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

f(x)=x2−bx+25x2−7x+10,  x≠5 f(x) is continuous atx=5, only if limx→5x2−bx+25x2−7x+10  is finiteNow x2−7x+10→0 when x→5,  then we must have x2−bx+25→0  for which b=10 Hence, limx→5x2−10x+25x2−7x+10=limx→5x−5x−2=0