If $\text{⨍ }\left(\mathrm{x}\right)\text{, g(x)}$ be twice differentiable function on [0,2] satisfying $\text{⨍''(x)=g''(x), ⨍ '(1)=2, g'(1)=4 and ⨍ (2)=3, g(2)=9, then ⨍ (x)-g(x) at x=4}$equals

$\begin{array}{l}Givenf\text{'}\text{'}\left(x\right)=g\text{'}\text{'}\left(x\right)\\ \Rightarrow \int f\text{'}\text{'}\left(x\right)dx=\int g\text{'}\text{'}\left(x\right)dx\\ \Rightarrow f\text{'}\left(x\right)=g\text{'}\left(x\right)+c\\ Putx=1\\ \Rightarrow f\text{'}\left(1\right)=g\text{'}\left(1\right)+c\\ \Rightarrow 2=4+c\\ c=-2\\ \therefore f\text{'}\left(x\right)=g\text{'}\left(x\right)-2\\ \Rightarrow \int f\text{'}\left(x\right)dx=\int g\text{'}\left(x\right)dx-2\int 1dx\\ \Rightarrow f\left(x\right)=g\left(x\right)-2x+k\end{array}$

$\begin{array}{l}putx=2\\ \Rightarrow f\left(2\right)=g\left(2\right)-4+k\\ \Rightarrow 3=9-4+k\\ \Rightarrow k=-2\\ \therefore f\left(x\right)=g\left(x\right)-2x-2\\ \Rightarrow f\left(x\right)-g\left(x\right)=-2x-2\\ Putx=4\\ \Rightarrow f\left(4\right)-g\left(4\right)=-8-2=-10\end{array}$