UPSC NDA I 2017 Mathematics Solved Paper

Concept:A function is continuous at a point if the left‑hand limit, right‑hand limit, and the function value at that point are all equal.Explanation:The given function is $f(x)=\begin{cases}2x+1, & -3 < x<-2 \\ x-1, & -2\le x<0 \\ x+2, & 0\le x<1\end{cases}$.Potential points of discontinuity are $x=-2$ and $x=0$ because the formula changes there.Check continuity at $x=-2$:Left‑hand limit: $\displaystyle\lim_{x\to -2^{-}}f(x)=\lim_{x\to -2^{-}}(2x+1)=2(-2)+1=-3$.Right‑hand limit: $\displaystyle\lim_{x\to -2^{+}}f(x)=\lim_{x\to -2^{+}}(x-1)=-2-1=-3$.Both limits are equal, so $f$ is continuous at $x=-2$.Check continuity at $x=0$:Left‑hand limit: $\displaystyle\lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{-}}(x-1)=0-1=-1$.Right‑hand limit: $\displaystyle\lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}(x+2)=0+2=2$.The left and right limits are different, so $f$ is discontinuous at $x=0$.Each piece is a polynomial, so $f$ is continuous everywhere else except at $x=0$.Answer:Option C: It is discontinuous at $x=0$ but continuous at every other point.