question_answer

A function\[y=f(x)\]has a second order derivative\[f'\,'=6(x-1)\]. 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

A) \[{{(x-1)}^{2}}\]       

B)                        \[{{(x-1)}^{3}}\]               

C)        \[{{(x+1)}^{3}}\]         

D)        \[{{(x+1)}^{2}}\]

Correct Answer: B

Solution :

Given, On integrating, we get                      ...(i) At the point (2, 1), the tangent to graph is Slope of tangent = 3   [from Eq. (i)]             From Eq. (i), \[f'(x)=3{{(x-1)}^{2}}\] On integrating, we get \[f(x)={{(x-1)}^{3}}+K\]              ...(ii) Since, graph passes through (2, 1). \[\therefore \]  \[1={{(2-1)}^{2}}+K\] \[\Rightarrow \]               \[K=0\] \[\therefore \] Equation of function is \[f(x)={{(x-1)}^{3}}\]