Question

Pls solve this question

Answer 1

$\large\bf{\underline{\purple{Given}}}$

• $\sf |x+1| \ge 4$

$\large\bf{\underline{\purple{Explanation}}}$

• Absolute value inequalities

✰ Case 1 ✰

➡ |x| < a or |x| ≤ a

• If |x| < a then -a < x < a
• If |x| < a then -a ≤ x ≤ a

✰ Case 2 ✰

➡ |x| > a or |x| ≥ a

• If |x| > a then x < -a or x > a
• If |x| ≥ a then x ≤ -a or x ≥ a

✰ Case 3 ✰

➡ |x| < -a or |x| ≤ -a

• There is always no solution in this case.

✰ Case 4 ✰

➡ |x| > -a or |x| ≥ -a

• There is always real number in this case.

$\large\bf{\underline{\purple{Solution}}}$

Now, according to the given condition

$\sf |x+1| \ge 4$

• This is case 2
• If |x| ≥ a then x ≤ -a or x ≥ a

$\qquad \qquad {\red{\boxed{\bf{\red{x + 1\:is\: negative}}}}}$

$\sf{x+1 \le - 4}$

$\sf{x+1-1 \le - 4-1}$

$\sf{x \le - 5}$

$\qquad \qquad \qquad \qquad {\bf{Or}}$

$\qquad \qquad {\red{\boxed{\bf{\red{x + 1\:is\:positive}}}}}$

$\sf{x+1 \ge 4}$

$\sf{x+1-1 \ge 4-1}$

$\sf{x \ge 3}$

• $\tt x \le -5\:or\:x \ge 3$
• Therefore D is correct option

Answer 2

In attachement, that's +* in case 1.

Step-by-step explanation:

Case 1: +(x + 1) ≥ 4 => x ≥ 3

Draw the straight line x = 3, which is a vertical line. The points which lie either to the right of the line x = 3 or on the line x = 3 satisfy the given inequality x ≥ 3. The solution region of the inequality x ≥ 3 is shown shaded.

Dark lines state that the line "x=3" comes under the solution region (as x ≥ 3).

Do the same with second case (which is -(x + 1) ≥ 4. => - 5 ≥ x).

The combination of both the cases defines the value of x(or where it can lie).

In form of equation, answer is (-∞, 5] U [3, ∞). Or, |x+1| ≥ 4 means,

- 4 ≤ x + 1 ≤ 4

- 4 - 1 ≤ x ≤ 4 - 1

- 5 ≤ x ≤ 3