台灣聯合大學系統112學年度學士班轉學生考試
科目:微積分
類組別:A2
甲、填充題:共８題，每題８分
解答:$$\lim_{x\to 0^+} \left({1\over x}-{1\over e^x-1} \right) =\lim_{x\to 0^+} {e^x-x-1\over x(e^x-1)} =\lim_{x\to 0^+} {(e^x-x-1)'\over (x(e^x-1))'} =\lim_{x\to 0^+} {e^x-1\over e^x-1+xe^x} \\ =\lim_{x\to 0^+} {(e^x-1)'\over (e^x-1+xe^x)'} =\lim_{x\to 0^+} {e^x\over 2e^x+xe^x} =\bbox[red, 2pt]{1\over 2}$$
解答:$${d\over dx}\sin x=\cos x= \sin(x+{\pi\over 2}) \Rightarrow {d^n\over d x^n}\sin x=\sin(x+{n\over 2}\pi) \\ {d\over dx}(x\sin x)=\sin x+ x\sin(x+{\pi \over 2}) \Rightarrow {d^2\over dx^2}(x\sin x)=2\sin(x+{1\over 2}\pi) +x\sin(x+{2\over 2}\pi) \\ \Rightarrow {d^n\over d x^n} (x\sin x)= n\sin (x+{n-1\over 2}\pi )+ x\sin(x+{n\over 2}\pi)\\ \Rightarrow {d^{2023}\over dx^{2023}} (x\sin x) =2023 \sin(x+{2022\over 2}\pi) +x\sin(x+{2023\over 2}\pi)=-2023\sin x+ x\cos(x+1011\pi) \\=\bbox[red, 2pt]{-2023\sin x-x\cos x}$$
解答:$$\int_{-1}^4 {dx\over \sqrt {|x|}} = \int_0^1{dx\over \sqrt x} +\int_0^4{dx\over \sqrt x}=2+4 = \bbox[red, 2pt]6$$
解答:$$\int_0^8 \int_{\sqrt[3]x}^2 {1\over y^4+1}\,dydx= \int_0^2 \int_0^{y^3} {1\over y^4+1}\,dx dy = \int_0^2 {y^3\over y^4+1}dy =\left. \left[{1\over 4}\ln(y^4+1) \right]\right|_0^2 = \bbox[red, 2pt]{{1\over 4}\ln 17}$$
解答:$$\int_1^e (\ln(x))^2\pi\,dx =\left. \left[\pi(x(\ln x)^2-2x\ln(x)+2x) \right] \right|_1^e =\bbox[red, 2pt]{\pi(e -2)}$$
解答:$$y=(\sin x)^x = e^{x\ln \sin x} \Rightarrow {dy\over dx} =(\ln \sin x+{x\cos x\over \sin x}) e^{x\ln \sin x} = \bbox[red, 2pt]{(\ln \sin x+x \cot x)(\sin x)^x}$$
解答:$$若a\gt 0 \Rightarrow \cases{\lim_{x\to \infty} {\sqrt{ax^2+4}\over x-b}=\sqrt a\\ \lim_{x\to -\infty} {\sqrt{ax^2+4}\over x-b}=-\sqrt a} \Rightarrow 有兩條水平漸近線\bbox[red,2pt]{y=\pm \sqrt a }$$
8. We say that the two commodities are substitute commodities if a decrease in the demand for one results in an increase in the demand for the other. Conversely, two commodities are referred to as complementary commodities if a decrease in the demand for on results in a decrease in the demand for the other as well. Suppose that the demand equations that relate the quantities demanded x and y to the unit prices p and q of the commodities A and B respectively are given by \( x=f(p,q)={q^2\over q+p^2}\) and \(y=q(p,q)= e^{-2q+p}\). Are A and B subtitute, complementary or neighter?
