The true set of values of x satisfying |x|-|x-3|=3, is

Question

The true set of values of x satisfying |x|-|x-3|=3, is (A) (-∞, 3] (B) [3, ∞) (C)  3  (D) (-∞, ∞)

Tardigrade Admin · Accepted Answer

text Case-I: x <0 - x + x -3=3 -3=3 text (no solution) text Case-II: 0 ≤ x <3 x+x-3=3 x=3 text (no solution) text Case-III: x ≥ 3 x-(x-3)=3 3=3 ∴ x ∈[3, ∞)

Q. The true set of values of $x$ satisfying $∣ x ∣ - ∣ x - 3∣ = 3$ , is

$(- \infty, 3]$

$[3, \infty)$

${3}$

$(- \infty, \infty)$

$Case-I : x < 0$
$- x + x - 3 = 3$
$- 3 = 3 (no solution)$
$Case-II : 0 \leq x < 3$
$x + x - 3 = 3$
$x = 3 (no solution)$
$Case-III : x \geq 3$
$x - (x - 3) = 3$
$3 = 3$
$∴ x \in [3, \infty)$

Q. The true set of values of x satisfying ∣x∣−∣x−3∣=3, is

(−∞,3]

[3,∞)

{3}

(−∞,∞)