Let $f(x)=\lim _{n \rightarrow \infty}\left(\frac{n^{n}(x+n)\left(x+\frac{n}{2}\right) \cdots\left(x+\frac{n}{n}\right)}{n!\left(x^{2}+n^{2}\right)\left(x^{2}+\frac{n^{2}}{4}\right) \cdots\left(x^{2}+\frac{n^{2}}{n^{2}}\right)}\right)^{\frac{x}{n}}$, for all $x>0$. Then

(B) $f\left(\frac{1}{3}\right) \leq f\left(\frac{2}{3}\right)$; (C) $f^{\prime}(2) \leq 0$

(A) $f\left(\frac{1}{2}\right) \geq f(1)$

(D) $\frac{f^{\prime}(3)}{f(3)} \geq \frac{f^{\prime}(2)}{f(2)}$

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Limits XII

Let

f (x) = lim_{n \to \infty} (\frac{n ^{n} ( x + n ) ( x + \frac{n}{2} ) \dots ( x + \frac{n}{n} )}{n ! ( x ^{2} + n ^{2} ) ( x ^{2} + \frac{n ^{2}}{4} ) \dots ( x ^{2} + \frac{n ^{2}}{n ^{2}} )})^{\frac{x}{n}}

, for all

x > 0

. Then

Answer: (B), (C)

Rewrite the limit as

f (x) = exp (x \int_{0}^{1} [ln (1 + x y) - ln (1 + x^{2} y^{2})] d y)

. Taking the derivative:

f^{'} (x) = f (x) ln (\frac{1 + x}{1 + x ^{2}})

. For

x \in (0, 1)

\frac{1 + x}{1 + x ^{2}} > 1

, so

f^{'} (x) > 0

and

f

is increasing. Thus

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

, confirming (B). For

x = 2

f^{'} (2) = f (2) ln (3/5) < 0

, confirming (C). Option (A) is false since

f (1/2) < f (1)

. For (D),

\frac{f ^{'} ( 3 )}{f ( 3 )} = ln (2/5)

and

\frac{f ^{'} ( 2 )}{f ( 2 )} = ln (3/5)

; since

ln (2/5) < ln (3/5)

, (D) is false.

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · One or more correct options