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Functions — JEE Maths practice question

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

Question
Domain of the definition f(x)=log⁡4(5−[x−1]−[x]2)x2+x−2f(x)=\dfrac{\log_{4}(5-[x-1]-[x]^{2})}{x^{2}+x-2} (where [⋅][\cdot] denotes greatest integer function) is:
A(−3,−2)∪(1,2)(-3,-2)\cup(1,2)
B(−4,−3)∪(2,3)∪(3,4)(-4,-3)\cup(2,3)\cup(3,4)
C(−43,−34)∪(23,23)\left(-\dfrac{4}{3},-\dfrac{3}{4}\right)\cup\left(\dfrac{2}{3},\dfrac{2}{3}\right)
D(−2,1)∪(1,2)(-2,1)\cup(1,2)correct
Solution
Step 1: For f(x)f(x) to be defined, two conditions must hold simultaneously. Condition 1: Denominator ≠0\neq 0:
x2+x−2≠0  ⇒  (x+2)(x−1)≠0  ⇒  x≠−2,  x≠1x^{2}+x-2\neq 0\;\Rightarrow\;(x+2)(x-1)\neq 0\;\Rightarrow\;x\neq -2,\;x\neq 1
 Condition 2: Argument of log⁡4\log_{4} must be >0>0:
5−[x−1]−[x]2>05-[x-1]-[x]^{2}>0
 Step 2: Note that [x−1]=[x]−1[x-1]=[x]-1 for all xx. Substituting:
5−([x]−1)−[x]2>0  ⇒  6−[x]−[x]2>0  ⇒  [x]2+[x]−6<05-([x]-1)-[x]^{2}>0\;\Rightarrow\;6-[x]-[x]^{2}>0\;\Rightarrow\;[x]^{2}+[x]-6<0
 Step 3: Solve [x]2+[x]−6<0[x]^{2}+[x]-6<0:
([x]+3)([x]−2)<0  ⇒  −3<[x]<2([x]+3)([x]-2)<0\;\Rightarrow\;-3<[x]<2
 So [x]∈{−2,−1,0,1}[x]\in\{-2,-1,0,1\}, giving x∈[−2,2)x\in[-2,2). Step 4: Combine with Condition 1 (x≠−2,  x≠1x\neq -2,\;x\neq 1):
x∈(−2,2)∖{1}=(−2,1)∪(1,2)x\in(-2,2)\setminus\{1\}=(-2,1)\cup(1,2)
 Answer: (4) (−2,1)∪(1,2)(-2,1)\cup(1,2)
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