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2024年10月9日 星期三

113年台綜大轉學考-微積分C詳解

 臺灣綜合大學系統113學年度學士班轉學生聯合招生考試

科目名稱:微積分C

解答:(a)limx2|6x17||6x7|3x6=limx2176x(6x7)3x6=limx22412x3x6=limx2(4)=4(b)L=[cos(2x)]x2lnL=lncos(2x)x2limx0lnL=limx0lncos(2x)x2=limx0(lncos(2x))(x2)=limx02sin(2x)cos(2x)2x=limx0sin(2x)xcos(2x)=limx0(sin(2x))(xcos(2x))=limx02cos(2x)cos(2x)2xsin(2x)=2limx0L=e2
解答:{x(t)=t3+1y(t)=t4+t{x(t)=3t2y(t)=4t3+1y=dydt=4t3+13t2y(0,0)=4t3+13t2|t=1=1
解答:y5+5xy+1=05y4y+5y+5xy=0(y4+x)y+y=0y=yy4+xy(0,1)=1y
解答:f(x)=x^8e^{1-x^2} \Rightarrow f'(x)=8x^7e^{1-x^2}-2x^9e^{1-x^2}=(8x^7-2x^9)e^{1-x^2}=0 \\ \Rightarrow x^7(8-2x^2)=0 \Rightarrow x=0,\pm 2 \Rightarrow \cases{f(0)=0\\ f(\pm 2)=256e^{-3}} \Rightarrow \text{ max}(f(x))= \bbox[red, 2pt]{256e^{-3}}
解答:I=\int_0^3 \sqrt{6x-x^2} \,dx =\int_0^3 \sqrt{9-(9-6x+x^2)} \,dx  =\int_0^3 \sqrt{9-(3-x)^2} \,dx \\ 取3\cos u=3-x \Rightarrow 3\sin u\,du=dx \Rightarrow I=3\int_0^{\pi/2} \sin u\sqrt{9-9\cos^2 u}\,du =9\int_0^{\pi/2} \sin^2 u\,du \\={9\over 2}\int_0^{\pi/2}(1-\cos 2u)\,du ={9\over 2}\left. \left[ u-{1\over 2}\sin 2u \right] \right|_0^{\pi/2} =\bbox[red, 2pt]{{9\over 4}\pi}
解答:R面積=\int_0^1 (x^2-x^7)\,dx ={5\over 24} \Rightarrow R質心的x坐標\bar x={\int_0^1 (x^3-x^8)\,dx\over 5/24} ={5/36\over 5/24} ={2\over 3} \\ \Rightarrow 質心與x=3的距離=3-{2\over 3}={7\over 3} \Rightarrow 欲求之體積={5\over 24}\cdot 2\pi\cdot {7\over 3} =\bbox[red, 2pt]{{35\over 36}\pi}
解答:a_n={n(x-3)^n\over 2^n(n^2+1)} \Rightarrow \lim_{n\to \infty}\left| {a_{n+1}\over a_n}\right| = \lim_{n\to \infty}\left| {(n+1)(x-3)^{n+1}\over 2^{n+1}((n+1)^2+1)} \cdot {2^n(n^2+1) \over n(x-3)^n}\right|\\ = \lim_{n\to \infty}\left| {(n+1)(x-3)(n^2+1)\over 2n(n^2+2n+2)}\right| = \left|{1\over 2}(x-3) \right|\lt 1 \Rightarrow -2\lt x-3\lt 2 \Rightarrow 1\lt x\lt 5\\ x=1 \Rightarrow \sum_{n=1}^\infty a_n= \sum_{n=1}^\infty (-1)^n{ n \over n^2+1}收斂(交錯級數審斂法) \\ x=5\Rightarrow \sum_{n=1}^\infty a_n= \sum_{n=1}^\infty{n\over n^2+1}發散 (積分審斂法)\\因此收斂區間為\bbox[red, 2pt]{[1,5)}
解答:f(x,y,z)=x^2+y^2-4xy+z+1 \Rightarrow \nabla f=(f_x,f_y,f_z)= (2x-4y,2y-4x,1) \\ \Rightarrow \nabla f=(2x-4y,2y-4x,1)=\lambda(1,1,2) \Rightarrow \lambda={1\over 2} \Rightarrow \cases{2x-4y=1/2\\ 2y-4x=1/2} \Rightarrow x=y=-{1\over 4} \\ \Rightarrow (a,b,c)=\bbox[red, 2pt]{(-{1\over 4},-{1\over 4},t),t\in \mathbb R}
解答:  \iint_D \cos(x+y)\,dA = \int_{-\pi/6}^0 \int_{-\pi/6-x}^{x+\pi/6} \cos(x+y)\,dydx +\int_0^{\pi/6}  \int_{x-\pi/6 }^{\pi/6-x} \cos(x+y)\,dydx  \\=\int_{-\pi/6}^0 \left( \sin(2x+{\pi\over 6})+{1\over 2} \right)\,dx +\int_0^{\pi/6} \left( {1\over 2}-\sin(2x-{\pi\over 6}) \right)\,dx ={\pi\over 12}+{\pi\over 12} =\bbox[red, 2pt]{\pi\over 6}另解:\cases{u= x+y\\ v=x-y} \Rightarrow \left|{\partial (u,v)\over \partial(x,y)} \right| =\begin{Vmatrix}1& 1\\ 1& -1 \end{Vmatrix} =2 \Rightarrow \iint_D \cos(x+y)\,dA= {1\over 2}\int_{-\pi/6}^{\pi/6} \int_{-\pi/6}^{\pi/6} \cos u\,dudv \\={1\over 2}\int_{-\pi/6}^{\pi/6}1\,dv={1\over 2}\cdot {\pi\over 3}=\bbox[red, 2pt]{\pi\over 6}
解答:
散度定理: \iint_S \mathbf F\cdot dS =\iiint_E \text{div }\mathbf F\,dV\\ \mathbf F=(9x^2z,8\sin(x^3)+e^{2z},x^5y\ln(x^2+1)) \Rightarrow \text{div }\mathbf F=18xz \\ \Rightarrow \iint_S \mathbf F\cdot dS =\int_0^3\int_0^2 \int_0^{6-3z}18xz\, dxdzdy =81\int_0^3\int_0^2 (z^3-4z^2+4z)dzdy =81\int_0^3 {4\over 3}dy\\ =\bbox[red, 2pt]{324}

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解題僅供參考,轉學考歷年試題及詳解

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