Now function $f(x)$ in continuous at $x=0$
$\therefore \lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0}\left(\frac{1+x}{2+x}\right)^{\frac{1}{x}}=b$
$\Rightarrow 0=b$
$\therefore g(x)=a x$
Now, for $\mathrm{x}>0$
$\mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \cdot\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{x}-1} \cdot \frac{1}{(2+\mathrm{x})^2}$
$+\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{x}} \cdot \ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right) \cdot\left(-\frac{1}{\mathrm{x}^2}\right)$
$\therefore \mathrm{f}^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right)$
$\text { And } \mathrm{f}(-1)=\mathrm{g}(-1)=-\mathrm{a}$
$\therefore \mathrm{a}=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9}$
$\therefore \mathrm{g}(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3}$
$=\ln \left(\frac{4}{9 \cdot \mathrm{e}^{1 / 3}}\right)$
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