113年台綜大轉學考-微積分B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B

 

解答:$$\textbf{(a)}\; f(x)=(x^2-x+1)^{100} \Rightarrow f'(x)=100(x^2-x+1)^{99}(2x-1) \Rightarrow f'(0)=100\cdot 1\cdot (-1) \\\quad =\bbox[red, 2pt]{-100} \\\textbf{(b)}\; f''(x)=9900(x^2-x+1)^{98}(2x-1)^2+ 200(x^2-x+1)^{99} \Rightarrow f''(0)= 9900+200\\ \quad = \bbox[red, 2pt]{10100}$$

解答:$$\lim_{x\to 0} {\sqrt{3+\cos 3x}-2 \over x^2} =\lim_{x\to 0} {(\sqrt{3+\cos 3x}-2)' \over (x^2)'} =\lim_{x\to 0}  {{-3\sin 3x\over 2\sqrt{3+\cos 3x}} \over 2x} = \lim_{x\to 0}  {{-3\sin 3x\over 4x\sqrt{3+\cos 3x}} }  \\= \lim_{x\to 0}  {{(-3\sin 3x )'\over (4x\sqrt{3+\cos 3x})'} }  = \lim_{x\to 0}  {-9\cos 3x\over 4\sqrt{3+\cos 3x}+{-6x \sin 3x\over \sqrt{3+\cos 3x}}}   = \bbox[red, 2pt]{-{9\over 8}}$$

解答:$$\textbf{(a)}\; f(x)=2^{x^2} =e^{x^2\ln 2} \Rightarrow f'(x)= 2x\ln 2 e^{x^2\ln 2} \Rightarrow f'(2)=4\ln 2\cdot 16=\bbox[red, 2pt]{64\ln 2} \\\textbf{(b)}\; g(x)=x^{2^x} =e^{2^x\ln x} \Rightarrow g'(x)=  (\ln 2 \cdot 2^x\ln x+{2^x\over x}) e^{2^x\ln x} \\ \quad \Rightarrow g'(2)=(4(\ln 2)^2+2) 2^{4} =\bbox[red, 2pt]{32+64(\ln 2)^2}$$

解答:$$\int_2^3 {x\over x^2-5x+4}\,dx =\int_2^3 {x\over (x-4)(x-1)}\,dx = \int_2^3 \left({4/3\over x-4} -{1/3 \over x-1} \right)\,dx  \\= \left. \left[ {4\over 3}\ln|x-4|-{1\over 3}\ln|x-1|\right] \right|_2^3 =-{1\over 3}\ln 2-{4\over 3}\ln 2 =\bbox[red, 2pt]{-{5\over 3}\ln 2}$$

解答:$$u=-\sqrt {3x} \Rightarrow du =-{3\over 2\sqrt {3x}}dx ={3\over 2u} dx \Rightarrow dx={2\over 3}u\,du \\ \Rightarrow \int e^{-\sqrt{3x}}\,dx = \int {2\over 3}ue^u\, du ={2\over 3}(ue^u-e^u) ={2\over 3}\left(-\sqrt{3x}e^{-\sqrt{3x}} -e^{-\sqrt{3x}}\right) \\ \Rightarrow  \int_3^\infty e^{-\sqrt{3x}}\,dx =\left. \left[ {2\over 3}\left(-\sqrt{3x}e^{-\sqrt{3x}} -e^{-\sqrt{3x}}\right) \right] \right|_3^\infty ={2\over 3}(3e^{-3}+e^{-3}) = \bbox[red, 2pt]{{8\over 3}e^{-3}}$$

解答:$$f(x)=\sqrt[5]{32+x} \Rightarrow \cases{f(0)=2\\ f'(x)={1\over 5 (32+x)^{4/5}}} \Rightarrow \cases{f'(0)= 1/80\\ f''(x)={-4\over 25(32+x)^{9/5}}} \\\Rightarrow \cases{f''(0)=-{1\over 3200} \\ f'''(x)={36\over 125(32+x)^{14/5}}} \Rightarrow f'''(0)= {9\over 512000}\\ \Rightarrow f(x)=f(0)+f'(0)x+ {1\over 2}f''(0)x^2 + {1\over 6}f'''(0)x^3+\cdots \\=2+{1\over 80}x-{1\over 6400}x^2+ {3\over 1024000}x^3+\cdots \Rightarrow \bbox[red, 2pt] {\cases{c_1=1/80\\ c_2=-1/6400\\ c_3= 3/1024000}}$$

