Question

A function $ y=f(x) $ has second order derivative $ f''(x)=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}} $

Answer 1

Answer: (B)

We have, $ f''\,(x)=6x-6 $ On integrating, we get $ f'(x)=3{{x}^{2}}-6x+c $ At $ x=2, $ $ f'(2)=3 $
$ \Rightarrow $ $ 12-12+c=3 $
$ \Rightarrow $ $ c=3 $
$ \therefore $ $ f'(x)=3{{x}^{2}}-6x+3 $
Again integrating we get $ f(x)={{x}^{3}}-3{{x}^{2}}+3x+d $
when $ x=2,\,\,y=1 $ $ f(2)=1={{(2)}^{3}}-3{{(2)}^{2}}+3(2)+d $
$ \Rightarrow $ $ d=-1 $
$ \therefore $ $ f(x)={{x}^{3}}-3{{x}^{2}}+3x-1 $
$={{(x-1)}^{3}} $