Question

STATEMENT-1 : If $f\left(x\right)=\{\begin{array}{cc}x,& ifxisrational\\ 1-x,& ifxisirrational\end{array}$, then ${lim}_{x\to 1/2}$ f(x) does not exist.
STATEMENT-2 : $x\to 1/2$ can be a rational or an irrational value.

• A. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
• B. STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
• C. STATEMENT-1 is True, STATEMENT-2 is False
• D. STATEMENT-1 is False, STATEMENT-2 is True

Answer 1

Answer: (D)

STATEMENT-1 is False, STATEMENT-2 is True

$f\left(x\right)=\{\begin{array}{c}x,xisrational\\ 1-x,xisirrational\end{array}$

Let $x\to \frac{1}{2}$ be rational

${lim}_{x\to \frac{1}{2}}x=\frac{1}{2}$

Let $x\to \frac{1}{2}$ be irrational

${lim}_{x\to \frac{1}{2}}(1-x)=1-\frac{1}{2}=\frac{1}{2}$

Similarly for $x\to {\frac{1}{2}}^{+}\&x\to {\frac{1}{2}}^{-}$

LHL$=$RHL $={lim}_{x\to \frac{1}{2}}f\left(x\right)=\frac{1}{2}$

$\therefore$ limit exists