$f(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\|x-1|, & x \geq 0\end{array} \text { and } g(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\1, & x \geq 0\end{array}\right. \text {. }\right.$
Then (gof) (x) is
$g(x)=\left\{\begin{array}{c} x +1, x < 0 \\ 1, x \geq 0\end{array}\right.$
$g(f(x))=\left\{\begin{array}{c} x +2, x < -1 \\ 1, x \geq-1\end{array}\right.$
$g(f(x))=\left\{\begin{array}{c} x +2, x < -1 \\1, x \geq-1\end{array}\right.$
$\therefore g ( f ( x ))$ is continuous everywhere
$g ( f ( x ))$ is not differentiable at $x =-1$
Differentiable everywhere else
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$f(t)=\left\{\begin{array}{cc}(-1)^{n+1} 2, & \text { if } t=2 n-1, n \in N , \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1 < t < 2 n+1, n \in N \end{array}\right.$
Define $g(x)=\int_1^x f(t) d t, x \in(1, \infty)$. Let $\alpha$ denote the number of solutions of the equation $g(x)=0$ in the interval $(1,8]$ and $\beta=\lim _{x \rightarrow 1+} \frac{g(x)}{x-1}$. Then the value of $\alpha+\beta$ is equal to. . . . . .