115年台北大學經濟系轉學考-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:經濟系日間學士班2、3年級
科目:微積分

解答:$$\textbf{(a) }\lim_{x\to 0}{e^{29x}-1 \over \sin(31x)} = \lim_{x\to 0}{{d\over dx}(e^{29x}-1) \over {d\over dx} \sin(31x)} =\lim_{x\to 0}{29e^{29x} \over 31\cos(31x)} = \bbox[red, 2pt]{29\over 31} \\ \textbf{(b) }\lim_{x\to -\infty}{41x^4-43x+47  \over 53x^4+59 x^2-61} =\lim_{x\to -\infty}{41-43/x^3+47/x^4  \over 53+59 /x^2-61/x^4} = \bbox[red, 2pt]{41\over 53} \\ \textbf{(c) }\lim_{x\to 0} {e^{3x}-1-3x \over x^2} = \lim_{x\to 0} {{d\over dx}(e^{3x}-1-3x )\over {d\over dx}x^2} = \lim_{x\to 0} {3e^{3x}-3 \over 2x} = \lim_{x\to 0} {{d\over dx}(3e^{3x}-3) \over{d\over dx} 2x} \\\qquad =\lim_{x\to 0} {9e^{3x} \over 2} = \bbox[red, 2pt]{9\over 2} \\ \textbf{(d) } \cases{\sqrt{7-x} \Rightarrow 7-x\ge 0 \Rightarrow x\le 7\\ \ln x \Rightarrow x\gt 0} \Rightarrow \bbox[red, 2pt]{0\lt x\le 7} \\\textbf{(e) } y=x^{11}+ \ln x \Rightarrow {dy\over dx}=11x^{10}+{1\over x} ={11x^{11}+1\over x} \Rightarrow {dx\over dy}= \bbox[red, 2pt]{x\over 11x^{11}+1} \\ \textbf{(f) }y=xe^{x^3} \Rightarrow {dy\over dx} ={d\over dx}x\cdot e^{x^3}+x{d\over dx}e^{x^3}=e^{x^3} +x\cdot3x^2e^{3x}= \bbox[red, 2pt]{e^{x^3}(1+3x^3)} \\ \textbf{(g) }\cases{z=7y^3\\ y=6x^2+8x} \Rightarrow  z=7(6x^2+8x)^3 \Rightarrow {dz\over dx}=21(6x^2+8x)^2(12x+8) =\bbox[red, 2pt]{336x^2(3x+4)^2(3x+2)} \\ \textbf{(h) }y={x^5+7e^x\over x^2+1} \Rightarrow {dy\over dx}={5x^4+7e^x\over x^2+1}-{(x^5+7e^x)(2x) \over (x^2+1)^2}= \bbox[red, 2pt]{3x^6+5x^4+7e^x(x-1)^2 \over (x^2+1)^2} \\ \textbf{(i) }z=x^2y+3xy^2+5y \Rightarrow \cases{{\partial z\over \partial x} =2xy+3y^2 \\ {\partial z\over \partial y} =x^2+6xy+5} \Rightarrow dz =\bbox[red, 2pt]{(2xy+3y^2)dx+(x^2+6xy+5)dy} \\\textbf{(j) } f(x)=|x-4| \Rightarrow f(x)=\begin{cases}x-4,& x\ge 4\\ 4-x, & x\lt 4 \end{cases} \Rightarrow \int_{-3}^6 f(x)\,dx =\int_{-3}^4(4-x)\,dx+ \int_4^6(x-4)\,dx \\\qquad = \left. \left[ 4x-{1\over 2}x^2 \right] \right|_{-3}^4+ \left. \left[ {1\over 2}x^2-4x \right] \right|_4^6= \left( 20+{9\over 2} \right)+ \left( 2 \right)=\bbox[red, 2pt]{53\over 2} \\ \textbf{(k) }\cases{u=x\\ dv=e^{-3x}\,dx} \Rightarrow \cases{du=dx\\ v=-e^{-3x}/3} \Rightarrow I= \int_0^\infty 101xe^{-3x}\,dx = 101 \left( \left. \left[ -{x\over 3}e^{-3x} \right] \right|_0^\infty+{1\over 3}\int_0^\infty e^{-3x}\,dx\right)  \\\qquad ={101\over 3}\int_0^\infty e^{-3x}\,dx ={101\over 3} \left. \left[ -{1\over 3}e^{-3x} \right] \right|_0^\infty = \bbox[red, 2pt]{101\over 9} \\ \textbf{(l) }\int_1^6 \int_0^3 (3x^2y+ 7xy^2+9x) \,dx \,dy =\int_1^6 \left. \left[ x^3y+{7\over 2}x^2y^2+{9\over 2}x^2 \right] \right|_0^3 \,dy = \int_1^6 \left( 27y+{63\over 2}y^2+{81\over 2} \right) \,dy \\ \qquad = \left. \left[ {27\over 2}y^2+{21\over 2}y^3+{81\over 2}y \right] \right|_1^6 = {1\over 2}(5994-129) = \bbox[red, 2pt]{5865\over 2}$$

