Given f′′(x)=g′′(x) for all x∈R.
Let h(x)=f(x)−g(x).
Taking the second derivative, h′′(x)=f′′(x)−g′′(x)=0.
Integrating with respect to x, h′(x)=c1.
Given f′(1)=4 and g′(1)=2, h′(1)=f′(1)−g′(1)=4−2=2.
Therefore, c1=2, which gives h′(x)=2.
Integrating again with respect to x, h(x)=2x+c2.
Given 3f(2)=9⇒f(2)=3 and g(2)=9, h(2)=f(2)−g(2)=3−9=−6.
Substituting x=2 in h(x), h(2)=2(2)+c2=−6⇒c2=−10.
Thus, h(x)=2x−10.
Substituting x=25, h(25)=f(25)−g(25)=2(25)−10=40.
Answer: 40