Function Discontinuity

Calculus Level 3

Consider the function

$\large{f\left( x \right) ={ \cot ^{ -1 }{ \left( \text{sgn}\left( \frac { \left\lfloor x \right\rfloor }{ 2x-\left\lfloor x \right\rfloor } \right) \right) } }}$

$\text{Statement 1}$ : $f(x)$ is discontinuous at $x=1$ .

$\text{Statement 2}$ : $f(x)$ is non-differentiable at $x=1$ .

Which of the following option is correct?

Notations : $\text{sgn}$ is signum function. And $\left\lfloor . \right\rfloor$ is floor function.

Statement 1 is incorrect and Statement 2 is correct Statement 1 and Statement 2 are correct Statement 1 and Statement 2 are incorrect Statement 1 is correct and Statement 2 is incorrect

1 solution

Akhil Bansal
Nov 15, 2015

$\large f(x) = \cot ^{ -1 }{ \left( \text{sgn}\left( \frac { \left\lfloor x \right\rfloor }{ 2x-\left\lfloor x \right\rfloor } \right) \right) }$ Checking function continuity at $x = 1$
Left hand limit : $\large \displaystyle \lim_{x \rightarrow 1^-} \cot ^{ -1 }{ \left( sgn\left( \frac {0 }{ 2 - 0 } \right) \right) } = \cot ^{-1}(\text{sgn}(0)) = \cot^{-1}(0)$ Right hand limit :

$\large \displaystyle \lim_{x \rightarrow 1^+} \cot ^{ -1 }{ \left( \text{sgn}\left( \frac {1 }{ 2 - 1 } \right) \right) } = \cot ^{-1}(\text{sgn}(1)) = \cot^{-1}(1)$

As, Left hand limit $\neq$ Right hand limit at $x = 1$ , therefore, function is discontinous at $x = 1$ .

A discontinuous function is always non- differentiable. Hence, both the statements are correct.

Function Discontinuity

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