Match List-I with List-II
[.] denotes the greatest integer function.
|
List-I |
List-II |
|
(A) $\int\limits_0^3[x]dx$ |
(I) $\frac{1}{2}$ |
|
(B) $\int\limits_0^1[2x]dx$ |
(II) 1 |
|
(C) $\int\limits_0^1 [3x]dx$ |
(III) $\frac{3}{2}$ |
|
(D) $\int\limits_0^1[4x]dx$ |
(IV) 3 |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
|
List-I |
List-II |
|
(A) $\int\limits_0^3[x]dx$ |
(IV) 3 |
|
(B) $\int\limits_0^1[2x]dx$ |
(I) $\frac{1}{2}$ |
|
(C) $\int\limits_0^1 [3x]dx$ |
(II) 1 |
|
(D) $\int\limits_0^1[4x]dx$ |
(III) $\frac{3}{2}$ |
Evaluating each integral:
(A) $\int_0^3 [x]\,dx$
$= \int_0^1 0\,dx + \int_1^2 1\,dx + \int_2^3 2\,dx = 0 + 1 + 2 = 3$
(A) → (IV)
(B) $\int_0^1 [2x]\,dx$
$2x \in [0, 2) \Rightarrow [2x] = 0$ on $[0, \frac{1}{2})$, $1$ on $[\frac{1}{2}, 1)$
$= \int_0^{1/2} 0\,dx + \int_{1/2}^{1} 1\,dx = 0 + \frac{1}{2} = \frac{1}{2}$
(B) → (I)
(C) $\int_0^1 [3x]\,dx$
$= \int_0^{1/3} 0\,dx + \int_{1/3}^{2/3} 1\,dx + \int_{2/3}^1 2\,dx$
$= 0 + \frac{1}{3} + \frac{2}{3} = 1$
(C) → (II)
(D) $\int_0^1 [4x]\,dx$
$= \int_0^{1/4} 0\,dx + \int_{1/4}^{1/2} 1\,dx + \int_{1/2}^{3/4} 2\,dx + \int_{3/4}^1 3\,dx$
$= 0 + \frac{1}{4} + \frac{1}{4} \cdot 2 + \frac{1}{4} \cdot 3 = \frac{1}{4} + \frac{1}{2} + \frac{3}{4} = \frac{3}{2}$
(D) → (III)