Question

If f(x) = `{{:(1 if x  "is rational"),(-1 if x  "is rational"):}` is continuous on ______.

पर्याय

• R

• `phi`   

• (–1, 1)

• (–1, 0)

Answer 1

If f(x) = `{{:(1 if x  "is rational"),(-1 if x  "is rational"):}` is continuous on `underlinebbphi`.

Explanation:

f(x) = `{{:(1 if x  "is rational"),(-1 if x  "is rational"):}`

Let 'a' is any rational number

⇒ f(a) = 1

Then, `lim_(x→a)f(x) = 1 = f(a)`

and, `lim_(x→a)f(x) = -1 ≠ f(a)`

⇒ f(x) is not continuous at any rational number.

Now, Let a ∈ Q^C

⇒ f(a) = –1

Then, \[\ce{lim_{\underset{x∈Q}{x→a}} f(x) = 1 ≠ f(a)}\]

and \[\ce{lim_{\underset{x∈Q^{C}}{x→a}} f(x) = -1 = f(a)}\]

⇒ f(x) is not continuous at any irrational number.

∴ The set of points of continuity = `phi`