$f(x) = \left\{ {\begin{array}{*{20}{r}}{1 + x;}&{x > 1}\\{2;}&{ - 1 \le x \le 1}\\{1 - x;}&{x < - 1}\end{array}} \right.$
Since $f(x) = 1 - x$ or $1 + x$ are polynomial functions and $f(x) = 2$ is a constant function.
$\therefore$ These are continuous at all points.....$(i)$
$\therefore$ $f(x)$ is differentiable at all the points, except at
$x = 1$ and at $x = - 1$.....$(ii)$
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$f(x)=\frac{\mathrm{P}(\mathrm{x})}{\sin (\mathrm{x}-2)}, \quad \mathrm{x} \neq 2$
$\quad \quad \quad \quad 7, \quad\quad\quad \mathrm{x}=2$
where $P(x)$ is a polynomial such that $P^{\prime \prime}(x)$ is always a constant and $P(3)=9$. If $f(x)$ is continuous at $x=2$, then $P(5)$ is equal to $.....$
$f(t)=\left\{\begin{array}{cc}(-1)^{n+1} 2, & \text { if } t=2 n-1, n \in N , \\ \frac{(2 n+1-t)}{2} f(2 n-1)+\frac{(t-(2 n-1))}{2} f(2 n+1), & \text { if } 2 n-1 < t < 2 n+1, n \in N \end{array}\right.$
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. . . . . .