Let $f(x)=(1-x)^{2} \sin ^{2} x+x^{2}$ for all $x \in \mathbb{R}$, and let $g(x)=\int_{1}^{x}\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t$ for all $x \in(1, \infty)$. Which of the following is true?

(B) $g$ is decreasing on $(1, \infty)$

(A) $g$ is increasing on $(1, \infty)$

(C) $g$ is increasing on $(1,2)$ and decreasing on $(2, \infty)$

(D) $g$ is decreasing on $(1,2)$ and increasing on $(2, \infty)$

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Application of Derivatives

Let

f (x) = (1 - x)^{2} sin^{2} x + x^{2}

for all

x \in R

, and let

g (x) = \int_{1}^{x} (\frac{2 ( t - 1 )}{t + 1} - ln t) f (t) d t

for all

x \in (1, \infty)

.

Which of the following is true?

Answer: (B)

By the Fundamental Theorem of Calculus:

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

For

x \in (1, \infty)

, we have

f (x) > 0

(since it's a sum of non-negative terms with at least one positive).

Let

h (x) = \frac{2 ( x - 1 )}{x + 1} - ln x

. Then:

h^{'} (x) = \frac{4}{( x + 1 ) ^{2}} - \frac{1}{x} = \frac{4 x - ( x + 1 ) ^{2}}{x ( x + 1 ) ^{2}} = \frac{- ( x - 1 ) ^{2}}{x ( x + 1 ) ^{2}} < 0

for

x > 1

. Since

h (1) = 0

and

h

is strictly decreasing on

(1, \infty)

, we have

h (x) < 0

for all

x > 1

.

Therefore

g^{'} (x) < 0

(1, \infty)

, so

g

is decreasing on

(1, \infty)

JEE Advanced · 2012 · 08 Apr 2012 · Paper 2 · Single correct option