Let f(x) and g (x) be twice differentiable functions satisfying f^′ ′(x)= g^′ ′(x) for all x ∈ R, f^′(1)=2 g^′(1)=4 and g (2)=3 f(2)=9. Then f(25)- g…

Let $\displaystyle f(x)$ and $\displaystyle \mathrm{g}(x)$ be twice differentiable functions satisfying $\displaystyle f^{\prime \prime}(x)=\mathrm{g}^{\prime \prime}(x)$ for all $\displaystyle x \in \mathbf{R}, f^{\prime}(1)=2 \mathrm{~g}^{\prime}(1)=4$ and $\displaystyle \mathrm{g}(2)=3 f(2)=9$. Then $\displaystyle f(25)-\mathrm{g}(25)$ is equal to :