Let y = f ( x ) be a thrice differentiable function in ( - 5, 5 ) ....

Given: 

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

$⇒ \dfrac{dy}{dx} = {f}^{'}\left(x\right)$

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

$⇒ {f}^{'}\left(1\right) = \dfrac{1}{\sqrt{3}} . . . \left(i\right)$

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

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

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

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

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

$z = {f}^{'} ( 3 ) = 1$

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

$⇒ I = {∫}_{\dfrac{1}{\sqrt{3}}}^{1}\left({z}^{2} + 1\right)dz$

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

$⇒ I = \left(\dfrac{1}{3} + 1\right) - \left(\dfrac{1}{3} \cdot \dfrac{1}{3\sqrt{3}} + \dfrac{1}{\sqrt{3}}\right)$

$⇒ I = \dfrac{4}{3} - \dfrac{10}{9\sqrt{3}}$

$⇒ I = \dfrac{4}{3} - \dfrac{10}{27}\sqrt{3}$

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

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

$⇒ \alpha = 36, \beta = - 10$

$⇒ \alpha + \beta = 36 - 10$

$⇒ \alpha + \beta = 26$