國立高雄師範大學 109 學年度學士班轉學生招生考試試題
系所別:電機工程學系 二年級
科 目:微積分(全一頁)
：$$\mathbf{(a)}\;\lim_{x\to 0}{2\sin x-\sin 2x\over x-\sin x} = \lim_{x\to 0}{(2\sin x-\sin 2x)' \over (x-\sin x)'} = \lim_{x\to 0}{2\cos x-2\cos 2x\over 1-\cos x} \\=\lim_{x\to 0}{ (2\cos x-2\cos 2x)' \over (1-\cos x)'} = \lim_{x\to 0}{-2\sin x+ 4\sin 2x\over \sin x} = \lim_{x\to 0}{ (-2\sin x+ 4\sin 2x)' \over (\sin x)'} \\= \lim_{x\to 0}{-2\cos x+ 8\cos 2x\over \cos x} = \bbox[red, 2pt] 6 \\ \mathbf{(b)}\; \lim_{x\to 0} {\sin \pi x\over \pi x} = \lim_{x\to 0} { (\sin \pi x)' \over (\pi x)'} = \lim_{x\to 0} {\pi \cos \pi x\over \pi } = \bbox[red, 2pt]1$$
解答：$$\mathbf{(a)}\; {d\over dx} \left.\left( {\cos x (1-x) \over 1+x} \right)\right|_{x=0} =\left.\left( {-\sin x (1-x) -\cos x\over 1+x} - {\cos x(1-x) \over (1+x)^2}\right)\right|_{x=0} ={-1}-1 =\bbox[red, 2pt]{-2} \\\mathbf{(b)}\; {d\over dx}a^x ={d\over dx}e^{\ln a^x} ={d\over dx}e^{x\ln a} =\ln a\cdot e^{x\ln a} =\ln a \cdot a^x \Rightarrow {d\over dx}a^x =a^x \ln a，\bbox[red,2pt]{故得證}$$
解答：$$f(x)=x^3+3x^2 -9x-13 \Rightarrow f'(x)=3x^2+6x-9 \Rightarrow f''(x)=6x+6\\ f'(x)=0 \Rightarrow 3(x+3)(x-1)=0 \Rightarrow x=1,-3 \\\Rightarrow \cases{f''(1)=12 \gt 0 \Rightarrow f(1)=-18為相對極小值\\ f''(-3)=-12 \lt 0 \Rightarrow f(-3)= 14為相對極大值} \\\Rightarrow f(x)的相對極值為\bbox[red,2pt]{-18(極小值),14(極大值)}$$
解答：$$\mathbf{(a)}\; y^2+2x^2y-x^4=7 \Rightarrow 2yy'+ 4xy +2x^2y'-4x^3=0 \Rightarrow (2y+2x^2)y'=4x^3-4xy\\ \Rightarrow y'= \bbox[red, 2pt]{{dy \over dx } ={4x^3-4xy\over 2y+2x^2}} \\\mathbf{(b)}\; 切線斜率:\left.{dy \over dx }\right|_{(1,2)} ={4-8\over 4+2} = \bbox[red, 2pt]{-{2\over 3}}$$
解答：$$\mathbf{(a)}\; 令\cases{u=\ln x \Rightarrow du ={1\over x}dx \\ dv=16x^3 dx \Rightarrow v=4x^4} \Rightarrow \int 16x^3 \ln(x)\,dx = \bbox[red, 2pt]{4x^4\ln x-\int 4x^3\,dx= 4x^4\ln x-x^4+C} \\\mathbf{(b)}\; \int_0^\pi \cos(2x) \sin(4x+\pi/4)\,dx = \int_0^\pi \cos(2x) (\sin(4x)\cos(\pi/4) +\sin(\pi/4)\cos(4x)) \,dx\\ = {1\over \sqrt 2}\int_0^\pi \cos(2x)\sin(4x) + \cos(2x)\cos(4x)\,dx= {1\over 2\sqrt 2}\int_0^\pi \sin (6x) +\sin(2x) +\cos(6x) +\cos(2x)\,dx \\={1\over 2\sqrt 2}\left. \left[ -{1\over 6}\cos (6x)-{1\over 2}\cos (2x)+ {1\over 6}\sin(6x)+ {1\over 2} \sin(2x) \right]\right|_0^\pi= \bbox[red, 2pt]0$$
解答：$$取u=t^2-9,則du = 2tdt \Rightarrow \int_3^5 t\sqrt{t^2-9}\,dt =\int_0^{16} {1\over 2} \sqrt{u}\,du = \left.\left [{1\over 3} u^{3/2}\right] \right|_0^{16} = \bbox[red, 2pt]{64\over 3}$$
解答：$$x=2t-1 \Rightarrow dx=2dt \Rightarrow \int_1^3 y\,dx = \int_1^2 (t^2+2)2\,dt =\left.\left[ {2\over 3} t^3 +4t\right] \right|_0^2 ={16\over 3}+8 =\bbox[red, 2pt]{40\over 3}$$
解答：$$\iint_R xe^{y^2}\,dA = \int_0^4 \int_0^{\sqrt y} xe^{y^2}\,dxdy =\int_0^4 {1\over 2}ye^{y^2}\,dy =\left.\left[ {1\over 4}e^{y^2} \right] \right|_0^4 =\bbox[red, 2pt]{{1\over 4}(e^{16}-1)}$$
解答：
$$\int_0^4 \int_\sqrt y^2 \sqrt{x^3+1}\,dxdy = \int_0^2 \int_0^{x^2} \sqrt{x^3+1}\,dydx =\int_0^2 x^2 \sqrt{x^3+1}\,dx \\ 令u=x^3+1 ,則du=3x^2 dx \Rightarrow \int_0^2 x^2 \sqrt{x^3+1}\,dx =\int_1^9 {1\over 3}\sqrt u\,du =\left.\left[{2\over 9} u^{3/2} \right] \right |_1^9 =6-{2\over 9} =\bbox[red, 2pt]{52\over 9}$$
解答：$$\vec r(t)=2\cos t \vec i+ (1+\sin t)\vec j \Rightarrow \vec v(t)={d\over dt}\vec r(t)=-2\sin t\vec i+ \cos t\vec j \\ \Rightarrow \vec v(\pi/3) = -\sqrt 3\vec i+{1\over 2}\vec j \Rightarrow 速度=|\vec v(\pi/3)| = \sqrt{3+ {1\over 4}} =\bbox[red, 2pt]{\sqrt{13}\over 2}$$
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解答：$$\vec r(t)=2\cos t \vec i+ (1+\sin t)\vec j \Rightarrow \vec v(t)={d\over dt}\vec r(t)=-2\sin t\vec i+ \cos t\vec j \\ \Rightarrow \vec v(\pi/3) = -\sqrt 3\vec i+{1\over 2}\vec j \Rightarrow 速度=|\vec v(\pi/3)| = \sqrt{3+ {1\over 4}} =\bbox[red, 2pt]{\sqrt{13}\over 2}$$
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