UPSC NDA 2 2016 Mathematics Solved Paper

Concept:We analyze composition of the floor function $f(x)=[x]$ and the sine function $g(x)=\sin x$ to verify three functional statements.Explanation:Statement 1: $(f \circ f)(x) = f(x)$$f(x)=[x]$ gives integer part.$(f \circ f)(x) = f(f(x)) = [[x]] = [x] = f(x)$ for any real $x$.Hence statement 1 is correct.Statement 2: $(g \circ g)(x) = g(x)$ only when $x=0$$(g \circ g)(x) = \sin(\sin x)$.For $(g \circ g)(x) = g(x)$ we need $\sin(\sin x) = \sin x$.This holds when $\sin x = 0$, i.e. $x = n\pi$ ($n \in \mathbb{Z}$).Thus it is true for many values, not just $x=0$.So statement 2 is incorrect.Statement 3: $g \circ (f \circ g)(x)$ can take only three valuesFirst, $(f \circ g)(x) = [\sin x]$ takes values $-1$, $0$, $1$ (since $\sin x \in [-1,1]$).Then $g \circ (f \circ g)(x) = \sin([\sin x])$.• If $[\sin x] = -1$ (when $-1 \le \sin x < 0$): value is $\sin(-1) = -\sin 1$.• If $[\sin x] = 0$ (when $0 \le \sin x < 1$): value is $\sin 0 = 0$.• If $[\sin x] = 1$ (when $\sin x = 1$): value is $\sin 1$.Thus only three distinct outputs: $-\sin 1$, $0$, $\sin 1$.So statement 3 is correct.Answer:Correct statements are 1 and 3 only. Option C.