Let $\mathrm{g}\left(\mathrm{x}\right)=\left[\begin{array}{l}\mathrm{a}\sqrt{\mathrm{x}+1},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{if} 0<\mathrm{x}<3\\ \mathrm{bx}+2,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \mathrm{if} 3\le \mathrm{x}<5\end{array}\right.$. If g(x) is differentiable on (0,5), then 25ab equals ______ .

${\mathrm{g}}^{\mathrm{\prime }}\left(3^{-}\right)=\underset{\mathrm{h}\to 0}{lim} \frac{\mathrm{g}(3-\mathrm{h})-\mathrm{g}\left(3\right)}{-\mathrm{h}}=\underset{\mathrm{h}\to 0}{lim} \frac{\mathrm{a}\sqrt{4-\mathrm{h}}-(3\mathrm{b}+2)}{-\mathrm{h}}......\left(1\right)$
For existence of limit $\underset{\mathrm{h}\to 0}{lim} =0$
$\therefore 2\mathrm{a}-3\mathrm{b}=2.......\left(2\right)$
Now, ${\mathrm{g}}^{\mathrm{\prime }}\left(3^{+}\right)=\underset{\mathrm{h}\to 0}{lim} \frac{\mathrm{b}(3+\mathrm{h})+2-(3\mathrm{b}+2)}{\mathrm{h}}=\mathrm{b}........\left(3\right)$
Substituting3b + 2 = 2a in equation (1), we get
${\mathrm{g}}^{\mathrm{\prime }}\left(3^{-}\right)=\underset{\mathrm{h}\to 0}{lim} \frac{\mathrm{a}\sqrt{4-\mathrm{h}}-2\mathrm{a}}{-\mathrm{h}}=\underset{\mathrm{h}\to 0}{lim} \left(\frac{(4-\mathrm{h})-4}{(-\mathrm{h})(\sqrt{4-\mathrm{h}}+2)}\right)=\frac{\mathrm{a}}{4}$
Hence, ${\mathrm{g}}^{\mathrm{\prime }}\left(3^{-}\right)={\mathrm{g}}^{\mathrm{\prime }}\left(3^{+}\right)$

$$\begin{gathered} \mathrm{b}=\frac{\mathrm{a}}{4}\Rightarrow \mathrm{a}=4\mathrm{b} \\ \mathrm{b}=\frac{2}{5}, \mathrm{a}=\frac{8}{5} \end{gathered}$$