Question

Let $g\left(x\right)=\underset{1-x}{\overset{1+x}{\int }}t|f^{\prime }\left(t\right)|dt$ where $f\left(x\right)$ does not behave like a constant in any interval $(a,b)$ and the graph of $y=f^{\prime }\left(x\right)$ is symmetrical about the line $x=1.$ Then

• A. $g\left(x\right)$ is increasing for all $x\in R$
• B. $g\left(x\right)$ is increasing only for $x>1$
• C. $g^{\prime }\left(x\right)$ is an even function
• D. $g^{\prime }\left(x\right)$ is symmetrical about the line $x=2$

Answer 1

Answer: (A), (C)

The correct options are
A $g\left(x\right)$ is increasing for all $x\in R$
C $g^{\prime }\left(x\right)$ is an even function

$g\left(x\right)=\underset{1-x}{\overset{1+x}{\int }}t|f^{\prime }\left(t\right)|dt$
Applying Leibnitz rule,
$g^{\prime }\left(x\right)=(1+x)|f^{\prime }(1+x)|+(1-x)|f^{\prime }(1-x)|$
As $y=f^{\prime }\left(x\right)$ is symmetrical about the line $x=1,$
$\therefore f^{\prime }(1+x)=f^{\prime }(1-x)$
$\Rightarrow g^{\prime }\left(x\right)=2|f^{\prime }(1+x)|>0$
$\Rightarrow g\left(x\right)$ is increasing for all $x\in R$

Also, $g^{\prime }(-x)=2|f^{\prime }(1-x)|=2|f^{\prime }(1+x)|=g^{\prime }\left(x\right)$
$\Rightarrow g^{\prime }\left(x\right)$ is an even function.