Let g: R → R be a non constant twice differentiable such that g ^ '...

Given: $f\left(x\right) = \dfrac{1}{2}\left[g\left(x\right) + g\left(2 - x\right)\right]$

$⇒ {f}^{'}\left(x\right) = \dfrac{1}{2}\left[{g}^{'}\left(x\right) - {g}^{'}\left(2 - x\right)\right]$

$⇒ {f}^{'}\left(\dfrac{1}{2}\right) = \dfrac{1}{2}\left[{g}^{'}\left(\dfrac{1}{2}\right) - {g}^{'}\left(\dfrac{3}{2}\right)\right]$

$⇒ {f}^{'}\left(\dfrac{1}{2}\right) = 0$

Also, ${f}^{'}\left(\dfrac{3}{2}\right) = \dfrac{1}{2}\left[{g}^{'}\left(\dfrac{3}{2}\right) - {g}^{'}\left(\dfrac{1}{2}\right)\right]$

$⇒ {f}^{'}\left(\dfrac{3}{2}\right) = 0$

Now, ${f}^{'}\left(1\right) = \dfrac{1}{2}\left[{g}^{'}\left(1\right) - {g}^{'}\left(1\right)\right]$

$⇒ {f}^{'}\left(1\right) = 0$

So, ${f}^{'}\left(x\right)$ has three roots and thus ${f}^{"}\left(x\right)$ will have atleast two roots in \left(0, 2\right)$