If f(x)=(x−1)(x−2)(x−3)  and a=0, b=4,  then c the using LMVT  (a,b) is 2±λ.  then value of λ  is

We have, f(x)=(x−1)(x−2)(x−3) x3−6x2+11x−6 ∴           f(a)=0−0+0−6=−6∴           f(b)=43−6.42+11.4−6                       =64−96+44−6=6 ∴f(b)−f(a)b−a=6−(−6)4−0=3 Also, f'(x)=3x2−12x+11 ∴             f'(c)=3c2−12c+11 From LMVT, f(b)−f(a)b−a=f'(c)⇒             3=3c2−12c+11∴   3c2−12c+8=0∴   c=12±(144−96)6=2±223 As both of the values of c lie in the open interval (0,4).∴     λ=223 Then, 4502λ=4502×223=600