A) \[\left[ -2,1 \right)\cup \left[ 2,3 \right)\]
B) \[\left[ -4,1 \right)\cup \left[ 2,3 \right)\]
C) \[\left[ 4,1 \right)\cup \left[ 2,3 \right)\]
D) \[\left[ 2,1 \right)\cup \left[ 2,3 \right)\]
Correct Answer: A
Solution :
| [a] We have |
| \[f(x)=lo{{g}_{e}}\{sgn(9-{{x}^{2}})\}+\sqrt{{{[x]}^{3}}-4[x]}\] |
| We must have, sgn \[(9-{{x}^{2}})>0\Rightarrow 9-{{x}^{2}}>0\] |
| \[\Rightarrow {{x}^{2}}-9<0\Rightarrow (x-3)(x+3)<0\Rightarrow -3<x<3\](i) |
| Also \[{{[x]}^{3}}-4[x]\ge 0\Rightarrow [x]({{[x]}^{2}}-4)\ge 0\] |
| \[\Rightarrow [x]([x]-2)([x]+2)\ge 0\] |
| \[\Rightarrow [x]\ge -2\] or [x] lies between -2 and 0 |
| i.e., \[[x]=-2,-1\] or 0 |
| Now, \[[x]\ge -2\Rightarrow x\ge 2\] (ii) |
| \[[x]=-2\Rightarrow -2\le x<-1;[x]=-1\Rightarrow -1\le x<0\] |
| \[[x]=0\Rightarrow 0\le x<1\] Hence \[[x]=-2,-1,0\] |
| \[\Rightarrow -2\le x<1\] |
| Hence \[{{D}_{f}}=[-2,\,\,1]\cup \left[ 2,3 \right).\] |
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