jee-advanced 2008 Q9

jee-advanced · India · paper2 Differentiating Transcendental Functions Higher-order or nth derivative computation

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

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

Paper Questions

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22