= $\mathop {Limit}\limits_{h \to 0} \frac{{3 - (1 + h) - 2}}{h}\,\, $
$= \, - 1$
$f ‘(1^-) =$ $\mathop {Limit}\limits_{h \to 0} \frac{{\frac{{{{(1 - h)}^2}}}{4}\,\, - \,\,\frac{3}{2}\,(1 - h)\, + \frac{{13}}{4}\, - 2}}{{ - \,\,h}}$
=$\mathop {Limit}\limits_{h \to 0} \frac{{{{(1 - h)}^2}\, - 6(1 - h)\, + 5}}{{ - 4h}}\,$
=$\mathop {Limit}\limits_{h \to 0} \frac{{{h^2} - 2h\, + 6h}}{{ - 4h}}\,\,$
$= - 1$
$\Rightarrow\,\, f$ is continuous at $x =1 $
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$f(x)=\left\{\begin{array}{l}\max \left\{t^{3}-3 t\right\} ; x \leq 2 \\ t \leq x \\ x^{2}+2 x-6 ; 2 < x < 3 \\ {[x-3]+9 ; 3 \leq x \leq 5} \\ 2 x+1 \quad ; \quad x > 5\end{array}\right\}$
Where $[t]$ is the greatest integer less than or equal to $t$. Let $m$ be the number of points where $f$ is not differentiable and $I =\int\limits_{-2}^{2} f( x ) dx$. Then the ordered pair $( m , I )$ is equal to
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
| $P(X)$ | $K^2$ | $2K$ | $K$ | $2K$ | $5K^2$ |
Then $\mathrm{P}(\mathrm{X}> 2)$ is equal to