Question

The solution set of the inequality ||x – 2| – 4| ≥ 1 is

Answer 1

Answer:

Answer

Correct option is

A

α+β+γ+δ=5

D

αβγ=0

x−2

1

−

x

1

−

x+2

2

≤0

x(x−2)(x+2)

x(x+2)−(x

2

−4)−2x(x−2)

≤0

x(x−2)(x+2)

x

2

+2x−x

2

+4−2x

2

+4x

≤0

x(x−2)(x+2)

−2x

2

+6x+4

≤0

x(x−2)(x+2)

x

2

−3x−4

≥0

x(x−2)(x+2)

x

2

−4x+x−4

≥0

(x−0)(x−2)(x+2)

(x−4)(x+1)

≥0

x∈(−2,−1)U(0,2)U(4,∞)

α=2

β=−1

γ=0

δ=4

α+β+γ+δ=4+2−1=5

αβγ=0

Answer 2

The solution set of the inequality is $-1 ≤ x ≤ 5 or x ≤ -3 or x ≥ 7$.

To solve the inequality, we can break it down into cases based on the sign of the expression inside the absolute value.

Case 1: $x ≤ 2$

In this case, |x-2| is negative or zero, so we have:

$||x-2|-4| ≥ 1$

$|-|x-2|-4| ≥ 1$

$-|x-2| ≥ -3$

$|x-2| ≤ 3$

This gives us the inequality $2 - 3 ≤ x ≤ 2$ + 3 or$-1 ≤ x ≤ 5.$

Case 2: $x > 2$

In this case,$|x-2|$ is positive, so we have:

$||x-2|-4| ≥ 1$

$|x-2|-4 ≥ 1 or |x-2|-4 ≤ -1$

Solving the first inequality, we ge$t |x-2| ≥ 5$, which gives$x ≤ -3 or x ≥ 7.$

Solving the second inequality, we get$|x-2| ≤ 3$, which gives us $2-3 ≤ x ≤ 2+3 or -1 ≤ x ≤ 5.$

Therefore, the solution set of the inequality is $-1 ≤ x ≤ 5 or x ≤ -3 or x ≥ 7$

For such more questions on inequality,

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