Let $g(x)=\log f(x)$ where $f(x)$ is a twice differentiable positive function on $(0, \infty)$ such that $f(x+1)=x f(x)$. Then, for $N=1,2,3 \ldots$, \[ g^{\prime \prime}\left(N+\frac{1}{2}\right)-g^{\prime \prime}\left(\frac{1}{2}\right)= \]

(A) $-4\left\{1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 N-1)^{2}}\right\}$

(B) $4\left\{1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 N-1)^{2}}\right\}$

(C) $-4\left\{1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 N+1)^{2}}\right\}$

(D) $4\left\{1+\frac{1}{9}+\frac{1}{25}+\cdots+\frac{1}{(2 N+1)^{2}}\right\}$

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Relations And Functions XII

Let

g (x) = lo g f (x)

where

f (x)

is a twice differentiable positive function on

(0, \infty)

such that

f (x + 1) = x f (x)

. Then, for

N = 1, 2, 3 \dots

g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) =

(A) $- 4 {1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{( 2 N - 1 ) ^{2}}}$ ✓ Correct
(B) $4 {1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{( 2 N - 1 ) ^{2}}}$
(C) $- 4 {1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{( 2 N + 1 ) ^{2}}}$
(D) $4 {1 + \frac{1}{9} + \frac{1}{25} + \dots + \frac{1}{( 2 N + 1 ) ^{2}}}$

Answer: (A)

From

f (x + 1) = x f (x)

, we get

g (x + 1) = lo g x + g (x)

. Differentiating twice:

g^{''} (x + 1) - g^{''} (x) = - \frac{1}{x ^{2}}

. For

x = \frac{1}{2}, \frac{3}{2}, \dots, N - \frac{1}{2}

, we sum:

g^{''} (\frac{3}{2}) - g^{''} (\frac{1}{2}) = - 4

g^{''} (\frac{5}{2}) - g^{''} (\frac{3}{2}) = - \frac{4}{9}

⋮

g^{''} (N + \frac{1}{2}) - g^{''} (N - \frac{1}{2}) = - \frac{4}{( 2 N - 1 ) ^{2}}

Summing all:

g^{''} (N + \frac{1}{2}) - g^{''} (\frac{1}{2}) = - 4 (1 + \frac{1}{9} + \dots + \frac{1}{( 2 N - 1 ) ^{2}})

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option