115年嘉義大學轉學考-微積分詳解

 國立嘉義大學 115 學年度轉學生招生考試

科目: 微積分 (每題 10 分，共 100 分)

解答:$$\lim_{x \to 0} {\sin 4x\over x} =\lim_{x \to 0} {{d\over dx}\sin 4x\over {d\over dx}x} = \lim_{x \to 0} {4\cos 4x\over 1} = \bbox[red, 2pt] 4$$

解答:$$f(x)=2x-3x^{2/3} \Rightarrow f'(x)=2-2x^{-1/3} \Rightarrow f''(x)={2\over 3}x^{-4/3} \\ f'(x)=0 \Rightarrow 2x^{-1/3}=2 \Rightarrow x^{-1/3}=1 \Rightarrow x=1 \Rightarrow f''(1)={2\over 3} \gt 0 \Rightarrow f(1)=-1 \text{ is a local minimum} \\ \Rightarrow \cases{f(-1)=-2-3=-5\\ f(3)=6-3^{5/3} \approx -0.2 \\f(0)=0} \Rightarrow \bbox[red, 2pt]{\cases{f(-1)=-5 \text{ global minimum}   \\ f(0)=0 \text{ global maximum}}}$$

解答:$$x^2(x^2 + y^2) = y^2 \Rightarrow x^4+x^2y^2=y^2 \Rightarrow  \frac{d}{dx}(x^4 + x^2y^2) = \frac{d}{dx}(y^2) \\ \Rightarrow  4x^3 + 2x^2y \frac{dy}{dx} + 2xy^2 = 2y \frac{dy}{dx} \Rightarrow   \frac{dy}{dx} = \frac{-2x^3 - xy^2}{x^2y - y} \Rightarrow \left. \frac{dy}{dx}\right|_{(\sqrt 2/2, \sqrt 2/2)} =3 \\ \Rightarrow \text{tangent line: } y - \frac{\sqrt{2}}{2} = 3\left(x - \frac{\sqrt{2}}{2}\right) \Rightarrow \bbox[red, 2pt]{y=3x-\sqrt 2}$$

解答:$$\int_0^{\pi/4} \sqrt{1+\cos 4x} \,dx =\int_0^{\pi/4} \sqrt{1+(2\cos^2 2x-1)} \,dx = \sqrt 2\int_0^{\pi/4} \cos 2x\,dx = \sqrt 2 \left. \left[ {1\over 2}\sin 2x \right] \right|_0^{\pi/4} \\=\sqrt 2\cdot {1\over 2} \cdot {1 } =\bbox[red, 2pt]{\sqrt 2\over 2}$$

解答:$$ a_n = \frac{(-1)^n(x+1)^n}{2^n} \Rightarrow  L = \lim_{n \to \infty} \left\vert{} \frac{a_{n+1}}{a_n} \right\vert{} = \left\vert{} \frac{(-1)^{n+1}(x+1)^{n+1}}{2^{n+1}} \cdot \frac{2^n}{(-1)^n(x+1)^n} \right\vert{} ={|x+1|\over 2}\\ L\lt 1 \Rightarrow |x+1|\lt 2 \Rightarrow -3\lt x\lt 1\\ x=1 \Rightarrow \sum_{n=0}^\infty a_n= \sum_{n=0}^\infty (-1)^n \text{ diverges} \\x=-3\Rightarrow \sum_{n=0}^\infty a_n= \sum_{n=0}^\infty 1^n \text{ diverges} \\ \text{The interval of convergence is } \bbox[red, 2pt]{(−3,1)}$$

解答:$$u= \ln(3x) \Rightarrow du ={dx\over x} \Rightarrow \int {1\over \ln(3x)} \cdot {1\over x}\,dx = \int {1\over u}\,du= \ln|u|+C =\bbox[red, 2pt]{ \ln\vert{}\ln(3x)\vert{} + C}$$

解答:$$y= \ln \sqrt{x+1\over x-1} = {1\over 2} \left[ \ln(x+1)-\ln(x-1) \right] \Rightarrow y'={1\over 2} \left[ {1\over x+1}-{1\over x-1} \right]= \bbox[red, 2pt]{1\over 1-x^2}$$

解答:$$ \lim_{x \to n^-} f(x) = \lim_{x \to n^-} \left( \frac{[x]}{2} + x \right) = \frac{n - 1}{2} + n = \frac{n - 1 + 2n}{2} = \frac{3n - 1}{2} \\ \lim_{x \to n^+} f(x) = \lim_{x \to n^+} \left( \frac{[x]}{2} + x \right) = \frac{n}{2} + n = \frac{3n}{2} \\ \frac{3n - 1}{2} \ne  \frac{3n}{2}, \text{ for all }n\in \mathbb Z \Rightarrow \text{$f(x)$ is discontinuous at} \bbox[red, 2pt]{ \text{all integer values of $x$}}$$

解答:$$ \int_{-\infty}^{0} \frac{1}{(1-2x)^{3/2}} dx = \lim_{t \to -\infty} \int_{t}^{0} (1-2x)^{-3/2} dx   = \lim_{t \to -\infty}  \left[ \frac{1}{\sqrt{1-2x}} \right]_{t}^{0} =  \lim_{t \to -\infty} \left( 1 - \frac{1}{\sqrt{1-2t}} \right) \\= \bbox[red, 2pt]1$$

解答:$$xy=1 \Rightarrow y={1\over x} \Rightarrow f(x,1/x)=x^2+ \left( {1\over x} \right)^2 =x^2+x^{-2} \ge 2\sqrt{x^2\cdot x^2}= \bbox[red, 2pt]2$$

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