Let denote the greatest integer . Consider the function . Then the value of the integral is
- A.
- B.
- C.
- D.
Show answer and solution
Correct option: C
Solution
On [0,2], [x] = 0 for x∈[0,1) and [x] = 1 for x∈[1,2), so 1+[x] = 1 on [0,1) and 2 on [1,2).
Compare x² with these values:
- On [0,1): x² ≤ 1, so f(x) = 1 (since 1+[x]=1 ≥ x² throughout, with equality only at x=1)
- On [1,2): x² vs 2: x²=2 when x=√2. For x∈[1,√2), x²<2 so f(x)=2; for x∈[√2,2), x²>2 so f(x)=x².
At x=2: [2]=2, so 1+[x]=3, x²=4, f(2)=4 (single point, no effect on integral).
Compute the integral:
∫₀¹ 1 dx = 1
∫₁^√2 2 dx = 2(√2−1)
∫_√22 x² dx = [x³/3]_√22 = 8/3 − 2√2/3
Total = 1 + 2√2 − 2 + 8/3 − 2√2/3
= −1 + 2√2 + 8/3 − 2√2/3
= (−3 + 8)/3 + (6√2 − 2√2)/3
= 5
Compare x² with these values:
- On [0,1): x² ≤ 1, so f(x) = 1 (since 1+[x]=1 ≥ x² throughout, with equality only at x=1)
- On [1,2): x² vs 2: x²=2 when x=√2. For x∈[1,√2), x²<2 so f(x)=2; for x∈[√2,2), x²>2 so f(x)=x².
At x=2: [2]=2, so 1+[x]=3, x²=4, f(2)=4 (single point, no effect on integral).
Compute the integral:
∫₀¹ 1 dx = 1
∫₁^√2 2 dx = 2(√2−1)
∫_√22 x² dx = [x³/3]_√22 = 8/3 − 2√2/3
Total = 1 + 2√2 − 2 + 8/3 − 2√2/3
= −1 + 2√2 + 8/3 − 2√2/3
= (−3 + 8)/3 + (6√2 − 2√2)/3
= 5
[JEE Main 2023 · 29 Jan · Shift 1]