台灣聯合大學系統113學年度學士班轉學生考試
科目:微積分
類組別:A2
一、填充題:共8題,每題8分,總計64分
解答:I=∫∞0∫∞yy2e−x2dxdy=∫∞0∫x0y2e−x2dydx=13∫∞0x3e−x2dx取u=x2,則du=2xdx⇒I=16∫∞0ue−udu=16[−ue−u−e−u]|∞0=16(0−(−1))=16解答:f(x)=exsinx⇒f′(x)=exsinx+excosx=ex(sinx+cosx)⇒f″(x)=2excosxf″(x)=0⇒cosx=0⇒x=π2,3π2⇒{f(π/2)=eπ/2f(3π/2)=−e3π/2⇒inflection points: (π2,eπ/2),(3π2,−e3π/2)
解答:limx→0xsinx1−cosx=limx→0(xsinx)′(1−cosx)′=limx→0sinx+xcosxsinx=limx→0(sinx+xcosx)′(sinx)′=limx→02cosx−xsinxcosx=2
解答:an=(−1)n(n+2)3n(x−3)n⇒limn→∞|an+1an|=limn→∞|(−1)n+1(n+3)(x−3)n+13n+1⋅3n(−1)n(n+2)(x−3)n|=limn→∞|(n+3)(x−3)3(n+2)|=13|x−3|⇒−1<13|x−3|<1⇒0<x<6⇒interval of convergence: (0,6)
解答:{{limx→0−f(x)=sin0=0limx→0+f(x)=b−3⇒b−3=0⇒b=3{limx→0−f′(x)=limx→0−2cos(2x)=2limx→0+f′(x)=limx→0+4x+a=a⇒a=2⇒(a,b)=(2,3)
解答:
圓錐體體積相當於直線y=−rhx+r繞x軸旋轉所得體積,即V=π∫h0(−rhx+r)2dx=π∫h0(r2h2x2−2r2hx+r2)dx=π[r23h2x3−r2hx2+r2x]|h0=13r2hπ
解答:f(x,y)=x3−12xy+8y3⇒{fx=3x2−12yfy=−12x+24y2⇒{fxx=6xfxy=−12fyy=48y⇒D(x,y)=fxxfyy−f2xy=288xy−144若{fx=0fy=0⇒{(x,y)=(0,0)(x,y)=(2,1)⇒{D(0,0)=−144<0⇒(0,0) is a saddle pointD(2,1)=432>0⇒fxx(2,1)=12>0⇒f(2,1)=−8⇒relative minimum: -8
解答:x2+xy=sinx⇒2x+y+xy′=cosx⇒y′=dydx=cosx−2x−yx
解答:假設M(t)為鹽含量⇒dMdt=2×4−M(t)20−(4−4)t×4=8−M(t)5⇒M′=8−15M,M(0)=0⇒M′+15M=8⇒et/5M′+15M′et/5=8et/5⇒(et/5M)′=8et/5⇒et/5M=∫8et/5dt=40et/5+c1⇒M(t)=40+c1e−t/5⇒M(0)=40+c1=0⇒c1=−40⇒M(t)=40−40e−t/5⇒M(10)=40(1−1e2)
解答:f(x)=g(x)⇒ex+6e−x=5⇒e2x−5ex+6=0⇒(ex−3)(ex−2)=0⇒x=ln3,ln2⇒面積=∫ln3ln2(f(x)−g(x))dx=∫ln3ln2(−ex−6e−x+5)dx=[−ex+6e−x+5x]|ln3ln2=5ln32−2
解答:f(x,y)=x3−12xy+8y3⇒{fx=3x2−12yfy=−12x+24y2⇒{fxx=6xfxy=−12fyy=48y⇒D(x,y)=fxxfyy−f2xy=288xy−144若{fx=0fy=0⇒{(x,y)=(0,0)(x,y)=(2,1)⇒{D(0,0)=−144<0⇒(0,0) is a saddle pointD(2,1)=432>0⇒fxx(2,1)=12>0⇒f(2,1)=−8⇒relative minimum: -8
解答:x2+xy=sinx⇒2x+y+xy′=cosx⇒y′=dydx=cosx−2x−yx
二、計算、證明題: 共3題,每題12分,總計36分
解答:By integral test, I=∫∞31x(lnx)pdx=∫∞ln31updu=[11−pu1−p]|∞ln3Case I p>1:I=−11−p(ln3)1−p⇒ convergentCases II p<1:I=∞⇒ divergentCase III p=1:I=∫∞ln31udu=ln∞−ln(ln3)=∞⇒ divergent⇒{convergent ,p>1divergent ,0≤p≤1解答:假設M(t)為鹽含量⇒dMdt=2×4−M(t)20−(4−4)t×4=8−M(t)5⇒M′=8−15M,M(0)=0⇒M′+15M=8⇒et/5M′+15M′et/5=8et/5⇒(et/5M)′=8et/5⇒et/5M=∫8et/5dt=40et/5+c1⇒M(t)=40+c1e−t/5⇒M(0)=40+c1=0⇒c1=−40⇒M(t)=40−40e−t/5⇒M(10)=40(1−1e2)
解答:f(x)=g(x)⇒ex+6e−x=5⇒e2x−5ex+6=0⇒(ex−3)(ex−2)=0⇒x=ln3,ln2⇒面積=∫ln3ln2(f(x)−g(x))dx=∫ln3ln2(−ex−6e−x+5)dx=[−ex+6e−x+5x]|ln3ln2=5ln32−2
======================== END ==========================
解題僅供參考, 其他歷年試題及詳解
沒有留言:
張貼留言