國立聯合大學108學年度暑假轉學生招生考試
科目:微積分
一、選擇題(單選題,每題5分)
解答:∫1011+x2dx=[tan−1x]|10=tan−11,故選(B)解答:∫4211−x2dx=∫421(1−x)(1+x)dx=12∫4211−x+11+xdx=12[−ln|1−x|+ln|1+x|]|42=12(−ln3+ln5−ln3)=12ln5−ln3,故無解
解答:利用複變的柯西積分:z=cosθ+isinθ=eiθ⇒{dz=ieiθdθ=izdθ⇒dθ=1izdzcosθ=(eiθ+e−iθ)/2=(z+1/z)/2因此∫2π02√2−cosθdθ=∮C2√2−(z+1/z)/2⋅1izdz=∮C4i(z−(√2−1))(z−(√2+1))dz,其中C為單位圓=4i−2×2πi=4π
解答:lim
解答:\lim_{x\to \pi/2} {\sec x \over 1+\tan x} =\lim_{x\to \pi/2} {1 \over \cos x+\sin x} =1,故選\bbox[red, 2pt]{(B)}
解答:{x^2\over 1}-{y^2\over 4}=1 為一雙曲線\Rightarrow {x^2\over 1}-{y^2\over 4}={z\over 3}為雙曲拋物面,故選\bbox[red, 2pt]{(D)}
解答:y={1\over x^2-4x+3} \Rightarrow y'=-{2x-4\over (x^2-4x+3)^2} \Rightarrow y'(0) = -{-4\over 9} ={4\over 9},故選\bbox[red, 2pt]{(C)}
解答:f(x,y,z) = 2x^2 +3y^2+z^2 \Rightarrow \cases{f_x=4x\\ f_y=6y\\ f_z=2z} \Rightarrow (f_x,f_y,f_z)|_{(2,1,3)} =(8,6,6) \\ \Rightarrow 方向導數=(8,6,6)\cdot {(1,0,-2)\over |(1,0,-2)|} ={1\over \sqrt 5}\left((8,6,6)\cdot (1,0,-2) \right) =-{4\over \sqrt 5},故選\bbox[red, 2pt]{(D)}
解答:f(x,y,z)=x^2+y^2+z^2 \Rightarrow \vec u=(f_x,f_y,f_z) =(2x,2y,2z) \\\Rightarrow {\vec u\over |\vec u|} ={(2x,2y,2z)\over \sqrt{4(x^2+y^2+ z^2)}} ={(2x,2y,2z)\over \sqrt{4 \cdot 9}} =({x\over 3}, {y\over 3},{z\over 3}),故選\bbox[red, 2pt]{(C)}
解答:(x+1){dy\over dx} =x(y^2+1) \Rightarrow \int {1\over y^2+1}dy = \int {x\over x+1}dx \Rightarrow \tan^{-1}y = x-\ln(x+1)+C\\ 將初始值y(0)=0 代入 \Rightarrow C=0 \Rightarrow y= \tan(x-\ln(x+1)),故選\bbox[red, 2pt]{(A)}
二、計算題(非選擇題,每題10分)
解答:令\cases{u=t^n \\ dv=e^t\,dt} \Rightarrow \cases{du =nt^{n-1}\,dt\\ v=e^t} \Rightarrow \int t^ne^t\,dt = t^ne^t-n\int t^{n-1}e^t\,dt\\ 因此\int t^3e^t\,dt = t^3e^t-3\int t^2e^t\,dt =t^3e^t-3(t^2e^t -2\int te^t\,dt) = t^3e^t -3t^2e^t+ 6\int te^t\,dt \\= t^3e^t -3t^2e^t+ 6(te^t-\int e^t\,dt) =\bbox[red, 2pt]{t^3e^t -3t^2e^t+ 6te^t-6e^t+C}解答:I=\int e^{ax}\cos(bx)\,dx = {1\over b}e^{ax}\sin(bx) -{a\over b}\int e^{ax}\sin(bx)\,dx \left( \because\cases{u=e^{ax} \Rightarrow du=ae^{ax}\\ dv=\cos(bx)dx \Rightarrow