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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Ans:
f'(0) = f'(-1) = f'(1) = 0
\Rightarrow f'(x) = ax(x-1)(x+1), a = constant
\int df(x) = \int ax(x-1)(x+1)dx
f(x) = a(\frac{x^{4}}{4}-\frac{x^{2}}{2}) + c, c = constant
f(0) = a(\frac{0^{4}}{4}-\frac{0^{2}}{2}) + c = 4
\Rightarrow c = 4
f(\pm 1) = a(\frac{(\pm 1)^{4}}{4}-\frac{(\pm 1)^{2}}{2}) + 4 = 3
\frac{-a}{4} = -1
a = 4
f(x) = x^{4}-2x^{2}+4
f'(x) = 4x^{3}-4x
f''(x) = 12x^{2}-4
f''(\sqrt{7}) = 12(\sqrt{7})^{2}-4 = 80
Thanks & Regards
Jitender Singh
IIT Delhi
askIITians Faculty
6 months ago

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