QJEEAdv16Ma4150: JEE Advanced 2016 Mathematics - Class 12 JEEAdv16Ma4150 Solution: Let for all Then

$f(x) = \mathop {\lim }\limits_{n \to \infty } {\left\{ {{{{n^{2n}}\left( {{x \over n} + 1} \right)\left( {{x \over n} + {1 \over 2}} \right)...\left( {{x \over n} + {1 \over n}} \right)} \over {n!{n^{2n}}\left( {{{{x^2}} \over {{n^2}}} + 1} \right)\left( {{{{x^2}} \over {{n^2}}} + {1 \over {{2^2}}}} \right)....\left( {{{{x^2}} \over {{n^2}}} + {1 \over {{n^2}}}} \right)}}} \right\}^{{x \over n}}}$

By taking logarithm we have

$\ln f(x) = \mathop {\lim }\limits_{n \to \infty } {x \over n}\left\{ {\sum\limits_{r = 1}^n {\ln \left( {1 + {{rx} \over n}} \right) - \sum\limits_{r = 1}^n {\ln \left( {1 + {{{r^2}{x^2}} \over {{n^2}}}} \right)} } } \right\}$

By definite integration, we can write it as

$\ln f(x) = x\int\limits_0^1 {\ln (1 + xy)dy - x\int\limits_0^1 {\ln (1 + {x^2}{y^2})dy} }$

Let xy = u

Then, $\ln f(x) = \int\limits_0^x {\ln (1 + u)du - \int\limits_0^x {\ln (1 + {u^2})du} }$

Using Newton-Leibnitz formula, we get

${1 \over {f(x)}}\,.\,f'(x) = log\left( {{{1 + x} \over {1 + {x^2}}}} \right)$ ...... (i)

Here, at x = 1,

${{f'(1)} \over {f(1)}} = \log (1) = 0$

$\therefore$ $f'(1) = 0$

Now, sign scheme of f'(x) is shown below

$\therefore$ At x = 1, function attains maximum. Since, f(x) increases on (0, 1).

$\therefore$ f(1) > f(1/2)

$\therefore$ Option (a) is incorrect.

f(1/3) < f(2/3)

$\therefore$ Option (b) is correct.

Also, f'(x) < 0, when x > 1

$\Rightarrow$ f'(2) < 0