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

解答：$$\lim_{t\to 0}{\tan t\sec(2t)\over 3t} =\lim_{t\to 0}{{d\over dt}\tan t\sec(2t)\over {d\over dt}3t} = \lim_{t\to 0}{\sec^2  t  \sec(2t) + \tan t\cdot 2\sec(2t)\tan (2t)\over 3} ={1\over 3}\\，故選\bbox[red, 2pt]{(C)}$$

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

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

解答：$$\int_0^{\pi/4} \int_0^{\cos \theta} 3r^2\sin \theta \,drd\theta = \int_0^{\pi/4} \left.\left[ r^3\sin \theta \right] \right|_0^{\cos\theta}\,d\theta = \int_0^{\pi/4} \cos^3\theta\sin \theta  \,d\theta = \int_0^{\pi/4} {1\over 2} \cos^2\theta\sin 2\theta  \,d\theta \\=\int_0^{\pi/4} {1\over 4} (\cos 2\theta+1)\sin 2\theta  \,d\theta =\int_0^{\pi/4} {1\over 8}  \sin 4\theta +{1\over 4}\sin 2\theta \,d\theta =\left. \left[ -{1\over 32}\cos 4\theta -{1\over 8}\cos 2\theta \right]\right|_0^{\pi/4}\\ ={1\over 32}-(-{1\over 32}-{1\over 8})={3\over 16}，故選\bbox[red, 2pt]{(D)}$$

解答：$$令\cases{u=x \Rightarrow du=dx\\ dv= e^{-x/2}\,dx \Rightarrow v=-2e^{-x/2}}，則\int_0^4 xe^{-x/2}\,dx = \left.\left[ -2xe^{-x/2} -4e^{-x/2} \right] \right|_0^{4} \\ =-8e^{-2}-4e^{-2} -(-4)=4-12e^{-2}，故選\bbox[red, 2pt]{(B)}$$

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

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

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

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