Let $g^{\prime}(1)=a$ and $g^{\prime \prime}(2)=b$ $\Rightarrow f(x)=x^{2}+a x+b$ Now, $f(1)=1+a+b ; f^{\prime}(x)=2 x+a ; f^{\prime \prime}(x)=2$ $g(x)=(1+a+b) x^{2}+x(2 x+a)+2$ $\Rightarrow g(x)=(a+b+3) x^{2}+a x+2$ \Rightarrow g^{\prime}(x)=2 x(a+b+3)+a \Rightarrow g^{\prime}(1)=2(a+b+3)$$+a=a $\Rightarrow a+b+3=0$ .......(1) $g^{\prime \prime}(x)=2(a+b+3)=b$ $\Rightarrow 2 a+b+6=0$ ........(2) Solving (i) and (ii), we get $a=-3$ and $b=0$ $f(x)=x^{2}-3 x$ and $g(x)=-3 x+2$ $f(4)=4$ and $g(4)=-12+2=-10$ $\Rightarrow f(4)-g(4)=16-2=14$