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 & RegardsJitender SinghIIT DelhiaskIITians Faculty

2 months ago

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