If the function $f(x)=a x^3+b x^2+11 x-6$ satisfies conditions of Rolle's theorem in [1, 3] and $f'\left(2+\frac{1}{\sqrt{3}}\right)=0$, then values of a and b are respectively

1, -6

f(1) = f(3)

a + b + 11 – 6 = 27a + 9b + 27

⇒ 26a + 8b + 22 = 0

⇒ 13a + 4b + 11 = 0

Also $f'\left(2+\frac{1}{\sqrt{3}}\right)=0$

$\Rightarrow 3 a\left(2+\frac{1}{\sqrt{3}}\right)^2+2 b\left(2+\frac{1}{\sqrt{3}}\right)+11=0$

⇒ a = 1, b = −6