" If "phi(x)=f(x)+f(2a-x)" and "f''(x)>0,a>0,0<=x<=2a

If phi (x)=f(x)+f(2a-x) and f'(x) gt a, a gt 0, 0 le x le 2a , then

Let phi(a)=f(x)+f(2ax) and f'(x)>0 for all x in[0,a]. Then ,phi(x)

If (x)=f(x)+f(1-x) and f'(x)<0;0<=x<=1g(x)=f(x)+f(1-x) and f'(x)<0;0<=x<=1 show that g(x) increases in [0,(1)/(2)) and decreases in ((1)/(2),1]

Let f''(x)gt0 and phi(x)=f(x)+f(2-x),x in(0,2) be a function then the function phi(x) is (A) increasing in (0,1) and decreasing (1,2) (B) decreasing in (0, 1) and increasing (1,2) (C) increasing in (0, 2) (D) decreasing in (0,2)

Let f : R to R be a function defined by f (x + y) = f (x).f (y) and f (x) ne 0 for andy x. If f '(0) exists, show that f '(x) = f (x). F'(0) AA x in R if f '(0) = log 2 find f (x).

Let f(x+y)=f(x)f(y) for all x,y in R and f(0)!=0 . Let phi(x)=f(x)/(1+f(x)^2) . Then prove that phi(x)-phi(-x)=0

Let f(x+y)=f(x)f(y) for all x,y in R and f(0)!=0. Let phi(x)=(f(x))/(1+f(x)^(2)). Then prove that phi(x)-phi(-x)=0