UPSC NDA I 2015 Mathematics 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 piecewise function is:$$f(x)=\begin{cases} -1 & x\le 0 \\ ax+b & 0 < x<1 \\ 1 & x\ge 1 \end{cases}$$Since $f(x)$ is continuous everywhere, it must be continuous at $x=0$.At $x=0$, the left-hand limit as $x\to0^-$ is $-1$.The right-hand limit as $x\to0^+$ is $a(0)+b = b$.For continuity at $x=0$: $\lim_{x\to0^-}f(x) = \lim_{x\to0^+}f(x) = f(0)$. But $f(0)=-1$.Thus, $b = -1$.Continuity at $x=1$ would give another equation, but it is not needed to find $b$.Answer:$b = -1$, which corresponds to option A.