Match the following List-I with List-II

$$\begin{aligned} & \text { (P) } \\ & f^{\prime}(x)=2 x+g^{\prime}(1), f^{\prime \prime}(x)=2 \\ & g^{\prime}(x)=2 x+f^{\prime}(2), g^{\prime \prime}(x)=2 \\ & \text { at } x=1, f^{\prime}(1)=2+g^{\prime}(1) \text { and } g^{\prime}(1)=2+f^{\prime}(2) \\ & \Rightarrow f^{\prime}(1)=4+f^{\prime}(2)\end{aligned}$$\text { (Q) } \therefore 2 \mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y})=\mathrm{f}(\mathrm{x}-\mathrm{y})+\mathrm{f}(\mathrm{x}+\mathrm{y})$Replacing x by y and y by x , then$2 f(y) f(x)=f(y-x)+f(y+x)$From equation (1) and (2), we get$f(x-y)=f(y-x)$Put$y=0$, then differentiate (R)$$\begin{aligned} & f^{\prime}(x)=3 x^2+2 x f^{\prime}(1)+f^{\prime \prime}(2) \\ & \Rightarrow f^{\prime}(1)^{\prime}=3+2 f^{\prime}(1)+f^{\prime \prime}(2) \\ & \Rightarrow f^{\prime}(1)+f^{\prime \prime}(2)=-3 \\ & \Rightarrow \text { and } f^{\prime \prime}(x)=6 x+2 f^{\prime}(1) \\ & \therefore f^{\prime \prime}(2)=12+2 f^{\prime}(1) \\ & \therefore-2 f^{\prime}(1)+f^{\prime \prime}(2)=12 \end{aligned}$$(S)$n(a+b)^{n-1}=n a^{n-1}+n b^{n-1}$