UPSC NDA I 2016 Mathematics Solved Paper

Concept: Express the given equation $x + |y| = 2y$ as a piecewise function using the definition of absolute value, then test domain, continuity, and differentiability at $x = 0$.Explanation:For $y \ge 0$: $|y| = y$, so $x + y = 2y$ gives $y = x$ (valid only when $x \ge 0$).For $y < 0$: $|y| = -y$, so $x - y = 2y$ gives $y = \frac{x}{3}$ (valid only when $x < 0$).Thus $y = f(x) = \begin{cases} \frac{x}{3}, & x < 0 \\ x, & x \ge 0 \end{cases}$, which is defined for all real $x$. Hence statement 1 is not correct.Check continuity at $x = 0$: $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{3} = 0$, $\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x = 0$, and $f(0)=0$. So $f$ is continuous at $x=0$. Statement 2 is not correct.Check differentiability: $f'(x) = \begin{cases} \frac{1}{3}, & x < 0 \\ 1, & x \ge 0 \end{cases}$. $\lim_{x \to 0^-} f'(x) = \frac{1}{3}$, $\lim_{x \to 0^+} f'(x) = 1$. Since left and right derivatives differ, $f$ is not differentiable at $x=0$. Statement 3 is not correct.All three statements are incorrect.Answer: D. 1, 2 and 3