A function $$y = f\left( x \right)$$  has a second order derivative $$f''\left( x \right) = 6\left( {x - 1} \right).$$    If its graph passes through the point (2,1) and at that point the tangent to the graph is $$y = 3x - 5,$$   then the function is

Solution :
$$\eqalign{ & f''\left( x \right) = 6\left( {x - 1} \right).\,{\text{Inegrating,we}}\,{\text{get}} \cr & f'\left( x \right) = 3{x^2} - 6x + c \cr & {\text{Slope}}\,{\text{at}}\,\left( {2,1} \right) = f'\left( 2 \right) = c = 3\,\,\left[ {\because \,{\text{slope}}\,{\text{of}}\,{\text{tangent}}\,{\text{at}}\,\left( {2,1} \right)\,{\text{is}}\,3} \right] \cr & \therefore f'\left( x \right) = 3{x^2} - 6x + 3 = 3{\left( {x - 1} \right)^2} \cr & {\text{Inegrating}}\,{\text{again,we}}\,{\text{get}}\,f\left( x \right) = {\left( {x - 1} \right)^3} + D \cr & {\text{The}}\,{\text{curve}}\,{\text{passes}}\,{\text{through}}\,\left( {2,1} \right) \cr & \Rightarrow 1 = {\left( {2 - 1} \right)^3} + D \Rightarrow D = 0 \cr & \therefore f\left( x \right) = {\left( {x - 1} \right)^3} \cr} $$