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Functions — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
The domain of the function log⁡e ⁣(sgn(9−x2))+[x]3−4[x]\log_{e}\!\left(\text{sgn}(9-x^{2})\right)+\sqrt{[x]^{3}-4[x]} (where sgn is signum function and [⋅][\cdot] is step function) is:
A[−2,1]∪[2,3][-2,1]\cup[2,3]
B[−2,1]∪[2,3)[-2,1]\cup[2,3)
C(−2,1]∪[2,3)(-2,1]\cup[2,3)
D[−2,1)∪[2,3)[-2,1)\cup[2,3)correct
Solution
Step 1: For log⁡e(sgn⁡(9−x2))\log_{e}(\operatorname{sgn}(9-x^{2})) to be defined: sgn⁡(9−x2)>0⇒9−x2>0⇒x2<9\operatorname{sgn}(9-x^{2})>0\Rightarrow 9-x^{2}>0\Rightarrow x^{2}<9 ⇒−3<x<3⇒x∈(−3,3)…(1)\Rightarrow -3<x<3\Rightarrow x\in(-3,3)\qquad \ldots(1) Step 2: For [x]3−4[x]\sqrt{[x]^{3}-4[x]} to be defined: [x]3−4[x]≥0[x]^{3}-4[x]\geq 0 ⇒[x]([x]2−4)≥0\Rightarrow [x]([x]^{2}-4)\geq 0 ⇒[x]([x]−2)([x]+2)≥0\Rightarrow [x]([x]-2)([x]+2)\geq 0 By wavy curve method: [x]∈[−2,0][x]\in[-2,0]\quad or [x]≥2\quad [x]\geq 2 Since [x][x] is integer, [x]∈{−2,−1,0}[x]\in\{-2,-1,0\}\quad or [x]≥2\quad [x]\geq 2 Step 3: If [x]∈{−2,−1,0}[x]\in\{-2,-1,0\}, then x∈[−2,1)x\in[-2,1). If [x]≥2[x]\geq 2, then x∈[2,∞)x\in[2,\infty). Combining with (1): x∈[−2,1)∪[2,3)x\in[-2,1)\cup[2,3) Hence, the required domain is [−2,1)∪[2,3)[-2,1)\cup[2,3) Answer: (4) [−2,1)∪[2,3)[-2,1)\cup[2,3)
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