarrow \lim_{x\to 0} |x^2+2x-1|=-x^2-2x+1 \\ \lim_{x\to 0} x^2+2x+1\gt 0 \Rightarrow \lim_{x\to 0} |x^2+2x+1|=x^2+2x+1}\\ \Rightarrow  \lim_{x \to 0} \frac{\vert{}x^2 + 2x - 1\vert{} - \vert{}x^2 + 2x + 1\vert{} + 4x}{x^2} = \lim_{x \to 0} \frac{-x^2 - 2x +1 - (x^2 + 2x + 1) + 4x}{x^2} \\= \lim_{x \to 0} {-2x^2\over x^2} =\bbox[red, 2pt]{-2}$$

解答:$$ H(x) = [f(e^{x^2})]^\alpha \Rightarrow  H'(x) = \alpha [f(e^{x^2})]^{\alpha - 1} \cdot \frac{d}{dx} \left( f(e^{x^2}) \right)   = \alpha [f(e^{x^2})]^{\alpha - 1} \cdot f'(e^{x^2}) \cdot \frac{d}{dx} \left( e^{x^2} \right) \\  = \alpha [f(e^{x^2})]^{\alpha - 1} \cdot f'(e^{x^2}) \cdot \left( e^{x^2} \cdot 2x \right) =  \bbox[red, 2pt]{2\alpha x e^{x^2} [f(e^{x^2})]^{\alpha - 1} f'(e^{x^2}) }$$

解答:$$\textbf{(a) }\frac{d}{dx}(\sin(xy)) = \frac{d}{dx}(x^2 - y^2) \Rightarrow  \cos(xy) \cdot \frac{d}{dx}(xy) = 2x - 2y \cdot y' \\ \Rightarrow  \cos(xy) \cdot (1 \cdot y + x \cdot y') = 2x - 2y y' \Rightarrow  y' (x \cos(xy) + 2y) = 2x - y \cos(xy) \Rightarrow \\y' = \bbox[red, 2pt]{\frac{2x - y \cos(xy)}{x \cos(xy) + 2y}} \\ \textbf{(b) } y = (\cos x)^{x^2} \Rightarrow \ln y= \ln (\cos x)^{x^2} =x^2 \ln \cos x \Rightarrow  \frac{d}{dx}(\ln y) = \frac{d}{dx}(x^2 \ln(\cos x)) \\ \Rightarrow  \frac{1}{y} \cdot y' = \frac{d}{dx}(x^2) \cdot \ln(\cos x) + x^2 \cdot \frac{d}{dx}( \ln(\cos x))\\ \Rightarrow \frac{1}{y} \cdot y' = 2x \ln(\cos x) + x^2 \cdot \left( \frac{1}{\cos x} \cdot (-\sin x) \right) \Rightarrow  \frac{1}{y} \cdot y' = 2x \ln(\cos x) - x^2 \tan x  \\ \Rightarrow y' = \bbox[red, 2pt]{(\cos x)^{x^2} \left( 2x \ln(\cos x) - x^2 \tan x \right)}$$

解答:$$\textbf{(a) } \text{for all }x\gt 0, \text{ we have }\ln(1+x)\lt x \Rightarrow 0\lt \ln(1+\sqrt{a_n}) \lt \sqrt{a_n} \\ \Rightarrow 0\lt [\ln(1+\sqrt{a_n})]^2 \lt (\sqrt{a_n})^2 =a_n \\ \text{Using Comparison Test,  } \sum_{n=1}^\infty a_n \text{ is convergent }\Rightarrow  \sum_{n=1}^\infty [\ln(1+\sqrt{a_n})]^2 \text{ is convergent }\bbox[red, 2pt]{QED.} \\ \textbf{(b) }a_n={1\over  n^2}\gt 0, \forall n\ge 1 \Rightarrow \sum_{n=1}^\infty {1\over n^2} \text{ is convergent} \\ \Rightarrow  \sum_{n=1}^{\infty} \ln(1+\sqrt{a_n}) = \sum_{n=1}^{\infty} \ln\left(1+\sqrt{\frac{1}{n^2}}\right) = \sum_{n=1}^{\infty} \ln\left(1+\frac{1}{n}\right) = \sum_{n=1}^{\infty} \left( \ln(n+1)-\ln n\right) \\=(\ln 2-\ln 1)+(\ln 3-\ln 2)+(\ln 4-\ln 3)+ \cdots=\infty \Rightarrow  \sum_{n=1}^{\infty} \ln(1+\sqrt{a_n}) \text{ is divergent} \\ \Rightarrow \bbox[red, 2pt]{\text{No, that won't be necessary.}}$$

解答:$$\textbf{(a) }  f_x(0,0) = \lim_{x \to 0} \frac{f(x,0) - f(0,0)}{x} = \lim_{x \to 0} \frac{x^2 \ln(x^2) - 0}{x} = \lim_{x \to 0} x \ln(x^2) =0\\ (x,y) \ne (0,0) \Rightarrow  f_x(x,y) = \frac{\partial}{\partial x} \left[ (x^2 + y^2) \ln(x^2 + y^2) \right] = 2x \ln(x^2 + y^2) + (x^2 + y^2) \cdot \frac{2x}{x^2 + y^2} \\= 2x \ln(x^2 + y^2) + 2x  \Rightarrow f_x(0,y\ne 0) =0 \Rightarrow  f_{xy}(0,0) = \lim_{y \to 0} \frac{f_x(0,y) - f_x(0,0)}{y} = \lim_{y \to 0} \frac{0 - 0}{y} = 0 \\ \text{At the same way, we have }f_y(0,0)=0 \text{ and } f_y(x\ne 0,0)=0. \text{Then }\\ f_{yx}(0,0) = \lim_{x \to 0} \frac{f_y(x,0) - f_y(0,0)}{x} = \lim_{x \to 0} \frac{0 - 0}{x} = 0 .\\ \text{Therefore, } \bbox[red, 2pt]{f_{xy}(0,0)=f_{yx}(0,0)=0} \\ \textbf{(b) } \text{In (a), we have $f_{xy}(0,0)=f_{yx}(0,0)=0$.} \\ (x,y) \ne (0,0) \Rightarrow \cases{f_x=2x\ln(x^2+y^2)+2x\\ f_y=2y\ln (x^2+ y^2)+ 2y} \\\Rightarrow \cases{ f_{xy}(x,y) = \frac{\partial}{\partial y} (2x \ln(x^2 + y^2) + 2x) = 2x \cdot \frac{2y}{x^2 + y^2} = \frac{4xy}{x^2 + y^2} \\f_{yx}(x,y) = \frac{\partial}{\partial x} (2y \ln(x^2 + y^2) + 2y) = 2y \cdot \frac{2x}{x^2 + y^2} = \frac{4xy}{x^2 + y^2} } \Rightarrow f_{xy}=f_{yx} \\ \lim_{x\to 0}f_{xy}(x,kx)= \lim_{x\to 0}{4kx^2 \over (k^2+1) x^2} = {4k  \over k^2+1 }\ne 0, \text{ for }k\ne 0 \\ \Rightarrow \text{$f_{xy}$ and $f_{yx}$ are not continuous at (0,0)} \quad \bbox[red, 2pt]{QED.}$$

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