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The function \mathrm{f(x)=\max |(1-x),(1+x), 2|, x \in(-\infty, \infty)}, is

Option: 1

discontinuous at all points.


Option: 2

Differentiable at all points.


Option: 3

Differentiable at all point except at x=1 and x=-1.


Option: 4

Continuous at all points except at x=1 and x=-1, where it is discontinuous.


Answers (1)

best_answer

We have,

\mathrm{\begin{aligned} f(x) =\max \{(1-x),(1+x), 2\} \\ =\left\{\begin{array}{cl} 1-x, \text { if } x \leq-1 \\ 2, \text { if }-1 \leq x \leq 1 \\ 1+x, \text { if } x \geq 1 \end{array}\right. \end{aligned}}

\mathrm{\Rightarrow \lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1} 1-x=2, \lim _{x \rightarrow-1^{+}} f(x)=\lim _{x \rightarrow-1} 2=2}

\mathrm{and, f(-1)=2 }

\mathrm{\therefore \quad \lim _{x \rightarrow-1^{-}} f(x)=\lim _{x \rightarrow-1^{+}} f(x)=f(-1) }

So, f(x) is continuous at x=-1.
It can be easily checked that f(x) is also continuous at x=1.
Since f(x) is a polynomial function for \mathrm{x \leq-1 } and \mathrm{x \geq 1 } and a constant function for \mathrm{-1 \leq x \leq 1 }. Hence, f(x) is continuous for all x.
We have,
\mathrm{ L f^{\prime}(-1)=-1, R f^{\prime}(-1)=0 \text { and } L f^{\prime}(1)=0, R f^{\prime}(1)=1 \text {. } }
So, f(x) is not differentiable at \mathrm{ x= \pm 1 }.

 

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Anam Khan

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