If  fx=x2-bx+25x2-7x+10  for  x≠5  is continuous at x = 5, then the value of f(5) is

fx=x2-bx+25x2-7x+10 ,  x≠5 fx  is   continuous  at  x=5  only  if  limx→5  x2-bx+25x2-7x+10  is  finite.Now,  x2-7x+10→0  when  x→5.Then  we  must  have  x2-bx+25→0  for  which  b=10.Hence,   limx→5  x2-10x+25x2-7x+10=   limx→5 x-5x-2=0.