107年海洋大學轉學考-微積分詳解

國立臺灣海洋大學107學年度轉學生入學招生考試

考試科目：微積分
系組名稱：商船二、輪機能源二(F)、輪機動力二(F)、電機二(B)、資工二(B)、通訊二(B)

解答：$$\mathbf{(1)}\; \lim_{x\to 0^+}\left({2^x+3^x \over 2} \right)^{1/x} =\lim_{x\to 0^+} e^{\ln(2^x+3^x)-\ln 2\over x} =\lim_{x\to 0^+} e^{(\ln(2^x+3^x)-\ln 2)'\over x'}\\ =\lim_{x\to 0^+} e^{\ln 2 \cdot 2^x+\ln 3\cdot 3^x \over 2^x+3^x} =\lim_{x\to 0^+} e^{\ln 2+\ln 3\over 2} =\sqrt 2\cdot \sqrt 3=\bbox[red, 2pt]{\sqrt 6}\\ \mathbf{(2)}\; \lim_{x\to 0^+} \sqrt x\ln x =\lim_{x\to 0^+} \sqrt {x(\ln x)^2} =\bbox[red, 2pt] 0\\ \mathbf{(3)}\; {d\over dx}[\ln |\tan x|] ={ (\tan x)'\over \tan x} ={\sec^2 x\over \tan x} =\bbox[red,2pt]{1\over \sin x\cos x} \\ \mathbf{(4)}\;{d\over dx}[\tan^{-1}(\sin 2x)] ={ (\sin 2x)'\over 1+\sin^2(2x)} = \bbox[red,2pt]{ 2 \cos 2x\over 1+\sin^2(2x)} \\\mathbf{(5)}\;{d\over dx}[{x^2+1\over x^3-1}] ={2x\over x^3-1} +{(x^2+1)(-3x^2)\over (x^3-1)^2} = \bbox[red, 2pt]{{2x\over x^3-1} -{3x^4 +3x^2\over (x^3-1)^2} }$$

解答：$$\mathbf{(1)}\;f'(0)= \lim_{h \to 0}{f(h)-f(0)\over h}= \lim_{h \to 0}{h^3\sin (1/h)\over h} =\lim_{h \to 0}{h^2\sin (1/h)\ } \\ 由於-h^2 \le h^2\sin(1/h)\le h^2 \Rightarrow \lim_{h \to 0}-h^2 \le \lim_{h \to 0}h^2\sin(1/h) \le \lim_{h \to 0} h^2 \\ \Rightarrow 0\le \lim_{h \to 0} h^2\sin(1/h)\le 0 \Rightarrow h^2\sin(1/h)=0 \Rightarrow \bbox[red,2pt]{f'(0)=0}\\ \mathbf{(2)}\;又{d\over dx}\left( x^3\sin{1\over x}\right) = 3x^2\sin{1\over x}-x\cos{1\over x}且\lim_{x\to 0}\left( 3x^2\sin{1\over x}-x\cos{1\over x} \right) =0\\ 因此\bbox[red,2pt]{f'(x)=3x^2\sin{1\over x}-x\cos{1\over x}}$$

解答：$$6+\int_a^x {f(t)\over t^2}\,dt = 2\sqrt x \Rightarrow {d\over dx}\left( 6+\int_a^x {f(t)\over t^2}\,dt \right) ={d\over dx} (2\sqrt x )\Rightarrow {f(x) \over x^2} ={1\over \sqrt x} \Rightarrow f(x)=x^{3/2} \\ \Rightarrow 6+\int_a^x {t^{3/2}\over t^2}\,dt =6+\int_a^x {1\over \sqrt t}\,dt = 6+\left.\left[ 2\sqrt t\right] \right|_a^x =6+2\sqrt x-2\sqrt a= 2\sqrt x \\ \Rightarrow \sqrt a=3 \Rightarrow a=9 \Rightarrow \bbox[red,2pt]{f(x)=x^{3/2},a=9}$$

