If 2-x-3-x- 5- then x - A- RB- R -C- 2- -D- -

The correct option is D ϕ2 |x|3+|x|≥5 ⇒ nbsp;2 |x|3+|x| 5≥ 0 ⇒ 13 6|x|3+|x|≥ 0 ⇒ nbsp; 13 6|x| ≥ 0 (∵ 3+|x| 0) ⇒ 6|x| ≥ 13 ⇒ |x| ≤ 136 Not possible as ...

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Byju's Answer

Standard XII

Mathematics

Inequalities Involving Modulus Function

If 2-|x|/3+|x...

Question

If $\frac{2 - | x |}{3 + | x |} \geq 5$ , then $x \in$

R

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R^{+}

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(2, \infty)

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ϕ

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Solution

The correct option is D $ϕ$
$\frac{2 - | x |}{3 + | x |} \geq 5 \Rightarrow \frac{2 - | x |}{3 + | x |} - 5 \geq 0 \Rightarrow \frac{- 13 - 6 | x |}{3 + | x |} \geq 0 \Rightarrow - 13 - 6 | x | \geq 0 (∵ 3 + | x | > 0)$
$\Rightarrow - 6 | x | \geq 13 \Rightarrow | x | \leq - \frac{13}{6}$
Not possible as $| x | \geq 0$

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If 2−|x|3+|x|≥5, then x∈

The correct option is D ϕ2−|x|3+|x|≥5⇒2−|x|3+|x|−5≥0⇒−13−6|x|3+|x|≥0⇒−13−6|x|≥0 (∵3+|x|>0) ⇒−6|x|≥13⇒|x|≤−136 Not possible as |x|≥0

If $\frac{2 - | x |}{3 + | x |} \geq 5$ , then $x \in$