A function $y=f(x)$ has a 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 $y=3 x-5$, then the function is

$(x-1)^3$

We have,

$f^{\prime \prime}(x)=6(x-1)$

$\Rightarrow f'(x)=3(x-1)^2+C$                     [On integrating]      ......(i)

It is given that $y=3 x-5$ is tangent to the curve $y=f(x)$ at the point (2, 1)

$\Rightarrow \left(\frac{d y}{d x}\right)_{(2,1)}$ = (Slope of the line $y=3 x-5$)

$\Rightarrow \left(\frac{d y}{d x}\right)_{(2,1)}=3 \Rightarrow\left\{f'(x)\right\}_{(2,1)}=3 \Rightarrow f'(2)=3$

Putting $x=2, f'(2)=3$ in (i), we get $C=0$

∴   $f'(x)=3(x-1)^2$               [Putting $c=0$ in (i)]

$\Rightarrow f(x)=(x-1)^3+C_1$       [On integrating]       ......(ii)

The curve $y=f(x)$ passes through $(2,1)$.

∴   $f(2)=1$

Putting $x=2, f(2)=1$ in (ii), we get $C_1=0$

Putting $C_1=0$ in (ii), we get $f(x)=(x-1)^3$