The favourable number of ways
= sum of coefficients of ${x^2},\,{x^4},\,.......$ in ${(x + {x^2} + {x^3})^4}$
= sum of coefficients of ${x^2},\,{x^4},\,......$ in ${x^4}{(1 + x + {x^2})^4}.$
Let ${(1 + x + {x^2})^4} = 1 + {a_1}x + {a_2}{x^2} + ..... + {a_8}{x^8}.$
Then ${3^4} = 1 + {a_1} + {a_2} + {a_3} + .... + {a_8}$, (On putting $x = 1)$
and $1 = 1 - {a_1} + {a_2} - {a_3} + ..... + {a_8}$, (On putting $x = - 1)$
$\therefore \,\,\,{3^4} + 1 = 2(1 + {a_2} + {a_4} + {a_6} + {a_8})$
$ \Rightarrow {a_2} + {a_4} + {a_6} + {a_8} = 41$
Thus sum of the coefficients of ${x^2},\,{x^4},\,...... = 41$
Hence the required probaility $ = \frac{{41}}{{81}}.$
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$f(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\|x-1|, & x \geq 0\end{array} \text { and } g(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\1, & x \geq 0\end{array}\right. \text {. }\right.$
મુજબ વ્યાખ્યાયિત વિધેયો છે. તો $(gof)(x)$ એ $........$