Let “g” be the function defined on the set of all real numbers by

g(x)=1, If x is rational and g(x)=exp(x) , if x is irrational,

Then the set of numbers at which “g”  is continuous is

(i)                  the empty set

(ii)                {0}

(iii)               {1}

(iv)              The set of rational numbers.

(v)                The set of irrational numbers.

If like this problem then rate me

very well explained.

hey what do u mean by exp(x)

plz explain

i m not getting the ques.

Given “g” be the function defined on the set of all real numbers by

 g(x)=1, If x is rational and g(x)=exp(x) , if x is irrational,

Now, at x = 0 since 0 is rational number g(x) =1

 

At h tending to 0 or h--->0,

Now RHL = Lim h--->0  g(0+h)

there are two possibilities

a) either h = rational, if so, Lim h--->0  g(0+h) = 1

b) or h = irrational, then Lim h--->0  g(0+h) = exp (h) = exp (0) = 1

Similarly we can prove that Left hand limit, LHL = Lim h--->0  g(0-h) = 1

thus we see that, RHL = LHL

Hence g(x) is continuous at 0.