解答:$$\textbf{(a)}\; f(x)=\sin(3x) \Rightarrow \cases{f'(x)=3\cos(3x)\\ f''(x)=-3^2\sin(3x) \\ f'''(x)=-3^3\cos(3x)\\ f^{[4]}(x)=3^4 \sin(3x)} \Rightarrow f^{[n]}(x)= \begin{cases}3^n \cos(3x) &n\equiv 1 \text{ mod } 4\\-3^n \sin(3x) &n\equiv 2 \text{ mod } 4\\-3^n \cos(3x) &n\equiv 3 \text{ mod } 4\\3^n \sin(3x) &n\equiv 0 \text{ mod } 4 \end{cases}\\\quad 111\equiv 3 \text{ mod }4 \Rightarrow f^{[111]}(0)=-3^{111} \cos 0=\bbox[red, 2pt]{-3^{111}} \\\textbf{(b)}\; \sin x= \sum_{n=0}^\infty (-1)^n {1\over (2n+1)!}x^{2n+1} \Rightarrow g(x)= \sin(x^3)= \sum_{n=0}^\infty (-1)^n {1\over (2n+1)!}x^{6n+3}  \\ n=18\Rightarrow g(x)的x^{111}係數為(-1)^{18} \cdot {1\over 37!} \Rightarrow g^{[111]}(0)=\bbox[red, 2pt]{111! \over 37!}$$

解答:$$f(x,y)=(4x^2+y^2)e^{-2x} \Rightarrow \cases{f_x= (8x-8x^2-2y^2)e^{-2x}\\ f_y= 2ye^{-2x}} \Rightarrow \cases{f_{xx}= (8-32x+16x^2+4y^2)e^{-2x} \\ f_{xy}=-4ye^{-2x}\\ f_{yy}=2e^{-2x}} \\ \Rightarrow d(x,y)=f_{xx}f_{yy}-f^2_{xy} =(16-64x+32x^2- 8y^2)e^{-4x}\\ \cases{f_x=0\\ f_y=0} \Rightarrow (x,y)=(0,0),(1,0) \Rightarrow \cases{d(0,0)=16\gt 0 \Rightarrow f_{xx}(0,0)=8\gt 0\\ d(1,0) =-16e^{-4}\lt 0} \\ \Rightarrow \bbox[red, 2pt]{\text{critical points: }(0,0,),(1,0), \text{ and local minimum: }f(0,0)=0, \text{saddle point: }(0,0)}$$

解答:$$\cases{P(x,y)= x^{1/3} y^{2/3} \\ g(x,y)=3x^{1/2}+ 5y^{1/2}-45} \Rightarrow \cases{P_x= \lambda g_x\\ P_y= \lambda g_y\\ g=0} \Rightarrow \cases{{1\over 3} x^{-2/ 3}y^{2/3} = \lambda \cdot {3\over 2}x^{-1/2} \cdots(1)\\ {2\over 3}x^{1/3}y^{-1/3} = \lambda \cdot {5\over 2}y^{-1/2} \cdots(2) \\ 3x^{1/2}+ 5y^{1/2}=45 \cdots(3)} \\ \Rightarrow {(1)\over (2)}={y\over 2x} ={3\sqrt y\over 5\sqrt x} \Rightarrow y^{1/2}={6\over 5}x^{1/2} 代入(3) \Rightarrow 3x^{1/2}+6x^{1/2} =45 \Rightarrow x^{1/2}=5 \\ \Rightarrow y^{1/2}=6 \Rightarrow P(25,36)= \bbox[red, 2pt]{25^{1/3} 36^{2/3}}$$

解答:$$改變積分順序: I=\int_0^4 \int_{\sqrt x}^2 \sqrt x\cdot {\sin(y^2)\over y^2} \,dy dx = \int_0^2 \int_0^{y^2}\sqrt x\cdot {\sin(y^2)\over y^2} \,dx dy = \int_0^2 {2\over 3}y \sin(y^2)\,dy \\ u=y^2 \Rightarrow du=2ydy \Rightarrow I= \int_0^4{1 \over 3}\sin u\,du =\left.\left[ -{1\over 3}\cos u\right] \right|_0^4 =\bbox[red, 2pt]{{1\over 3}(1-\cos 4)}$$

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