A function $ y=f\left( x \right) $ has a second order derivative $ f''\left( x \right)=6\left( x-1 \right) $ . If its graph passes through a point $ \left( 2,1 \right) $ and at that point the tangent to the graph is $ y=3x-5 $ , then the function is
(a) $ {{\left( x-1 \right)}^{2}} $
(b) $ {{\left( x+1 \right)}^{2}} $
(c) $ {{\left( x+1 \right)}^{3}} $
(d) $ {{\left( x-1 \right)}^{3}} $

Hint: First, we will take integration of second order derivative which will be $ \int{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}=\int{6\left( x-1 \right)} $ . In solving this we will get an equation for first order derivatives. Then we will find the value of the constant term c by putting point $ \left( 2,1 \right) $ in the equation we got. Also, that constant term will be equal to the slope of the tangent equation $ y=3x-5 $ which will be obtained by comparing with $ y=mx+c $ . Then, again we will integrate the first order equation we got in order to get the equation in the form of function y. Then again by putting points we will get constant terms and on further simplification we will get the answer. Formula use are $ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c} $ , $ \int{1dx=x+c} $ .

Complete step-by-step answer:
Here, it is given that the second order derivative is $ f''\left( x \right)=6\left( x-1 \right) $ . It is denoted as $ \dfrac{{{d}^{2}}y}{d{{x}^{2}}} $ . So, we can write it as
 $ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=6\left( x-1 \right) $
Now, in order to get the first derivative, we will take integration on both sides of the equation. We get as
 $ \int{\dfrac{{{d}^{2}}y}{d{{x}^{2}}}}=\int{6\left( x-1 \right)} $
We know that $ \int{{{x}^{n}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c} $ , $ \int{1dx=x+c} $ and integration of second order derivative will be first order derivative. So, we get as
 $ \dfrac{dy}{dx}=\int{6x-\int{6}} $
 $ \dfrac{dy}{dx}=\dfrac{6{{x}^{2}}}{2}-6x+c $
 $ \Rightarrow \dfrac{dy}{dx}=3{{x}^{2}}-6x+c $ ……………………(1)
Now, it is given that graph of this equation passes through point $ \left( 2,1 \right) $ we can substitute x=2 and we get as,
 $ \Rightarrow \dfrac{dy}{dx}=3{{\left( 2 \right)}^{2}}-6\left( 2 \right)+c $
On further solving, we get as
 $ \Rightarrow \dfrac{dy}{dx}=12-12+c=c $ ………………(2)
We are given a tangent equation $ y=3x-5 $ . So, the slope of this tangent and constant term in (2) will be equal. To find slope, we will compare the tangent equation with the general equation i.e. $ y=mx+c $ where m is slope and c is constant.
Thus, we get slope form equation as $ m=3 $ .
So, we can say that the constant term in equation (1) is the same as slope. So, we get equation as
 $ \Rightarrow \dfrac{dy}{dx}=3{{x}^{2}}-6x+3 $
Now, we are given that $ y=f\left( x \right) $ . For this we will again take integration on the above equation. We will get as
 $ \int{\dfrac{dy}{dx}}=\int{3{{x}^{2}}-6x+3} $
Here, integration of $ \dfrac{dy}{dx} $ will be equal to function y. So, on solving we will get as
 $ y=\dfrac{3{{x}^{3}}}{3}-\dfrac{6{{x}^{2}}}{2}+3x+c $
 $ y={{x}^{3}}-3{{x}^{2}}+3x+c $
Now, here we will again put point $ \left( 2,1 \right) $ to get a constant term. So, we get as
 $ 1={{\left( 2 \right)}^{3}}-3{{\left( 2 \right)}^{2}}+3\left( 2 \right)+c $
 $ 1=8-12+6+c $
 $ 1-2=c\Rightarrow c=-1 $
So, we get function y as
 $ y={{x}^{3}}-3{{x}^{2}}+3x-1 $
Thus, this is an expansion of $ {{\left( x-1 \right)}^{3}} $ .
Hence, option (d) is the correct answer.

Note: Another approach to get an answer is by taking the option method and performing a second derivative to check which option after the second order derivative gives the answer $ f''\left( x \right)=6\left( x-1 \right) $ . If we take option (a): $ {{\left( x-1 \right)}^{2}} $ . On expanding we get as $ y={{x}^{2}}-2x+1 $. Now, on differentiating, we get as $ \dfrac{dy}{dx}=2x-2 $. On the second derivative we get as $ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=2 $ . Thus, it is not the same as $ f''\left( x \right)=6\left( x-1 \right) $. So, this is not the answer. Similarly, on checking this way we will get the option (d) correct.