In the functionf(x)=ax3+bx2+11x−6  satisfies conditions of Rolle’s Theorem in [1,3]  andf'(2+13)=0,  then value of a and b are respectively

∵  f(1)=f(3)⇒a+b+11−6=27a+9b+33−6 ⇒   13a+4b=−11 And       f'(x)=3ax2+2bx+11        .....(i) ⇒f'(2+13)=3a(2+13)2+2b(2+13)+11=0⇒3a(4+13+43)+2b(2+13)+11=0           ......(ii) From Eqs, (i) and (ii), we get a=1,   b=−6.