If the greatest and least values of the function f(x)=x3−6x2+9x+1 on [0,2]  are λ and μ,  then the value of λ4+μ8 must be

Given, f(x)=x3−6x2+9x+1f′(x)=3x2−12x+9=3(x−1)(x−3)For maxima or minima f '(x) = 0x =1, 3 but  3∉[0,2]Only one critical point in [0, 2] is x = 1Greatest value λ=max{f(0),f(1),f(2)}=max{1,5,3}=5and least value μ=min{f0,f1,f2}     =min 1,5,3=1 λ4+μ8=54+18=625+1=626