Let y = f ( x ) be a thrice differentiable function in ( - 5 , 5 ) . Let the tangents to the curve y = f ( x…

Given: 

$y=f\left(x\right)$

$\Rightarrow \frac{dy}{dx}=f^{\text{'}}\left(x\right)$

${\left(\frac{dy}{dx}\right)}_{\left(1,f\left(1\right)\right)}=f^{\text{'}}\left(1\right)=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$

$\Rightarrow f^{\text{'}}\left(1\right)=\frac{1}{\sqrt{3}} ...\left(i\right)$

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{\left(3,\mathrm{f}\left(3\right)\right)}=f^{\text{'}}\left(3\right)=\tan \frac{\pi }{4}=1$

$\Rightarrow f^{\text{'}}\left(3\right)=1 ...\left(ii\right)$

It is given that, $27{\int }_1^3\left({\left(f^{\text{'}}\left(t\right)\right)}^2+1\right)f^{"}\left(t\right)dt=\alpha +\beta \sqrt{3}$

Let, $I={\int }_1^3\left({\left(f^{\text{'}}\left(t\right)\right)}^2+1\right)f^{"}\left(t\right)dt$

Putting, $f^{\text{'}}\left(t\right)=z\Rightarrow f^{"}\left(t\right)dt=dz$

$z=f^{\text{'}}\left(3\right)=1$

$z=f^{\text{'}}\left(1\right)=\frac{1}{\sqrt{3}}$

$\Rightarrow I={\int }_{\frac{1}{\sqrt{3}}}^1\left(z^2+1\right)dz$

$\Rightarrow I={\left(\frac{z^3}{3}+z\right)}_{\frac{1}{\sqrt{3}}}^1$

$\Rightarrow I=\left(\frac{1}{3}+1\right)-\left(\frac{1}{3}·\frac{1}{3\sqrt{3}}+\frac{1}{\sqrt{3}}\right)$

$\Rightarrow I=\frac{4}{3}-\frac{10}{9\sqrt{3}}$

$\Rightarrow I=\frac{4}{3}-\frac{10}{27}\sqrt{3}$

$\Rightarrow \alpha +\beta \sqrt{3}=27\left(\frac{4}{3}-\frac{10}{27}\sqrt{3}\right)$

$\Rightarrow \alpha +\beta \sqrt{3}=36-10\sqrt{3}$

$\Rightarrow \alpha =36, \beta =-10$

$\Rightarrow \alpha +\beta =36-10$

$\Rightarrow \alpha +\beta =26$