Question

Let $\underset{x\to a}{lim}f\left(x\right)$ exists but it is not equal to f (a). Then f(x) is discontinuous at x$=$ a and a is called a removable discontinuity. If $\underset{x\to a^{-}}{lim}f\left(x\right)=land\underset{x\to a^{+}}{lim}f\left(x\right)=m$ exist but $l\ne m.$ Then a is called a jump discontinuity. If one of the limits (left hand limit or right hand limit ) does not exist, then a is called an infinite discontinuity.

Let f(x) be defined by f(x) $=\{\begin{array}{cc}2x^2& x\text{is rational}\\ 1-x,& x\text{is irrational}\end{array}$ Then $f$ is

• A. discontinuous at $x=\frac{1}{2}$
• B. continuous at $x=\frac{1}{2}$
• C. continuous everywhere
• D. discontinuous everywhere

Answer 1

The correct option is B continuous at $x=\frac{1}{2}$

$f:R\to R$ is defined by
$f\left(x\right)=\{\begin{array}{cc}2x^2,& xisrational\\ 1-x,& xisirrational\end{array}$
${lim}_{x\to \frac{1}{2}}f\left(x\right)$ ($x\to \frac{1}{2}$ through rational )
${lim}_{x\to \frac{1}{2}}2x^2=2\times \frac{1}{4}=\frac{1}{2}$
$limitoff\left(x\right)atx=\frac{1}{2}$ though irrational numbers $=1-\frac{1}{2}=\frac{1}{2}$

Also, $f\left(\frac{1}{2}\right)=\frac{1}{2}$
$\therefore$ the function is continuous at $x=\frac{1}{2}$
clearly , it is not continuous everywhere .