FunctionsmediumFree

Functions: Domain Definition Denotes Greatest Integer Function

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)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Functions · easy
The range of the function y=[x2]−[x]2y=[x^{2}]-[x]^{2}, x∈[0,2]x\in[0,2] (where [⋅][\cdot] denotes the integral part) is:
Functions · easy
If [x][x] denotes the integral part of xx, then domain of the function f(x)=3−x(x−1)(x−2)(x−3)+sin⁡−1 ⁣[3x−22]f(x)=\dfrac{\sqrt{3-x}}{(x-1)(x-2)(x-3)}+\sin^{-1}\!\left[\dfrac{3x-2}{2}\right] is:
Functions · easy
The number of integers in the domain of f(x)=1ln⁡cos⁡−1xf(x) = \dfrac{1}{\sqrt{\ln\cos^{-1}x}}.
Functions · easy
Let [⋅][\cdot] denote the greatest integer function. If the domain of f(x)=cos⁡−1(4x+2[x]3)f(x)=\cos^{-1}\left(\dfrac{4x+2[x]}{3}\right) is [α,β][\alpha,\beta], then 12(α+β)12(\alpha+\beta) is equal to

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath