國立臺灣大學 115 學年度碩士班招生考試試題
科目: 微積分(D)
解答:$$\textbf{(a) }\text{Let }\cases{u(x)=f(x)-2e^{3x} \sin(4x) \\ v(x)=g(x)-2e^{3x} \cos(4x)} \text{, then we have }H(x)=u(x)^2+ v(x)^2, \text{ and}\\\quad \cases{u'(x)= f'(x)-(6e^{3x} \sin(4x)+ 8e^{3x} \cos(4x))=3f(x)+4g(x)-(6e^{3x} \sin(4x)+ 8e^{3x} \cos(4x)) \\v'(x)=g'(x)-(6e^{3x}\cos(4x)-8e^{3x}\sin (4x)) =-4f(x)+3g(x)-(6e^{3x}\cos(4x)-8e^{3x}\sin (4x))} \\ \Rightarrow \cases{u'(x)=(3(f(x)-2e^{3x}\sin (4x)))+ 4(g(x)- 2e^{3x}\cos(4x)) =3u(x)+4v(x) \\ v'(x)=-4(f(x)-2e^{3x} \sin(4x))+ 3(g(x)-2e^{3x} \cos(4x)) =-4u(x)+3v(x)} \\ \Rightarrow H'(x)=2u(x)u'(x)+2v(x)v'(x) =2u(x) \left( 3u(x)+4v(x) \right) +2v(x)(-4u(x)+3v(x)) \\\qquad =6u(x)^2+ 6v(x)^2=6H(x) \Rightarrow H'(x)=6H(x) \Rightarrow \bbox[red, 2pt]{\text{ the constant }k=6} \\\textbf{(b) }H'(x)=6H(x) \Rightarrow H(x)=Ce^{6x} \Rightarrow H(0)= (f(0)-2e^0\sin 0)^2+ (g(0)-2e^0 \cos 0)^2= 0=C \\ \quad \Rightarrow H(x)=0 \Rightarrow u(x)^2+ v(x)^2=0 \Rightarrow \cases{u(x)=0\\ v(x)=0} \Rightarrow \bbox[red, 2pt]{\cases{f(x)=2e^{3x} \sin(4x) \\g(x)=2e^{3x} \cos(4x)}}$$
解答:$$\textbf{(a) } f(x)={1\over 27+x^3} ={1\over 27}\cdot {1\over 1-(-x^3/27)} ={1\over 27} \sum_{k=0}^\infty \left( -{x^3\over 27} \right)^k \Rightarrow \bbox[red, 2pt]{f(x) =\sum_{k=0}^\infty (-1)^k{x^{3k} \over 27^{k+1}}}\\ \quad \left| -{x^3\over 27}\right| \lt 1 \Rightarrow |x^3|\lt 27 \Rightarrow |x|\lt 3 \Rightarrow \bbox[red, 2pt]{R=3} \\\textbf{(b)(i) } I_m= \int_0^3 {x^m\over 27+x^3}\,dx = \int_0^3 x^m \left( \sum_{k=0}^\infty (-1)^k {x^{3k}\over 27^{k+1}}\right) \,dx = \int_0^3 \left( \sum_{k=0}^\infty (-1)^k {x^{3k+m}\over 27^{k+1}}\right) \,dx \\\quad =\sum_{k=0}^\infty (-1)^k {1\over 27^{k+1}} \left. \left[ {x^{3k+m+1 }\over 3k+m+1} \right] \right|_0^3 =\sum_{k=0}^\infty (-1)^k {1\over 27^{k+1}} \cdot {3^{3k+m+1} \over 3k+m+1} \\\quad =\sum_{k=0}^\infty (-1)^k{3^{m-2} \over 3k+m+1} \Rightarrow \bbox[red, 2pt]{a_{k,m} ={3^{m-2} \over 3k+m+1}} \\ \textbf{(b)(ii)} S= \sum_{k=0}^\infty (-1)^k{108k+45\over (3k+1)(3k+2)} = \sum_{k=0}^\infty (-1)^k \left( {9\over 3k+1}+{27\over 3k+2} \right)\\\quad = \sum_{k=0}^\infty (-1)^k \left( 81a_{k,0}+ 81a_{k,1} \right) =81 \int_0^3 \left( {1\over 27+x^3} +{x\over 27+x^3} \right)\,dx \\\quad = 3 \int_0^3 \left( {-2\over x+3}+{2x-3\over x^2-3x+9}+{18\over (x-3/2)^2+(3\sqrt 3/2)^2} \right) \,dx \\\quad =3 \left. \left[ -2\ln|x+3| + \ln|x^2-3x+9| +4\sqrt 3 \tan^{-1} {2x-3\over 3\sqrt 3}\right] \right|_0^3 = \bbox[red, 2pt]{-6\ln 2+4\sqrt 3 \pi}$$
解答:$$\textbf{(a)} \cases{u=x+y\\ v=x-y} \Rightarrow \cases{x=(u+v)/2\\ y=(u-v)/2} \Rightarrow |J|= \begin{Vmatrix} 1/2& 1/2\\ 1/2&-1/2 \end{Vmatrix} ={1\over 2} \\ \Rightarrow \iint_R (x+y)^2 \,dA = \int_{-1}^1 \int_{-1}^1 u^2\cdot {1\over 2}\,du\,dv = \bbox[red, 2pt]{2\over 3} \\ \textbf{(b) }\iiint_D (x+y+z)^2 \,dV =\iiint_D x^2\,dV + \iiint_D y^2\,dV+ \iiint_D z^2\,dV +\iiint_D 2(xy+yz+zx) \,dV \\\quad =\iiint_D z^2\,dV +\iiint_D z^2\,dV +\iiint_D z^2\,dV + 0=\iiint_D 3z^2\,dV \\\quad = 8\iiint_{D_1} 3z^2\,dV ,\text{ where }D_1=\{(x,y,z) \mid x+y+z\le 1, x\ge 0, y\ge 0,z\ge 0\} \\ \quad = 24 \int_0^1 \int_0^{1-z} \int_0^{1-y-z}z^2 \,dx\,dy\,dz = \bbox[red, 2pt]{2\over 5}$$
解答:$$\textbf{(a) }\text{The constraints define a closed and bounded, and the objective function}\\\quad f\text{ is continuous. By the Extreme Valuse Theorem, }f\text{ has its maximum and minimum values.} \\\textbf{(b) }\text{Suppose the optimal solution occurs at }P_1x_1^* + P_2x_2^* +P_3x_3^* \lt I. \text{ Because }{\partial U\over \partial x_1}\gt 0,\\ \quad \text{We can silightly increase }x_1\text{ by a small amount without violating the budget constraint.} \\\quad \text{This slight increase would yield a higher utility value than }U(x_1^*,x_2^*,x_3^*). \text{ This contradicts} \\\quad \text{ the assumption that }(x_1^*,x_2^*, x_3^*) \text{ is the global maximizer. Therefore, at the maximum,} \\\quad \text{ the budget constraint must hold with equality.} \\\textbf{(c) } L= U(x_1,x_2,x_3)+ \lambda(I-P_1x_1-P_2x_2-P_3x_3) \\\quad \text{According to the Envelope Theorem, we have } \cases{\displaystyle {\partial U^*\over \partial I}= {\partial L\over \partial I}= \lambda^*\\ \displaystyle {\partial U^*\over \partial P_i} = {\partial L\over \partial P_i}=- \lambda^*x_i} \Rightarrow \\\quad \text{Increasing I expands the feasible region, which can only increase or maintain the maximum} \\\text{ possible value of }U. \text{ Conversely, increasing }P_i \text{ will shrinks the feasible region, which can }\\\text{only decrease or maintain the maximum possible value of }U\\ \text{For example, in real-life,} \cases{U: \text{ the consumer's level of satisfaction}\\ x_i: \text{ the quantity consumed of various goods} \\P_i: \text{ price of }x_i\\ I: \text{the consumer's total income or budget}}$$
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解題僅供參考，碩士班歷年試題及詳解
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