112年普考-微積分詳解

 112年公務人員普通考試試題

類 科：天文、氣象
科 目：微積分

解答：$$f(x)=2x^3-3x^2-12x+2 \Rightarrow f'(x)=6x^2-6x-12 \Rightarrow f''(x)=12x-6\\ 若f'(x)=0 \Rightarrow 6(x-2)(x+1)=0 \Rightarrow x=2,-1 \Rightarrow \cases{f''(2)=18 \gt 0\\ f''(-1)=-18 \lt 0} \Rightarrow \cases{f(2)=-18\\ f(-1)=9}\\ 邊界點\cases{f(-3)=-43\\ f(3)=-7} \Rightarrow \bbox[red, 2pt]{ \cases{最大值=9\\ 最小值=-43}}$$

解答：$$\mathbf{(一)}\;\lim_{n\to \infty}\left| a_{n+1} \over a_n \right| =\lim_{n\to \infty}\left| n+1　 \over n　 \right| = 1\Rightarrow 收斂半徑=\bbox[red, 2pt]1 \\ 又|x-2| \lt 1 \Rightarrow 1\lt x\lt 3 \\ 考慮邊界點：\begin{cases}x=3 \Rightarrow  & \sum_{n=1}^\infty n(x-2)^n= \sum_{n=1}^\infty n 發散\\x=1 \Rightarrow  & \sum_{n=1}^\infty n(x-2)^n=-1+ 2-3+4-\cdots發散\end{cases}\\ \Rightarrow 收斂區間=\bbox[red, 2pt]{(1,3)} \\ \mathbf{(二)}\; f(x)=\sqrt{1+x} =(x+1)^{1/2} \Rightarrow f'(x)={1\over 2}(x+1)^{-1/2} \Rightarrow f''(x)=-{1\over 4}(x+1)^{-3/2}\\ \Rightarrow f'''(x)={3\over 8}(x+1)^{-5/2} \Rightarrow f^{[4]}=-{15\over 16}(x+1)^{-7/2}\\ \Rightarrow \cases{f(0)=1\\ f'(0)=1/2\\ f''(0)=-1/4\\ f'''(0)=3/8 \\ f^[4](0)=-15/16} \Rightarrow f(x)=\sum_{n=0}^\infty {f^{[n]}(0)\over n!}x^n=\bbox[red,2pt]{ 1+{1\over 2}x-{1\over 8}x^2+{1\over 16}x^3-{5\over 128}x^4}+O(x^5)$$

解答：$$\mathbf{(一)}\; f(x,y)=xe^{xy} \Rightarrow \cases{f_x=e^{xy}+xye^{xy}\\ f_y=x^2e^{xy}} \Rightarrow \cases{f_{xx}=2ye^{xy}+ xy^2e^{xy}\\ f_{xy}= 2xe^{xy}+ x^2ye^{xy}\\ f_{yy}=x^3e^{xy}} \\ 一階偏導數：\bbox[red, 2pt] {\cases{f_x=e^{xy}+xye^{xy}\\ f_y=x^2e^{xy}}}，二階偏導數：\bbox[red, 2pt]{\cases{f_{xx}=2ye^{xy}+ xy^2e^{xy}\\ f_{xy}= 2xe^{xy}+ x^2ye^{xy}\\ f_{yy}=x^3e^{xy}}} \\ \mathbf{(二)}\; f(x,y)=x^2-y^2+xy-x+y \Rightarrow \nabla f =(f_x,f_y)= (2x+y-1, -2y+x+1) \\\Rightarrow \nabla f(1,2)=(3, -2) \Rightarrow 方向導數D_{\vec u}f(1,2) =\nabla f(1,2) \cdot {\vec u\over |\vec u|}=(3,-2)\cdot ({3\over 5},{4\over 5})=\bbox[red, 2pt]{1\over 5}$$

解答：$$\mathbf{(一)}\;令\cases{u=x \\ dv=e^{2x}\, dx} \Rightarrow \cases{du=dx \\ v={1\over 2}e^{2x}} \Rightarrow \int xe^{2x}\,dx ={1\over 2}xe^{2x}-{1\over 2}\int e^{2x}\,dx = {1\over 2}xe^{2x}-{1\over 4}e^{2x}+C\\ \Rightarrow \int_0^2 xe^{2x}\,dx =\left. \left[ {1\over 2}xe^{2x}-{1\over 4}e^{2x} \right] \right|_0^2 =\bbox[red, 2pt]{{1\over 4}(3e^4+1)} \\ \mathbf{(二)}\cases{x=r \cos \theta\\ y=r\sin \theta} \Rightarrow \int_0^1 \int_0^{1-x^2} \sqrt{x^2+y^2}\,dydx =\int_0^{\pi/2} \int_0^1 r^2\,drd\theta = \int_0^{\pi/2} {1\over 3}d\theta =\bbox[red, 2pt]{\pi\over 6}$$