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If \mathrm{f(x)=[x-11 \cdot([x]-[-x])}, then (where [.] represents greatest integer function)

 

Option: 1

\mathrm{f(x)} is continuous and differentiable at x=1


Option: 2

\mathrm{f(x)} is discontinuous at x=1


Option: 3

f(x) is continuous at x=2


Option: 4

\mathrm{f(x)} is continuous but non-differentiable at x=1


Answers (1)

best_answer

\mathrm{ f(x)=|x-1| \cdot([x]-[-x])}

\mathrm{[x]-[-x] is \: discontinuous\: at\: x=1}

\mathrm{But \: g(x)=L x-11\: is\: continuous\: and\: g(1)=0\\\\ }

\mathrm{Hence\: f(x) is\: continuous\: at\: x=1}

\mathrm{\text { Also } \begin{aligned} f(x) & = \begin{cases}(1-x)(0-(-1)), & x \in(0,1) \\ 0(1-(-1)), & x=1 \\ (x-1)(1-(-2)), & x \in(1,2)\end{cases} \\ & = \begin{cases}1-x, & x \in(0,1] \\ 3(x-1), & x \in(1,2)\end{cases} \end{aligned}}

Clearly \mathrm{f(x)} is continuous at x=1 and non-differentiable at \mathrm{x=1 \text { as } f^{-}\left(1^{-}\right)=-1 \text { and } f^{\prime}\left(1^\times\right)=3}

 

Posted by

Deependra Verma

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