TIFR-2011-Maths-A-25

Question

The function

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

is not continuous anywhere on the real line.

Answer 1

False ;

It is continuous at x=0 since |f(x₀)-0|<a for all a , whenever |x-x₀|<a.

Answer 2

Suppose, f is conti. at x = 0

As "0" is rational number f(x)= f(0)= 0

For any rational number x there exist a seq. Of Irrational xn converges to x .

Here, f is conti. f(xn) converges to f(x)

But f(xn) never converges to 0 so f is nowher conti.

Ans. True