The correct option is C g(110)<g(1110)
g(x)=2f(x2)+f(2−x)
⇒g′(x)=f′(x2)−f(2−x)
Given, f′′(x)<0, ∀ x∈(0,2)
⇒f′(x) is decreasing in (0,2).
Case 1: x2<2−x i.e., x<43
f′(x2)>f′(2−x)
⇒g′(x)>0
∴g(x) is increasing in (0,43)
Case 2: x2>2−x i.e., x>43
f′(x2)<f′(2−x)
⇒g′(x)<0
∴g(x) is decreasing in (43,2)
Since g(x) is increasing in (0,43) and 110<1110
∴g(110)<g(1110)
g′(x)>0 for x∈(0,43) and g′(x)<0 for x∈(43,2).
So, g(x) has a local maximum at x=43