UPSC NDA 2 2016 Mathematics Solved Paper

Concept:A function $g(x)$ is continuous at $x = a$ if the left-hand limit, right-hand limit, and the function value at $x = a$ are all equal.Explanation:The given piecewise function is:$$g(x) = \begin{cases} -x, & -3 \le x \le 0 \\ 0, & 0 < x \le 2 \\ 2x - 4, & 2 < x \le 3 \end{cases}$$Check continuity at $x = 0$:$g(0) = -0 = 0$Left-hand limit: $\lim_{h \to 0} g(0 - h) = \lim_{h \to 0} g(-h) = \lim_{h \to 0} -(-h) = \lim_{h \to 0} h = 0$Right-hand limit: $\lim_{h \to 0} g(0 + h) = \lim_{h \to 0} g(h) = \lim_{h \to 0} 0 = 0$Since LHL = RHL = $g(0)$, $g(x)$ is continuous at $x = 0$.Check continuity at $x = 2$:$g(2) = 2(2) - 4 = 4 - 4 = 0$Left-hand limit: $\lim_{h \to 0} g(2 - h) = \lim_{h \to 0} [2(2 - h) - 4] = \lim_{h \to 0} (4 - 2h - 4) = \lim_{h \to 0} (-2h) = 0$Right-hand limit: $\lim_{h \to 0} g(2 + h) = \lim_{h \to 0} [2(2 + h) - 4] = \lim_{h \to 0} (4 + 2h - 4) = \lim_{h \to 0} (2h) = 0$Since LHL = RHL = $g(2)$, $g(x)$ is continuous at $x = 2$.Check continuity at $x = -1$:$g(-1) = -(-1) = 1$Left-hand limit: $\lim_{h \to 0} g(-1 - h) = \lim_{h \to 0} -(-1 - h) = \lim_{h \to 0} (1 + h) = 1$Right-hand limit: $\lim_{h \to 0} g(-1 + h) = \lim_{h \to 0} -(-1 + h) = \lim_{h \to 0} (1 - h) = 1$Since LHL = RHL = $g(-1)$, $g(x)$ is continuous at $x = -1$.Thus, all three statements (1, 2, and 3) are correct.Answer:All three statements are correct. The correct code is D.