Mathematics - Functions Question with Solution | TestHub

(i) $f(x)$ is defined if (i) $(4-|x|)>0$ (iii) $[x^2]>0$ but $[x^2]\ne 1$.

Now, (i) $\Rightarrow |x|<4\Rightarrow -4<x<4$ ............(A)

From (iii), $[x^2]>0\Rightarrow [x^2]=1,2,3,\dots$

But $[x^2]\ne 1$.

$\therefore [x^2]=2,3,4,\dots$ i.e. $[x^2]\ge 2$.

$\Rightarrow x^2\ge 2$; $\because [f(x)]\ge n\Rightarrow f(x)\ge n$.

$\Rightarrow x\le -\sqrt{2}$ or $x\ge \sqrt{2}$.

Combining (A) and (B), we find that $-4<x\le -\sqrt{2}$ or $\sqrt{2}\le x<4$. …(B) Hence the domain of the given function is $(-4,-\sqrt{2}]\cup [\sqrt{2},4)$.

(ii) The function is defined if (i) $sgn(x^2)>0$ and (ii) $[x]>0$ but $[x]\ne 1$. We know that

$$sgn(x^2)=\{\begin{array}{ll}1& if{x}^2>0\\ 0& if{x}^2=0\\ -1& if{x}^2<0\end{array}$$

(i) Since $sgn(x^2)$ is non-negative, we have $x^2>0\Rightarrow x\in \mathbb{R}-\{0\}$. …(A)

(ii) $\Rightarrow [x]=2,3,4,\dots \therefore x\in [2,\infty )$. …(B)

Hence, $D_f=A\cap B=[2,\infty )$.