If the functionf(x)=ax3+bx2+11x−6 satisfies condition of Rolle’s theorem in  [1,3]  for  x=2+13,  then value of a and b are respectively

f(x)=ax3+bx2+11x−6 Satisfies condition of Rolle’s theorem in [1,3] ⇒f(1)=f(3) ⇒a+b+11−6=27a+9b+33−6 ⇒13a+4b=−11                            (1) And f'(x)=3ax2+2bx+11 ⇒f'(2+13)=3a(2+13)2+2b(2+13)+11                                =0⇒3a(4+13+43)+2b(2+13)+11=0   ----   (2)From (1) and (2), we get a=1, b=−6