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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)=cos−1⁡(4x+2[x]3) is [α,β], then 12(α+β) is equal to:

The argument of the inverse cosine function, cos−1⁡(y), must lie in the interval [−1,1].

● Video Walkthrough
Step 1: Establish the domain condition for the inverse cosine function✦ Active

For the function f(x)=cos−1⁡(4x+2[x]3) to be defined, the argument of the inverse cosine must be within the interval [−1,1]. Thus, we must have:

−1≤4x+2[x]3≤1

Multiplying by 3, we get:

−3≤4x+2[x]≤3
Step 2: Substitute x=[x]+{x} and determine possible values for [x]○ Expand

Let x=[x]+{x}, where [x] is the greatest integer less than or equal to x, and {x} is the fractional part of x, such that 0≤{x}<1. Substitute this into the inequality:

−3≤4([x]+{x})+2[x]≤3

Simplify the expression:

−3≤6[x]+4{x}≤3

Since 0≤{x}<1, we know that 0≤4{x}<4. We analyze the two parts of the inequality:

1. 6[x]+4{x}≥−3⟹6[x]≥−3−4{x}. Since 4{x}<4, −4{x}>−4. So, 6[x]>−3−4=−7. This implies [x]>−7/6≈−1.16, so [x]≥−1 (as [x] is an integer).

2. 6[x]+4{x}≤3⟹6[x]≤3−4{x}. Since 4{x}≥0, −4{x}≤0. So, 6[x]≤3−0=3. This implies [x]≤3/6=1/2, so [x]≤0 (as [x] is an integer).

Combining these, the possible integer values for [x] are −1 and 0.

💡 Teacher's Secret Hint

Remember to use the full range of the fractional part, 0≤{x}<1, to establish the tightest bounds for [x].

Step 3: Determine the intervals for x and calculate 12(α+β)○ Expand

Case 1: [x]=−1. Substitute into −3≤6[x]+4{x}≤3:

−3≤6(−1)+4{x}≤3⟹−3≤−6+4{x}≤3

Adding 6 to all parts: 3≤4{x}≤9⟹3/4≤{x}≤9/4. Intersecting with 0≤{x}<1, we get 3/4≤{x}<1. Since x=[x]+{x}=−1+{x}, this gives −1+3/4≤x<−1+1, so −1/4≤x<0.

Case 2: [x]=0. Substitute into −3≤6[x]+4{x}≤3:

−3≤6(0)+4{x}≤3⟹−3≤4{x}≤3

Dividing by 4: −3/4≤{x}≤3/4. Intersecting with 0≤{x}<1, we get 0≤{x}≤3/4. Since x=[x]+{x}=0+{x}, this gives 0≤x≤3/4.

The domain of f(x) is the union of these two intervals: [−1/4,0)∪[0,3/4]=[−1/4,3/4]. Therefore, α=−1/4 and β=3/4.

Finally, we calculate 12(α+β):

12(α+β)=12(−14+34)=12(24)=12(12)=6
💡 Teacher's Secret Hint

Carefully combine the intervals obtained from each case. The union of [a,b) and [b,c] is [a,c] if the intervals are contiguous.

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