Match the following List-I with List-II  

	COLUMN-I		COLUMN-II
P)	Let $f\left(x\right)=x^2-xg\text{'}\left(1\right)+g\text{'}\text{'}\left(2\right)$ and $g\left(x\right)=x^2+xf\text{'}\left(2\right)+f\text{'}\text{'}\left(3\right),$ then $2(f\text{'}\left(2\right)-f\text{'}\left(1\right))$ is equal to	1)	2
Q)	If $f(x-y),f\left(x\right),f\left(y\right)$ and $f(x+y)$ are in A.P for all x, y and $f\left(0\right)\ne 0,$ then $f\text{'}\left(2\right)+f\text{'}(-2)$ is equal to	2)	1
R)	If $f\left(x\right)=x^3+x^2.f\text{'}\left(1\right)+xf\text{'}\text{'}\left(2\right)+f\text{'}\text{'}\text{'}\left(3\right)$ for all x, then $f\left(0\right)+f\left(3\right)+1$ is equal to	3)	4
S)	Let $f\left(x\right)=x^n,$ n being a positive integer, then value of n for which the equality $f\text{'}(a+b)=f\text{'}\left(a\right)+f\text{'}\left(b\right)$ is valid for all a, b>0 is	4)	0

$$\begin{gathered} \left(P\right)f^{\text{'}}\left(x\right)=2x+g^{\text{'}}\left(1\right),f^{\text{'}\text{'}}\left(x\right)=2 \\ {g}^{\text{'}}\left(x\right)=2x+f^{\text{'}}\left(2\right),g^{\text{'}\text{'}}\left(x\right)=2 \\ At x=1, f^{\text{'}}\left(1\right)=2+g^{\text{'}}\left(1\right) and g^{\text{'}}\left(1\right)=2+f^{\text{'}}\left(2\right) \\ \Rightarrow f^{\text{'}}\left(1\right)=4+f^{\text{'}}\left(2\right) \\ \left(Q\right)\therefore 2f\left(x\right)f\left(y\right)=f(x-y)+f(x+y) \mathrm{....}\left(1\right) \end{gathered}$$

Replacing x by y and y by x, then
$2f\left(x\right)f\left(x\right)=f(y-x)+f(y+x)\mathrm{.....}\left(2\right)$        
 From equation (1) and (2), we get $f(x-y)=f(y-x)$
 Put  $y=0$, then differentiate
$$\begin{gathered} \left(R\right)f^{\text{'}}\left(x\right)=3x^2+2x{f}^{\text{'}}\left(1\right)+f^{\text{'}\text{'}}\left(2\right) \\ \Rightarrow f^{\text{'}}\left(1\right)=3+2f^{\text{'}}\left(1\right)+f^{\text{'}\text{'}}\left(2\right) \\ \Rightarrow f^{\text{'}}\left(1\right)+f^{\text{'}\text{'}}\left(2\right)=-3 \\ \Rightarrow \text{and\hspace{0.17em}\hspace{0.17em}}{f}^{\text{'}\text{'}}\left(x\right)=6x+2f^{\text{'}}\left(1\right) \\ \therefore f^{\text{'}\text{'}}\left(2\right)=12+2f^{\text{'}}\left(1\right) \\ \therefore -2f^{\text{'}}\left(1\right)+f^{\text{'}\text{'}}\left(2\right)=12 \\ \left(S\right)n{(a+b)}^{n-1}=n{a}^{n-1}+n{b}^{n-1} \end{gathered}$$