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