Question

Let $f\left(x\right)=(1-x{)}^2{\sin }^{2}x+x^2$ for all $x\in$ IR, and let $g\left(x\right)={\int }_1^x(\frac{2(t-1)}{t+1}-\ln t)f\left(t\right)dt$ for all $x\in (1,\infty )$.
Which of the following is true?

• A. $g$ is increasing on $($1, $\infty )$
• B. $g$ is decreasing 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, $\infty )$

Answer 1

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

$g^{\prime }\left(x\right)=[\frac{2(x-1)}{x+1}-\ln x]f\left(x\right)$,

For $xϵ(1,\infty ),f\left(x\right)>0$
(Using Newton-Leibnitz rule)
Let $h\left(x\right)=(\frac{2(x-1)}{x+1}-\ln x)$

$\Rightarrow h^{\prime }\left(x\right)=[\frac{4}{(x+1{)}^2}-\frac{1}{x}]$
$=\frac{-(x-1{)}^2}{(x+1{)}^2}<0$
Also $h\left(1\right)=0$.

So, $h\left(x\right)<0\forall x>1$
$\Rightarrow g\left(x\right)$ is decreasing on $(1,\infty )$