The tangent to the graph of the function y = f(x) at the point with abscissa x = 1 form an angle of π/6 and at the point x = 2 an angle of π/3 and at the point x = 3 an angle of π/4. The value of $\int_1^3f'(x)f''(x)dx+\int_2^3f''(x)dx$.

None of these

$\frac{dy}{dx}|_{x=1}=\tan\frac{π}{6}=\frac{1}{\sqrt{3}},\,\frac{dy}{dx}|_{x=2}=\tan\frac{π}{3}=\sqrt{3}$

$\frac{dy}{dx}|_{x=3}=\tan\frac{π}{4}=1$

$I_1=\int\limits_1^2f'(x)f''(x)dx$

Let t = f'(x) ⇒ dt = f''(x) dx

x = 1, $t = f'(1) = \frac{1}{\sqrt{3}}$ and x = 2, $f'(2) = \sqrt{3}$

$⇒I_1=\int\limits_{\frac{1}{\sqrt{3}}}^{\sqrt{3}}t\,dt=\frac{1}{2}(3-\frac{1}{3})=\frac{+4}{3},I_2=\int\limits_2^3f''(x)dx=\int\limits_2^3d[f'(x)]=f'(3)-f'(2)=1-\sqrt{3}$

So $(\frac{4}{3})+(1+\sqrt{3})=\frac{7-3\sqrt{3}}{3}$