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: $$\textbf{(a)}\; \lim_{x\to 2}{|6x-17|-|6x-7|\over 3x-6} =\lim_{x\to 2}{17-6x-(6x-7)\over 3x-6}  =\lim_{x\to 2}{24-12x\over 3x-6}  \\\quad =\lim_{x\to 2} (-4)= \bbox[red, 2pt]{-4} \\\textbf{(b)}\; L=[\cos(2x)]^{x^{-2} } \Rightarrow \ln L={\ln \cos(2x) \over x^2} \Rightarrow \lim_{x\to 0}\ln L= \lim_{x\to 0}{\ln \cos(2x) \over x^2} = \lim_{x\to 0}{(\ln \cos(2x))' \over (x^2)'} \\\quad = \lim_{x\to 0}{{-2\sin(2x)\over \cos(2x)}\over 2x} = \lim_{x\to 0}{-\sin(2x)\over x\cos(2x)}  = \lim_{x\to 0}{(-\sin(2x))'\over (x\cos(2x))'}   = \lim_{x\to 0}{-2\cos(2x)\over \cos(2x)-2x\sin(2x)} \\\quad =-2 \Rightarrow \lim_{x\to 0} L= \bbox[red,2pt]{e^{-2}}$$

解答: $$\cases{x(t)=t^3+1\\ y(t)=t^4+t} \Rightarrow \cases{x'(t)=3t^2\\ y'(t)= 4t^3+1} \Rightarrow y'={dy\over dt}={4t^3+1\over 3t^2} \\ \Rightarrow y'(0,0)= \left. {4t^3+1\over 3t^2}\right|_{t=-1} =\bbox[red, 2pt]{-1}$$

解答: $$y^5+5xy+1=0 \Rightarrow 5y^4y'+5y+5xy'=0 \Rightarrow (y^4+x)y'+y=0 \Rightarrow y'=-{y\over y^4+x}  \\ \Rightarrow y'(0,-1)= 1 \Rightarrow y''=-{y'\over y^4+x} +{y(4y^3y'+1)\over (y^4+x)^2} \Rightarrow y''(0,-1)=-1+3=\bbox[red, 2pt]2$$

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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