UPSC NDA II 2015 Mathematics Paper

Concept:For a function to be continuous at a point, left-hand limit, right-hand limit, and function value must be equal.Explanation:The piecewise function is continuous everywhere, so we check the two boundary points.At $x = \frac{\pi}{2}$:Right-hand limit: $\lim_{x \to \pi/2^+} f(x) = \cos\left(\frac{\pi}{2}\right) = 0$Left-hand limit: $\lim_{x \to \pi/2^-} f(x) = A\sin\left(\frac{\pi}{2}\right) + B = A + B$Equating: $A + B = 0$ ...(1)At $x = -\frac{\pi}{2}$:Right-hand limit: $\lim_{x \to -\pi/2^+} f(x) = A\sin\left(-\frac{\pi}{2}\right) + B = -A + B$Left-hand limit: $\lim_{x \to -\pi/2^-} f(x) = -2\sin\left(-\frac{\pi}{2}\right) = -2(-1) = 2$Equating: $-A + B = 2$ ...(2)Solve (1) and (2): $A + B = 0$ and $-A + B = 2$. Adding gives $2B = 2 \Rightarrow B = 1$. Then $A = -1$.Answer:$A = -1$ (Option C)