台灣聯合大學系統112學年度學士班轉學生考試
科目:微積分
類組別:A3/A4/A6
甲、填充題:共8題,每題8分
解答:L=(cos(2√x))2/x⇒lnL=2lncos(2√x)x⇒limx→0+lnL=limx→0+2lncos(2√x)x=limx→0+ddx(2lncos(2√x))ddxx=limx→0+−2sin(2√x)/√xcos(2√x)=limx→0+−2tan(2√x)√x=limx→0+ddx(−2tan(2√x))ddx√x=limx→0+−2sec2(2√x)/√x1/2√x=limx→0+−4sec2(2√x)=−4⇒limx→0+lnL=−4⇒limx→0+L=e−4=1e4
解答:g(x)=tan−1(x2)+∫x21sec(t−1)dt⇒g′(x)=2x1+x4+2xsec(x2−1)⇒{g(−1)=tan−1(1)+0=π/4g′(−1)=−1−2⋅1=−3⇒f′(x)=g′(x)eg(x)⇒f′(−1)=g′(−1)eg(−1)=−3eπ/4
解答:{x(t)=ln(sect+tant)−sinty(t)=cost⇒{x′(t)=sect−costy′(t)=−sint⇒(x′(t))2+(y′(t))2=sec2t−2+cos2+sin2t=sec2t−1 =tan2t⇒曲線長=∫π/30√(x′(t))2+(y′(t))2dt=∫π/30tantdt=[−ln(cost)]|π/30=−ln12=ln2
解答:∫e1(ln(x))2πdx=[π(x(lnx)2−2xln(x)+2x)]|e1=π(e−2)
解答:位置r(t)=(cos3t,sin3t,3t)⇒速度r′(t)=(−3sin3t,3cos3t,3)⇒{r(π/3)=(−1,0,π)r′(π/3)=(0,−3,3)又f(x,y,z)=xyz⇒∇f=(fx,fy,fz)=(yz,xz,xy)=(3tsin3t,3tcos3t,(sin6t)/2))⇒∇f(π/3)=(0,−π,0)⇒∇f(π/3)⋅r′(π/3)‖r′(π/3)‖=(0,−π,0)⋅(0,−1√2,1√2)=√2π2
解答:∫80∫23√x1y4+1dydx=∫20∫y301y4+1dxdy=∫20y3y4+1dy=[14ln(y4+1)]|20=14ln17
解答:
積分區域如上圖著色區域,令{x=rcosθy=rsinθ⇒x2+(y−1)2=1⇒r2=2rsinθ⇒r=2sinθ因此∬R√x2+y2dA=2∫π/20∫22sinθr2drdθ=2∫π/20[13r3]|22sinθdθ=163∫π/20(1−sin3θ)dθ=163[θ−112(cos3θ−9cosθ)]|π/20=163(π2+112(−8))=83π−329
解答:∫4−1dx√|x|=∫10dx√x+∫40dx√x=2+4=6
乙、計算、證明題: 共3題,每題12分,共36分
解答:(a)令bn=lnnn−lnn⇒limn→∞bn=limn→∞ddnlnnddn(n−lnn)=limn→∞1n−1=0 又ddnbn=1−lnn(n−lnn)2<0,∀n≥3;因此由交錯級數審定法知該級數收斂現在要判斷是絕對收斂或條件收斂: 由於lnnn−lnn>1n−lnn>1n而調和級數發散,因此原級數絕對值發散,即條件收斂(b)limn→∞|an+1an|=limn→∞|(x+4)n+1(n+1)3n+1⋅n3n(x+4)n|=limn→∞|x+43⋅nn+1|=|x+43|<1⇒−7<x<−1x=−1⇒∞∑n=1(x+4)nn3n=∞∑n=11n為調和級數⇒發散x=−7⇒∞∑n=1(x+4)nn3n=∞∑n=1(−1)n1n⇒{limn→∞1n=01n>1n+1,交錯級數審定法⇒收斂 因此級數收斂範圍為−7≤x<−1
解答:(a)取y=mx⇒lim(x,y)→(0,0)f(x,y)=limx→0x2⋅mxx3+m3x3=limx→0m1+m3=m1+m3≠f(0,0)⇒f(0,0)不連續,故得證 (b)∂f∂x=2xyx3+y3−x2y⋅3x2(x3+y3)2=2xy4−x4y(x3+y3)2(c)∂f∂x=limh→0f(0+h,0)−f(0,0)h=limh→00h=0
解答:f(x,y)=x2+3y2+2y⇒{fx=2xfy=6y+2⇒{fxx=2fxy=0fyy=6⇒d(x,y)=fxxfyy−f2xy=12>0若{fx=0fy=0⇒(x,y)=(0,−1/3)⇒f(0,−1/3)=−13為極小值 再利用 Lagrange 算子求極大值:{f(x,y)=x2+3y2+2yg(x,y)=x2+y2−1⇒{fx=λgxfy=λgyg=0⇒{2x=λ(2x)6y+2=λ(2y)x2+y2=1⇒{x=0⇒y=±1λ=1⇒y=−1/2⇒x=±√3/2⇒{f(0,1)=5f(0,−1)=1f(±√3/2,−1/2)=1/2⇒最大值5⇒最大值5,最小值−13
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第7題是(8/3)pi-(32/9)吧,不是+
回覆刪除已更正,謝謝!
刪除後面可以直接用wallis公式
回覆刪除第三題似乎是-ln2
回覆刪除應該還是ln 2,曲線長度不會負值
刪除5的r(π/3) 有誤 應為(-1,0,π)
回覆刪除已更正,謝謝
刪除