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

Option: 1

f(x) is continuous and differentiable at x=1


Option: 2

f(x)  is discontinuous at x=1


Option: 3

f(x)  is continuous at x=2


Option: 4

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 \mathrm{x=1}

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

Hence \mathrm{f(x)}  is continuous at \mathrm{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 \mathrm{x=1} and non-differentiable at \mathrm{x=1} as \mathrm{f\left(1^{-}\right)=-1\, and \, f^{\prime}\left(1^{*}\right)=3}

Posted by

Divya Prakash Singh

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