f(x)=f(x)+f(2−x) Now, differentiate w.r.t. x f′(x)=f′(x)−f′(2−x) For f(x) to be increasing f′(x)>0 ⇒f′(x)−f′(2−x)>0 ⇒f′(x)>f′(2−x) But f"(x)>0⇒f′(x) is an increasing function Then, f′(x)>f′(2−x)>0 ⇒x>2−x ⇒x>1 Hence, f(x) is increasing on (1,2) and decreasing on (0,1)