Mathematics - Application of Derivative Question with Solution | TestHub

$\because \mathrm{f}(1)=\mathrm{f}(3)$

$\begin{array}{llc}\Rightarrow & a+b+11-6=27a+9b+33-6\Rightarrow & 13a+4b=-11\end{array}$

and $f^{\prime }(x)=3a{x}^2+2bx+11$..........(i)

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

$\Rightarrow 3a\left(4+\frac{1}{3}+\frac{4}{\sqrt{3}}\right)+2b\left(2+\frac{1}{\sqrt{3}}\right)+11=0$........(ii)

From eqs. (i) and (ii), we get $\mathrm{a}=1,\mathrm{b}=-6$.