114年台北大學轉學考-微積分詳解

 國立臺北大學114學年度日間學士班轉學生招生考試

學制系級:經濟學系日間學士班2、3年級
科目:微積分

1.(50%)Answer the following questions.

解答:$$\lim_{x\to 0}{\sin 3x\over x} =\lim_{x\to 0}{3\cos 3x\over 1} =\bbox[red, 2pt] 3$$

解答:$$\lim_{x\to -\infty} {5^x+7^x\over 2\cdot 5^x-3\cdot 7^x} =\lim_{x\to -\infty} {1+(7/5)^x\over 2\cdot 1 -3\cdot (7/5)^x} = \bbox[red, 2pt]{1\over 2}$$

解答:$$f(x)=\sqrt x-\ln x \Rightarrow f'(x)={1\over 2\sqrt x}-{1\over x} =0 \Rightarrow x=4 \Rightarrow f(4)=2-2\ln 2 \\ \Rightarrow \text{range}(f)= \bbox[red, 2pt]{[2-2\ln 2, \infty)}$$

解答:$$z=\ln (1+x^2y^3) \Rightarrow {dz\over dx} ={{d\over dx}(1+x^2y^3)\over 1+x^2y^3} = \bbox[red, 2pt]{{2xy^3\over 1+x^2y^3}}$$

解答:$$y=x^3+\tan x \Rightarrow {dy\over dx} =3x^2+\sec^2 x \Rightarrow {dx\over dy} = \bbox[red, 2pt]{1\over 3x^2+\sec^2 x}$$

解答:$$y={e^x\over x^2+1} \Rightarrow {dy\over dx}={e^x\over x^2+1}-{2xe^x\over (x^2+1)^2} = \bbox[red, 2pt]{(x-1)^2e^x\over (x^2+1)^2}$$

解答:$$f(x)=x^3+x =2  \Rightarrow f(1)=2 \Rightarrow f^{-1}(2)=1\\ y=f^{-1}(x) \Rightarrow f(y)=x \Rightarrow y'f'(y)=1 \Rightarrow \left( f^{-1}(x)\right)'\cdot f'(f^{-1}(x)) =1\\ \Rightarrow \left( f^{-1}(x)\right)' ={1\over f'(f^{-1}(x))} \Rightarrow\left( f^{-1}(2)\right)' ={1\over f'(f^{-1}(2))} ={1\over f'(1)} ={1\over 3\cdot 1^2+1} = \bbox[red, 2pt]{1\over 4}$$

解答:$$\int_2^8 {1\over x+1}\,dx = \left. \left[ \ln(x+1)\right] \right|_2^8 =\ln 9-\ln 3= \bbox[red, 2pt]{\ln 3}$$

解答:$$\cases{u=x^2 \\ dv=e^x\,dx} \Rightarrow \cases{du=2x\,dx\\ v=e^x} \Rightarrow \int x^2e^x\,dx =x^2e^x-\int 2xe^x\,dx =x^2e^x-2(xe^x-e^x)+C\\ \Rightarrow \int_0^1 x^2e^x\,dx = \left. \left[ x^2e^x-2xe^x+2e^x\right] \right|_0^1 = \bbox[red, 2pt]{e-2}$$

解答:$$\int_0^a \int_0^b (x-a)(x-b)dxdy =\int_0^a \int_0^b (x^2-(a+b)x+ ab)dxdy = \int_0^a \left. \left[ {1\over 3 }x^3-{1\over 2}(a+b)x^2+abx  \right] \right|_0^b \,dy \\=\int_0^a \left( {1\over 3}b^3-{1\over 2}b^2(a+b) +ab^2\right)\,dy = \bbox[red, 2pt]{{1\over 3}ab^3-{1\over 2}ab^2(a+b) +a^2b^2}$$

2.(50%)Answer the following questions.

解答:$$y={x_1^2+x_2^2 \over x_1+x_2} \Rightarrow dy={\partial \over \partial x_1} \left({x_1^2+x_2^2 \over x_1+x_2} \right)\cdot dx_1 +{\partial \over \partial x_2} \left({x_1^2+x_2^2 \over x_1+x_2} \right)\cdot dx_2 \\= \bbox[red, 2pt] {{x_1^2+2x_1x_2-x_2^2 \over (x_1+x_2)^2}dx_1 +{-x_1^2+2x_1x_2+x_2^2 \over (x_1+x_2^2)}dx_2}$$

解答:$${\partial \over \partial x}(xye^z+xz^2-y)=0 \Rightarrow ye^z+xy{\partial z\over \partial x}e^z+z^2+2xz{\partial z\over \partial x}=0 \Rightarrow {\partial z\over \partial x}= -{ye^z+z^2 \over xye^z+2xz} \\ \Rightarrow {\partial z\over \partial x}(1,1,0)=-{1\over 1} \Rightarrow \bbox[red, 2pt]{{\partial z\over \partial x}(1,1,0)=-1} \\{\partial z\over \partial y}(xye^z+xz^2-y)=0 \Rightarrow xe^z+xy{\partial z\over \partial y}e^z+2xz{\partial z\over \partial y}-1=0 \Rightarrow {\partial z\over \partial y}={1-xe^z\over xy+2xz} \\ \Rightarrow \bbox[red, 2pt]{{\partial z\over \partial y}(1,1,0)=0}$$


解答:$$f(x)=\ln x \Rightarrow f'(x)={1\over x} \Rightarrow f''(x)=-{1\over x^2} \Rightarrow f'''(x)={2\over x^3} \\ \Rightarrow f(x) \approx f(1)+f'(1)(x-1)+{1\over 2}f''(1)(x-1)^2+{1\over 6}f'''(1)(x-1)^3 \\\qquad =0+(x-1)+{1\over 2}(-1)(x-1)^2+{1\over 6}\cdot 2(x-1)^3\\ \Rightarrow \bbox[red, 2pt]{f(x) \approx (x-1)-{1\over 2}(x-1)^2+{1\over 3}(x-1)^3}$$

解答:$$\bbox[cyan,2pt]{試題有疑義}:應該是求最小值，而非最大值;\\ x+2y=6 \Rightarrow x=6-2y \Rightarrow f(6-2y,y) =(6-2y)^2+y^2=5y^2-24y+36 \\ \Rightarrow y={12\over 5} \Rightarrow x=6-{24\over 5}={6\over 5} \Rightarrow f\bbox[red, 2pt]{(6/5,12/5)}={36\over 5}$$

解答:$$f(x)=x^4-8x^2+16 \Rightarrow f'(x)=4x^3-16x =4x(x^2-4) =4x(x-2)(x+2) =0 \\ \Rightarrow x=0,\pm 2 \Rightarrow \text{critical points: }(0,f(0))=(0,16), (2,f(2)) =(2,0), (-2,f(-2)) =(-2,0) \\ \Rightarrow \bbox[red, 2pt]{\text{critical points: }(0,16), (\pm 2,0) }\\ f''(x)=12x^2-16 \Rightarrow \cases{f''(0) \lt 0\\ f''(\pm 2) \gt 0} \Rightarrow \bbox[red, 2pt]{\cases{\text{local maximum: }16\\ \text{local minimum: }0}}$$

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