FunctionsmediumFree

Functions: Denotes Integral Part Domain

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

Question
If [x][x] denotes the integral part of xx, then the domain of f(x)=cos⁡−1(x+[x])f(x)=\cos^{-1}(x+[x]) is:
A(0,1)(0,1)
B[0,1)[0,1)correct
C[0,1][0,1]
D[−1,1][-1,1]
Solution
Step 1: For cos⁡−1\cos^{-1} to be defined: −1≤x+[x]≤1-1\le x+[x]\le 1. Step 2: Case x∈Zx\in\mathbb{Z}: Let x=Rx=R (integer). Then x+[x]=2Rx+[x]=2R. Need −1≤2R≤1-1\le 2R\le 1, so R=0R=0. Thus x=0x=0. Step 3: Case x∉Zx\notin\mathbb{Z}: Let x=R+αx=R+\alpha where R=[x]R=[x] (integer) and 0<α<10<\alpha<1. Then x+[x]=2R+αx+[x]=2R+\alpha. Need −1≤2R+α≤1-1\le 2R+\alpha\le 1. From 2R+α≥−12R+\alpha\ge -1: Since α>0\alpha>0, 2R≥−1−α>−22R\ge -1-\alpha>-2, so R≥−1R\ge-1. But α>0\alpha>0 means 2R+α>−12R+\alpha>-1 when R≥0R\ge 0. From 2R+α≤12R+\alpha\le 1: α≤1−2R\alpha\le 1-2R. If R=0R=0: 0<α≤10<\alpha\le 1, but α<1\alpha<1, so 0<α<10<\alpha<1, giving x∈(0,1)x\in(0,1). If R≥1R\ge 1: α≤1−2R≤−1<0\alpha\le 1-2R\le-1<0. Impossible. If R=−1R=-1: α≤3\alpha\le 3 (always true) and 2(−1)+α≥−1  ⇒  α≥12(-1)+\alpha\ge-1\;\Rightarrow\;\alpha\ge 1. But α<1\alpha<1. Impossible. Step 4: Combining: x=0x=0 (from integer case) and x∈(0,1)x\in(0,1) (from non-integer case). Domain =[0,1)=[0,1). Answer: (2) [0,1)[0,1)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Functions · medium
Let f(x)+3f ⁣(π2−x)=sin⁡xf(x)+3f\!\left(\dfrac{\pi}{2}-x\right)=\sin x, x∈Rx\in\mathbb{R}. Let α\alpha be the maximum value of ff. If the area bounded by g(x)=x2g(x)=x^2 and h(x)=βx3h(x)=\beta x^3, β>0\beta>0, equals α2\alpha^2, then 30β330\beta^3 equals
Functions · medium
A={1,2,3}A = \{1,2,3\}, B={1,3,5,7,9}B = \{1,3,5,7,9\}. Then No. of one-one functionsNo. of strictly monotonic functions\dfrac{\text{No. of one-one functions}}{\text{No. of strictly monotonic functions}}.
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 · expert
Let [k][k] and {k}\{k\} denote the greatest integer and fractional part of kk. A real-valued function ff is defined for all real x≠−1x \neq -1 by f(x)=([log⁡e(1+{x})]∫0x2+1sin⁡(et) dt+2026−x5x5+1)1/5.f(x) = \left(\left[\log_e(1 + \{x\})\right]\int_{0}^{x^2+1}\sin(e^t)\,dt + \frac{2026 - x^5}{x^5 + 1}\right)^{1/5}. Find the value of f−1(200)−f(200)+f(f(2026))f(f(2)).\frac{f^{-1}(200) - f(200) + f(f(2026))}{f(f(2))}.

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