臺灣綜合大學系統108學年度學士班轉學生聯合招生考試
科目名稱：微積分B
解答：$$\mathbf{(a)}\lim_{x\to 0}{1\over \cos x} ={1\over 1}=\bbox[red, 2pt]1 \\\mathbf{(b)}\lim_{x\to 0}{1-\cos x\over e^x-1-x} = \lim_{x\to 0}{(1-\cos x)' \over (e^x-1-x)'} =\lim_{x\to 0}{ \sin x\over e^x-1 } =\lim_{x\to 0}{ \cos x\over e^x } =\bbox[red, 2pt]1$$
解答：$$V={4\over 3}\pi r^3 \Rightarrow {dV\over dt} 4\pi r^2 {dr\over dt} \Rightarrow 5=4\pi\cdot 2^2\cdot {dr\over dt} \Rightarrow {dr\over dt}={5\over 16\pi }\\ \Rightarrow 速度為\bbox[red, 2pt]{{5\over 16\pi }{cm/sec}}$$
解答：$$(x^2+y^2)'=((2x^2+ 2y^2-x)^2)' \Rightarrow 2x+2yy' = 2(2x^2+ 2y^2-x)(4x+4yy'-1)\\ 將y(0)={1\over 2}代入上式\Rightarrow y'(0)=2\cdot {1\over 2}(2y'(0)-1) \Rightarrow y'(0)=1\\ \Rightarrow 過(0,{1\over 2})之切線:\bbox[red, 2pt]{y=x+{1\over 2}}$$
解答：$$f(x) = \int_1^{\sqrt x} {e^t\over t}\,dt \Rightarrow f'(x)={e^{\sqrt x} \over 2x} \gt 0,\forall x \in [1,\infty) \Rightarrow f(x)在區間[1,\infty)為嚴格遞增函數 \\ \Rightarrow \bbox[red,2pt]{最小值=f(1)=0}$$
解答：$$底面為單位圓:x^2+y^2=1 \Rightarrow x^2=1-y^2，\\只考慮第一象限與z軸所圍體積:dV=2xx' = 2(1-y^2)dy\\ \Rightarrow \int dV= \int_0^1 2(1-y^2)\,d y ={4\over 3} \Rightarrow 全部體積=4\times {4\over 3}=\bbox[red, 2pt]{16\over 3}$$
解答：$$\cases{u=\tan^{-1}x \Rightarrow du ={dx\over 1+x^2}\\ dv= {xdx\over (1+x^2)^2} \Rightarrow v=-{1\over 2(1+x^2)}} \Rightarrow \int_0^\infty {x\tan^{-1} x\over (1+x^2)^2}\,dx= \left. -{\tan^{-1}x\over 2(1+x^2)}\right|_0^\infty +{1\over 2}\int {1\over (1+x^2)^2} \,dx \\= \left. -{\tan^{-1}x\over 2(1+x^2)} +{1\over 4}\left( {x\over 1+x^2} +\tan^{-1} x\right) \right|_0^\infty =\bbox[red, 2pt]{\pi\over 8}\\註:v=\tan^{-1}x \Rightarrow \cases{dv={dx\over 1+x^2}\\ 1+x^2=1+\tan^2 u = \sec^2 v} \Rightarrow \int{1\over (1+x^2)^2}dx = \int {dv\over \sec^2 v} =\int \cos^2 v\,dv \\= {1\over 2} \int \cos 2v+1\,dv ={1\over 2}({1\over 2}\sin 2v+v)={1\over 2}(\sin v\cos v+v)={1\over 2}({x\over 1+x^2} +\tan^{-1}x)$$
解答：$$\lim_{n\to \infty}\left| {a_{n+1} \over a_n}\right| =\lim_{n\to \infty}\left| {(-2)^{n+1} \cdot x^{n+1}\over (n+1)^2} \cdot {n^2 \over (-2)^n x^n}\right| = 2|x| \lt 1 \Rightarrow |x|\lt {1\over 2} \\\Rightarrow \bbox[red,2pt]{\cases{收斂半徑=1/2\\ 收斂區間=(-1/2,1/2)}}$$
解答：$$\cases{f_x(0,0)= \lim_{h\to 0}{f(h,0)-f(0,0)\over h}= \lim_{h\to 0}{f(h,0) \over h}= \lim_{h\to 0}{0/h^2\over h} =0 \\ f_y(0,0)= \lim_{h\to 0}{f(0,h)-f(0,0)\over h}= \lim_{h\to 0}{f(0,h) \over h}= \lim_{h\to 0}{h/2\over h}= {1\over 2} } \Rightarrow \bbox[red,2pt]{\cases{f_x(0,0)=0\\ f_y(0,0)={1\over 2}}}$$
解答：$$\cases{f(x,y,z)= xyz\\ g(x,y,z)= xy+ 2xz+ 2yz-12} \Rightarrow \cases{f_x =\lambda g_x\\ f_y =\lambda g_y\\ f_z =\lambda g_z\\ g=0} \Rightarrow \cases{yz =\lambda (y+2z) \cdots(1)\\ xz = \lambda (x+2z) \cdots(2)\\ xy = \lambda(2x+2y) \cdots(3)\\ xy+ 2xz+ 2yz=12 \cdots(4)} \\ 因此\cases{{(1)\over (2)} \Rightarrow {y\over x}= {y+2z\over x+2z} \Rightarrow x=y \\ {(2)\over (3)} \Rightarrow {z\over y}= {x+2z\over 2x+2y} \Rightarrow y=2z\\ {(1)\over (3)} \Rightarrow {z\over x}= {y+2z\over 2x+2y} \Rightarrow x=2z} 代入(4) \Rightarrow 4z^2+4z^2+ 4z^2 =12 \Rightarrow z^2=1 \Rightarrow z=\pm 1\\ \Rightarrow \cases{A=(2,2,1)\\ B=(-2,-2,-1)} \Rightarrow \cases{f(A)= 4\\ f(B)=-4} \Rightarrow 極值為\bbox[red, 2pt]{\pm 4}$$
解答：$$令\cases{x=r\cos \theta\\ y=r\sin \theta} \Rightarrow \int_{-a}^a \int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} \cos(x^2+y^2)\,dy dx = \int_0^{2\pi} \int_0^a \cos(r^2)r\,drd\theta \\ =\int_0^{2\pi} \left.\left[ {1\over 2} \sin(r^2)\right]\right|_0^a\,d\theta =\int_0^{2\pi} {1\over 2}\sin (a^2)\,d\theta= \bbox[red,2pt]{\pi \sin(a^2)}$$
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