Let f(x) be a cubic polynomial with leading coefficient unity such that f(a) = b  and f′(a)=f′′(a)=0. Suppose g(x)=f(x)−f(a)+(a−x)f′(x)+3(x−a)2 for which  conclusion of Rolle's theorem in [a,b] holds at x=2, where 2∈(a,b) The value of f′′(2) , is  The value of definite integral ∫ab f(x)dx is λμ (where λ&μ are relatively prime numbers), then :

We have, f(x)=(x−a)3+b----1 Since,  Rolle's theorem   is applicable to g(x) at x=2 So, g(a)=g(b) and g′(2)=0⇒0=f(b)−f(a)+(a−b)f′(b)+3(b−a)2 and (a−2)f′′(2)+6(2−a)=0  ⇒ f′′(2)=6,(a≠2) We have, f(x)=(x−a)3+b----1 Since,  Rolle's theorem _ is applicable to g(x) at x=2 So, g(a)=g(b) and g′(2)=0⇒0=f(b)−f(a)+(a−b)f′(b)+3(b−a)2 and (a−a)f′′(2)+6(2−a)=0f′(b)=f(b)−f(a)(b−a)+3(b−a)→2 and f′′(2)=6,(a≠2) from (i), f′′(x)=6(x−a) and f′′(2)=6⇒6(2−a)=6⇒a=1 also, from (1) and (2) (b−a)=3/2⇒b=5/2∴∫15/2 (x−1)3+5/2dx=32164