114年台綜大轉學考-微積分A詳解

 臺灣綜合大學系統114學年度學士班轉學生聯合招生考試

科目名稱:微積分A

解答:$$f(x)=\log_2 (x^2) ={\ln x^2\over \ln 2} \Rightarrow f'(x)={2\over x\ln 2} \Rightarrow \bbox[red, 2pt]{f'(2)={2\over \ln 2}} \\g(x)=\log_x (2^x) ={\ln 2^x\over \ln x} \Rightarrow g'(x)={\ln 2(\ln x-1) \over (\ln x)^2} \Rightarrow g'(2)={\ln 2(\ln 2-1) \over (\ln 2)^2} \Rightarrow \bbox[red, 2pt]{g'(2)=1-{1\over \ln 2}}$$

解答:$$\lim_{x\to 0}\frac{\sin(2 \sin^{-1}(3x))-6x}{x^3} =\lim_{x\to 0}\frac{2\sin(\sin^{-1}(3x)) \cos(\sin^{-1}(3x))-6x}{x^3} =\lim_{x\to 0}\frac{6x \cos(\sin^{-1}(3x)) -6x}{x^3} \\=\lim_{x\to 0}\frac{6\cos(\sin^{-1}(3x))- 6}{x^2} =\lim_{x\to 0}\frac{-6\sin(\sin^{-1}(3x))\cdot{3\over \sqrt{1-9x^2}}}{2x}  =\lim_{x\to 0}{-27 \over \sqrt{1-9x^2}} =\bbox[red, 2pt]{-27}$$

解答:$$f(x)=x\sqrt{1+2x^2} \Rightarrow f(2)=6 \Rightarrow f^{-1}(6)=2 \\ f'(x)=\sqrt{1+2x^2}+{2x^2  \over \sqrt{1+2x^2}} \Rightarrow f'(2)={17\over 3} \Rightarrow g'(6) =(f^{-1})'(6) ={1\over f'(f^{-1}(6))} =\bbox[red, 2pt]{3\over 17}$$

解答:$$L=\left[ {2 \arctan(2n) \over \pi}\right]^n \Rightarrow \ln L=n \ln{2 \arctan(2n) \over \pi} \Rightarrow \lim_{n\to \infty} \ln L=\lim_{n\to \infty} \frac{\displaystyle \ln{2 \arctan(2n) \over \pi}}{1/n} \\=\lim_{n\to \infty}\frac{2\over (4n^2+1) \arctan(2n)}{-1/n^2} =\lim_{n\to \infty}{-2n^2\over (4n^2+1) \arctan(2n)} ={-1\over 2 \cdot {\pi\over 2}} =-{1\over \pi} \\ \Rightarrow \lim_{n\to \infty}   L = \bbox[red, 2pt]{e^{-1/\pi}}$$

解答:$$\int_0^{\pi/3} (3+\tan \theta \sec \theta)^2\,d\theta =\int_0^{\pi/3} (9+ 6\tan \theta \sec \theta + \tan^2 \theta \sec^2\theta)\,d\theta \\= \int_0^{\pi/3}9\,d \theta+ 6\int_0^{\pi/3} \tan \theta \sec \theta\,d\theta +\int_0^{\pi/3} \tan^2 \theta \sec^2 \theta\,d\theta \\= \left. \left[ 9\theta \right] \right|_0^{\pi/3} + 6\left. \left[ \sec \theta \right] \right|_0^{\pi/3}+  \left. \left[{1 \over 3}\tan^3 \theta \right] \right|_0^{\pi/3} =3\pi+6(2-1)+{1\over 3}\cdot 3\sqrt 3 = \bbox[red, 2pt]{3\pi+6+\sqrt 3}$$

解答:$$e^x =\sum_{n=0}^\infty {x^n\over n!} \Rightarrow e^{-4x} =\sum_{n=0}^\infty {(-1)^nx^{4n}\over n!}  \Rightarrow f(x) =(1-x)^3 e^{-4x} \\=(1-3x+3x^2-x^3) \left(1-{x^4} +\cdots-{1\over 505!}x^{2020} +{1\over 506!}x^{2024}-\cdots \right) =\sum_{n=0}^{\infty} a_nx^n \\ \Rightarrow a_{2025} =(-3)\cdot {1\over 506!} \Rightarrow f^{[2025]}(0)= -{3\over 506!}\cdot 2025! = \bbox[red, 2pt]{-{3\cdot 2025!\over 506!}}$$

