QJEEAdv12Mab0e0: JEE Advanced 2012 Mathematics - Class 12 JEEAdv12Mab0e0 Solution: Let for all and let for all . Which of the following is true?

Given,

$\begin{aligned}g(x) & =\int_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right) f(t) d t \\\text { and } f(x) & =(1-x)^2 \sin ^2 x+x^2 \\\Rightarrow g(x) & =\int_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right)\left((1-t)^2 \sin ^2 t+t^2\right) d t\end{aligned}$

On differentiating the above equation w.r.t. $x$

$\begin{aligned}& \Rightarrow g^{\prime}(x)=\left(\frac{2(x-1)}{x+1}-\ln x\right)\left((1-x)^2 \sin ^2 x+x^2\right) \\& \Rightarrow g^{\prime}(x)=\left(2-\frac{4}{x+1}-\ln x\right)\left((1-x)^2 \sin ^2 x+x^2\right)\end{aligned}$

Here (1-x)^2 \sin ^2 x+x^2>0 for all $x \in \mathrm{R}$

Now, draw the graph of $y=2-\frac{4}{x+1}$ and $y=-\ln x$

For x>0

$\text { Add the graph of } y=-\ln x \text { and } y=2-\frac{4}{x+1}$

\Rightarrow g^{\prime}(x)<0 \text { for } x \in(1, \infty)

Hence, $g(x)$ is decreasing function for $x \in(1, \infty)$