UPSC NDA I 2018 Mathematics Solved Paper

Concept:The function is $f(x) = |x+1|$. We test each given property by substituting into the function.Explanation:We check option A: $f(x^{2}) = |f(x)|^{2}$.$f(x^{2}) = |x^{2}+1|$ and $|f(x)|^{2} = (|x+1|)^{2} = (x+1)^{2}$.Since $|x^{2}+1| \neq (x+1)^{2}$ for all $x$ (e.g., $x = -2$ gives $5 \neq 1$), option A is false.Check option B: $f(|x|) = |f(x)|$.$f(|x|) = \lvert |x|+1 \rvert$ and $|f(x)| = |x+1|$.For $x = -2$, $f(|-2|) = f(2) = |2+1| = 3$, but $|f(-2)| = |-2+1| = 1$. They are not equal, so option B is false.Check option C: $f(x+y) = f(x) + f(y)$.$f(x+y) = |x+y+1|$ and $f(x)+f(y) = |x+1| + |y+1|$.These are not equal in general (e.g., $x=1, y=2$ gives $|1+2+1| = 4$ and $|1+1| + |2+1| = 2+3 = 5$), so option C is false.Since none of A, B, or C holds, the correct choice is D.Answer:D. None of the above