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(B) Given $y=f(x)$,the slope of the tangent is $\frac{dy}{dx} = f'(x)$.At $x=1$,$f'(1) = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.At $x=3$,$f'(3) = \

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$26$

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(B) Given $y=f(x)$,the slope of the tangent is $\frac{dy}{dx} = f'(x)$.At $x=1$,$f'(1) = 	an(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$.At $x=3$,$f'(3) = 	an(\frac{\pi}{4}) = 1$.Consider the integral $I = \int_1^3 ((f'(t))^2 + 1) f''(t) dt$.Let $z = f'(t)$,then $dz = f''(t) dt$.When $t=1$,$z = f'(1) = \frac{1}{\sqrt{3}}$.When $t=3$,$z = f'(3) = 1$.Substituting these into the integral:$I = \int_{1/\sqrt{3}}^1 (z^2 + 1) dz = \left[ \frac{z^3}{3} + z ight]_{1/\sqrt{3}}^1$$I = (\frac{1}{3} + 1) - (\frac{1}{3} \cdot \frac{1}{3\sqrt{3}} + \frac{1}{\sqrt{3}})$$I = \frac{4}{3} - (\frac{1}{9\sqrt{3}} + \frac{3}{3\sqrt{3}}) = \frac{4}{3} - \frac{10}{9\sqrt{3}} = \frac{4}{3} - \frac{10\sqrt{3}}{27}$.Now,$27I = 27(\frac{4}{3} - \frac{10\sqrt{3}}{27}) = 36 - 10\sqrt{3}$.Comparing with $\alpha + \beta\sqrt{3}$,we get $\alpha = 36$ and $\beta = -10$.Therefore,$\alpha + \beta = 36 - 10 = 26$.

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