Q233 / 270 · NDA 2016 (Phase 2) · Mathematics

Paragraph: Let $f(x)=[x]$, where $[$.$] is the greatest integer function and $g(x)=\sin x$ be two real valued functions over $R$.<br><strong>Question:</strong> Which of the following statements are correct? 1. $(f \circ f)(x)=f(x)$. 2. $(\operatorname{gog})(x)=g(x)$ only when $x=0$. 3. (go (fog)) (x) can take only three values. Select the correct answer using the code given below:

  • A
    1 and 2 only
  • B
    2 and 3 only
  • Correct answer
    C
    1 and 3 only
  • D
    1,2 and 3

Solution

$\begin{aligned} & (f \circ f)(x)=[[x]] \text { and } f(x)=[x] \\ & \text { Suppose } x=0.2 \\ & \Rightarrow(f \circ f)(x)=[[0.2]]=[0]=0 \\ & f(x)=[0.2]=0 \\ & \Rightarrow f \circ f(x)=f(x) \end{aligned}$ Now, $(\operatorname{gog})(x)=\sin \sin x$ and $g(x)=\sin x$ At $x=0$ $\begin{aligned} & (\operatorname{gog})(x)=\sin \sin 0=\sin 0=0 \\ & g(x)=\sin 0=0 \\ & \Rightarrow(\operatorname{gog})(x)=g(x) \text { at } x=0 \end{aligned}$ and this is true for $\mathrm{x}=\mathrm{np}$, where $\mathrm{n}=0,1,2,3,4 \ldots \ldots$. $\begin{aligned} & \because(f \circ g)(x)=[\sin x] \\ & (g o(f \circ g))(x)=\sin [\sin x] \\ & \because \sin x \text { has value from }-1 \text { to } 1 \\ & \text { If }-1 \leq \sin x \lt 0 \\ & (g o(f \circ g))(x)=\sin (-1)-\sin (-1) \\ & \text { If } 0 \leq \sin x \lt 1 \\ & \therefore(\operatorname{go}(f \circ g))(x)=\sin (0)=\sin 0 \text { and } \\ & \text { If } \sin x=1 \\ & \therefore(\operatorname{go}(f \circ g))(x)=\sin (1)=\sin 1 \end{aligned}$

Hence, the correct answer is option: 'C'.

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