If - x -3- - 10-thenA- x -13-7-B- x- -13-7C- x -13 -7- -D- x -13 -7- -

The correct option is D xϵ ( ∞ , 13) ∪ [7,∞) Since |x+3|≥ 10 ⇒ x+3 ≤ 10 or x+3 ≥10 ⇒ x ≤ 13 or x≥7 ⇒ x ϵ ( ∞, 13) ∪ [7,∞) ...

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Byju's Answer

Standard XII

Mathematics

Inequalities Involving Modulus Function

If | x +3| ≥ ...

Question

If $| x + 3 | \geq 10$ ,then

$x ϵ (- 13, 7]$

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$x ϵ (- 13, 7)$

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$x ϵ (- \infty, - 13) \cup (7, \infty)$

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$x ϵ (- \infty, - 13) \cup [7, \infty)$

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Solution

The correct option is D
$x ϵ (- \infty, - 13) \cup [7, \infty)$

$S i n c e | x + 3 | \geq 10 \Rightarrow x + 3 \leq - 10 o r x + 3 \geq 10 \Rightarrow x \leq - 13 o r x \geq 7 \Rightarrow x ϵ (- \infty, - 13) \cup [7, \infty)$

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Similar questions

Q. Mark the correct alternative in each of the following:
If

|x + 3|

\geq

10, then
(a) x

\in

(

-

13, 7]
(b) x

\in

(

-

13, 7)
(c) x

\in

(

-

\infty

-

13)

\cup

(7,

\infty

)
(d) x

\in

(

-

\infty

-

13]

\cup

[7,

\infty

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If $| x + 2 | \leq 9$ ,then

Q. If

\frac{5 - \sqrt{3}}{2 + \sqrt{3}} = x + y \sqrt{3}

, then

(a) x = 13, y = −7

(b) x = −13, y = 7

(c) x = −13, y = −7

(d) x = 13, y = 7

Q. Solve for

| 2 x + 3 | + x = 10

, 'x' is a real number

Q. If

x > 3

and

y = 3 x + 4

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If |x+3|≥10,then

The correct option is D xϵ(−∞,−13)∪[7,∞) Since |x+3|≥10⇒ x+3≤−10 or x+3 ≥10⇒ x≤−13 or x≥7⇒ x ϵ (−∞,−13)∪[7,∞)

If $| x + 3 | \geq 10$ ,then

The correct option is D
$x ϵ (- \infty, - 13) \cup [7, \infty)$

$S i n c e | x + 3 | \geq 10 \Rightarrow x + 3 \leq - 10 o r x + 3 \geq 10 \Rightarrow x \leq - 13 o r x \geq 7 \Rightarrow x ϵ (- \infty, - 13) \cup [7, \infty)$