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Let g(x) = ln f(x) where f(x) is a twic...

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

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))}`

Text Solution

Verified by Experts

The correct Answer is:
A
Since, `f(x) = e^(g(x)) rArr e^(g(x+1))=f(x+1) = xf(x) = xe^(g(x)) `
`and" "g(x+1) = log x + g (x) `
i.e.`" " g(x+1) - g(x) = log x ` …(i)
Replacing x by ` x - 1/2 `, we get
`g(x+1/2)-g(x-1/2) = log(x-1/2) = log (2x-1) - log 2`
`:. g''(x+1/2)-g''(x-1/2) = (-4)/((2x-1)^(2))` ....(ii)
On substituting, x = 1,2,3, ...,N in Eq. (ii) and adding, we get
`g''(N+1/2)-g''(1/2) =- 4{1+1/9+1/25+...+1/((2N-1)^(2))}`,
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