Statement-I: |3x - 5| 9 ⇒ x ∈(-∞, (-4/3)) ∪ ((14/3), ∞). Statement-II: The region containing all the solutions of an inequality is called the solution region.

Question

Statement-I: |3x - 5| > 9 ⇒ x ∈(-∞, (-4/3)) ∪ ((14/3), ∞). Statement-II: The region containing all the solutions of an inequality is called the solution region. (A) If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I (B) If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I (C) If Statement-I is true but Statement-II is false (D) If Statement-I is false and Statement-II is true

Tardigrade Admin · Accepted Answer

We have, |3x - 5| > 9 ⇒ 3 x -5 < -9 or 3 x -5> 9 ⇒ 3x < - 4 or 3x > 14 ⇒ x < (-4/3) or x > (14/3) ∴ x ∈(-∞, (-4/3)) ∪ ((14/3), ∞)

Q. Statement-I: $∣3 x - 5∣ > 9 \Rightarrow x \in (- \infty, \frac{- 4}{3}) \cup (\frac{14}{3}, \infty)$ .
Statement-II: The region containing all the solutions of an inequality is called the solution region.

If both Statement-I and Statement-II are true and Statement-II is the correct explanation of Statement-I

If both Statement-I and Statement-II are true but Statement-II is not the correct explanation of Statement-I

If Statement-I is true but Statement-II is false

If Statement-I is false and Statement-II is true

We have, $∣3 x - 5∣ > 9$
$\Rightarrow 3 x - 5 < - 9$ or $3 x - 5 > 9$
$\Rightarrow 3 x < - 4$ or $3 x > 14$
$\Rightarrow x < \frac{- 4}{3}$ or $x > \frac{14}{3} ∴ x \in (- \infty, \frac{- 4}{3}) \cup (\frac{14}{3}, \infty)$

Q. Statement-I: ∣3x−5∣>9⇒x∈(−∞,3−4​)∪(314​,∞). Statement-II: The region containing all the solutions of an inequality is called the solution region.