UPSC NDA 2 2016 Mathematics Solved Paper

Concept:For a function to be differentiable at a point, its left-hand derivative and right-hand derivative must be equal.Explanation:First, define the given function $f(x)$ piecewise:$f(x) = \begin{cases} -2, & -3 \le x \le 0 \\ x - 2, & 0 < x \le 3 \end{cases}$Now, $f(|x|)$ is formed by replacing $x$ with $|x|$, which for $x \le 0$ becomes $-x$:$f(|x|) = \begin{cases} -x - 2, & -3 \le x \le 0 \\ x - 2, & 0 < x \le 3 \end{cases}$Next, $|f(x)|$ is obtained by taking the absolute value:$|f(x)| = \begin{cases} 2, & -3 \le x \le 0 \\ 2 - x, & 0 < x \le 2 \\ x - 2, & 2 < x \le 3 \end{cases}$Since $g(x) = f(|x|) + |f(x)|$, add the two piecewise functions:$g(x) = \begin{cases} (-x - 2) + 2 = -x, & -3 \le x \le 0 \\ (x - 2) + (2 - x) = 0, & 0 < x \le 2 \\ (x - 2) + (x - 2) = 2x - 4, & 2 < x \le 3 \end{cases}$Check differentiability at $x=0$:Left-hand derivative: $\lim_{h \to 0} \frac{g(0+h) - g(0)}{h} = \lim_{h \to 0} \frac{0 - 0}{h} = 0$Right-hand derivative: $\lim_{h \to 0} \frac{g(0) - g(0-h)}{h} = \lim_{h \to 0} \frac{0 - [ - (-h) ]}{h} = \lim_{h \to 0} \frac{-h}{h} = -1$Since $0 \neq -1$, $g$ is not differentiable at $x=0$. Statement 1 is false.Check at $x=2$:Left-hand derivative: $\lim_{h \to 0} \frac{g(2+h) - g(2)}{h} = \lim_{h \to 0} \frac{[2(2+h)-4] - 0}{h} = \lim_{h \to 0} \frac{2h}{h} = 2$Right-hand derivative: $\lim_{h \to 0} \frac{g(2) - g(2-h)}{h} = \lim_{h \to 0} \frac{0 - 0}{h} = 0$Since $2 \neq 0$, $g$ is not differentiable at $x=2$. Statement 2 is false.Answer:Both statements 1 and 2 are false. The correct option is D. Neither 1 nor 2.