The correct option is D g′(x)<0 for all x∈(12,∞)
f′′(x)≤0 ∀ x∈R
⇒f′ is decreasing in R
g′(x)=f′(x)−f′(1−x) ⋯(1)
If x>1−x i.e., x>12
then f′(x)≤f′(1−x)
⇒f′(x)−f′(1−x)≤0
⇒g′(x)≤0 for all x∈(12,∞)
If x<1−x i.e., x<12
then f′(x)≥f′(1−x)
⇒g′(x)≥0 for all x∈(−∞,12)
By first derivative test,
g(x) has local maximum at x=12
g(x)=f(x)+f(1−x)
Replacing x by 1−x, we get
g(x)=g(1−x)
Again, replacing x by x+12
g(12+x)=g(12−x)
⇒ Graph of y=g(x) is symmetric about x=12
Given, g′(14)=0
By symmetricity, g′(34)=0
Also, from (1), g′(12)=0
By Rolle's theorem, g′′(x) vanishes for at least two values in [0,1]