Answer: Option B
:
B
Since f"(x)=6(x-1)
$\Rightarrow f^{\prime }\left(x\right)=3(x-1{)}^2+c$ (integrating)        ----(i)
Also, at the point (2,1), the tangent to the graph is y =3x-5and slope of thetangent = 3
$\Rightarrow f^{\prime }\left(2\right)=3$
$3(2-1{)}^3+c=3$ [from eq(i)
$\Rightarrow 3+c=3\Rightarrow c=0$
From Eq (i) we have
$f^{\prime }\left(x\right)=3(x-1{)}^2$
$\Rightarrow f\left(x\right)=(x-1{)}^3+k$ (Integrating)----(ii)
$\therefore 1=(2-1{)}^3+k\Rightarrow k=0$
Hence the equation of the function is $f\left(x\right)=(x-1{)}^3$.