Mathematics - Functions Question with Solution | TestHub

Given $f(x)={\log }_e\left\{\mathrm{sgn}\left(9-x^2\right)\right\}+\sqrt{[x{]}^3-4[x]}=y_1+y_2$ (say)

Now, $y_1$ is defined if $\mathrm{sgn}\left(9-x^2\right)>0$

But $\mathrm{sgn}\mathrm{x}=1$ (i.e. $>0$ ) if $\mathrm{x}>0$ $\therefore \mathrm{sgn}\left(9-x^2\right)>0\Rightarrow 9-x^2>0\Rightarrow x^2-9<0\Rightarrow (x-3)(x+3)<0\Rightarrow -3<x<3\dots \left(A\right)$ Again, $y_2$ is defined if $[x{]}^3-4[x]\ge 0\Rightarrow [x]\left\{[x{]}^2-4\right\}\ge 0\Rightarrow [x]([x]-2)([x]+2)\ge 0$. Following the wavy curve method, we find

Thus $[x]\ge 2$ or $[x]$ lies between - 2 and 0, i.e. $[\mathrm{x}]=-2,-1$ or 0

Now, $[x]\ge 2\Rightarrow x\ge 2$ .........(B)

$[\mathrm{x}]=-2\Rightarrow -2\le \mathrm{x}<1$

$[\mathrm{x}]=-1\Rightarrow -1\le \mathrm{x}<0$

$[\mathrm{x}]=0\Rightarrow 0\le \mathrm{x}<1$.

Hence $[\mathrm{x}]=-2,-1,0\Rightarrow -2\le x<1$

$\therefore (B)\cup (C)=(x\ge 2)\text{ or }(-2\le x<1)$ .............(C)

Hence $D_f=(A)\cup (C)=[-2,1)\cup [2,3)$.