臺灣綜合大學系統106學年度學士班轉學生聯合招生考試
科目名稱:微積分A
解答:limx→0(3x)=limx→0(sin(4x)+a−2b)⇒a−2b=0⇒a=2b⋯(1)又limx→0(sin(4x)+a−2b)′(3x)′=limx→04cos(4x)3=43=2a+b⋯(2)將(1)代入(2)⇒5b=43⇒b=415⇒a=815⇒(a,b)=(815,415)
解答:f(x)=tan−1(2x)⇒f′(x)=21+4x2=2(1−4x2+16x4−64x6+⋯)=⇒f″(x)=2(−8x+64x3−384x5+⋯)⇒f‴(x)=2(−8+192x2−1920x4+⋯)⇒{f(0)=0f′(0)=2f″(0)=0f‴(0)=−16⇒f(x)的泰勒級數=f(0)+f′(0)x+12f″(0)x2+16f‴(0)x3+⋯=2x−83x3+⋯又g(x)=11−x⇒f(x)/g(x)=2x−83x3+⋯1−x=2x+2x2−23x3+⋯
解答:
積分區域如上圖,改變積分順序,則∫10∫1√xxcos(y5+2)dydx=∫10∫y20xcos(y5+2)dxdy=∫10[12x2cos(y5+2)]|y20dy=∫1012y4cos(y5+2)dy令u=y5+2⇒du=5y4dy,則上式變為∫32110cosudu=110[sinu]|32=110(sin3−sin2)
解答:ex2+y2sin(2x)=4y⇒2xex2+2y2cos(2x)+2yy′sin(2x)=4y′⇒(4−2ysin(2x))y′=2xex2+2y2cos(2x)⇒y′=dydx=2xex2+2y2cos(2x)4−2ysin(2x)
解答:3a−1≤1⇒a≤23
解答:I=∫∞0e−4x2dx⇒2I=∫∞−∞e−4x2dx⇒(2I)2=4I2=∫∞−∞∫∞−∞e−4(x2+y2)dydx令{x=rcosθy=rsinθ⇒∫∞−∞∫∞−∞e−4(x2+y2)dydx=∫2π0∫∞0re−4r2drdθ=∫2π0[−18e−4r2]|∞0dθ=∫2π018dθ=π4⇒4I2=π4⇒I=√π4
解答:令{A(−2,0)B(4,0)C(3,3)D(−1,3),則ABCD為一梯形,面積為(6+4)×32=15依格林定理∮C→F⋅d→r=∮C(4y+6ye2x)dx+(6x+3e2x)dy=∬R∂∂x(6x+3e2x)−∂∂y(4y+6ye2x)dA=∬R((6+6e2x)−(4+6e2x))dA=∬R2dA=2×15=30
解答:F=ex+y2+cosz⇒∇F=(Fx,Fy,Fz)=(ex+y2+cosz,2yex+y2+cosz,−sinzex+y2+cosz)⇒∇F(0,0,0)=(e,0,0)⇒遞減最快的方向=−∇F(0,0,0)|∇F(0,0,0)|=(−1,0,0)遞減率為−|∇F|=−e
解答:折疊後的盒子體積f(x)=(3−2x)2x⇒f′(x)=3(2x−1)(2x−3)=0⇒x=1/2,3/2⇒{f(1/2)=2f(3/2)=0⇒最大積體為2
解答:3a−1≤1⇒a≤23
解答:I=∫∞0e−4x2dx⇒2I=∫∞−∞e−4x2dx⇒(2I)2=4I2=∫∞−∞∫∞−∞e−4(x2+y2)dydx令{x=rcosθy=rsinθ⇒∫∞−∞∫∞−∞e−4(x2+y2)dydx=∫2π0∫∞0re−4r2drdθ=∫2π0[−18e−4r2]|∞0dθ=∫2π018dθ=π4⇒4I2=π4⇒I=√π4
解答:令{A(−2,0)B(4,0)C(3,3)D(−1,3),則ABCD為一梯形,面積為(6+4)×32=15依格林定理∮C→F⋅d→r=∮C(4y+6ye2x)dx+(6x+3e2x)dy=∬R∂∂x(6x+3e2x)−∂∂y(4y+6ye2x)dA=∬R((6+6e2x)−(4+6e2x))dA=∬R2dA=2×15=30
解答:F=ex+y2+cosz⇒∇F=(Fx,Fy,Fz)=(ex+y2+cosz,2yex+y2+cosz,−sinzex+y2+cosz)⇒∇F(0,0,0)=(e,0,0)⇒遞減最快的方向=−∇F(0,0,0)|∇F(0,0,0)|=(−1,0,0)遞減率為−|∇F|=−e
解答:折疊後的盒子體積f(x)=(3−2x)2x⇒f′(x)=3(2x−1)(2x−3)=0⇒x=1/2,3/2⇒{f(1/2)=2f(3/2)=0⇒最大積體為2
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