Let $f(x)$ be non-constant, twice differentiable function define on $\mathbb R$ such that $y=f(x)$ is symmetric about line $x=1$ and $f(-1)=f'(\frac{1}{4})=f'(\frac{1}{2})=0$ then which of the following statements is\are correct
(A) $f''(x)=0$ has atleast four roots in the interval $(0,2)$
(B) $\int_{-2}^{2} (x^5+x^3)f(1+x)<\int_0^2 \frac{1}{1+2^{f(x)}}$
(C) There exist at least one $c\in (\frac{3}{2},\frac{7}{4})$, such that $f'(c)+cf''(c)=0$
(d) for some $c\in(1,\frac{3}{2}), f'(c)=cf''(c)$
My Approach:
Option (A)
since $f(x)$ is symmetric about $x=1$
$\implies$ $f(1+x)=f(1-x)$$\quad$.......$(1)$
and by property of continuous symmetric function $f'(1)=0$
Now Differentiating equation $(1)$
$f'(1+x)=-f'(1-x)$$\quad$.... $(2)$
After substituting $x=\frac{1}{2}$ and $x=\frac{3}{4}$ in above equation $(2)$ I obtained
$\implies$ $f'(\frac{3}{2})=-f'(\frac{1}{2})=0$ and $f'(\frac{7}{4})=-f'(\frac{1}{4})=0$
So overall
$\implies$ $f'(1)=f'(\frac{1}{2})=f'(\frac{3}{2})=f'(\frac{1}{4})=f'(\frac{7}{4})=0$
Since $f'(x)=0$ has minimum $5$ roots in $(0,2)$ so $f''(x)=0$ will have at least four roots in $(0,2)$
Option (B)
Clearly Left side Integral is $0$ because it is an odd function and right side integral will be positive
Option (C)
for $c\in (\frac{3}{2},\frac{7}{4})$ $f''(c)=0$ at least once but $f'(c)$ may not be zero
Option (D)
I am not sure how to solve this option.
I tried using LMVT but I didn't arrive at the result.
My doubts
(1) How to solve option (D)
(2) Is my approach for option (B) and option (C) correct?