Consider the function, f(x)=ax3+bx2+11x−6 Above satisfies the Rolle's theorem in [1,3] so, f(1)=0‌ and ‌f(3)=0 This implies, a+b+5=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(I) And, 27a+9b+27=0 9a+3b+9=0 3a+b+3=0‌‌‌‌‌⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅⋅(II) From equation (I) and (II), a+b=−5