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Functions — JEE Maths practice question

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)
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