UPSC NDA 2 2016 Mathematics Solved Paper

Concept:A function is continuous on a set if it is continuous at every point in that set.The given function simplifies to a piecewise constant function on its domain.Explanation:Given $f(x) = \frac{x + |x|}{x}$ with domain $A = \mathbb{R}\setminus\{0\}$.We write $|x| = \begin{cases} -x, & x < 0 \\ x, & x \ge 0 \end{cases}$.Thus $f(x) = \begin{cases} \frac{x - x}{x} = 0, & x < 0 \\ \frac{x + x}{x} = 2, & x \ge 0 \end{cases}$.For $x$ in $A$ (nonzero reals), $f(x)$ is constant on $x<0$ and constant on $x>0$, so it is continuous on each open interval.However, at $x=0$ (which is not in $A$), the left-hand limit is $0$ and the right-hand limit is $2$, so $f$ is neither defined nor continuous at $0$.Therefore, $f$ is continuous on $A$ (all nonzero reals) because it is continuous at every point of $A$.Set $B = \{x \in \mathbb{R} : x \ge 0\}$ includes $0$, so $f$ is not continuous on $B$.Set $C = \{x \in \mathbb{R} : x \le 0\}$ also includes $0$, so not continuous on $C$.Set $D = \mathbb{R}$ includes $0$ and also points outside domain, so not continuous on $D$.Hence, the only set on which $f$ is continuous is $A$.Answer:Option A.