The tangent of the curve $$y = f\left( x \right)$$   at the point with abscissa $$x = 1$$  form an angle of $$\frac{\pi }{6}$$ and at the point $$x = 2$$  an angle of $$\frac{\pi }{3}$$ and at the point $$x = 3$$  an angle of $$\frac{\pi }{4}.$$ If $$f''\left( x \right)$$  is continuous, then the value of $$\int\limits_1^3 {f''\left( x \right)f'\left( x \right)dx} + \int\limits_2^3 {f''\left( x \right)dx} $$       is :

Solution :
According to question
$$\eqalign{ & f'\left( 1 \right) = \tan \frac{\pi }{6} = \frac{1}{{\sqrt 3 }} \cr & f'\left( 2 \right) = \tan \frac{\pi }{3} = \sqrt 3 \cr & {\text{and }}f'\left( 3 \right) = \tan \frac{\pi }{4} = 1 \cr & {\text{So, }}\int\limits_1^3 {f''\left( x \right)f'\left( x \right)dx} + \int\limits_1^3 {f''\left( x \right)dx} \cr & = \left[ {\frac{{{{\left\{ {f'\left( x \right)} \right\}}^2}}}{2}} \right]_1^3 + \left[ {f'\left( x \right)} \right]_2^3 \cr & = \frac{1}{2}\left[ {1 - \frac{1}{3}} \right] + \left[ {1 - \sqrt 3 } \right] \cr & = \frac{4}{3} - \sqrt 3 \cr} $$