Consider a differentiable f:R→R for which f(1) = 2 and fx+y=2xfy+4yfx∀x,y∈R.The value of f(4) isThe minimum value of f(x) isThe number of solutions of  f(x) = 2 is

fx+y=2xfy+4yfxInterchanging  x  and  y,  we get⇒fx+y=2yfx+4xfy⇒fx4x-2x=fy4y-2y=k⇒fx=k4x-2x      f1=2⇒k=1.Hence,  fx=4x-2x              f4=44-24=240       fx+y=2xfy+4yfxInterchanging  x  and  y,  we get⇒fx+y=2yfx+4xfy⇒fx4x-2x=fy4y-2y=k⇒fx=k4x-2x      f1=2⇒k=1.Hence,  fx=4x-2x                            =2x2-2x=2x-1/22-1/4Thus  fx  has  least  value  -1/4fx+y=2xfy+4yfxInterchanging  x  and  y,  we get⇒fx+y=2yfx+4xfy⇒fx4x-2x=fy4y-2y=k⇒fx=k4x-2x      f1=2⇒k=1.Hence,  fx=4x-2x Also  4x-2x=2⇒2x2-2x-2=0⇒2x-22x+1=0⇒2x-2=0⇒x=1