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Differential Calculus

Cool goIITian

Joined:	27 Nov 2010
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7 Jan 2012 21:33:46 IST

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If f(x) is a polynomial function which satisfy the relation (f(x))² f"(x)=(f"(x))³ f'(x), f'(0)=f'(1)=f'(-1)=0,f(0)=4, f(±1)=3, then f"(i) (where i=√7) is equal to

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saurav

Cool goIITian

Joined: 30 May 2010

Posts: 48

8 Jan 2012 02:33:16 IST

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Let the degree of polynomial be n.according to first equation the degrees of resulting polynomials on lhs and rhs must be equal.this gives 2n+n-2=3(n-2)+n-1.so n=5.so its derivative is a fourth degree polynomial whose roots are 0,1and-1and let the remaining one be @.f'(x)=x(x^2-1)(x-@).integrate to get f(x) with a constant of integration c.use f(0)=4 and f(+_1)=3 to get @ and c.then differentiate it twice to get f''(x) and place x=whatever you want :-)

Arjun Virmani

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Joined: 1 Jun 2009

Posts: 1294

12 Jan 2012 13:46:15 IST

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awesome as answered :)

a small correction:

f'(x) = k * x (x^2 - 1) (x-@)

now u'll have 3 eq. and 3 unknwn @ , c and k !!!

after integration!!!

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