If Z and R denote set of integers and set of real numbers respectively, then match List I with List II.
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List-I |
List-II |
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(A) $5x-3<3x + 1, x ∈ Z$ |
(I) $x ∈ (-∞,-3]$ |
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(B) $3x+17≤ 2(1-x),x ∈ R$ |
(II) $x ∈ (-∞,-1)$ |
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(C) $13x + 17 <2(1-x),x ∈R$ |
(III) $\{........-4,-3,...,0,1\}$ |
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(D) $\frac{2x + 3}{5}-2>\frac{3(x-2)}{5},x ∈Z$ |
(IV) $\{....... -4, -3, -2\}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A)-(III), (B)-(I), (C)-(II), (D)-(IV)
|
List-I |
List-II |
|
(A) $5x-3<3x + 1, x ∈ Z$ |
(III) $\{........-4,-3,...,0,1\}$ |
|
(B) $3x+17≤ 2(1-x),x ∈ R$ |
(I) $x ∈ (-∞,-3]$ |
|
(C) $13x + 17 <2(1-x),x ∈R$ |
(II) $x ∈ (-∞,-1)$ |
|
(D) $\frac{2x + 3}{5}-2>\frac{3(x-2)}{5},x ∈Z$ |
(IV) $\{....... -4, -3, -2\}$ |
Matching List-I with List-II with solutions:
(A) $5x - 3 < 3x + 1$, $x \in \mathbb{Z}$
Simplify: $5x - 3 < 3x + 1 \Rightarrow 2x < 4 \Rightarrow x < 2$
Integer solutions: $\{..., -4, -3, -2, -1, 0, 1\}$ → (III)
(B) $3x + 17 \le 2(1 - x)$, $x \in \mathbb{R}$
$3x + 17 \le 2 - 2x \Rightarrow 5x \le -15 \Rightarrow x \le -3$ → (I)
(C) $13x + 17 < 2(1 - x)$, $x \in \mathbb{R}$
$13x + 17 < 2 - 2x \Rightarrow 15x < -15 \Rightarrow x < -1$ → (II)
(D) $(2x + 3)/5 - 2 > 3(x - 2)/5$, $x \in \mathbb{Z}$
Simplify: $(2x + 3)/5 - 2 > (3x - 6)/5 \Rightarrow (2x + 3 - 10) > 3x - 6 \Rightarrow 2x - 7 > 3x - 6 \Rightarrow -x > 1 \Rightarrow x < -1$
Integer solutions: $\{..., -4, -3, -2\}$ → (IV)