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Postati aici intrebari legate de problemele din aceasta varianta.
--- Pitagora,
Pro-Didactician
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La 2 b si c va rog.
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La varianta 70 punctl 2.c
terbuie sa detreminam h(x),si avem:
se observa ca toti termeni se reduc ramanand doar:
acum se calc volumul:
si rezolvi restul.ti a ramas doar de calc integralei aplicand formula.....
--- We can still be happy !
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la varianta 70 subiectul III 2. c avem:
se observa ca f este descrescatoare pe 0 si plus infinit deci avem:
din obs astea rezulta ca prima egaliatate este rezolvate mai precis f(0)este mai mare sau egal cu (1) (0 fiind punctul de maxim al f)
ramane de rezolvat cealalta egaliatate adica ln3 mai mic decat f0 care este clar deci demonstram doar ln3 mai mare sau egal cu f(1) si avem:
Edit Admin: v-am corectat o mica greseala in codul LaTeX care genera acele mesaje de eroare.
--- We can still be happy !
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f(x)=1/(x+1) + 1/(x+2)
-f(x+1)=-1/(x+2) - 1/(x+3)
f(x+2)=1/(x+3) + 1/(x+4)
-f(x+3)=-1/(x+4) - 1/(x+5)
......................
f(x+2006)=1/(x+2007) + 1/(x+2008)
-f(x+2007)=-1/(x+2008) - 1/(x+2009)
f(x+2008)=1/(x+2009) + 1/(x+2010)
__________________________________ +
f(x)-f(x+1)+...+f(2008)= 1/(x+1) +1/(x+2010)
h(x)=f(x)-f(x+1)+...+f(2008)-1/(x+1)=1/(x+1) +1/(x+2010)-1/(x+1) = 1/(x+2010)
V=pi*int(h(x)^2)dx de la 0..1
int(h(x)^2)dx de la 0..1 = int[1/((x+2010)^2)]de la 0..1 = -1/(x+2010) bara 0..1=
=-1/2011 + 1/2010=1/(2010*2011)
V=pi/(2010*2011)
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[Citat] f(x)=1/(x+1) + 1/(x+2)
-f(x+1)=-1/(x+2) - 1/(x+3)
f(x+2)=1/(x+3) + 1/(x+4)
-f(x+3)=-1/(x+4) - 1/(x+5)
......................
f(x+2006)=1/(x+2007) + 1/(x+2008)
-f(x+2007)=-1/(x+2008) - 1/(x+2009)
f(x+2008)=1/(x+2009) + 1/(x+2010)
__________________________________ +
f(x)-f(x+1)+...+f(2008)= 1/(x+1) +1/(x+2010)
h(x)=f(x)-f(x+1)+...+f(2008)-1/(x+1)=1/(x+1) +1/(x+2010)-1/(x+1) = 1/(x+2010)
V=pi*int(h(x)^2)dx de la 0..1
int(h(x)^2)dx de la 0..1 = int[1/((x+2010)^2)]de la 0..1 = -1/(x+2010) bara 0..1=
=-1/2010 + 1/2011=1/(2010*2011)
V=pi/(2010*2011) |
da intradevar acolo este 2010 nu 2000 mam grabit putin,bn mai.
--- We can still be happy !
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2 b
derivezi functia f'(x)=(-2x^2-6x-5)/(x^2+3x+2)^2
f'(x)<0 pe interval [0,+inf) deoarece are rad complexe
este descrescatoare pe [0,1]
=> f(1)<f(x)<f(0) pe [0.1] => 5/6<f(x)<3/2
integram relatia
5/6<intf(x)dx de la 0..1<3/2
la 2 a integrala a dat ln(3) => 5/6<ln(3)<3/2
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ce frumos ies pana la urma, va multumesc
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