Let f:ℝ→ℝ and g:ℝ→ℝ be functions satisfying f(x+y)=f(x)+f(y)+f(x)f(y) and f(x)=xg(x) for all x,y∈ℝ . If limx→0 g(x)=1, then which of the following statements is/are TRUE?

∵ Put x=y=0 is given relation ⇒f(0)=f(0)+f(0)+f2(0)⇒f(0)=0 or −1∵f(x+y)=f(x)+f(y)+f(x)⋅f(y)⇒f(x+y)−f(x)y=f(y)(1+f(x))y⇒limy→0 f(x+y)−f(x)y=limy→0 (1+f(x))⋅f(y)y  sincelimy→0fyy=limy→0gy=1 ⇒f′(x)=1+f(x)⇒f′(0)=1+f(0)⇒f′(0)=1+0⇒f′(0)=1  Again f′(x)1+f(x)=1⇒∫f′(x)dx1+f(x)dx=∫dx⇒ln⁡(1+f(x))=x+c⇒ln⁡[1+f(x)]=x         since f0=0⇒1+f(x))=ex⇒f(x)=ex−1⇒f′(x)=ex   ⇒fx is differentiable for all x.and f′(1)=e g(x)=f(x)x=ex−1x         If g(0)=1 then since g′(0)=limh→0 g(0+h)−g(0)hg′0+=limh→0 eh−1h−1h=limh→0 eh−1−hh2=12g′0−=limh→0 g(0−h)−g(0)−h=limh→0 e−h−1-h-1-hlimh→0 e−h−1+hh2=12g(x) is differentiable.