$$x = {e^{y + {e^{y + .....\infty }}}}\,\, \Rightarrow x = {e^{y + x}}.$$

Taking log.

$$\log \,\,x = y + x$$

$$ \Rightarrow {1 \over x} = {{dy} \over {dx}} + 1$$

$$ \Rightarrow {{dy} \over {dx}} = {1 \over x} - 1 = {{1 - x} \over x}$$

$$f\left( x \right) = a{x^2} + bx + c$$

$$f\left( 1 \right) = f\left( { - 1} \right)$$

$$ \Rightarrow a + b + c = a - b + c$$

or $$b = 0$$

$$\therefore$$ $$f\left( x \right) = a{x^2} + c$$

or $$f'\left( x \right) = 2ax$$

Now $$f'\left( a \right);f'\left( b \right);$$

and $$f'\left( c \right)$$ are $$2a\left( a \right);2a\left( b \right);2a\left( c \right)$$

i.e.$$\,2{a^2},\,2ab,\,2ac.$$

$$ \Rightarrow $$ If $$a,b,c$$ are in $$A.P.$$ then

$$f'\left( a \right);f'\left( b \right)$$ and

$$f'\left( c \right)$$ are also in $$A.P.$$

$$f\left( x \right) = {x^n} \Rightarrow f\left( 1 \right) = 1$$

$$f'\left( x \right) = n{x^{n - 1}} \Rightarrow f'\left( 1 \right) = n$$

$$f''\left( x \right) = n\left( {n - 1} \right){x^{n - 2}}$$

$$ \Rightarrow f''\left( 1 \right) = n\left( {n - 1} \right)$$

$$\therefore$$ $${f^n}\left( x \right) = n!$$

$$ \Rightarrow {f^n}\left( 1 \right) = n!$$

$$ = 1 - {n \over {1!}} + {{n\left( {n - 1} \right)} \over {2!}}{{n\left( {n - 1} \right)\left( {n - 2} \right)} \over {3!}}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + .... + {\left( { - 1} \right)^n}{{n!} \over {n!}}$$

$$ = {}^n\,{C_0} - {}^n\,{C_1} + {}^n\,{C_2} - {}^n\,{C_3}$$

$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + ...... + {\left( { - 1} \right)^n}\,{}^n{C_n} = 0$$

$$F\left( t \right) = \int\limits_0^t {f\left( {t - y} \right)g\left( y \right)} dy$$

$$ = \int\limits_0^t {{e^{t - y}}} ydy = {e^t}\int\limits_0^t {{e^{ - y}}} \,ydy$$

$$ = {e^t}\left[ { - y{e^{ - y}} - {e^{ - y}}} \right]_0^t$$

$$ = - {e^t}\left[ {y{e^{ - y}} + {e^{ - y}}} \right]_0^t$$

$$ = - {e^t}\left[ {t\,{e^{ - t}} + {e^{ - t}} - 0 - 1} \right]$$

$$ = - {e^t}\left[ {{{t + 1 - {e^t}} \over {{e^t}}}} \right]$$

$$ = {e^t} - \left( {1 + t} \right)$$