UPSC NDA 2 2016 Mathematics Solved Paper

Concept:The limits of composites $f(g(x))$ and $g(f(x))$ as $x\to0$ are compared using left and right hand limits.Explanation:Let $f(x) = [x]$ (greatest integer) and $g(x) = \sin x$.Then $(fog)(x) = [\sin x]$ and $(gof)(x) = \sin ([x])$.For $x \to 0^-$, $\sin x$ is negative near $0$, so $[\sin x] = -1$.For $x \to 0^+$, $\sin x$ is positive small, so $[\sin x] = 0$.Hence $\lim_{x\to0^-}(fog)(x) = -1$ and $\lim_{x\to0^+}(fog)(x) = 0$, not equal so $\lim_{x\to0}(fog)(x)$ does not exist.For $x \to 0^-$, $[x] = -1$, so $\sin([x]) = \sin(-1) = -\sin 1$.For $x \to 0^+$, $[x] = 0$, so $\sin([x]) = \sin 0 = 0$.Thus $\lim_{x\to0^-}(gof)(x) = -\sin 1$ and $\lim_{x\to0^+}(gof)(x) = 0$, not equal so $\lim_{x\to0}(gof)(x)$ does not exist.Now check the given options:Option A and B are false because both limits do not exist.Option C: $\lim_{x\to0^-}(fog)(x) = -1$ and $\lim_{x\to0^-}(gof)(x) = -\sin 1$, not equal.Option D: $\lim_{x\to0^+}(fog)(x) = 0$ and $\lim_{x\to0^+}(gof)(x) = 0$, they are equal.So option D is correct.Answer:D. $\mathop {\lim }\limits_{x \to {0^ + }} \left( {{\rm{fog}}} \right)\left( x \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {{\rm{gof}}} \right)\left( x \right)$