QJEEAdv08Ma6f5c: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Ma6f5c Solution: Let , where is a twice differentiable positive function on (0,…

Given that,

$g(x) = \log f(x)$

$\Rightarrow g(x + 1) = \log f(x + 1)$

$\Rightarrow g(x + 1) = \log xf(x)$

[ $\because$ $f(x + 1) = xf(x)$ ]

$g(x + 1) = \log x + \log f(n)$

$\Rightarrow g(x + 1) - g(x) = \log (x)$

$g'(x + 1) - g'(x) = {1 \over x}$

$\Rightarrow g''(x + 1) - g''(x) = {{ - 1} \over {{x^2}}}$

Putting, $x = x - {1 \over 2}$ , we get

$g''\left( {x + {1 \over 2}} \right) - g''\left( {x - {1 \over 2}} \right) = - {1 \over {{{\left( {x - {1 \over 2}} \right)}^2}}} = {{ - {{(2)}^2}} \over {{{(2x - 1)}^2}}}$

Putting $x = 1,2,3$ .........., N we get

$g''\left( {x + {1 \over 2}} \right) - g''\left( {x - {1 \over 2}} \right) - = {1 \over {{{\left( {x - {1 \over 2}} \right)}^2}}} = {{ - {{(2)}^2}} \over {{{(2x - 1)}^2}}}$

Putting $x = 1,2,3$ , .........., N we get

$g''\left( {{3 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = {{ - {2^2}} \over {{1^2}}}$ .... (i)

$g''\left( {{5 \over 2}} \right) - g''\left( {{3 \over 2}} \right) = {{ - {2^2}} \over {{1^2}}}$ ..... (ii)

$g''\left( {{7 \over 2}} \right) - g''\left( {{5 \over 2}} \right) = {{ - {2^2}} \over {{5^2}}}$ ..... (iii)

$g''\left( {N + {1 \over 2}} \right) - g''\left( {N - {1 \over 2}} \right) = {{ - {2^2}} \over {{{(2N - 1)}^2}}}$ ..... (iv)

Adding all the above equations, we get

$g''\left( {N + {1 \over 2}} \right) - g''\left( {{1 \over 2}} \right) = 4$

$\left[ {1 + {1 \over {{3^2}}} + {1 \over {{5^2}}}\, + \,...\, + \,{1 \over {{{(2N - 1)}^2}}}} \right]$