f(x) = 1 + $\underset{0}{\overset{\mathrm{x}}{\int }}(\mathrm{x}-\mathrm{t})\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}\Rightarrow$ f '(x) = $\underset{0}{\overset{\mathrm{x}}{\int }}\mathrm{f}\left(\mathrm{t}\right)\mathrm{dt}$
f "(x) = f (x)
f ''(x) + f ' (x) = f (x) + f ' (x)
$\frac{\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)+\mathrm{f}\text{'}\left(\mathrm{x}\right)}{\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\text{'}\left(\mathrm{x}\right)}$ = 1
On integrating, we get
ln(f(x) + f'(x)) = x + C
Now    f (0) = 1  and  f ' (0) = 0    ⇒    C = 0
Hence    f (x) + f ' (x) = e^x
$\therefore \mathrm{\varphi }$ (x) = f (x) + f ' (x) = e^x
$\frac{\mathrm{\varphi }\text{'}\left(\mathrm{x}\right)}{\mathrm{\varphi }\left(\mathrm{x}\right)}=\frac{{\mathrm{e}}^{\mathrm{x}}}{{\mathrm{e}}^{\mathrm{x}}}=1\Rightarrow {\frac{\mathrm{\varphi }\text{'}\left(\mathrm{x}\right)}{\mathrm{\varphi }\left(\mathrm{x}\right)}|}_{\mathrm{x}=2}=1$       [twice differential f(x) ]