JEE MAIN - Mathematics (2022 - 24th June Evening Shift - No. 5)
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 :
Explanation
$$f(x) = \left\{ {\matrix{ {{{\sin (x - [x])} \over {x[x]}}} & , & {x \in ( - 2, - 1)} \cr {\max \{ 2x,3[|x|]\} } & , & {|x| < 1} \cr 1 & , & {otherwise} \cr } } \right.$$
$$f(x) = \left\{ {\matrix{ {{{\sin (x + 2)} \over {x + 2}}} & , & {x \in ( - 2, - 1)} \cr 0 & , & {x \in ( - 1,0]} \cr {2x} & , & {x \in (0,1)} \cr 1 & , & {otherwise} \cr } } \right.$$
It clearly shows that f(x) is discontinuous
At x = $$-$$1, 1 also non differentiable
and at $$x = 0$$, $$L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}} = 0$$
$$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h} = 2$$
$$\therefore$$ f(x) is not differentiable at x = 0
$$\therefore$$ m = 2, n = 3
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