Mathematics - Application of Derivative Question with Solution | TestHub

$g(x)={\int }_1^x\left(\frac{2(t-1)}{t+1}-\ln t\right)f(t)dt$, $\therefore g^{\prime }(x)=\left[\frac{2(x-1)}{x+1}-\ln x\right]f(x)$ Here $f(x)>0,\forall x\in (1,\infty )$ Let $h(x)=\frac{2(x-1)}{x+1}-\ln x$ $\therefore h^{\prime }(x)=\frac{4}{(x+1{)}^2}-\frac{1}{x}=\frac{-(x-1{)}^2}{(x+1{)}^{2}x}<0,x\in (1,\infty )$ $\Rightarrow h(x)$ is decreasing function. Hence, for $x>1,h(x)<h(1)\Rightarrow h(x)<0\forall x>1$ $\Rightarrow g^{\prime }(x)<0\forall x\in (1,\infty )$ Therefore, $g(x)$ is decreasing on $(1,\infty )$.