A function `y=f(x)`has a second order derivative `f''(x)=6(x-1)dot` 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

Options

A. `(x+1)^(2)`
B. `(x-1)^(3)`
C. `(x+1)^(3)`
D. `(x-1)^(2)`

Answer

B

Explanation

We have,
`f'(x)=6(x-1)`
`rArr" "f'(x)=3(x-1)^(2)+C" [On integrating] …(i)"`
It is given that `y=3x-5` is tangent to the curve y = f(x) at the point (2, 1).
`rArr" "((dy)/(dx))_("(2,1)")=("Slope of the line y"=3x-5)`
`rArr" "((dy)/(dx))_("(2,1)")=3rArr{f'(x)}_("(2,1)")=3rArrf'(2)=3`
Putting x = 2, f'(2) = 3 in (i), we get C = 0
`therefore" "f'(x)=3(x-1)^(2)" [Putting c = 0 in (i)]"`
`rArr" "f(x)=(x-1)^(3)+C_(1)" [On integrating ...(ii)]"`
The curve y = f(x) passes though (2,1).
`therefore" "f(2)=1`
Putting x = 2, f(2) = 1 in (ii), we get `C_(1)=0`
Putting `C_(1)=0` in (ii), we get `f(x)=(x-1)^(3)`