$f(x)=a x^3+b x^2+26 x-24$ ...(i) on $[2,4]$
$\because f(x)$ satisfies the Role's theorem
$\begin{array}{ll}\Rightarrow & f(2)=f(4) \\ \Rightarrow & a(2)^3+b\left(2^2\right)+26(2)-24\end{array}$
$=a(4)^3+b(4)^2+26(4)-24$
$\Rightarrow \quad 8 a+4 b+28=64 a+16 b+80$
$\Rightarrow \quad 56 a+12 b+52=0$
$14 a+3 b+13=0$ ...(ii)
On differentiating Eq. (i) w.r.t. $x$
$\begin{aligned} & f^{\prime}(x)=\frac{d}{d x}\left(a x^3+b x^2+26 x-24\right) \\ & f^{\prime}(x)=3 a x^2+2 b x+26\end{aligned}$
At, $x=3+\frac{1}{\sqrt{3}}$
$f^{\prime}\left(3+\frac{1}{\sqrt{3}}\right)=3 a\left(3+\frac{1}{\sqrt{3}}\right)^2+2 b\left(3+\frac{1}{\sqrt{3}}\right)+26$
$\begin{aligned} & 0=3 a\left(9+\frac{1}{3}+\frac{6}{\sqrt{3}}\right)+6 b+\frac{2 b}{\sqrt{3}}+26 \\ & 0=28 a+6 b+6 \sqrt{3} a+\frac{2 b}{\sqrt{3}}+26\end{aligned}$
$\Rightarrow \quad(28+6 \sqrt{3}) a+\left(\frac{6 \sqrt{3}+2}{\sqrt{3}}\right) b+26=0$ ...(iii)
$(28 a+6 b+26)+(18 a+2 b) \frac{1}{\sqrt{3}}=0+0 \frac{1}{\sqrt{3}}$
On comparing rational and irrational part,
$28 a+6 b+26=0$ and $18 a+2 b=0$
and $54 a+6 b=0$
$\therefore \quad 28 a-54 a+26=0$ and $6 b=-54 a$
$\begin{aligned} \Rightarrow & & -26 a+26 & =0 \\ \Rightarrow & & a & =1\end{aligned}$
$b=\frac{-54 \times 1}{6}$
$\begin{aligned} \Rightarrow & b & =-9 \\ \therefore & a b & =-9\end{aligned}$