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  • Let <img alt="\mathrm{ f(x)= \begin{cases}\frac{\sin (x-[x])}{x-[x]} & , x \in(-2,-1) \\ \max \{2 x, 3[|x|]\} & ,|x|<1 \\ 1 & , \text { otherwise }\end{cases}}" src="https://ent

Let \mathrm{ f(x)= \begin{cases}\frac{\sin (x-[x])}{x-[x]} & , x \in(-2,-1) \\ \max \{2 x, 3[|x|]\} & ,|x|<1 \\ 1 & , \text { otherwise }\end{cases}} 

where [t] denotes greatest integer \leq 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:

Option: 1

(3,3)


Option: 2

(2,4)


Option: 3

(2,3)


Option: 4

(3,4)


Answers (1)

\mathrm{f(x)=\left\{\begin{array}{l} \frac{\sin (x-[x])}{(x-[x])} ; \quad x \in(-2,7) \\ \max \{2 x, 3[|x|]\} ;|x|<1 \end{array}\right. }\\

\mathrm {f(x)=\left\{\begin{array}{cl} \frac{\sin (x+2)}{x+2} ; & x \in(-2,-1) \\ 0 & ;(-1,0] \\ 2 x & ;[0,1) \\ 1 & ; \text { atherisise } \end{array}\right. }

Discontinuous at \mathrm{x=-1,1}

Non differentiable at \mathrm{x=-1,1,-2 }\\

\therefore \mathrm {m =2 }\\

\mathrm {n =3}\\

\mathrm {(m, n) \equiv(2,3)}\\

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Kshitij

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