解答:$$f(p,q)={q^2\over q+p^2} \Rightarrow \cases{f_p={-2pq^2 \over (q+p^2)^2} \lt 0\\ f_q={2q\over q+p^2}-{q^2\over (q+p^2)^2} ={q^2+2p^2q\over (q+p^2)^2}\gt 0} \\ g(p,q)=e^{-2q+p} \Rightarrow \cases{g_p = e^{-2q+p} \gt 0\\ g_q =-2e^{-2q+p}\lt 0}\\ 因此我們有\cases{f_q\gt 0\\ g_p\gt 0} \Rightarrow \text{A and B are } \bbox[red, 2pt]{\text{substitue}}$$
乙、計算、證明題: 共３題，每題１２分，共３６分
解答:$$\mathbf{(a)}\;0\le \left| x\sin({1\over x})\right| \le |x| \Rightarrow 0\le \lim_{x\to 0}\left| x\sin({1\over x})\right| \le \lim_{x\to 0}|x| =0 \Rightarrow 0\le \lim_{x\to 0}\left| x\sin({1\over x})\right| \le0\\ 由夾擠定理得\lim_{x\to 0}\left| x\sin({1\over x})\right|=0 \Rightarrow \lim_{x\to 0} x\sin({1\over x}) =0 \Rightarrow 在x=0,f是連續的,\bbox[red, 2pt]{故得證}\\ \mathbf{(b)}\; x\ne 0 \Rightarrow f'(x)=\sin({1\over x})+x\cos({1\over x})\cdot (-{1\over x^2}) = \bbox[red, 2pt]{\sin({1\over x})-{1\over x}\cos({1\over x})}\\ \mathbf{(c)}\; \lim_{h\to 0}{f(0+h)-f(0)\over h} = \lim_{h\to 0}{h\sin (1/h)-0\over h} = \lim_{h\to 0}\sin({1\over h})不存在\\ \quad \Rightarrow f在x=0不可微,\bbox[red, 2pt]{故得證}$$
解答:$$\mathbf{(a)}\;\int_1^\infty {x\over x^2+4}\, dx = \left. \left[{1\over 2}\ln(x^2+4) \right] \right|_1^\infty =\infty \Rightarrow \bbox[red,2pt]{\text{diverge}}\\ \mathbf{(b)}\; \lim_{n\to \infty} \left|{a_{n+1}\over a_n} \right| =\lim_{n\to \infty} \left|{(\sqrt{n+2}-\sqrt {n+1}) (x-3)^{n+1}\over (\sqrt{n+1}-\sqrt n)(x-3)^n} \right| =|x-3|\lt 1 \Rightarrow 2\lt x\lt 4\\ x=4 \Rightarrow \sum_{n=1}^\infty(\sqrt{n+1}-\sqrt n)(x-3)^n =\sum_{n=1}^\infty(\sqrt{n+1}-\sqrt n),\\利用\text{integral test}: \int_1^\infty \sqrt{x+1}-\sqrt x\,dx = \infty \Rightarrow 發散\\ x=2 \Rightarrow \sum_{n=1}^\infty(\sqrt{n+1}-\sqrt n)(x-3)^n =\sum_{n=1}^\infty(\sqrt{n+1}-\sqrt n)(-1)^n\\ 利用\text{Alternating Series Test: }\cases{\lim_{n\to \infty}(\sqrt{n+1}-\sqrt n)=0\\ (\sqrt{n+1}-\sqrt n)-(\sqrt{n+2}-\sqrt{n+1})\ge 0} \Rightarrow 收斂\\ 因此\bbox[red,2pt]{2\le x\lt 4},級數收斂$$
解答:$$此題相當於求:在條件0\le 200x+300y\le 60000下,求f(x,y)=100x^{3/4}y^{1/4}的最大化\\利用\text{Lagrange }算子求極值: \cases{f(x,y)=100x^{3/4}y^{1/4}\\ g(x,y)=200x+300y-60000} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y \\ g=0} \\ \Rightarrow \cases{75x^{-1/4}y^{1/4} =\lambda\cdot 200\\ 25x^{3/4}y^{-3/4} =\lambda\cdot 300} \Rightarrow {3y\over x}={2\over 3} \Rightarrow y={2\over 9}x \Rightarrow 200x+300\cdot {2\over 9}x=6000\\ \Rightarrow x=225 \Rightarrow y=50 \Rightarrow 當\bbox[red, 2pt]{\cases{x=225\\y=50}}時,f(x,y)有最大值$$ ==================== END =========================
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