解答:$$\cases{x=r \cos \theta=e^{2\theta} \cos \theta={\sqrt 2\over 2}e^{\pi/2} \\ y=r\sin \theta=e^{2\theta }\sin \theta={\sqrt 2\over 2}e^{\pi/2}} \Rightarrow {dy\over dx } ={{dr \over d\theta}\sin \theta+ r\cos \theta\over {dr\over d\theta}\cos \theta -r\sin \theta} ={2e^{2\theta} \sin \theta+e^{2\theta} \cos \theta \over 2e^{2\theta} \cos \theta- e^{2\theta}\sin \theta} \\={2\sin \theta+\cos \theta\over 2\cos \theta-\sin \theta} ={\sqrt 2+\sqrt 2/2\over \sqrt 2-\sqrt 2/2} =3 \Rightarrow \\ r=e^{2\theta} \Rightarrow r'=2e^{2\theta} \Rightarrow L=\int_0^{\pi/2} \sqrt{r^2+(r')^2}\,d\theta =\int_0^{\pi/2} \sqrt{e^{4\theta}+ 4e^{4\theta}}\,d \theta =\int_0^{\pi/2} \sqrt 5 e^{2\theta}\,d\theta \\=\left. \left[ {\sqrt 5\over 2}e^{2\theta}\right] \right|_0^{\pi/2} = \bbox[red, 2pt]{{\sqrt 5\over 2}\left(e^\pi-1 \right)}$$

解答:$$\cases{F_1(x,y,z)=xy+yz+zx+14\\ F_2(x,y,z) =x^2+y^2+z^2-29} \Rightarrow \cases{\nabla F_1=(y+z,x+z,y+x) \\ \nabla F_2=(2x,2y,2z)} \Rightarrow \cases{\nabla F_1(2,3,-4)=(-1,-2,5) \\ \nabla F_2(2,3,-4) =(4,6,-8)} \\ \Rightarrow \vec u=\nabla F_1\times \nabla F_2 =(-14,12,2) \parallel (a,b,1) \Rightarrow \bbox[red, 2pt]{\cases{a=-7\\ b=6}}$$

解答:$$\cases{f(x,y) = \sqrt{x^2+y^2} \\g(x,y) =9x^2+16xy+21y^2-125} \Rightarrow \cases{f_x= \lambda g_x\\ f_y= \lambda g_y\\ g=0} \Rightarrow \cases{{x\over \sqrt{x^2+y^2}}=\lambda (18x+16y) \\ {y\over \sqrt{x^2+y^2}} =\lambda(16x+42y) \\9x^2+16xy+21y^2=125} \\ \Rightarrow {x\over y}={18x+16y\over 16x+42y} \Rightarrow 2x^2+3xy-2y^2=0 \Rightarrow (2x-y)(x+2y) =0 \\ \Rightarrow \cases{2x=y \Rightarrow x^2=1 \Rightarrow x=\pm 1 \Rightarrow (x,y)=(1,2),(-1,-2)\\ x=-2y \Rightarrow y^2=5 \Rightarrow y=\pm\sqrt 5 \Rightarrow (x,y) =(-2\sqrt 5,\sqrt 5),(2\sqrt 5, -\sqrt 5)} \\ \Rightarrow \cases{f(\pm 1,\pm 2) =\sqrt 5\\ f(\pm 2\sqrt 5, \mp\sqrt 5) =5} \Rightarrow \bbox[red, 2pt]{\cases{\text{ the shortest distance: }\sqrt 5\\ \text{the longest distance: }5}}$$

解答:$$\cases{u=x^2/y \\ v=y^2/x} \Rightarrow \cases{x=u^{2/3}v^{1/3}\\ y= u^{1/3}v^{2/3}} \Rightarrow {\partial(x,y) \over \partial(u,v)} =\begin{vmatrix} {2\over 3}u^{-1/3}v^{1/3} &{1\over 3} u^{2/3} v^{-2/3} \\ {1\over 3}u^{-2/3}v^{2/3} & {2\over 3}u^{1/3}v^{-1/3}\end{vmatrix} ={1\over 3} \\ \Rightarrow \iint_D {y^2\over x^4+3y^2}\,dA= \int_{u=1}^3 \int_{v=1}^3 {u^{2/3}v^{4/3} \over u^{8/3}v^{4/3} +3u^{2/3}v^{4/3}} \cdot {1\over 3}\,dvdu= {1\over 3}\int_{u=1}^3 \int_{v=1}^3 {1\over u^2+3} \,dvdu \\= {2\over 3}\int_{u=1}^3 {1\over u^2+3}\,du= {2\over 9}\int_{u=1}^3 {1\over (u/\sqrt 3)^2+1}\,du ={2\over 9} \left. \left[ \sqrt 3\tan^{-1}{u\over \sqrt 3}\right] \right|_1^3 \\={2\sqrt 3\over 9} \left( {\pi\over  3}-{\pi\over 6} \right) = \bbox[red, 2pt]{{\sqrt 3\over 27} \pi}$$

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解題僅供參考，轉學考歷年試題及詳解