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Let \mathrm{g(x)=\log (f(x)) } where f(x) is a twice differentiable positive function on \mathrm{(0, \infty) } such that \mathrm{f(x+1)=x\, f(x) }. Then, for N=1,2,3, \mathrm{\ldots }..
\mathrm{g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=k\left\{\frac{1}{1}+\frac{1}{9}+\frac{1}{25}+\ldots \ldots+\frac{1}{(2 N-1)^2}\right\}, }
then the value of k is

Option: 1

-4


Option: 2

4


Option: 3

1


Option: 4

-1


Answers (1)

best_answer

\mathrm{ g(x)=\log (f(x)), }
\mathrm{\begin{aligned} & g(x+1)=\log (f(x+1)) \\ & =\log (x(f(x))=\log x+\log (f(x))=\log x+g(x) \\ & \Rightarrow g(x+1)-g(x)=\log x . \end{aligned} }
Differentiating twice w.r.t. x, we obtain
\mathrm{g^{\prime \prime}(x+1)-g^{\prime \prime}(x)=-\frac{1}{x^2} }
Change x to \mathrm{x-\frac{1}{2} } to get
\mathrm{g^{\prime \prime}\left(x+\frac{1}{2}\right)-g^{\prime \prime}\left(x-\frac{1}{2}\right)=-\frac{1}{\left(x-\frac{1}{2}\right)^2}=-\frac{4}{(2 x-1)^2} }
Set x=1,2, \mathrm{\ldots } . N in succession, we get
\mathrm{ \begin{aligned} & g^{\prime \prime}\left(1+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-\frac{4}{1^2} \\ & g^{\prime \prime}\left(2+\frac{1}{2}\right)-g^{\prime \prime}\left(1+\frac{1}{2}\right)=-\frac{4}{3^2} \\ & g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(N-\frac{1}{2}\right)=-\frac{4}{(2 N-1)^2} \end{aligned} }
Adding them all we obtain,
\mathrm{\begin{aligned} & g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)=-4\left[1+\frac{1}{9}+\ldots . .+\frac{1}{\text { Activ }(2 N / / m 1)^2}\right] \\ & \therefore \quad k=-4 \end{aligned} }
 

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