If (3x - 4/2) ge (x + 1/4) - 1, then x∈

Question

If (3x - 4/2) ge (x + 1/4) - 1, then x∈ (A) [1, ∞) (B) (1, ∞) (C) (-5, 5) (D) [-5, 5]

Tardigrade Admin · Accepted Answer

We have (3x - 4/2) ge (x + 1/4) - 1 or, (3x - 4/2) ge (x -3/4) or, 2(3x - 4) ge(x -3 ) or, 6x - 8 ge x - 3 or, 5x ge 5 or x ge 1 Thus all real numbers which are greater than or equal to 1 is the solution set of the given inequality ∴ x ∈ [1, ∞)

Q. If $\frac{3 x - 4}{2} \geq \frac{x + 1}{4} - 1$ , then $x \in$

$[1$ , $\infty)$

$(1$ , $\infty)$

$(- 5$ , $5)$

$[- 5$ , $5]$

Q. If 23x−4​≥4x+1​−1, then x∈

[1, ∞)

(1, ∞)

(−5, 5)

[−5, 5]

We have 23x−4​≥4x+1​−1 or, 23x−4​≥4x−3​ or, 2(3x−4)≥(x−3) or, 6x−8≥x−3 or, 5x≥5 or x≥1 Thus all real numbers which are greater than or equal to 1 is the solution set of the given inequality ∴x∈[1, ∞)

Q. If $\frac{3 x - 4}{2} \geq \frac{x + 1}{4} - 1$ , then $x \in$

$[1$ , $\infty)$

$(1$ , $\infty)$

$(- 5$ , $5)$

$[- 5$ , $5]$