108年宜蘭大學轉學考-微積分詳解

國立宜蘭大學108學年度暑假轉學招生考試微積分試題

解答： $$\lim_{t\to 0}{ \tan t\sec 2t\over 3t} =\lim_{t\to 0}{ \sin t\over 3t \cos t \cos(2t)} = \lim_{t\to 0}{ 2\sin t\over 3t(\cos(3t)+\cos t )} =  \lim_{t\to 0}{ (2\sin t)'\over (3t(\cos(3t)+\cos t ))'} \\=   \lim_{t\to 0}{ 2\cos t\over 3(\cos(3t)+\cos t ) -3t(3\sin(3t)+\sin(t))} = {2\over 6} ={1\over 3}，故選\bbox[red, 2pt]{(C)}$$

解答： $$f(x)={x-4\over x+4} =1-{8\over x+4} \Rightarrow f'(x)={8\over (x+4)^2} \Rightarrow f'(3)={8\over 49}，故選\bbox[red, 2pt]{(B)}$$

解答： $$\lim_{x \to \infty}{3x+2\over \sqrt{2x^2 +1}} =\lim_{x \to \infty} {3+2/x\over \sqrt{2+1/x^2}} ={3\over \sqrt 2}，故選\bbox[red, 2pt]{(C)}$$

解答： $$f(x)= x^3-3x^2-24x+2 \Rightarrow f'(x)=3x^2-6x-24 \Rightarrow f''(x)=6x-6\\ 因此f'(x)=0 \Rightarrow x^2-2x-8=(x-4)(x+2)=0 \Rightarrow x=4,-2 \Rightarrow \cases{f''(4)=18 \gt 0\\ f''(-2)= -18\lt 0} \\ \Rightarrow \cases{f(-2)=30 \\ f(-3)= 20\\ f(1)= -24} \Rightarrow 絕對最大值=30，故選\bbox[red, 2pt]{(B)}$$

解答： $$g(2)=2^3 = a\cdot 2^2 \Rightarrow a=2，故選\bbox[red, 2pt]{(E)}$$

解答： $$\int_0^{\pi/4} \int_0^{\cos \theta} 3r^2 \sin \theta \,drd\theta =\int_0^{\pi/4} \cos^3\theta \sin \theta \,d\theta =\int_1^{\sqrt 2/2} -u^3\,du (取u=\cos\theta \Rightarrow du=-\sin\theta d\theta) \\ =\left.\left[ -{1\over 4}u^4 \right] \right|_1^{\sqrt 2/2} =-{1\over 16}+{1\over 4} ={3\over 16}，故選\bbox[red, 2pt]{(D)}$$

解答： $$取\cases{u=x \Rightarrow du=dx\\ dv=e^{-x/2}dx \Rightarrow v=-2e^{-x/2}} \Rightarrow \int_0^4 xe^{-x/2}\,dx = \left.\left[ -2xe^{-x/2} \right]\right|_0^4 +2\int_0^4 e^{-x/2}\,dx\\ =\left.\left[ -2xe^{-x/2} -4e^{-x/2}\right]\right|_0^4 =-12e^{-2}+4，故選\bbox[red, 2pt]{(B)}$$

解答： $$取\cases{u=x \Rightarrow du=dx\\ dv=\cos xdx \Rightarrow v=\sin x} \Rightarrow \int_0^{\pi/2} x\cos x\,dx = \left.\left[ x\sin x \right]\right|_0^{\pi/2} -\int_0^{\pi/2} \sin x\,dx\\ =\left.\left[ x\sin x+\cos x\right]\right|_0^{\pi /2} ={\pi \over 2}-1，故選\bbox[red, 2pt]{(C)}$$

解答： $$\int_0^2 \int_0^{\sqrt{2x-x^2}} xy\,dydx =\int_0^2 {1\over 2}x(2x-x^2)\,dx =\left.\left[ {1\over 3}x^3 - {1\over 8}x^4\right] \right|_0^2 ={8\over 3}-2={2\over 3}，故選\bbox[red, 2pt]{(D)}$$

解答： $$\sum_{n=1}^\infty {2\over 4n^2-1} =\sum_{n=1}^\infty {2\over (2n-1)(2n+1)} =\sum_{n=1}^\infty \left({1\over 2n-1}-{1\over 2n+1} \right)\\ = (1-{1\over 3}) +({1\over 3}-{1\over 5}) +({1\over 5}-{1\over 7}) +\cdots  =1，故選\bbox[red, 2pt]{(C)}$$

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