Question

The solution set of inequality \( |x+2|<3 \) is

• A. \( -1<x<5 \)
• B. \( -5<x<1 \)
• C. \( -1<x<3 \)
• D. \( 1<x<3 \)
• E. \( -1<x<1 \)

Hint:

For inequalities of the form \( |A|<k \), always convert them into the double inequality \( -k<A<k \). This is the fastest and safest method.

Answer 1

The Correct Option is B

Step 1: Recall the basic property of modulus inequality.
For any real expression \( A \), the inequality \[ |A|<k \] where \( k>0 \), means \[ -k<A<k \] Here, the given inequality is \[ |x+2|<3 \] So we can write: \[ -3<x+2<3 \]

Step 2: Separate the inequality into a double inequality.
The modulus sign has now been removed, and we have the equivalent compound inequality: \[ -3<x+2<3 \] Now we solve this for \( x \).

Step 3: Subtract \( 2 \) from all three parts.
Subtracting \( 2 \) from the left side, middle term, and right side, we get: \[ -3-2<x+2-2<3-2 \] \[ -5<x<1 \]

Step 4: Interpret the result on the number line.
This means \( x \) can take any real value strictly between \( -5 \) and \( 1 \).
The endpoints \( -5 \) and \( 1 \) are not included because the inequality is strict: \[ |x+2|<3 \] and not \[ |x+2|\leq 3 \]

Step 5: Verify by checking the endpoints.
If \( x=-5 \), then \[ |x+2|=|-5+2|=|-3|=3 \] which does not satisfy \( |x+2|<3 \).
If \( x=1 \), then \[ |x+2|=|1+2|=3 \] which also does not satisfy the inequality.
So both endpoints must be excluded.

Step 6: Match with the given options.
The interval obtained is \[ -5<x<1 \] Among the given options, this matches: \[ \boxed{(2)\ -5<x<1} \]

Step 7: Final conclusion.
Hence, the solution set of the inequality is \[ \boxed{-5<x<1} \] Therefore, the correct option is \[ \boxed{(2)\ -5<x<1} \]