If x2 -6-x-5 - 0 and -x-2- 3- then x -

X^2 6|x|+5 ≤ 0 ⇒ |x|^2 6|x| +5 ≤ 0 ⇒ (|x| 1)(|x| 5) ≤ 0 ⇒ |x| ∈ [1,5] ⇒ |x| ≥ 1 and |x| ≤ 5 ⇒ x ∈ ( ∞, 1] ∪ [1,∞) and x ∈ [ 5,5] ⇒ x ∈ [ 5, 1] ∪ [1,5] ⋯ ...

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Standard XII

Mathematics

Inequalities Involving Modulus Function

If x2 -6|x|+5...

Question

If $x^{2} - 6 | x | + 5 \leq 0$ and $| x + 2 | < 3,$ then $x \in$

[- 5, - 1] \cup [1, 5]

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[- 5, - 1]

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(- 5, - 1]

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(- 5, - 1] \cup [1, 5)

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Solution

The correct option is C $(- 5, - 1]$
$x^{2} - 6 | x | + 5 \leq 0$
$\Rightarrow | x |^{2} - 6 | x | + 5 \leq 0$
$\Rightarrow (| x | - 1) (| x | - 5) \leq 0$
$\Rightarrow | x | \in [1, 5]$
$\Rightarrow | x | \geq 1$ and $| x | \leq 5$
$\Rightarrow x \in (- \infty, - 1] \cup [1, \infty)$ and $x \in [- 5, 5]$
$\Rightarrow x \in [- 5, - 1] \cup [1, 5] \dots (1)$

$| x + 2 | < 3$
$\Rightarrow - 3 < x + 2 < 3$
$\Rightarrow - 5 < x < 1 \dots (2)$

From $(1)$ and $(2)$ ,
$x \in (- 5, - 1]$

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If x2−6|x|+5≤0 and |x+2|<3, then x∈

The correct option is C (−5,−1]x2−6|x|+5≤0 ⇒|x|2−6|x|+5≤0 ⇒(|x|−1)(|x|−5)≤0 ⇒|x|∈[1,5] ⇒|x|≥1 and |x|≤5 ⇒x∈(−∞,−1]∪[1,∞) and x∈[−5,5] ⇒x∈[−5,−1]∪[1,5] ⋯(1) |x+2|<3 ⇒−3<x+2<3 ⇒−5<x<1 ⋯(2) From (1) and (2), x∈(−5,−1]

If $x^{2} - 6 | x | + 5 \leq 0$ and $| x + 2 | < 3,$ then $x \in$