Let g(x) = log f(x), where f(x) is a twice differentiable positive function on `(0, oo)` such that `f(x + 1) = xf(x)`. Then, for `N = 1, 2, 3, ......, g''(N+(1)/(2))-g''((1)/(2))` is equal to

Options

A. `-4{1+(1)/(9)+(1)/(25)+...+(1)/((2N-1)^(2))}`
B. `4{1+(1)/(9)+(1)/(25)+...+(1)/((2N-1)^(2))}`
C. `-4{1+(1)/(9)+(1)/(25)+...+(1)/((2N+1)^(2))}`
D. `4{1+(1)/(9)+(1)/(25)+...+(1)/((2N+1)^(2))}`

Answer

A

Explanation

We have,
`g(x)=logf(x)" and "f(x+1)=xf(x)`
`implies" "g(x+1)=logf(x+1)`
`implies" "g(x+1)=log(xf(x))" "[becausef(x+1)=xf(x)]`
`implies" "g(x+1)=logx+logf(x)`
`implies" "g(x+1)-g(x)=logx`
`implies" "g'''(x+1)-g''(x)=-(1)/(x^(2))`
Putting `x=(1)/(2),(3)/(2),(5)/(2),(7)/(2),...,(2N-1)/(2),`we get
`g''(1+(1)/(2))-g''((1)/(2))=-4`
`g''(2+(1)/(2))-g''(1+(1)/(2))=-(4)/(9)`
`g''(3+(1)/(2))-g''(2+(1)/(2))=-(4)/(25)`
`{:(vdots," "vdots," "vdots),(vdots," "vdots," "vdots):}`
`g''(N+(1)/(2))-g''(N-(1)/(2))=-(4)/((2N-1)^(2))`
Adding all these, we get
`g''(N+(1)/(2))-g''((1)/(2))=-4(1+(1)/(9)+(1)/(25)+(1)/((2N-1)^(2)))`