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The function $f(x)$ defined by  

$$f(x)= \begin{cases}  0 & \text{if x is rational }  \\  x & \text{if } x\text{ is irrational }   \end{cases}$$

  1. is not continuous at any point
  2. is continuous at every point
  3. is continuous at every rational number
  4. is continuous at $x=0$
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2 Answers

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A- False , Because irrational numbers are dense in real numbers so there would be some irrational number which are contiguous so f(x) may be continuous at some point and also see reason for option-D.

https://proofwiki.org/wiki/Irrationals_are_Everywhere_Dense_in_Reals

B- False, Because between every two rational number there is an irrational number, so f(x) is dis-continuous at almost all rational number except 0 ( For this see option D).

https://proofwiki.org/wiki/Between_two_Rational_Numbers_exists_Irrational_Number

C- False, Reason same as B.

D- True , Because at x=0, all values are very small so x-> +0 and x - > - 0, irrational numbers are also very close to 0, hence f(x) is continuous at x=0.
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x is discontinuous at points of rational numbers other than point 0,

x is continuous if there are more than 1 irrational number in contiguous form and also at point 0

A) False. If there are more than one irrational no in contiguous form then it is continuous

B) False. If there are rational number in between two irrational no , it is not continuous

C) False . Other than 0 no other rational number makes f(x) continuous

D)True. x=0 is rational number and x->-0 and x-> +0 are irrational number where f(x) is continuous

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