115年台北大學資工轉學考-微積分詳解

 國立臺北大學115學年度日間學士班轉學招生考試

學制系級:資訊工程學系日間學士班2年級
科目:微積分

解答:$$\textbf{(a) }E(x)=f(x)+f(-x) \Rightarrow E(-x)=f(-x)+f(x) \Rightarrow E(-x)= E(x) \Rightarrow E(x)\text{ is even}\; \bbox[red, 2pt]{QED.} \\ \textbf{(b) }O(x)=f(x)-f(-x) \Rightarrow O(-x) =f(-x)-f(x) =-O(x) \Rightarrow O(-x)=-O(x) \\ \qquad \Rightarrow O(x)\text{ is odd.}\; \bbox[red, 2pt]{QED. } \\ \textbf{(c)} \cases{E_1(x)=(f(x)+f(-x))/2 \text{ is even} \\ O_1(x)=(f(x)-f(-x))/2 \text{ is odd}} \Rightarrow f(x)=E_1(x)+O_1(x) \quad \bbox[red, 2pt]{QED.} \\ \textbf{(d) }f(x)=2^x+ (x-3)^2 =2^x+x^2-6x+9 \Rightarrow f(-x)=2^{-x}+x^2+6x+9 \\ \Rightarrow E(x)= {1\over 2}(f(x)+f(-x))={1\over 2}\left( 2^x+2^{-x} \right)+x^2+9 \\ \qquad \Rightarrow O(x)={1\over 2}(f(x)-f(-x))={1\over 2}(2^x-2^{-x})-6x \Rightarrow E(x)+O(x)=2^x+x^2-6x+9 \\\qquad \Rightarrow \bbox[red, 2pt]{f(x)=E(x)+O(x), \text{ where} \cases{E(x)=(2^x+2^{-x})/2+x^2+9 \text{ is even} \\ O(x)=(2^x-2^{-x})/2-6x \text{ is odd}}}$$

解答:$$\textbf{(a) } \textbf{Method 1: }m = \lim_{x \to 1} \frac{(x^3 + 1) - 2}{x - 1}  = \lim_{x \to 1} \frac{x^3 - 1}{x - 1} = \lim_{x \to 1} \frac{(x - 1)(x^2 + x + 1)}{x - 1} \\\qquad  = \lim_{x \to 1} (x^2 + x + 1) = 1^2 + 1 + 1 = \bbox[red, 2pt]3 \\ \textbf{Method 2: }m = \lim_{h \to 0} \frac{f(1+h) - f(1)}{h} = \lim_{h \to 0} \frac{((1+h)^3 + 1) - 2}{h}\\ \qquad  = \lim_{h \to 0} \frac{(1 + 3h + 3h^2 + h^3 + 1) - 2}{h}  = \lim_{h \to 0} \frac{h^3 + 3h^2 + 3h}{h} = \lim_{h \to 0} (h^2 + 3h + 3) \\ \qquad  = 0^2 + 3(0) + 3 = \bbox[red, 2pt]3 \\ \textbf{(b) } y=3(x-1)+2 \Rightarrow \bbox[red, 2pt]{y=3x-1}$$

解答:$$\int_0^1 (u+2)(u-3) du = \int  \left(  \frac{1}{3}u^3 - \frac{1}{2}u^2 - 6u  \right)\,du=  \left[ \frac{1}{3}u^3 - \frac{1}{2}u^2 - 6u \right]_0^1 =\bbox[red, 2pt]{- {37\over 6}}$$

解答:$$\textbf{(a) }\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \text{circle: }(x-a)^2+y^2=a^2 \Rightarrow x^2-2ax+y^2=0  \Rightarrow r=2a\cos \theta\\ \cases{\overleftrightarrow{OB}: y=x\tan \theta \\ B \in x=2a} \Rightarrow B=(2a,2a \tan \theta) \Rightarrow \overline{OB} = \sqrt{4a^2+4a^2\tan^2 \theta}=2a\sec \theta\\ \overline{OA}=2a\cos \theta \Rightarrow \overline{AB}= \overline{OB}-\overline{OA}=2a\sec \theta-2a\cos \theta = 2a\sin \theta \tan \theta\\ \Rightarrow P=(\overline{OP}\cos \theta, \overline{OP}\sin \theta)= (2a\sin \theta \tan \theta \cos \theta, 2a\sin^2\theta \tan \theta) =(2a\sin^2 \theta, 2a\sin^2\theta \tan \theta)\\ \Rightarrow \bbox[red, 2pt]{\left\{ (2a\sin^2\theta, 2a\sin^2\theta \tan \theta) \mid -{\pi\over 2}\lt \theta\lt {\pi\over 2}\right \} } \\\textbf{(b)}$$

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