解答：$$\mathbf{(1)}\;u=x^2 \Rightarrow du=2xdx \Rightarrow \int {x\over 1+x^4}\,dx ={1\over 2}\int {1\over 1+u^2}\,du ={1\over 2}\tan^{-1}u +C\\ \Rightarrow \int_0^\sqrt 3{x\over 1+x^4}\,dx =\left.\left[ {1\over 2}\tan^{-1}x^2 \right]\right|_0^\sqrt 3 ={1\over 2} \tan^{-1}\sqrt 3={1\over 2}\cdot {\pi\over 3} =\bbox[red,2pt]{\pi\over 6} \\\mathbf{(2)}\; \int_{-\pi}^\pi \sin^2 x\,dx =2\int_0^\pi \sin^2 x\,dx = \int_0^\pi 1-\cos 2x\,dx =\left.\left[ x-{1\over 2}\sin 2x \right] \right|_0^\pi = \bbox[red,2pt]\pi \\\mathbf{(3)}\; u=\ln x \Rightarrow du={1\over x}dx \Rightarrow e^u\,du=dx \Rightarrow \int \cos(\ln x)\,dx =\int e^u\cos u\,du \\ ={1\over 2}e^u(\sin u+\cos u)+C =\bbox[red, 2pt]{{1\over 2}x(\sin (\ln x)+\cos(\ln x))+C} \\\mathbf{(4)}\;\int_0^1\int_y^1 \sin x^2\,dxdy =\int_0^1 \int_0^x \sin x^2 \,dydx =\int_0^1 x\sin x^2\,dx =\left.\left[ -{1\over 2}\cos x^2\right]\right|_0^1 =\bbox[red, 2pt]{{1\over 2}(1+\cos 1)} \\\mathbf{(5)}\; \int_0^{\pi/2} \int_0^{1-\cos \theta} (\sin \theta)r\,drd\theta =\int_0^{\pi/2} {1\over 2}(1-\cos\theta)^2\sin\theta \,d\theta  ={1\over 2}\int_0^{\pi/2} \sin\theta -2\sin\theta\cos \theta+\cos^2\theta \sin \theta \,d\theta\\ ={1\over 2}\left(\left.\left[ -\cos\theta+ {1\over 2}\cos 2\theta -{1\over 3} \cos^3\theta \right] \right|_0^{\pi/2}\right) = \bbox[red, 2pt]{1\over 6}$$

解答：

$$\cases{y=4x-x^2 \Rightarrow x=2-\sqrt{4-y}\\ y=x^2 \Rightarrow x=\sqrt y} \Rightarrow 繞x=4旋轉體積=\pi \int_{0}^4 (2+ \sqrt{4-y})^2-(4-\sqrt y)^2\,dy \\ =\pi \int_{0}^4 8\sqrt y+4\sqrt{4-y}-2y-8\,dy =\pi \left.\left[ {16\over 3} y^{3/2} -{8\over 3} (4-y)^{3/2}-y^2-8y \right] \right|_0^4=\bbox[red,2pt]{16\pi}$$

解答：$$\mathbf{(1)}\;利用\bbox[red,2pt]{比值審斂法}:\lim_{n\to \infty} \left|{a_{n+1}\over a_n} \right| =\lim_{n\to \infty} \left|{n+1 \over 2^{n+1}} \cdot {2^n\over n} \right| = =\lim_{n\to \infty}  {n+1\over 2n} ={1\over 2}\lt 1 \Rightarrow \bbox[red,2pt]{收斂} \\\mathbf{(2)}\;利用\bbox[red,2pt]{交錯級數判別法}:令a_n={\sqrt n\over n+2} \Rightarrow \lim_{n\to \infty}a_n=0 且  a_n-a_{n+1} ={(n+3)\sqrt n-(n+2)\sqrt{n+1}\over (n+2)(n+3)} \\ 由於((n+3)\sqrt n)^2-((n+2)\sqrt{n+1})^2 =n^2+n-4 \gt 0, n\ge 2 \Rightarrow a_n \gt a_{n+1}, n\ge 2\\ \Rightarrow \bbox[red,2pt]{收斂}$$

========================== END ============================

解題僅供參考，其他轉學考試題及詳解