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# It is given that a function $f{x} = x^3 - 6x^2 + ax+ b$ on [1,3] , Rolle's theorem holds with $c = 2 + \frac{1}{\sqrt3}$. Find the value of a and b if f(1) = f(3) = 0

$(A)\;a = 11 , b = -6$
$(B)\;a = 10 , b = -5$
$(C)\;a = -11 , b = 6$
$(D)\;a = -10 , b = 5$

(i.e) $1^3 -6 \times 1 + a +b = 3^3 - 6\times 3^2 +3a + b = 0$
$\implies a+b-5=0$ and $3a+b-27=0$
\begin{align*}f'(C) & = 3 (2 + \frac{1}{\sqrt 3})^2 + 12 (2 + \frac{1}{\sqrt 3}) + 11 \\ & = 3 [1 + \frac{1}{3} + \frac{4}{\sqrt 3}] + [24 + \frac{12}{\sqrt 3}] +11 \\ & = 12 + \frac{12}{\sqrt 3} + 1 - 24 -\frac{12}{\sqrt 3} + 11 \\ & = 0\end{align*}
$\therefore$ a = 11 and b = -6