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Let \mathrm{f(x)=2|[x] x| for -1 \leq x \leq 2}, then number of points where f(x) is discontinuous, is

Option: 1

1


Option: 2

2


Option: 3

4


Option: 4

2.5


Answers (1)

best_answer

\mathrm{ f(x)=\left\{\begin{array}{l}-2 x,-1 \leq x<0 \\ 0,0 \leq x<1 \\ 2 x, 1 \leq x<2 \\ 8,2 \leq x<3 \text { or } x=2 \text { (as given) }\end{array}\right. }
Now, \mathrm{\lim _{x \rightarrow 0^{-}} f(x)=0, \lim _{x \rightarrow 0^{+}} f(x)=0. Also, f(0)=0 \therefore f(x)\, \, is \, \, continuous \, \, at\, \, x=0 }
Again, \mathrm{\lim _{x \rightarrow 1^{-}} f(x)=0, \lim _{x \rightarrow 1^{+}} f(x)=2 \therefore \lim _{x \rightarrow 1^{+}} f(x) \neq \lim _{x \rightarrow 1^{-}} f(x) \therefore \quad f(x) is \, \, discontinuous \, \, at \, x=1. }\mathrm{\lim _{x \rightarrow 1^{-}} f(x)=0, \lim _{x \rightarrow 1^{+}} f(x)=2 \therefore \lim _{x \rightarrow 1^{+}} }
\mathrm{f(x) \neq \lim _{x \rightarrow 1^{-}} f(x) \therefore \quad f(x) is \, \, discontinuous \, \, at \, x=1. }
Again, \mathrm{\lim _{x \rightarrow 2^{-}} f(x) \neq(f(x))_{x=2}=8 }
\mathrm{ \therefore \quad } f is discontinuous at x=2.

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