Let $f\left(x\right)$ be a non-constant, twice differentiable function defined on R such that $y=f\left(x\right)$ is symmetric about line x = 1 and  $f(-1)=f\text{'}\left(\frac{1}{4}\right)=f\text{'}\left(\frac{1}{2}\right)=0$, then which of following statement(s) is (are) TRUE?

$$\begin{gathered} f(1-x)=f(1+x) \Rightarrow f\text{'}(1-x)+f\text{'}(1+x)=0 \\ put, x=0\Rightarrow f\text{'}\left(1\right)=0 \\ Given, f\text{'}\left(\frac{1}{4}\right)=0\Rightarrow f\text{'}\left(\frac{7}{4}\right)=0 \\ f\text{'}\left(\frac{1}{2}\right)=0\Rightarrow f\text{'}\left(\frac{3}{2}\right)=0 \end{gathered}$$

applying Rolle’s theorem in $(\frac{1}{4},\frac{1}{2}),(\frac{1}{2},1),(1,\frac{3}{2}),(\frac{3}{2},\frac{7}{4})$ 
$f"\left(x\right)=0$  has at least 4 roots in (0, 2)
$\Rightarrow \left(A\right)$

$$\begin{gathered} I=\underset{-2}{\overset{2}{\int }}(x^5+x^3)f(1+x)dx \\ I=\underset{-2}{\overset{2}{\int }}({(-x)}^5+{(-x)}^3)f(1-x)dx\Rightarrow I+I=\underset{-2}{\overset{2}{\int }}f(1+x)\times 0dx=0 \end{gathered}$$

$\Rightarrow I=0<\underset{0}{\overset{2}{\int }}\frac{1}{1+2^{f\left(x\right)}}dx\left(B\right)$

Suppose$F\left(x\right)=\frac{f\text{'}\left(x\right)}{x}$

$F\left(1\right)=F\left(\frac{3}{2}\right)=0$

Applying Rolles theorem
$F\text{'}\left(c\right)=0$ for at least one c in  $(1,\frac{3}{2})$
$\Rightarrow c.f"\left(c\right)-f\text{'}\left(c\right)=0 \Rightarrow c.f"\left(c\right)=f\text{'}\left(c\right)$

for at least one c in  $(1,\frac{3}{2})$
Suppose $G\left(x\right)=x.f\left(x\right)$

$$\begin{gathered} G\left(\frac{3}{2}\right)=G\left(\frac{7}{4}\right)=0 \\ G\text{'}\left(x\right)=1.f\text{'}\left(x\right)+xf"\left(x\right) \end{gathered}$$

Applying Rolle’s theorem in  $(\frac{3}{2},\frac{7}{4})$
$\Rightarrow f\text{'}\left(c\right)+cf"\left(c\right)=0$ for at least one c in  $(\frac{3}{2},\frac{7}{4})$