111年普考-微積分詳解

111年公務人員普通考試試題

類 科： 氣象
科 目： 微積分

解答： $$\lim_{x\to 0} {\sin x\over x} =\lim_{x\to 0} {\cos x\over 1} =1 \Rightarrow \lim_{x\to 0}f(x)=f(0)=1 \Rightarrow \bbox[red, 2pt]{f(x)是連續的, x\in \mathbb{R}}\\ 又f'(x)=\cases{{x\cos x-\sin x\over x^2},x\ne 0\\ 0,x=0}而 \lim_{x\to 0}{x\cos x-\sin x\over x^2} = \lim_{x\to 0}{-\sin x\over 2} =0=f'(0)\\ 因此\bbox[red,2pt]{f(x)是可微分的, x\in \mathbb{R}}$$

解答： $$f(x,y)=\tan^{-1}{y\over x} \Rightarrow \cases{f_x= \cfrac{-y}{x^2+y^2} \Rightarrow \cases{\bbox[red, 2pt]{f_{xx}= \cfrac{2xy}{(x^2+y^2)^2}} \\ \bbox[red,2pt]{f_{xy}= \cfrac{y^2-x^2}{(x^2+y^2)^2}}}\\ f_y= \cfrac{x}{x^2+y^2} \Rightarrow \cases{ \bbox[red,2pt]{f_{yx} = \cfrac{y^2-x^2}{(x^2+y^2)^2}}\\ \bbox[red,2pt]{f_{yy}= \cfrac{-2xy}{(x^2+y^2)^2}}}}$$

解答： $$假設 \vec f(t)=(f_1(t),f_2(t),f_3(t)) ;令u=\sqrt{t+1} \Rightarrow du=\cfrac{1}{2\sqrt{t+1}} dt ={1\over 2u}dt \Rightarrow dt=2udu \\\Rightarrow f_1(t)=\int \cfrac{t}{t+1-\sqrt{t+1}}dt = \int {u^2-1\over u^2-u} \cdot 2udu =\int 2(u+1)\,du =(u+1)^2 =(\sqrt{t+1}+1)^2+C\\ =t+ 2\sqrt{t+1}+C'; 再由f_1(0)=3 \Rightarrow 2+C'=3 \Rightarrow C'=1 \Rightarrow f_1(t)=t+ 2\sqrt{t+1}+1;\\ 同理,f_2(t)= \int {1\over \sqrt t(t+1)}\,dt=2\tan^{-1}\sqrt t+C; 再由f_2(0)=2 \Rightarrow C=2 \Rightarrow  f_2(t)=2\tan^{-1}\sqrt t+2;\\ f_3(t) =\int {1\over (t^2+1)^2}\,dt={1\over 2}\left( {t\over t^2+1} +\tan^{-1} t\right)+C; 再由f_3(0)=1 \Rightarrow C=1\\ \Rightarrow f_3(t)={1\over 2}\left( {t\over t^2+1} +\tan^{-1} t\right)+1\\ 因此\vec f(t)=\bbox[red, 2pt]{(t+ 2\sqrt{t+1}+1, 2\tan^{-1}\sqrt t+2,{1\over 2}\left( {t\over t^2+1} +\tan^{-1} t\right)+1)}$$

解答： $$f(x)=\ln x \Rightarrow f'(x)={1\over x} \Rightarrow f''(x)=-{1\over x^2} \Rightarrow f'''(x)={2\over x^3} \Rightarrow f^{[n]}=(-1)^{n-1}{(n-1)!\over x^n}\\ \Rightarrow f(x)=f(a) +{f'(a)\over 1!}(x-a) + {f''(a)\over 2!}(x-a)^2 +\cdots\\ \qquad \quad = \ln(a)+{1\over a}(x-a)-{1\over 2a^2}(x-a)^2+{1\over 3a^3}(x-a)^3-\cdots+ {1\over na^n}\cdot (-1)^{n-1}(x-a)^n+\cdots\\ \Rightarrow f(x)=\bbox[red,2pt]{\ln x= \sum_{k=1}^\infty {(-1)^{k-1}\over ka^k}(x-a)^k}\\ \lim_{n\to \infty}\left| {a_{n+1}\over a_n}\right| =\lim_{n\to\infty}\left| {n\over a(n+1)}(x-a)\right| =\left| {x-a\over a }\right|\lt 1 \Rightarrow |x-a|\lt a \Rightarrow 收斂半徑=\bbox[red,2pt]{a}$$

解答： $$f(x,y)=x^3+y^3-3x-3y^2 \Rightarrow \cases{f_x=3x^2-3\\ f_y=3y^2-6y} \Rightarrow \cases{f_{xx}=6x\\ f_{xy}=0\\ f_{yy}= 6y-6} \Rightarrow D(x,y)=f_{xx}f_{yy}-f_{xy}^2\\ \Rightarrow 判別式D(x,y)=36x(y-1)\\ 因此\cases{f_x=0\\ f_y=0}\Rightarrow \cases{x=\pm 1\\ y=0,2} \Rightarrow 臨界點(1,0),(1,2) (只考慮0\le x,y\le 3)\\ \Rightarrow \cases{D(1,0)=-36\lt 0 \Rightarrow (1,0)為鞍點\\ D(1,2)=36 \gt 0又 f_{xx}(1,2)=6 \gt 0 \Rightarrow (1,2)為極小值} \Rightarrow f(1,2)=-6\\ 考慮邊界點\cases{f(3,3)=18\\ f(3,0)=18 \\ f(0,3)=0\\ f(0,0)= 0}，因此\bbox[red,2pt]{\cases{最小值=f(1,2)=-6\\ 最大值=f(3,0)=f(3,3)=18}}$$

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