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Let $\mathrm{f},\mathrm{g} : \mathrm{R} \rightarrow \mathrm{R}$ be functions defined by
$\mathrm{f(x)}=\left\{\begin{array}{ll} {[x],} & x<0 \\ |1-x|, & x \geqslant 0 \end{array} \text { and } g(x)= \begin{cases}\mathrm{e}^{x}-x, & x<0 \\ (x-1)^{2}-1, & x \geqslant 0\end{cases}\right.$
where $\mathrm{[x]}$ denote the greatest integer less than or equal to $\mathrm{x}$ . Then, the function fog is discontinuous at exactly :
Option: 1
One point

Option: 2
Two points

Option: 3
Three points

Option: 4
Four points

Answers (1)

best_answer

$\mathrm{g\left(0^{-}\right)=e^{0}-0=1} \\$

$\mathrm{g(0)=g\left(0^{+}\right)=1-1=0}\\$

$\mathrm{\therefore g(x)\: is \: discontinuous\: at \: x=0}$

$\mathrm{f\left(0^{-}\right)=-1 }\\$

$\mathrm{f(0)= f\left(0^{+}\right)=1\\ }$

$\mathrm{\therefore \: f(x)\: is\: discontinuous \: at \: x=0}$

$\mathrm{\: \therefore\: we\: need \: to\: check \: fog \: continuity\: at \: x=0, at \: x\: where \: g(x)=0}$

$\mathrm{\text { At } x=0 }\\$

$\mathrm{f o g(0) =f(0)=1} \\$

$\mathrm{f o g\left(0^{+}\right) =f\left(0^{-}\right)=-1 }\\$

$\mathrm{\therefore\: Discontinuous\: at \: x=0}$

$\\\mathrm{At \: x=2\: } \\f o g\left(2^{+}\right)=f\left(0^{+}\right)=1$

$\mathrm{f o g\left(2^{-}\right)=f\left(0^{-}\right)=-1}$

$\mathrm{\therefore\: Discontinuous \: at \: x=2}$

Hence the correct answer is option 2.

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