UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The function is defined as $f(x) = (|x| - |x - 1|)^2$.For $x > 1$, both absolute values simplify to positive expressions, allowing the function to reduce to a constant.The derivative of a constant is zero.Explanation:Step 1: For $x > 1$, the absolute value $|x|$ equals $x$.Step 2: For $x > 1$, the absolute value $|x - 1|$ equals $x - 1$.Step 3: Substitute into $f(x)$: $f(x) = (x - (x - 1))^2$.Step 4: Simplify inside the parentheses: $x - (x - 1) = x - x + 1 = 1$.Step 5: Thus $f(x) = (1)^2 = 1$, a constant function for all $x > 1$.Step 6: The derivative of a constant is $0$, so $f'(x) = 0$.Answer:Option A: $0$ is correct.