解答:$$\textbf{(a) } \lim_{x\to \infty} \left( \sqrt{x^2+6}-x \right) = \lim_{x\to\infty} \left( \sqrt{x^2 + 6x} - x \right) \cdot \frac{\sqrt{x^2 + 6x} + x}{\sqrt{x^2 + 6x} + x}  = \lim_{x\to\infty} \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} \\ \qquad =  \lim_{x\to\infty} \frac{6x}{x\left(\sqrt{1 + \frac{6}{x}} + 1\right)}  = \lim_{x\to\infty} \frac{6}{\sqrt{1 + \frac{6}{x}} + 1} = \bbox[red, 2pt]3 \\\textbf{(b) }  f(x) = x^3 - 36x^2 + 429x - 37 \Rightarrow  f'(x) = 3x^2 - 72x + 429 =0 \Rightarrow 3(x-11)(x-13) =0 \\ \qquad x=11,13 \Rightarrow \bbox[red, 2pt]{\text{The critical points are $x = 11$ and $x = 13$}} \\ \textbf{(c) }f(x,y)=x^2+13y-12xy \Rightarrow \cases{f_x= 2x-12y=0 \Rightarrow x=6y\\ f_y=13-12x =0 \Rightarrow x=13/12} \Rightarrow \cases{x=13/12\\ y=13/72} \\ \qquad \Rightarrow f \left( {13\over 12},{13\over 72} \right) = \left(\frac{13}{12}\right)^2 + 13 \left(\frac{13}{72}\right) - 12\left(\frac{13}{12}\right)\left(\frac{13}{72}\right) = \frac{169}{144} \approx 1.17 \\ \text{boundary }\cases{y=0\\ x=10\\ y=x} \Rightarrow \text{ vertices }\cases{(0,0)\\(10,0)\\ (10,10) \\(13/22, 13/22)} \Rightarrow \cases{f(0,0)=0\\ f(10,0)=100 \\ f(10,10)= -970\\ f(13/12, 13/12)= 169/44 \approx3.8} \\ \Rightarrow \bbox[red, 2pt]{\cases{\text{absolute maximum point: }(10, 0) \\\text{absolute minimum point: }(10, 10)}} \\\textbf{(d) }  f(x,y) = (x+y)^{1/2} \Rightarrow \cases{f_x(x,y) = \frac{1}{2}(x+y)^{-\frac{1}{2}} \\f_y(x,y) = \frac{1}{2}(x+y)^{-\frac{1}{2}} } \Rightarrow \cases{f_{xx}(x,y) = -\frac{1}{4}(x+y)^{-\frac{3}{2}} \\ f_{xy}(x,y) = -\frac{1}{4}(x+y)^{-\frac{3}{2}} \\ f_{yy}(x,y) = -\frac{1}{4}(x+y)^{-\frac{3}{2}}} \\ \qquad \Rightarrow \cases{f(1,0) = 1 \\ f_x(1,0) = \frac{1}{2} \\f_y(1,0) = \frac{1}{2} \\f_{xx}(1,0) = -\frac{1}{4} \\f_{xy}(1,0) = -\frac{1}{4} \\f_{yy}(1,0) = -\frac{1}{4}} \\ \Rightarrow P_2(x,y) = 1 + \frac{1}{2}(x-1) + \frac{1}{2}(y) + \frac{1}{2}\left[ -\frac{1}{4}(x-1)^2 + 2\left(-\frac{1}{4}\right)(x-1)(y) - \frac{1}{4}(y)^2 \right] \\ \qquad = \bbox[red, 2pt]{ P_2(x,y) = 1 + \frac{1}{2}(x-1) + \frac{1}{2}y - \frac{1}{8}(x-1)^2 - \frac{1}{4}(x-1)y - \frac{1}{8}y^2 }$$

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