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Domain of arcsin with Greatest Integer: α²+β² = 5 | JEE Main 2026

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

Question
Let [⋅][\cdot] denote the greatest integer function. If the domain of the function f(x)=sin⁡−1(x+[x]3)f(x)=\sin^{-1}\left(\dfrac{x+[x]}{3}\right) is [α,β)[\alpha,\beta), then α2+β2\alpha^2+\beta^2 is equal to
A22
B55correct
C1010
D1313
Solution
Step 1: sin⁡−1(u)\sin^{-1}(u) requires −1≤u≤1-1\le u\le1. With u=x+[x]3u=\dfrac{x+[x]}{3}:
−1≤x+[x]3≤1 ⇒ −3≤x+[x]≤3.-1\le\dfrac{x+[x]}{3}\le1\ \Rightarrow\ -3\le x+[x]\le3.
 Step 2: On x∈[n,n+1)x\in[n,n+1), [x]=n[x]=n and x+[x]=x+n∈[2n,2n+1)x+[x]=x+n\in[2n,2n+1). Test each nn against −3≤x+n≤3-3\le x+n\le3: ∙\bullet n=−2n=-2 (x∈[−2,−1)x\in[-2,-1)): x−2∈[−4,−3)⇒x−2<−3x-2\in[-4,-3)\Rightarrow x-2<-3 ∴ invalid. ∙\bullet n=−1n=-1 (x∈[−1,0)x\in[-1,0)): x−1∈[−2,−1)⊂[−3,3]x-1\in[-2,-1)\subset[-3,3] ∴ valid. ∙\bullet n=0n=0 (x∈[0,1)x\in[0,1)): x∈[0,1)⊂[−3,3]x\in[0,1)\subset[-3,3] ∴ valid. ∙\bullet n=1n=1 (x∈[1,2)x\in[1,2)): x+1∈[2,3)⊂[−3,3]x+1\in[2,3)\subset[-3,3] ∴ valid. ∙\bullet n=2n=2 (x∈[2,3)x\in[2,3)): x+2∈[4,5)⇒x+2>3x+2\in[4,5)\Rightarrow x+2>3 ∴ invalid. Step 3: ⇒x∈[−1,0)∪[0,1)∪[1,2)=[−1,2)\Rightarrow x\in[-1,0)\cup[0,1)\cup[1,2)=[-1,2). ⇒α=−1, β=2\Rightarrow\alpha=-1,\ \beta=2.
∴α2+β2=(−1)2+22=1+4=5.\therefore\alpha^2+\beta^2=(-1)^2+2^2=1+4=5.
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