Let $$f(x) = \left\{ {\matrix{ {{{\sin (x - [x])} \over {x - [x]}}} & {,\,x \in ( - 2, - 1)} \cr {\max \{ 2x,3[|x|]\} } & {,\,|x| < 1} \cr 1 & {,\,otherwise} \cr } } \right.$$

where [t] denotes greatest integer $\le$ t. If m is the number of points where $f$ is not continuous and n is the number of points where $f$ is not differentiable, then the ordered pair (m, n) is :

<p>$$f(x) = \left\{ {\matrix{ {{{\sin (x - [x])} \over {x[x]}}} & , & {x \in ( - 2, - 1)} \cr {\max \{ 2x,3[|x|]\} } & , & {|x| < 1} \cr 1 & , & {otherwise} \cr } } \right.$$</p> <p>$$f(x) = \left\{ {\matrix{ {{{\sin (x + 2)} \over {x + 2}}} & , & {x \in ( - 2, - 1)} \cr 0 & , & {x \in ( - 1,0]} \cr {2x} & , & {x \in (0,1)} \cr 1 & , & {otherwise} \cr } } \right.$$</p> <p>It clearly shows that f(x) is discontinuous</p> <p>At x = $-$1, 1 also non differentiable</p> <p>and at $x = 0$, $L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}} = 0$</p> <p>$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h} = 2$</p> <p>$\therefore$ f(x) is not differentiable at x = 0</p> <p>$\therefore$ m = 2, n = 3</p>