UPSC NDA I 2016 Mathematics Solved Paper

Concept:Continuity at a point $x = a$ requires that the left-hand limit, right-hand limit, and the function's value at $a$ are all equal.Explanation:The function is given as:$f(x)= \begin{cases} x+\pi & \text{for } x\in[-\pi,0) \\ \pi\cos x & \text{for } x\in[0,\frac{\pi}{2}] \\ (x-\frac{\pi}{2})^2 & \text{for } x\in(\frac{\pi}{2},\pi] \end{cases}$Check continuity at $x = 0$:$f(0) = \pi\cos 0 = \pi$Left-hand limit (LHL): $\displaystyle\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} (x+\pi) = \pi$Right-hand limit (RHL): $\displaystyle\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+} \pi\cos x = \pi$Since LHL = RHL = $f(0) = \pi$, the function is continuous at $x = 0$.Check continuity at $x = \frac{\pi}{2}$:$f\left(\frac{\pi}{2}\right) = \pi\cos\frac{\pi}{2} = 0$LHL: $\displaystyle\lim_{x\to \frac{\pi}{2}^-} f(x) = \lim_{x\to \frac{\pi}{2}^-} \pi\cos x = \pi\cos\frac{\pi}{2} = 0$RHL: $\displaystyle\lim_{x\to \frac{\pi}{2}^+} f(x) = \lim_{x\to \frac{\pi}{2}^+} \left(x-\frac{\pi}{2}\right)^2 = 0$Since LHL = RHL = $f\left(\frac{\pi}{2}\right) = 0$, the function is continuous at $x = \frac{\pi}{2}$.Thus, both statements are correct.Answer:Option C: Both 1 and 2