Match List-I with List-II
Choose the correct answer from the options given below: |
(A)-(IV), (B)-(III), (C)-(II), (D)-(I) (A)-(III), (B)-(IV), (C)-(II), (D)-(I) (A)-(III), (B)-(II), (C)-(I), (D)-(IV) (A)-(III), (B)-(IV), (C)-(I), (D)-(II) |
(A)-(III), (B)-(IV), (C)-(I), (D)-(II) |
The correct answer is Option (4) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)
List-I and List-II Matching: (A) $\displaystyle \int_0^4 |x|\, dx$ For $x \in [0, 4]$, $|x| = x$ $\Rightarrow \int_0^4 x\, dx = \left[\frac{x^2}{2}\right]_0^4 = \frac{16}{2} = 8$ ⇒ (A) → (III) (B) $\displaystyle \int_{-2}^2 |x|\, dx$ Even function: $= 2 \int_0^2 x\, dx = 2 \cdot \left[\frac{x^2}{2}\right]_0^2 = 2 \cdot \frac{4}{2} = 4$ ⇒ (B) → (IV) (C) $\displaystyle \int_0^3 [x]\, dx$ $[x]$ is the greatest integer ≤ x
Total = $0 + 1 + 2 = 3$ ⇒ (C) → (I) (D) $\displaystyle \int_{-1}^1 [x]\, dx$ On $[-1, 0)$: $[x] = -1$ → Interval length = 1 → Area = $-1 \cdot 1 = -1$ On $[0,1)$: $[x] = 0$ → Area = $0$ Total = $-1 + 0 = -1$ ⇒ (D) → (II) |