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Q.The solution of the inequality 3x−2≤03x - 2 \le 0 is:

(a) [3,∞)[3, \infty)
(b) (−∞,23)\left(-\infty, \frac{2}{3}\right)
(c) (−∞,23]\left(-\infty, \frac{2}{3}\right]
(d) [23,∞)\left[\frac{2}{3}, \infty\right)
Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2023MCQ· 1mImportance★★★★★
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3x−2≤03x - 2 \le 0 solves to x≤23x \le \frac{2}{3}, i.e. the closed interval (−∞,23]\left(-\infty, \frac{2}{3}\right].

Start with 3x−2≤03x - 2 \le 0.

Add 22 to both sides: 3x≤23x \le 2.

Divide both sides by 33 (positive, so the inequality direction is unchanged): x≤23x \le \frac{2}{3}.

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