臺灣綜合大學系統112學年度學士班轉學生聯合招生
科目名稱:微積分B
解答:∫211x2cos(πx)dx=[−1πsin(πx)]|21=−1π
解答:limx→−∞ln(8x+9x)3x=limx→−∞∂∂xln(8x+9x)∂∂x3x=13limx→−∞8xln8+9xln98x+9x=13limx→−∞ln8+(9/8)xln91+(9/8)x=13ln8=ln2
解答:f(x,y)=x2+4xy+y3+5⇒{fx=2x+4yfy=4x+3y2⇒(fx(2,−1),fy(2,−1))=(0,11)⇒切線方程式:11(y+1)=0⇒y=−1
解答:f(x)=x(x−1)2⇒f′(x)=−x+1(x−1)3⇒{f(x)≥0−1≤x<1f(x)≤0−∞<x≤−1,x>1⇒{遞增區間:[−1,1)遞減區間:(−∞,−1]∪(1,∞)
解答:∫1012x+10√x+12dx=∫1012(√x+3)(√x+2)dx=12∫101√x+2−1√x+3dx=∫10uu+2−uu+3du(其中u=√x)=∫101−2u+2−1+3u+3du=∫103u+3−2u+2du=[3ln(u+3)−2ln(u+2)]|10=8ln2−5ln3
解答:
兩積分區域可合併,並將積分順序對調⇒∫π/20∫2yysinxxdx+∫ππ/2∫πysinxxdx=∫π0∫xx/2sinxxdydx=∫π0sinxx⋅x2dx=12∫π0sinxdx=12[−cosx]|π0=1
解答:令{f(x,y,z)=x−y+zg(x,y,z)=2x2+4y2+4z2−9⇒{fx=λgxfy=λgyfz=λgzg=0⇒{1=λ(4x)−1=λ(8y)1=λ(8z)⇒{x=−2yz=−y代入g=0⇒8y2+4y2+4y2−9=0⇒y2=916⇒{y=3/4⇒{x=−3/2z=−3/4y=−3/4⇒{x=3/2z=3/4⇒{f(−3/2,3/4,−3/4)=−3f(3/2,−3/4,3/4)=3⇒最小值=−3
解答:f(x,y,z)=x2zey+xz2⇒∇f(x,y,z)=(2xzey+z2,x2zey,x2ey+2xz)⇒∇f(1,ln2,2)=(12,4,6)⇒令→u=∇f(1,ln2,2)‖∇(1,ln2,2)‖=(67,27,37)⇒在方向(67,27,37)最化最大⇒最大變化率=(12,4,6)⋅(67,27,37)=14
解答:|x|\lt {3\over 2} \Rightarrow \left|{2x\over 3} \right|\lt 1\\ f(x)={2x^5\over 16x^4+81} ={2\over 81}x^5 \cdot {1\over 1-(-({2\over 3}x)^4)} \\= {2\over 81}x^5\left(1- ({2\over 3}x)^4 +({2\over 3}x)^8-({2\over 3}x)^{12} +({2\over 3}x)^{16}-({2\over 3}x)^{20}+ \cdots \right) \\ \Rightarrow x^{25}的係數=-{2\over 81}\cdot ({2\over 3})^{20} =\bbox[red, 2pt]{-{2^{21}\over 3^{24}}}
解答:令{f(x,y,z)=x−y+zg(x,y,z)=2x2+4y2+4z2−9⇒{fx=λgxfy=λgyfz=λgzg=0⇒{1=λ(4x)−1=λ(8y)1=λ(8z)⇒{x=−2yz=−y代入g=0⇒8y2+4y2+4y2−9=0⇒y2=916⇒{y=3/4⇒{x=−3/2z=−3/4y=−3/4⇒{x=3/2z=3/4⇒{f(−3/2,3/4,−3/4)=−3f(3/2,−3/4,3/4)=3⇒最小值=−3
解答:f(x,y,z)=x2zey+xz2⇒∇f(x,y,z)=(2xzey+z2,x2zey,x2ey+2xz)⇒∇f(1,ln2,2)=(12,4,6)⇒令→u=∇f(1,ln2,2)‖∇(1,ln2,2)‖=(67,27,37)⇒在方向(67,27,37)最化最大⇒最大變化率=(12,4,6)⋅(67,27,37)=14
解答:|x|\lt {3\over 2} \Rightarrow \left|{2x\over 3} \right|\lt 1\\ f(x)={2x^5\over 16x^4+81} ={2\over 81}x^5 \cdot {1\over 1-(-({2\over 3}x)^4)} \\= {2\over 81}x^5\left(1- ({2\over 3}x)^4 +({2\over 3}x)^8-({2\over 3}x)^{12} +({2\over 3}x)^{16}-({2\over 3}x)^{20}+ \cdots \right) \\ \Rightarrow x^{25}的係數=-{2\over 81}\cdot ({2\over 3})^{20} =\bbox[red, 2pt]{-{2^{21}\over 3^{24}}}
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解答僅供參考,其他歷年試題及詳解
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