台灣聯合大學系統108學年度學士班轉學生考試
科目:微積分
類組別:A2
甲、填充題:共8題,每題8分,共64分
{△ACD=1△BDE=2⇒平均值=2−13−(−1)=14
解答:令{u=x⇒du=dxdv=sec2xdx⇒v=tanx⇒∫xsec2xdx=xtanx−∫tanxdx=xtanx+ln|cosx|+C
解答:∫20∫1y/2yex3dxdy=∫10∫2x0yex3dydx=∫102x2ex3dydx=[23ex3]|10=23(e−1)
解答:{limx→0+f(x)=limx→0+2sin2xx=0limx→0−f(x)=limx→0−ax+bcosx=b⇒b=0⇒f′(x)={2sin(2x)x−2sin2xx2,if x>0a,if x≤0⇒{limx→0+f′(x)=limx→0+2sin(2x)x−2sin2xx2=2limx→0−f(x)=limx→0−a=a⇒a=2⇒a=2,b=0
解答:∂∂x(x2cos2y−siny)=0⇒2xcos2y−2x2y′cosysiny−y′cosy=0將(0,π)代入上式⇒y′=0⇒切線為一水平線且經過(0,π),即y=π
解答:
解答:∫20∫1y/2yex3dxdy=∫10∫2x0yex3dydx=∫102x2ex3dydx=[23ex3]|10=23(e−1)
解答:{limx→0+f(x)=limx→0+2sin2xx=0limx→0−f(x)=limx→0−ax+bcosx=b⇒b=0⇒f′(x)={2sin(2x)x−2sin2xx2,if x>0a,if x≤0⇒{limx→0+f′(x)=limx→0+2sin(2x)x−2sin2xx2=2limx→0−f(x)=limx→0−a=a⇒a=2⇒a=2,b=0
解答:∂∂x(x2cos2y−siny)=0⇒2xcos2y−2x2y′cosysiny−y′cosy=0將(0,π)代入上式⇒y′=0⇒切線為一水平線且經過(0,π),即y=π
解答:
兩圖形{y=f(x)=−x2+4xy=g(x)=x2交於{O(0,0)A(2,4)⇒相交區域如上圖著色區域因此繞x軸旋轉體積為∫20(f(x)−g(x))2πdx=π∫20(−2x2+4x)2dx=π∫204x4−16x3+16x2dx=π[45x5−4x4+163x3]|20=6415π
解答:limx→0+(sinx)x=limx→0+exln(sinx)=1
解答:g(x)={x2sin1x,if x≠00,if x=0.⇒g′(x)={2xsin1x−cos1x,if x≠00,if x=0.⇒{g′(0)=0⇒g′(0)存在limx→0g′(x)=limx→0(2xsin1x−cos1x)不存在,故得證
解答:a.limn→∞|an+1an|=limn→∞|e−(n+1)2e−n2|=limn→∞|en2e(n+1)2|=limn→∞|1e2n+1|=0⇒收斂b.limn→∞sin(1/n)1/n=limn→∞(sin(1/n))′(1/n)′=limn→∞(1/n)′cos(1/n)(1/n)′=limn→∞cos(1/n)=1由於∞∑n=11n發散,因此∞∑n=1sin1n發散(極限比較審斂法)
解答:∞∑n=3ln(1+1n)(lnn)ln(n+1)=∞∑n=3ln(n+1n)(lnn)ln(n+1)=∞∑n=3ln(n+1)−lnn(lnn)ln(n+1)=∞∑n=3(1lnn−1ln(n+1))=(1ln3−1ln4)+(1ln4−1ln5)+(1ln5−1ln6)+⋯=1ln3
乙、計算、證明題:共3題,每題12分,共36分
解答:假設三邊長度分別為x,y,z,則{容量=xyz費用3xy+2yz+2zx=36;由費用可知z=36−3xy2(x+y)代入容量⇒f(x,y)=xy⋅36−3xy2(x+y)⇒{fx=0⇒x2+2xy=12fy=0⇒y2+2xy=12⇒x=y=2⇒z=3因此三邊長分別為2,2,3解答:g(x)={x2sin1x,if x≠00,if x=0.⇒g′(x)={2xsin1x−cos1x,if x≠00,if x=0.⇒{g′(0)=0⇒g′(0)存在limx→0g′(x)=limx→0(2xsin1x−cos1x)不存在,故得證
解答:a.limn→∞|an+1an|=limn→∞|e−(n+1)2e−n2|=limn→∞|en2e(n+1)2|=limn→∞|1e2n+1|=0⇒收斂b.limn→∞sin(1/n)1/n=limn→∞(sin(1/n))′(1/n)′=limn→∞(1/n)′cos(1/n)(1/n)′=limn→∞cos(1/n)=1由於∞∑n=11n發散,因此∞∑n=1sin1n發散(極限比較審斂法)
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