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(B) Given that $f^{\prime \prime}(x)=g^{\prime \prime}(x)$.Integrating both sides with respect to $x$, we get $f^{\prime}(x)=g^{\prime}(x)+C_1$.Substi

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$-4$

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(B) Given that $f^{\prime \prime}(x)=g^{\prime \prime}(x)$.Integrating both sides with respect to $x$, we get $f^{\prime}(x)=g^{\prime}(x)+C_1$.Substituting $x=1$, we have $f^{\prime}(1)=g^{\prime}(1)+C_1$.Given $f^{\prime}(1)=4$ and $g^{\prime}(1)=6$, so $4=6+C_1$, which implies $C_1=-2$.Thus, $f^{\prime}(x)=g^{\prime}(x)-2$.Integrating again with respect to $x$, we get $f(x)=g(x)-2x+C_2$.Substituting $x=2$, we have $f(2)=g(2)-2(2)+C_2$.Given $f(2)=3$ and $g(2)=9$, so $3=9-4+C_2$, which implies $3=5+C_2$, so $C_2=-2$.Thus, $f(x)=g(x)-2x-2$.Rearranging the terms, we get $f(x)-g(x)=-2x-2$.For $x=1$, $f(1)-g(1)=-2(1)-2=-4$.

If $f(x)$ and $g(x)$ are twice differentiable functions on $(0,3)$ satisfying $f^{\prime \prime}(x)=g^{\prime \prime}(x)$, $f^{\prime}(1)=4$, $g^{\prime}(1)=6$, $f(2)=3$, and $g(2)=9$, then $f(1)-g(1)$ is

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