IIT JEE Advanced 2008 Paper 2

Since g(x) = logF(x), we get
g (x + 1) = log F (x + 1)
= log [xF (x)]
= log x + log F (x)
= log x + g (x)
g (x + 1) - g (x) = log x
Differentiating on both sides, we get
g' (x + 1) - g' (x) =

1

x

Differentiating again, we get
g" (x + 1) - g" (x) -

1

x2

Substituting x = x -

1

2

, we get
g" (x+

1

2

)−g"(x−

1

2

) = −

−1

(x− 1 2 )2

⇒ g"(x+

1

2

)−g"(x−

1

2

) = −

−4

(2x−1)2

Substituting x = 1, 2, 3, …, N, we get
g"(1+

1

2

)−g"(1−

1

2

) =

−4

1

g"(2+

1

2

)−g"(2−

1

2

) =

−4

9

g"(3+

1

2

)−g"(3−

1

2

) =

−4

25

g"(N+

1

2

)−g"(N−

1

2

) =

−4

(2N−1)2

Hence, we get
g"(N+

1

2

)−g"(

1

2

) = −4[+

11

9

+

1

25

+...+

1

(2N−1)2

]