v={1\over b}\sin(bx)}\right)\\ ={1\over b}e^{ax}\sin(bx) -{a\over b}\left(-{1\over b}e^{ax}\cos (bx)+ {a\over b}I\right)\left( \because\cases{u=e^{ax} \Rightarrow du=ae^{ax}\\ dv=\sin (bx)dx \Rightarrow v=-{1\over b}\cos(bx)}\right) \\ ={1\over b}e^{ax}\sin(bx) +{a\over b^2} e^{ax}\cos(bx)-{a^2\over b^2}I \Rightarrow (1+{a^2\over b^2}) I= {1\over b}e^{ax}\sin(bx) +{a\over b^2} e^{ax}\cos(bx) \\ \Rightarrow I= {b^2\over a^2+b^2}\left({1\over b}e^{ax}\sin(bx) +{a\over b^2} e^{ax}\cos(bx)\right) =\bbox[red, 2pt]{{b\over a^2+b^2} e^{ax}\sin (bx)+ {a\over a^2+b^2} e^{ax}\cos(bx)+C}
解答:{dy\over dx}+3x^2y= 6x^2 \Rightarrow {dy\over dx}= 3x^2(2-y) \Rightarrow \int {1\over 2-y}dy = \int 3x^2 dx \Rightarrow -\ln |2-y|=x^3+C_1\\ \Rightarrow 2-y=C_2e^{-x^3} \Rightarrow y=2-C_2e^{-x^3},將y(0)=1 代入左式\Rightarrow 1=2-C_2 \Rightarrow C_2=1\\ \Rightarrow \bbox[red, 2pt]{y=2-e^{-x^3}}
解答:\lim_{x\to 0}{\tan x-x\over x^3} =\lim_{x\to 0}{(\tan x-x)'\over (x^3)'} =\lim_{x\to 0}{\sec^2 x-1\over 3x^2} = \lim_{x\to 0}{\tan^2 x \over 3x^2} = \lim_{x\to 0}{(\tan^2 x)' \over (3x^2)'}\\ =\lim_{x\to 0}{2\tan x\sec^2 x \over 6x} = \lim_{x\to 0}{(\tan x\sec^2 x)' \over (3x)'} = \lim_{x\to 0}{ \sec^4x +2\tan^2 x\sec^2 \over 3} \\={1\over 3} \lim_{x\to 0} \left({1\over \cos^4 x} +{2\sin^2 x\over \cos^4 x}\right) =\bbox[red, 2pt]{1\over 3}
解答:\iint_D x^3dydz +x^2ydzdx + x^2zdxdy \Rightarrow \mathbf{F}=(x^3,x^2y,x^2z),再利用散度定理\\ \Rightarrow \iint_D x^3dydz +x^2ydzdx + x^2zdxdy = \iiint\nabla \cdot \mathbf F\,dV= \iiint (3x^2+x^2+x^2)\,dV =\iiint 5x^2\,dV\\ 利用圓柱坐標: \cases{x=r\cos \theta\\ y= r\sin \theta \\ z=z},其中\cases{0\le r\le 2\\ 0\le \theta \le 2\pi\\ 0\le z\le 6}; \\因此\iiint 5x^2\,dV = \int_0^6 \int_0^{2\pi} \int_0^2 5r^3\cos^2\theta \,drd\theta dz = 20\int_0^6 \int_0^{2\pi}\cos^2\theta \,d\theta dz\\ =20\int_0^6 \int_0^{2\pi}{\cos 2\theta +1\over 2} \,d\theta dz = 20\pi \int_0^6 dz = \bbox[red,2pt]{120\pi}
===================== END ======================
解題僅供參考
沒有留言:
張貼留言