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JEE Main 202622 JanuaryEvening Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2026 · 22 January · Evening Shift · Mathematics · Definite Integration

Let f(x)=[x]^ 2 -[x+3]-3, x ∈ R , where [] is the greatest integer...

Let f(x)=[x]2−[x+3]−3,x∈Rf(x)=[x]^{2}-[x+3]-3, x \in \mathbf{R}, where [] is the greatest integer funtion. Then

Options

  1. A∫02f(x)dx=−6\int_{0}^{2} f(x) \mathrm{d} x=-6
  2. Bf(x)<0f(x)<0 only for x∈[−1,3)x \in[-1,3)
  3. Cf(x)>0f(x)>0 only for x∈[4,∞)x \in[4, \infty)
  4. Df(x)=0f(x)=0 for finitely many values of xx

Correct answer

B. f(x)<0f(x)<0 only for x∈[−1,3)x \in[-1,3)

Step-by-step solution

For x∈[n,n+1)x \in [n, n+1) where nn is an integer, [x]=n[x] = n and [x+3]=n+3[x+3] = n+3.
Thus f(x)=n2−(n+3)−3=n2−n−6=(n−3)(n+2)f(x) = n^2 - (n+3) - 3 = n^2 - n - 6 = (n-3)(n+2).
For f(x)<0f(x) < 0, we need (n−3)(n+2)<0(n-3)(n+2) < 0, which gives −2<n<3-2 < n < 3, i.e., n∈{−1,0,1,2}n \in \{-1, 0, 1, 2\}.
This corresponds to x∈[−1,3)x \in [-1, 3).
We can verify:
For n=−1n = -1: f=1−2−3=−4<0f = 1 - 2 - 3 = -4 < 0
For n=0,1n = 0,1: f=−6<0f = -6 < 0
For n=2n = 2: f=4−5−3=−4<0f = 4 - 5 - 3 = -4 < 0
For n=3n = 3: f=9−6−3=0f = 9 - 6 - 3 = 0
For n≥4n \geq 4, (n−3)(n+2)>0(n-3)(n+2) > 0.
Therefore f(x)<0f(x) < 0 only for x∈[−1,3)x \in [-1, 3).
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