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Maths Question 1 – JEE-MAIN 2026

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Let [] denote the greatest integer function. If the domain of the function f(x)=sin−1⁡(x+[x]3) is [α,β], then α2+β2 is equal to:

Recall that for sin−1⁡(y) to be defined, the argument y must satisfy −1≤y≤1.

Step 1: Apply the domain condition for the inverse sine function✦ Active

For the function f(x)=sin−1⁡(x+[x]3) to be defined, the argument of the inverse sine function must lie in the interval [−1,1]. Thus, we must have:

−1≤x+[x]3≤1

Multiplying by 3, we get:

−3≤x+[x]≤3
Step 2: Analyze the inequality using the greatest integer function property○ Expand

Let [x]=n, where n is an integer. By definition, n≤x<n+1. Substituting [x]=n into the inequality from Step 1:

−3≤x+n≤3

Since n≤x<n+1, we can write 2n≤x+n<2n+1. Combining this with the inequality above:

From x+n≤3, we have 2n≤x+n≤3⟹2n≤3⟹n≤1.5. From −3≤x+n, we have −3≤x+n<2n+1⟹−3<2n+1⟹−4<2n⟹n>−2. Therefore, the possible integer values for n=[x] are −1,0,1.

💡 Teacher's Secret Hint

Remember that x=[x]+{x}, where 0≤{x}<1. This helps in bounding x+[x].

Step 3: Determine the domain and calculate α2+β2○ Expand

We examine the inequality −3≤x+[x]≤3 for each possible integer value of [x]:

1. **If [x]=−1**: Then −1≤x<0. The inequality becomes −3≤x−1≤3⟹−2≤x≤4. The intersection of [−1,0) and [−2,4] is [−1,0). 2. **If [x]=0**: Then 0≤x<1. The inequality becomes −3≤x+0≤3⟹−3≤x≤3. The intersection of [0,1) and [−3,3] is [0,1). 3. **If [x]=1**: Then 1≤x<2. The inequality becomes −3≤x+1≤3⟹−4≤x≤2. The intersection of [1,2) and [−4,2] is [1,2).

The domain of f(x) is the union of these intervals: [−1,0)∪[0,1)∪[1,2)=[−1,2). Comparing this with the given domain [α,β], we have α=−1 and β=2. We need to calculate α2+β2:

α2+β2=(−1)2+(2)2=1+4=5
💡 Teacher's Secret Hint

Carefully combine the intervals obtained for each case of [x] to form the complete domain.

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